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ПРОФ. РЕДАКТУРУ И МОЖЕТ СОДЕРЖАТЬ ОШИБКИ.

МЕХАНИКА • СЛЕПКОВ АЛЕКСАНДР ИВАНОВИЧ СЛЕДИТЕ ЗА ОБНОВЛЕНИЯМИ НА VK.COM/TEACHINMSU.

ФИЗИЧЕСКИЙ
ФАКУЛЬТЕТ
МГУ ИМЕНИ
М.В. ЛОМОНОСОВА

THE PRINCIPLES OF QUANTUM

FIELD THEORY

АРУТЮНОВ
ГЛЕБ ЭДУАРДОВИЧ

ФИЗФАК МГУ

КОНСПЕКТ ПОДГОТОВЛЕН
СТУДЕНТАМИ, НЕ ПРОХОДИЛ
ПРОФ. РЕДАКТУРУ И МОЖЕТ
СОДЕРЖАТЬ ОШИБКИ.
СЛЕДИТЕ ЗА ОБНОВЛЕНИЯМИ
НА VK.COM/TEACHINMSU.

ЕСЛИ ВЫ ОБНАРУЖИЛИ
ОШИБКИ ИЛИ ОПЕЧАТКИ,
ТО СООБЩИТЕ ОБ ЭТОМ,
НАПИСАВ СООБЩЕСТВУ
VK.COM/TEACHINMSU.
БЛАГОДАРИМ ЗА ПОДГОТОВКУ КОНСПЕКТА
СТУДЕНТА ФИЗИЧЕСКОГО ФАКУЛЬТЕТА МГУ
СОЛОВЫХ АЛЕКСАНДРА АЛЕКСЕЕВИЧА
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Contents
1 Lecture 1. Classical Fields and Symmetries 5
1.1 The history and main challenge of QFT . . . . . . . . . . . . . . . . . . . . 5
1.2 Discussion of symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Classical fields and symmetries . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Lecture 2. Tensor Fields, Euler-Lagrange Equations,

Lorentz Transformations 16
2.1 Classical fields in many dimensional case . . . . . . . . . . . . . . . . . . . 16
2.2 Tensors and tensor fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Action principle and Euler-Lagrange equations . . . . . . . . . . . . . . . . 19
2.4 Lorentz and Poincare groups . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.5 Boosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 Lecture 3. Lorentz and Poincare Groups, Noether’s Theorem 34

3.1 Repeating of the rotation group . . . . . . . . . . . . . . . . . . . . . . . . 34
3.2 Lorentz boosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.4 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Lecture 4. Applications of Noether’s Theorem. Conservation Laws and

Symmetries 52
4.1 Internal symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.2 Energy-momentum tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.3 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.4 Lagrangians of Wess-Zumino“ . . . . . . . . . . . . . . . . . . . . . . . . 61
”
4.5 Improvement procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.6 Casimir operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5 Lecture 5. Conserved Charges as Symmetry Generators. Klein-Gordon

Field. Mass-Shell Condition 73
5.1 Conserved charges as the symmetry generators . . . . . . . . . . . . . . . . 73
5.2 Klein-Gordon field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
5.3 How to solve Klein-Gordon equation? . . . . . . . . . . . . . . . . . . . . . 88

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6 Lecture 6. More on Klein-Gordon equation. Canonical Quantization 91

6.1 How to solve Klein-Gordon equation? . . . . . . . . . . . . . . . . . . . . . 91
6.2 Canonical quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

7 Lecture 7. Second Quantization. Commutation and Green’s Functions,

Pauli-Jordan Function 107
7.1 Second quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
7.2 Commutation and Greens functions . . . . . . . . . . . . . . . . . . . . . . 114

8 Lecture 8. Retarded and Advanced Green’s Functions. Feynman Propagator.

Yukawa Force 123
8.1 Retarded Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
8.2 Advanced Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
8.3 Feynman propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
8.4 Yukawa force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

9 Lecture 9. Yukawa Force. Dirac Equation 137

9.1 Yukawa force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
9.2 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

10 Lecture 10. The Dirac Equation and Lorentz Transformations 151

10.1 Lorentz transformations of spinors . . . . . . . . . . . . . . . . . . . . . . . 160

11 Lecture 11. Lorentz Boosts. Parity 166

11.1 Spatial rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
11.2 Lorentz boosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
11.3 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

12 Lecture 12. Time Reversal. Weyl Spinors and Weyl Equations 181
12.1 Time reversal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
12.2 Weyl spinoes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

13 Lecture 13. Solution of the Dirac Equation 196

13.1 Helicity operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

14 Lecture 14. Charge Conjugation and Anti-Particles 212

14.1 Majorana spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

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Lecture 1. Classical Fields and Symmetries

The history and main challenge of QFT

Quantum field theory (QFT), perhaps, is one of the greatest achievements of theoretical
physicists of 20s century. It is arose from an attempt to combine two already existent and
experimentally verified theories: theory of special relativity and quantum mechanics. QFT
is the unique tool which is used to describe the physics of elementary particles. QFT is
characterized by numerous precision experiments or tests such as collider experiments
like, for instance, those which people carry out in the Large Hadron Collider at Cern or
at Fermi National Accelerator Laboratory in USA.
Methods of QFT has an universal character and this theory is not only applied in
sphere of particle physics, but also widely used in, for example, condensed matter theory.
Thus, QFT applied for dynamical systems with large or infinite degrees of freedom.
In spite of well developed state of QFT, starting from 1928 discovery of the electron
by Dirac, it has many undiscovered peaces of knowledge.
The fact that in QFT scientists deal with dynamical systems with large or infinite
degrees of freedom leads to a list of well-known problems. Some of the physical observables
or quantities which you would like to measure or predict will be given by divergent
integrals. It is a so-called problem of infinities. Perhaps, most sharply the problem of
infinities shows itself when people try to merge QFT with gravitational theory.
All of today’s experience shows that, perhaps, the degrees of freedom that can be
obtained in QFT are not a suitable starting point for building a quantum theory. This
means that, probably, there is some hidden theory on the next level where degrees of
freedom should be reconsidered or recovered and understood by another point of view. One
of such next-level theory is a string theory. There is a theory of one-dimensional extended
objects and it offers very intriguing possibility how to think about quantum fields in a
very different way and, hopely, to construct the quantum approach to avoid the problem
of infinities. The problem of infinities is also known under the name renormalization.
Concerning particle physics, the main goal is to be able to ultimately compute and
predict physical observables such as, for instance, cross-ratios in scattering experiments,
decays of unstable particles and the spectra of bound states. All of these are very difficult
theoretical questions because there is no direct access to local losses during particle
interaction. It is unknown how to access this loss, but as one of the possible ways is

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to build a theoretical model of interaction loss between elementary particles and then test
this predictions, for instance, in colliding experiments, when cross-ratios or decay rates of
unstable particles can be experimentally measured.

Discussion of symmetries
Symmetries play an extremely important role not in a QFT, but in physics in general
and in our lives. It essentially means that the loss of nature is not arbitrary from one
space-time point to another. If you have a certain knowledge of measure at one space-time
point and you move away and you made measurements in the other points of space-time,
you better find something which in agreement with your previous experiments, but not
something arbitrary subjected to new loss. So, symmetries provide a certain order in our
world, allowing us to talk about universal laws. That is why the first theme will be about
symmetries and their relations to conservation laws. This will be the so-called Noether’s
first theorem. he transition will be made to the traditional or canonical quantization of
3 basic fields: scalar, spinor and electromagnetic, and then to the scattering matrix (or
S-matrix). So, the gotten knowledge about quantized fields will be applied to evaluation
of the scattering matrix of the quantum electrodynamics (QED) theory, which is used to
describe electrons and photons.
The lecturer suggests people to read some books for better understanding of the course.
There are:

1) Michael E. Pestin and Daniel V. Schroeder An introduction to quantum field

”
theory“ (canonical book, which is used in many universities)

2) A. Zee Quantum Field Theory in a Nutshell“

”
3) L.D. Landau and E.M. Lifshits Theoretical physics. Quantum electrodynamics“
”
4) N.N. Bogolyubov and D.V. Shirkov Quantum fields“
”
5) Silvan S. Schweber An Introduction to Relativistic Quantum Field Theory“
”
6) L.A. Takhtajyan Quantum mechanics for mathematicians“
”
7) Brian C. Hall Quantum Theory for Mathematicians“
”
8) Faddeev and Yakubovsky Lectures on quantum mechanics for mathematics students“
”

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Classical fields and symmetries

The discussion of classical fields and symmetries includes lagrangian description of
classical systems with an infinite number of degrees of freedom and Noether’s first theorem
that allows to construct dynamical invariants or quantities that remain invariants during
dynamical evolution of the system.
Let’s remind how in classical mechanics the transition from discrete to continuum
takes place. To describe continuous systems like, for instance, solid, it is necessary to
make a transition from finite number of degrees of freedom to infinite. This means that
it is necessary to specify at least coordinates of all points and the number of this points
is infinite. By the fact continuous can be reached by taking appropriate limit of a system
with a finite number of discrete coordinates. For instance, let’s consider an elastic rod of
fixed length which undergoes small longitudinal vibrations (fig. 1.1).

Fig. 1.1. An elastic rod of fixed length 𝑙

Let’s describe this system in the framework of classical mechanics. The rod can be
approximated as a system of equal mass and particles with equal distances between
neighbor points.
If the rod is represented as a system of 𝑛 particles connected with each other by a
springs (fig. 1.2), then the distance between each particle will be ∆𝑎 = 𝑙/𝑛, where we
denoted a distance between particles in equilibrium position as ∆𝑎.

Fig. 1.2. The rod is represented as a system of 𝑛 particles connected with each other by
a springs

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From fig. 2 it can be seen that system has 𝑛 + 1 springs and that is why the length of
rod can be written as:

𝑙 = (𝑛 + 1)∆𝑎 . (1.1)

Each spring has such characteristic as Hooke’s constant 𝑘.

The displacement of 𝑖𝑡ℎ particle from it’s equilibrium position will be described by the
coordinate 𝜑𝑖 . Displacement means that the particle is taken and moved a little bit away
from it’s equilibrium position.
The kinetic energy of all particles can be easily found from:
𝑛
∑︁ 𝑚
𝑇 = 𝜑˙ 2 . (1.2)
𝑖=1
2

It is also known that the presented system has a potential energy stored in springs,
and such an energy may be computed. For this reason, the coordinate system should be
introduced.
The left end of 𝑖𝑡ℎ particle will have coordinate:

𝑥𝑖 = 𝑖∆𝑎 + 𝜑𝑖 . (1.3)

At the same time the right end of 𝑖𝑡ℎ particle will have coordinate:

𝑥𝑖+1 = (𝑖 + 1)∆𝑎 + 𝜑𝑖+1 . (1.4)

The potential energy stored in the 𝑖𝑡ℎ spring will be:

𝑘
𝑈𝑖 = (𝑥𝑖+1 − 𝑥𝑖 )2 . (1.5)
2
With the use of (1.3) and (1.4), (1.5) can be simplified to:
𝑘
𝑈𝑖 = (∆𝑎 + 𝜑𝑖+1 − 𝜑𝑖 )2 . (1.6)
2
The total potential energy of the system can be gotten by summarizing of impacts
from each spring:
𝑛
∑︁ 𝑘
𝑈= (∆𝑎 + 𝜑𝑖+1 − 𝜑𝑖 )2 . (1.7)
𝑖=0
2
Formula (1.7) can be simplified by opening the brackets:
𝑛 𝑛
𝑘 ∑︁ 𝑘 ∑︁
𝑈 == (𝜑𝑖+1 − 𝜑𝑖 )2 + ∆𝑎2 (𝑛 + 1) + 𝑘∆𝑎 (𝜑𝑖+1 − 𝜑𝑖 ) , (1.8)
2 𝑖=0 2 𝑖=0

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where the last expression represents cross member.

Let’s associate
𝜑0 = 𝜑𝑛+1 = 0 , (1.9)

because the left end of 0𝑡ℎ spring and the right end of (𝑛 + 1)𝑡ℎ spring are not movable.
Because of the statement (1.9) and the fact that all displacements will compensate
each other, the last term of the (1.8) will be equal to zero.
It is known that in classical mechanics the constant shift of the potential plays no role,
because the force is given by the minus gradient of the potential energy:
𝜕𝑈𝑖
𝐹𝑖 = − . (1.10)
𝜕𝜑𝑖
Such a way formula (1.9) can be simplified to:
𝑛
1 ∑︁
𝑈= 𝑘 (𝜑𝑖+1 − 𝜑𝑖 )2 . (1.11)
2 𝑖=0

If we place (1.11) into (1.10), we will get that

𝐹𝑖 = 𝑘 (𝜑𝑖+1 + 𝜑𝑖−1 − 2𝜑𝑖 ) . (1.12)

The force 𝐹𝑖 presented in (1.12) consists of difference between forces that acts on 𝑖𝑡ℎ
particle from the right and from the left sides. It can be seen from:

𝐹𝑖 = 𝑘 (𝜑𝑖+1 − 𝜑𝑖 ) − 𝑘 (𝜑𝑖 − 𝜑𝑖−1 ) . (1.13)

⏟ ⏞ ⏟ ⏞
from the right from the left

This means that kinetic and potential energies of the system are known and, therefore,
the lagrangian of the system may be formed. From course of theoretical mechanics it is
well known that:
L=𝑇 −𝑈 , (1.14)

Then it is necessary to place (1.2) and (1.11) into (1.14):

𝑛 𝑛
∑︁ 𝑚 2 𝑘 ∑︁
L= 𝜑˙𝑖 − (𝜑𝑖+1 − 𝜑𝑖 )2 . (1.15)
𝑖=1
2 2 𝑖=0

As can be seen from (1.15), we have a system with finite number of degrees of freedom
and these degrees of freedom can be described by 𝜑𝑖 .
Then it is necessary to take a continuum limit by sending the number of particles to
infinity. At the same time, the number of springs also will tend to infinity. The distance
between particles in the opposite side will tend to zero:

𝑛 → ∞ , ∆𝑎 → 0 . (1.16)

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From (1.1) it can be clearly seen that 𝑛 and 𝐷𝑎 changes with the same speed and that
is why 𝑙 remains constant. Despite this fact that there is an another problem which is
linked with the fact that each particle in system has a mass. With increasing of particles
number it will tend to infinity and therefore it is necessary to keep it finite. This target
can be reached if mass of each particle 𝑚 will decrease with the same speed as ∆𝑎. Let’s
define a mass density 𝜇:
𝑚
𝜇= . (1.17)
∆𝑎
From (1.17) it can be seen that the mass density is constant.
Besides the mass of each particle it is also necessary to keep constant forces between
particles. For this goal the Hook’s constant with the ∆𝑎 change will be scaled as follows:

𝑘∆𝑎 = 𝑐𝑜𝑛𝑠𝑡 = κ , (1.18)

where 𝑘 is now a function of ∆𝑎 and

κ
𝑘∼ . (1.19)
∆𝑎
Then it is necessary to rewrite lagrangian in terms of ∆𝑎:
𝑛 𝑛 (︂ )︂2
1 ∑︁ (︁ 𝑚 )︁ 2 1 ∑︁
˙ 𝜑𝑖+1 − 𝜑𝑖
L= ∆𝑎 𝜑𝑖 − ∆𝑎 (𝑘∆𝑎) . (1.20)
2 𝑖=1 ∆𝑎 2 𝑖=0 ∆𝑎

Using (1.17) and (1.18), formula (1.20) can be written as:

𝑛 𝑛 (︂ )︂2
1 ∑︁ ˙ 2 1 ∑︁ 𝜑𝑖+1 − 𝜑𝑖
L= ∆𝑎𝜇𝜑𝑖 − ∆𝑎κ . (1.21)
2 𝑖=1 2 𝑖=0 ∆𝑎

When the limit will be taken, ∆𝑎 will tend to zero. Due to this fact the discrete index
𝑖 of displacement variable can be replaced by a continuum variable 𝑥. In other worlds, the
variable 𝜑𝑖 will be replaced by 𝜑 (𝑖∆𝑎). Due to the replacement of 𝑖, 𝑖∆𝑎 may be changed
by 𝑥 and as a result:
𝜑𝑖 → 𝜑 (𝑥) . (1.22)

It should be noticed, that with 𝑛 → ∞ 𝜑(𝑥) becomes a continuous field.

After the transition to the continuum
𝜑𝑖+1 − 𝜑𝑖 𝜑 (𝑥 + ∆𝑎) − 𝜑(𝑥)
→ −→ 𝜕𝑥 𝜑 (𝑥) . (1.23)
∆𝑎 ∆𝑎 𝑛→∞

Using expressions (1.22) and (1.23), (1.21) can be written as:

1 𝑙 [︁ ˙ 2
∫︁ ]︁
L= d𝑥 𝜇𝜑 (𝑥) − κ (𝜕𝑥 𝜑 (𝑥))2 , (1.24)
2 0

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where sums from 𝑖 = 0 or 1 to 𝑛 were replaced by integrals from 𝑥 = 0 to 𝑙.

The limiting procedure with equations of motion can also be performed. For this task
it is necessary to get them from lagrangian with finite 𝑛 and then go to the limit. The
equations of motion for discrete system can be written as:
𝑚 ¨ 𝜑𝑖+1 + 𝜑𝑖−1 − 2𝜑𝑖
𝜑𝑖 − 𝑘∆𝑎 =0. (1.25)
∆𝑎 (∆𝑎)2

The second expression in (1.25) can be replaced by second derivative when ∆𝑎 tends
to zero:
𝜑𝑖+1 + 𝜑𝑖−1 − 2𝜑𝑖 𝜕 2𝜑
lim = = 𝜕𝑥𝑥 𝜑 . (1.26)
Δ𝑎→0 (∆𝑎)2 𝜕𝑥2
As a result the final view of equation of motion (1.25) will be:

𝜇𝜑¨ (𝑥) − κ𝜕𝑥𝑥 𝜑 (𝑥) = 0 . (1.27)

Equation (1.27) also can be gotten from (1.24), but for this target it is necessary to
represent (1.27) in the next view:
∫︁ 𝑙
L= d𝑥 L , (1.28)
0

where L is a lagrangian density and equal

1 (︁ ˙ 2 )︁
L= 𝜇𝜑 − κ (𝜕𝑥 𝜑)2
(1.29)
2
for the presented situation.
In theoretical mechanics there is one more important term such as action, which can
be defined as an integral of the lagrangian with respect to time. By this moment it is not
needed to pass the general picture, which says that the case is being conducted with an
action for our continuum dynamical system.
The action will be indicated as 𝑆 and it will be a functional of displacement field
𝜑 (𝑥, 𝑡): ∫︁ 𝑡2
𝑆 [𝜑] = Ld𝑡 . (1.30)
𝑡1

Lagrangian at the same time can be written as presented at (1.29) and, therefore,
(1.30) will have a view:
∫︁ 𝑡2 ∫︁ 𝑙 (︁ )︁
𝑆 [𝜑] = d𝑡 d𝑥 × L 𝜑 (𝑥, 𝑡) , 𝜑˙ (𝑥, 𝑡) , 𝜕𝑥 𝜑 (𝑥, 𝑡) . (1.31)
𝑡1 0

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To obtain equations of motion directly from lagrangian (1.28) it is needful to understand

how the action changes under an infinitesimal change of the field. For this purpose it
is necessary to replace 𝜑(𝑥, 𝑡) with 𝜑(𝑥, 𝑡) + 𝛿𝜑(𝑥, 𝑡). This procedure is called variation
because the field is varied to observe how the action changes.
The derivatives of 𝜑 can be also written as:
⎧
⎨ 𝜕 𝜑 (𝑥, 𝑡) → 𝜕 𝜑 (𝑥, 𝑡) + 𝜕
𝛿𝜑 (𝑥, 𝑡)
𝜕𝑡 𝜕𝑡 𝜕𝑡
(1.32)
⎩ 𝜕 𝜑 (𝑥, 𝑡) → 𝜕
𝜑 (𝑥, 𝑡) + 𝜕
𝛿𝜑 (𝑥, 𝑡)
𝜕𝑥 𝜕𝑥 𝜕𝑥

Then it is needed to see how the action reacts on the infinitesimal change of the field.
So the result was calculated as follows:

𝛿𝑆 [𝜑] = 𝑆 [𝜑 + 𝛿𝜑] − 𝑆 [𝜑] . (1.33)

Formula (1.33) gives the following:

∫︁ 𝑡2 ∫︁ 𝑙 [︂ ]︂
𝜕L 𝜕L 𝜕L
𝛿𝑆 [𝜑] = d𝑡 d𝑥 𝛿𝜑 + 𝜕𝑡 𝛿𝜑 + 𝜕𝑥 𝛿𝜑 . (1.34)
𝑡1 0 𝜕𝜑 𝜕 𝜑˙ 𝜕 (𝜕𝑥 𝜑)

Integral in (1.34) can be evaluated by parts:

∫︀ 𝑡2 ∫︀ 𝑙 [︁ ]︁
𝜕L
𝛿𝑆 [𝜑] = 𝑡1
d𝑡 0
d𝑥 𝜕𝜑
− 𝜕𝑡 𝜕L
𝜕 𝜑˙
− 𝜕 𝜕L
𝑥 𝜕(𝜕𝑥 𝜑) 𝛿𝜑+

⃒𝑡=𝑡2 ⃒𝑥=𝑙 (1.35)

∫︀ 𝑙 ⃒ ⃒
𝜕L
∫︀ 𝑡2 𝜕L
+ 0 d𝑥 𝜕(𝜕𝑡 𝜑) 𝛿𝜑⃒ + 𝑡1 d𝑡 𝜕(𝜕𝑥 𝜑) 𝛿𝜑⃒
⃒ ⃒
⃒ ⃒
𝑡=𝑡1 𝑥=0

To find the trajectories (get equations of motion) of the dynamical systems it is

necessary to minimize the action which means that it is needed put to zero the variational
derivative:
𝛿𝑆 [𝜑]
=0. (1.36)
𝛿𝜑
It should be noticed that it was needed to minimize the action under the condition
that values of the field in the initial and finite times are unaffected. In other words:

𝛿𝜑 (𝑥, 𝑡1 ) = 𝛿𝜑 (𝑥, 𝑡2 ) = 0 , (1.37)

that means that initial and final profile of the field must be kept in tact.
Expression (1.37) can be clearly understood if return to classical mechanics. In classical
case there is a particle which moves from one point to another and it makes some actual

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trajectory. The particle moves in the some potential and interacts with this one which
lead to formation of the definite trajectory. In the variational principle, the fixed initial
and final positions of the particle cannot be changed, and all other trajectories that are
in the vicinity of the actual trajectory are tried. For all of this trajectories the principle
of least action telling us that the value of the action must be largen than the value of the
action on the actual trajectory (fig. 1.3). But comparison of values of the action on the
virtual and the actual trajectories is made under the assumption that the initial and the
final points taken by a particle keep fixed.

Fig. 1.3. Possible trajectories of the particle

The same thing happens for a field. Suppose that there is an initial moment of time
that equals to 𝑡1 and the 𝜑 function can be drawn as represented on (fig. ??). Then when
time grows, the field starts to move and at the time moment equals to 𝑡2 has another
profile. Such a way trajectory of the field forms a surface.

Fig. 1.4. The field evolution from 𝑡 = 𝑡1 to 𝑡 = 𝑡2

Due to formula (1.37) the integral

⃒𝑡=𝑡2
∫︁ 𝑙
𝜕L ⃒
d𝑥 (1.38)
⃒
𝜕 (𝜕𝑡 𝜑) 𝛿𝜑 ⃒
⃒
0
𝑡=𝑡1

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will be equal to zero.

Then it is necessary to find the third integral in formula (1.35). This term can be put
to zero or not put to zero due to physical grounds. For instance, if we say that we are
dealing with the rod which is clamped between two walls which means that the profile
of the field at the left and right points is fixed (fig. 1.5), the 𝛿𝜑 will equal to zero and
therefore third integral in formula (1.35) will be equal to zero.

Fig. 1.5. The profile of the field of the rod which is clamped between two walls

Such a way in formula (1.35) only one expression stayed. For the actual trajectory
the vanishing of 𝛿𝑆 for any variation 𝛿𝜑 by the basic lemma of a variational calculus
represents the equation of motion:
(︂ )︂ (︂ )︂
𝜕 𝜕L 𝜕 𝜕L 𝜕L
+ − =0. (1.39)
𝜕𝑡 𝜕 (𝜕𝑡 𝜑) 𝜕𝑥 𝜕 (𝜕𝑥 𝜑) 𝜕𝜑
If we apply equations of motion (1.39) to the concrete problem and place the lagrangian
density (1.29) into (1.39), we will get equations of motion similar to (1.27):

𝜑¨ − 𝑐2 𝜕𝑥𝑥 𝜑 = 0 , (1.40)

where 𝑐 is a next ratio between 𝜇 and κ:

√︂
κ
𝑐= . (1.41)
𝜇
The value 𝑐 can interpreted as a speed which characterizes propagation of vibrations
through the rod.
Equations of motions gotten for the presented dynamical system are very easy because
of their linearity. As it well known, for the linear equations the superposition principle
works. So it is possible to solve equations by creating of the next linear combination:

𝜑 (𝑥, 𝑡) = 𝑒𝑖𝑝𝑥 𝑎𝑝 (𝑡) + 𝑒−𝑖𝑝𝑥 𝑏𝑝 (𝑡) . (1.42)

Then it is necessary to apply boundary conditions for the gotten solution. For the left
wall of the rod, the boundary condition will give:

𝜑 (0, 𝑡) = 0 ⇒ 𝑎𝑝 (𝑡) + 𝑏𝑝 (𝑡) = 0 → 𝑏𝑝 (𝑡) = −𝑎𝑝 (𝑡) . (1.43)

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Expression (1.43) leads to the simplification of the (1.42):

𝜑 (𝑥, 𝑡) = 𝑎𝑝 (𝑡) 𝑒𝑖𝑝𝑥 − 𝑒−𝑖𝑝𝑥 → 𝑎𝑝 (𝑡) sin (𝑝𝑥) ,

(︀ )︀
(1.44)

where the multiplier 2𝑖 was entered in 𝑎𝑝 (𝑡).

At the same time for the right border of the rod there will be the following expression:

𝜑 (𝑙, 𝑡) = 𝑎𝑝 (𝑡) sin (𝑝𝑙) = 0 . (1.45)

In expression (1.45) it is necessary to equate to zero one of two multipliers. It can be

clearly seen that 𝑎𝑝 (𝑡) = 0 corresponds to 𝜑 (𝑥, 𝑡) = 0. That is why it is necessary to
equate to zero sin (𝑝𝑙) and get the condition for 𝑝-value.
𝜋𝑛
sin (𝑝𝑙) = 0 ⇒ 𝑝𝑙 = 𝜋𝑛 → 𝑝 = 𝑝𝑛 = , 𝑛∈Z. (1.46)
𝑙
Then it is possible to take the gotten anzac and plug it into equations of motion. In
this way there will be an equation which will describe a time evolution of the 𝑎𝑝 coefficient:

¨𝑝 + 𝑐2 𝑝2 𝑎𝑝 (𝑡) = 0 .
𝑎 (1.47)

The solution of the (1.47) is well known and equal to:

𝑎𝑝 (𝑡) = 𝑒𝑖𝜔𝑝 𝑡 𝑎𝑝 , (1.48)

where 𝜔𝑝 is equals to:

𝜔𝑝 = ±𝑐𝑝 . (1.49)
For any 𝑝𝑛 the solution of equation of motion can be found, but for general solution
𝜑 (𝑥, 𝑡) it is required to find a sum of all possible 𝑛 values:
∞
∑︁
𝜑 (𝑥, 𝑡) = sin (𝑝𝑛 𝑥) (𝐴𝑛 cos (𝜔𝑛 𝑡) + 𝐵𝑛 sin (𝜔𝑛 𝑡)) , (1.50)
𝑛=0

where sine and cosine replace exponents +𝑖𝜔𝑛 𝑡 and −𝑖𝜔𝑛 𝑡, because 𝜔𝑝 can be positive or
negative due to formula (1.49). That is why it is enough to sum only from 𝑛 = 0 to ∞. It
should be noticed that there is a zero-solution when 𝑛 = 0.
Formula (1.50) has two undetermined coefficients 𝐴𝑛 and 𝐵𝑛 which can be determined
if specify the initial value of the field and it’s first derivative at the time moment 𝑡 = 0.
It should be noticed that the initial conditions must be represented as a sum of cosines
and sines of 𝜔𝑛 𝑡 and then we can equate coefficients between this cosines and sines with
coefficients 𝐴𝑛 and 𝐵𝑛 from (1.50).
Summarizing the previous information, a description of the continuum system in
classical mechanics was made and the expression for the displacement field was found.

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Lecture 2. Tensor Fields, Euler-Lagrange Equations,

Lorentz Transformations

Classical fields in many dimensional case

The concept of the displacement field can be further generalized to the case of multidimensional
systems. For instance, in two dimensions it is possible to consider two-dimensional lattice
of springs. In such a way it is necessary to present a plane of particles with mass 𝑚 and
distance between neighbors ∆𝑎 (fig. 2.1).

Fig. 2.1. A two-dimensional lattice of springs

Then it is necessary to allow longitudinal fluctuations of particles connected with the

springs existence. In this case a natural quantity to discuss is the displacement field 𝜑
depending on coordinates of a point at the equilibrium position. This coordinates can be
written as 𝑖 and 𝑗 for 𝑥 and 𝑦 respectively. In such a way the displacement field 𝜑 will be
a vector with the following notation:
⃗ (𝑖,𝑗) .
𝜑 (2.1)

The continuum approximation of the presented system is called a membrane. In this

case a longitudinal fluctuations of the membrane will be searched.
Further a transmission to three-dimensional case can be completed. For this purpose
it is required to represent the system as it shown on fig. 2.2.
⃗ (𝑖,𝑗,𝑘) coordinates
The displacement field of such system will be a three-dimensional vector 𝜑
of which can be specified by indexes 𝑖, 𝑗 and 𝑘 for 𝑥, 𝑦 and 𝑧 respectively.

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Fig. 2.2. A three-dimensional lattice of springs

Then it is necessary to complete absolutely the same derivation procedure which have
been already done for the case of the one-dimensional system. At the end it is needed
to end up with the following differential equation which describes the evolution of the
⃗ Any component of the displacement field vector will undergo
three-dimensional vector 𝜑.
the following dynamics:

⃗¨ − 𝑐1 𝜕𝑥𝑥 𝜑
𝜑 ⃗ − 𝑐2 𝜕𝑦𝑦 𝜑
⃗ − 𝑐3 𝜕𝑧𝑧 𝜑
⃗ − 𝑐4 𝜕𝑥𝑦 𝜑
⃗ − 𝑐5 𝜕𝑦𝑧 𝜑
⃗ − 𝑐6 𝜕𝑥𝑧 𝜑
⃗=0, (2.2)

where 𝜑¨ is the second time derivative and coefficients 𝑐1 , . . . , 𝑐6 describe a properties of

the observed solid body. It also can be seen that many quantities which appear in the
description are fields depending on space-time variables and this fields behaves themselves
as tensors or more generally tensor fields.
Therefore it is very natural to introduce and discuss the concept of a tensor field from
the beginning.

Tensors and tensor fields

As it well known tensors in any coordinate system represent a set of numbers which
transforms with changing from one coordinate system to another in a definite way.
For tensor fields we are asking about transformation of the set of numbers under general
coordinate transformations. Consider a transformation of coordinates of this type:

𝑥𝜇 → 𝑥′𝜇 (𝑥𝜈 ) . (2.3)

Such transformations are assumed to be invertible and this transformations belong to

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the class which is directly defined as general coordinate transformations. Let’s assume
that the work is done in 𝑑 dimensions. That is why index 𝜇 can take values from 1 to 𝑑.
Then a tensor field
𝜑𝜇𝜈11...𝜈
...𝜇𝑝
𝑞
(𝑥) (2.4)

is a tensor field of rank (𝑝, 𝑞).

Under general coordinate transformations it transforms in the following way. In the
new coordinate system it is described as
𝜕𝑥′𝜇1 𝜕𝑥′𝜇𝑝 𝜕𝑥𝜌1 𝜕𝑥𝜌𝑞 𝜆1 ...𝜆𝑝
𝜑′𝜇1 ...𝜇𝑝 ′
𝜈1 ...𝜈𝑞 (𝑥 ) = . . . . . . 𝜑 (𝑥) , (2.5)
𝜕𝑥𝜆1 𝜕𝑥𝜆𝑝 𝜕𝑥′𝑛𝑢1 𝜕𝑥′𝜈𝑞 𝜌1 ...𝜌𝑞
where over indexes 𝜆1 . . . 𝜆𝑝 and indexes 𝜌1 . . . 𝜌𝑞 are used for summation due to Einstein’s
convention.
Partial derivatives
𝜕𝑥′𝜇
= 𝐽𝜇𝜆 (2.6)
𝜕𝑥𝜆
can be represented as elements of the matrix which is called Jacobian matrix. This matrix
is non degenerate and therefore it is the element of the group 𝐺𝐿 (𝑑, R), where 𝑑 means
dimension and R means that the Jacobian matrix is real. The group 𝐺𝐿 (𝑑, R) is a group
of all invertible 𝑑 × 𝑑 matrices. 𝑑 is a positive integer number which specifies the range of
the coordinates
𝜇, 𝜆, 𝜈, 𝜌 = 1, . . . , 𝑑 . (2.7)

Further restrictions on possible transformations of coordinates can be imposed by

physical requirements. For instance, if you are talking about Galilean invariants then you
restrict general coordinate transformations to that of the rotation group. The transition
from 𝑥 to 𝑥′ in this situation will be only rotation.
Analogously, if you implement Einstein’s relativity principle, which allows only for
Lorentz transformation, than in this case you will restrict the general coordinate transformation
to that of Lorentz transformations.
A simple example of a tensor is given by a scalar. Scalar in this case is an object which
transform under general coordinate transformations according to the formula:

𝜑′ (𝑥′ ) = 𝜑 (𝑥) . (2.8)

The scalar does not have any indexes and therefore the Jacobian matrices do not
appear. It is a tensor of rank (0, 0).

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It’s important to realize that the point in different coordinates can be specified by the
coordinate 𝑥 and 𝑥′ , but physically it is the one point.
Summarizing, the scalar field is a function, which under the change of coordinates
behaves itself in the way written by formula (2.8).
It should be noticed that the scalar field is not something which behaves like (2.8). It is
not true. If you take any function like, for instance, sin 𝑥 and then make a next coordinate
transformation:
𝑥 → 𝑥′2 (2.9)

you will get

sin 𝑥 → sin 𝑥′2 . (2.10)

But, of course, now it is necessary to realize that the written in (2.10) expression will
not the old function depending on 𝑥′ , because if you write

𝜑 (𝑥) = sin 𝑥 (2.11)

and complete a transformation written in (2.9), the 𝜑 (𝑥′ ) will be described by formula:

𝜑 (𝑥′ ) = sin 𝑥′ . (2.12)

That is why it is needed to make transformed in the (2.10) expression equal to 𝜑′ (𝑥′ ).
Next to the scalar field is a vector field, which carries one index up. An example of a
vector field would be, for instance, a velocity vector which can be also made as a vector
field. For instance, if you discuss the flow of a liquid in hydrodynamics, then you need
tensor field of velocities of volume elements of the liquid.
There are also some examples of a co-vector, which means a tensor with a lower index.
This tensor would be a gradient of a function. More generally you might have, for instance,
symmetric on the manifold which would be a tensor with two lower indices and so on. It
will also be seen in the development of the course, a few other examples of tensor fields.

Action principle and Euler-Lagrange equations

Then it is necessary to return back to the action principle in general. Action principle
that was already discussed plays an important role in describing of the dynamics.
In the action principle the work takes place with with the functional 𝑆, which is a
functional of field 𝜑. This functional has the following form:
∫︁
𝑆 [𝜑] = d𝑥L (𝑥, 𝜑𝑖 , 𝜕𝜇 𝜑𝑖 ) , (2.13)

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where the d𝑥 can be defined as

d𝑥 := d𝑥0 d𝑥1 . . . d𝑥𝑑−1 , (2.14)

where 𝑑 is a number of space-time dimensions, where 𝑥0 signifies the time direction.

The 𝑥0 element can be represented by the next formula:

𝑥0 = 𝑐𝑡 , (2.15)

where 𝑐 is the speed of light.

Other directions such as 𝑥1 , . . . , 𝑥𝑑−1 are spatial directions.
Finally, action is an integral over 𝑑 dimensional space-time coordinates of the Lagrangian
density, which is considered in general to be a function of 𝑥, fields 𝜑𝑖 , and the first
derivatives.
It should be noticed that Lagrangian density depends on only first derivatives because
non-degenerate or what is called unitary theories is considered, where equations of motion,
which would be derived from this section, contain only second derivatives in time of the
field 𝜑 and second derivatives in the spatial directions. The theories with derivatives of
the third and higher orders of the field 𝜑 are much more complicated in the classical and
especially the quantum case.
An example of the theory limited by the second order is a very natural generalization
of the Newtonian mechanics, because in Newtonian mechanics deriving equations have a
second order in time. It means that to specify the trajectory of a particle it is needed
to specify the initial coordinate and initial velocity and it is enough to solve the Koshi
problem for such dynamical system which linked with the fact that a unique solution is
needed. This principle was inherited from standard Newtonian mechanics, and our goal
is to apply it to fields as well.
Further, it will also be required that there is no explicit dependence on 𝑥. All dependence
on 𝑥 will come only through the field 𝜑 and its derivatives:
∫︁
𝑆 [𝜑] = d𝑥L (𝜑𝑖 , 𝜕𝜇 𝜑𝑖 ) , (2.16)

where the index 𝑖 has double meaning. It will combine labels of different fields when the
discussion will be about fields of different nature and also it will be uniform label for
all possible tensor indices of one field. In formula (2.16) it is possible to consider the
Lagrangian not only for a scalar field, but, for instance, Lagrangian for a vector field, as it
can be, for instance, in electrodynamics where the electromagnetic field is a vector. Also

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by index 𝑖, for example, it is possible to mean the index of a vector of the one vector field,
but it is also possible to mean a collection of scalars by index 𝑖.
From the action 𝑆 described by formula (2.16), equations of motion are derived by
using the principle of the least action. In fact, an example how this is done with the help
of variational principle where variational derivative was used has already been seen. As it
was described in the previous lecture the variational derivative is a variation of the action
with respect to variation of the field 𝜑:
(︂ )︂
𝛿𝑆 𝜕L d 𝜕L
𝜀𝑖 = = − 𝜇 =0. (2.17)
𝛿𝜑𝑖 𝜕𝜑𝑖 d𝑥 𝜕 (𝜕𝜇 𝜑𝑖 )

where the gotten expression that is known as Euler-Lagrange equations. Solving of the
Euler-Lagrange equations it is possible to find the actual dynamical trajectories of the
system.
Then it is necessary to differentiate the Lagrangian derivative with respect to 𝑥𝜇 in
an explicit way by taking into account that Lagrangian density is a function of 𝜑𝑖 and its
derivative. It is possible to write for this Lagrangian derivative:

𝜕L 𝜕 2L 𝜕 2L
− 𝜑𝑗,𝜇 − 𝜑𝑗,𝜇𝜈 = 0 , (2.18)
𝜕𝜑𝑖 𝜕𝜑𝑖,𝜇 𝜕𝜑𝑗 𝜕𝜑𝑖,𝜇 𝜕𝜑𝑗,𝜈

where
𝜕𝜑𝑖
𝜑𝑖,𝜇 = . (2.19)
𝜕𝑥𝜇
Then it is important to realize that in fact Lagrangian is not uniquely defined, which
leads to the one and the same equations of motion. It is practicable to have different
Lagrangians, which lead to the one and the same equations of motion. This acts because
the Lagrangian density may always be changed by adding to it total derivative:

L → L + 𝜕 𝜇 Λ𝜇 , (2.20)

where Λ𝜇 is considered as a function

Λ𝜇 = Λ𝜇 (𝜑𝑖 ) . (2.21)

which depends on 𝜑𝑖 but not of their derivatives, because if you also assume that the
Λ𝜇 depends on the derivative of 𝜑𝑖 , then it would lead to the appearance of the second
derivative of 𝜑 in the Lagrangian. This is something which was prohibited because it was
required that the Lagrangian depends on the first derivative of 𝜑.

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Then it is necessary to find out why the equations of motion will not be changed under
the addition. This acts because of the next expression:

𝜇 𝜕Λ𝜇
𝜕𝜇 Λ = 𝜕𝜇 𝜑 , (2.22)
𝜕𝜑
where it was considered that we deal with one scalar field 𝜑.
Then it is possible to trace how under this addition the Lagrangian derivative changes.
It is known that Lagrangian derivative is given by expression (2.17). If we want to observe
how the Lagrangian derivative changes, we need to substitute the additional term instead
of Lagrangian in (2.17) and see what happens. This means that the Lagrangian derivative
will get an extra contribution depending on Λ:

𝜀 → 𝜀 + 𝛿𝜀 (Λ) , (2.23)

where the extra contribution 𝛿𝜀 (𝐿) will be given by:

𝜕 𝜕Λ𝜆 𝜕2
(︂ 𝜆 )︂
𝜕2
(︂ )︂ (︂ 𝜆 )︂
𝜕Λ 𝜕Λ
𝛿𝜀 (𝐿) = 𝜕𝜆 𝜑 − 𝜑,𝜆 𝜑,𝜇 − 𝜑,𝜆 𝜑,𝜇𝜈 = 0 . (2.24)
𝜕𝜑 𝜕𝜑 𝜕𝜑𝜕𝜑,𝜇 𝜕𝜑 𝜕𝜑,𝜇 𝜕𝜑,𝜈 𝜕𝜑
It can be seen from (2.24) that the additional contribution 𝛿𝜀 (𝐿) is zero by itself. First
of all, if you look at the last term in (2.24), you will see that the Λ𝜆 depends on 𝜑 we
have its first derivative. If we differentiate the expression in the brackets
(︂ 𝜆 )︂
𝜕Λ
𝜑,𝜆 (2.25)
𝜕𝜑
with respect to derivatives, since we have two derivatives before brackets, the result of
the differentiation will be zero, because of the fact that expression in the brackets involve
the first derivative of 𝜑 only once. Remaining two terms will cancel each other and it
can be checked by opening the brackets and evaluating of the derivatives. Finally, the
contribution to the Euler-Lagrangian equations from the total derivative presented in
(2.24) is actually zero. That is why an additional term does not contribute to equations
of motion.

Lorentz and Poincare groups

After the general remarks presented in previous section it is now possible to come to
the discussion of Lorentz and Poincare groups.
As it was already mentioned in the first lecture, symmetries play an extremely important
role in the description of nature. For instance, the translational symmetry implies that if

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we know physical loss at one space-time point, then we know them at any other which
means that in fact, the physical losses are not random from point to point. Similar remarks
concern, for instance, rotational symmetry, which means that losses in different directions
can be related to each other. Among all possible symmetries, there is one which has a
universal character and this is Poincare symmetry, which combines Lorentz symmetry or
Lorentz transformations with shifts of space-time coordinates.
From mathematical point of view, these transformations have a structure of a Lie
group. First of all, before describing of the Poincare symmetry, it is needed to introduce
the notion of the Lie group which will be defined as 𝐺. The group 𝐺 is a set of elements
of any nature, which satisfies the following set of axioms:

1) For any two elements 𝐺1 and 𝐺2 , which belong to group 𝐺, one can define their
product, 𝐺1 · 𝐺2, which also belongs to the group 𝐺. The product of 𝐺1 and 𝐺2 is
associative, which means that it does not matter how to put the brackets:

(𝑔1 · 𝑔2 ) · 𝑔3 = 𝑔1 · (𝑔2 · 𝑔3 ) . (2.26)

2) Group 𝐺 must include a unit element 𝑒 such that:

𝑔·𝑒=𝑒·𝑔 =𝑔 , (2.27)

where a unit element can be also called as identity.

3) For any element 𝑔 from the group 𝐺, there exists an inverse element 𝑔 −1 such that:

𝑔 · 𝑔 −1 = 𝑔 −1 · 𝑔 = 𝑒 . (2.28)

An important class of groups constitute Lie groups. A smooth manifold 𝐺 of dimension

𝑛 is called Lie group, if 𝐺 is supplied with a structure of a group and compatible with a
structure of a smooth manifold, which means that the group operations are smooth.
A physical important example of Lie group, which can be well known from courses
on classical mechanics is a rotation group. The rotation group is the group of rotational
matrices in three dimensions. The explicit discussion of this group will be held a little bit
later, because it is a subgroup of the Lorentz group.
Another example, which is primarily important, is a Lorentz group. Now consider a 𝑑
dimensional Minkowski space R1,𝑑−1 . The Lorentz transformations by definition are linear
coordinate transformations of Minkowski space of the form:

𝑥′𝜇 = Λ𝜇𝜈 𝑥𝜈 , (2.29)

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where over index 𝜈 there is a summation in the Einstein’s term.

These transformations of a Minkowski space are such, that they preserve the interval
between two events in a Minkowski space. The infinitesimal relativistic interval 𝑑𝑠 is
defined through the quadratic form:
(︀ )︀2 (︀ )︀2 (︀ )︀2
𝑑𝑠2 = 𝜂𝜇𝜈 d𝑥𝜇 d𝑥𝜈 = 𝑐2 d𝑡2 − d𝑥1 − d𝑥2 − d𝑥3 , (2.30)

where 𝜂𝜇𝜈 is Minkowski metric, which is a diagonal matrix and indexes 𝜇 and 𝜈 are running
in general from 0 to 3, if the four dimensional Minkowski space is discussing:

𝜂𝜇𝜈 = diag (+1, −1, −1, −1) , 𝜇 , 𝜈 = 0, 1, 2, 3 . (2.31)

Now it can be seen that 𝑥𝜇 is a vector with components 𝑥0 , 𝑥1 , 𝑥2 , 𝑥3 :

𝑥𝜇 = 𝑥0 , 𝑥 1 , 𝑥 2 , 𝑥 3 .
(︀ )︀
(2.32)

If you treat 𝑥𝜇 as a vector, the Lorentz transformations can be written in the matrix
form
𝑥′ = Λ𝑥 , (2.33)

cover vector 𝑥 and it transforms passing from one Lorentz frame into another, simply with
matrix Λ like:
Λ = |Λ𝜇𝜈 | , 𝜇, 𝜈 = 0, 1, 2, 3 . (2.34)

The requirement of Λ to preserve the quadratic form is an explicit form written in the
following way:
𝜂𝜇𝜈 Λ𝜇 𝛼Λ𝜈𝛽 = 𝜂𝛼𝛽 . (2.35)

In other words, Lorentz transformations are transformations which leave our Minkow’s
metric invariant. Formula (2.35) defines the class of matrices Λ, which can be associated
with Lorentz transformations.
The transformation from formula (2.35), which preserves Minkowsky metric can also
be written in the matrix form:
Λ𝑡 𝜂Λ = 𝜂 , (2.36)

where 𝜂 and Λ is a four by four matrix.

Definition presented in (2.36) is a generalization of the standard condition

Λ𝑡 Λ = 1 , (2.37)

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which is known for orthogonal matrices representing the rotation group. Because of the
fact that Minkowsky metric is used in expression (2.36), the condition in (2.36) can be
called as pseudoorthogonality condition arising from the fact that now the work with
Minkowsky space takes place rather than the work with Euclidean 𝑑 dimensional space.
It can be showed that matrices, which satisfies condition (2.36), form a group. It is clear
that as a product in this group, it is possible to consider the usual product of matrices.
It should be also noticed that condition (2.36) works only for Minkowsky space. It
is a condition which is directly related with the definition of something that is meant
by Lorentz transformations. Lorentz transformations are transformations which preserve
Minkowsky metric 𝜂. It is also practicable to define a group which preserves this metric
in general relativity, or in special relativity, such a group is called a group of isometrics.
Group of isometrics is a group which preserves a given metric. That is why it is possible to
say in this respect that the Lorentz group is a group of isometrics of Minkowsky metric. If
we want a different constant metric instead of 𝜂, it is possible to use, if we have a different
manifold, not Minkowsky space. Then it is necessary to consider what a different space
with a different metric and also define a group of isometrics in a similar way, which was
completed before.
Let’s take two matrices, Λ1 and Λ2 , which satisfy the condition (2.36) and then consider
the product of these two matrices. It is needed also to show that the condition (2.36) works
for Λ1 and Λ2 : ⎧
⎨Λ𝑡 𝜂Λ1 = 𝜂
1
(2.38)
⎩Λ𝑡 𝜂Λ = 𝜂
2 2

meaning that the expression

(Λ1 Λ2 )𝑡 𝜂Λ1 Λ2 = 𝜂 (2.39)

will be satisfied.
Expression (2.39) would be also a Lorentz transformation. It is easy to proof that
(2.39) is right:
Λ𝑡2 Λ𝑡1 𝜂Λ1 Λ2 = Λ𝑡2 𝜂Λ2 = 𝜂 . (2.40)
⏟ ⏞
=𝜂

Finally, from (2.40) it is clearly seen that the product of two Lorentz transformations
is a Lorentz transformation. Identity matrix would be a trivial Lorentz transformation,
which is a identity matrix. Then it is needed to show that any Λ has an inverse, which
is also a Lorentz transformation. To complete this it is possible to take and compute the

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determinant of both sides:

det Λ𝑡 𝜂Λ = (det Λ)2 det 𝜂 ,

(︀ )︀
(2.41)

where the rule that under transposition the determinant is unchanged was used.
On the other hand, by the definition of Lorentz transformation, expression (2.41) is the
same as the determinant of 𝜂. The determinant of 𝜂 is not equal to zero by the definition
and equal to minus one. That is why

(det Λ)2 = 1 . (2.42)

From formula (2.42), it was concluded that all matrices of Lorentz transformations
always have
det Λ = ±1 . (2.43)

But this means that matrix of Lorentz transformation is non-degenerate and therefore
it is invertible. The inverse is easy to find from the defining relation of the Lorentz
transformations:

Λ𝑡 𝜂Λ = 𝜂 ⇒ Λ𝑡 𝜂 = 𝜂Λ−1 ⇒ 𝜂Λ𝑡 𝜂 = 𝜂 2 Λ−1 = Λ−1 . (2.44)

Finally, we got that

Λ−1 = 𝜂Λ𝑡 𝜂 . (2.45)

Therefore to show that Λ−1 is also Lorentz transformation, it is necessary to check the
following:
)︀𝑡
Λ−1 𝜂Λ−1 = 𝜂 .
(︀
(2.46)

It may be done by substituting the expression for Λ−1 :

(︀ −1 )︀𝑡 −1 (︀ 𝑡 )︀𝑡
Λ 𝜂Λ = 𝜂Λ 𝜂 𝜂𝜂 Λ𝑡 𝜂 = 𝜂 Λ𝜂Λ𝑡 𝜂 = 𝜂 𝜂𝜂 = 𝜂 , (2.47)
⏟ ⏞ ⏟ ⏞ ⏟ ⏞
1 𝜂 1

where the statement that

Λ𝜂Λ𝑡 = 𝜂 (2.48)

was used. Expression (2.48) follows from the defining relation

Λ𝑡 𝜂Λ = 𝜂 (2.49)

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Expression (2.48) means that Λ𝑡 is also a matrix of Lorentz transformation. By this

moment it is clear that if Λ is a matrix of Lorentz transformations, Λ𝑡 is also a matrix of
Lorentz transformation. And this can be gotten again from the same expression for Λ−1 :

Λ−1 = 𝜂Λ𝑡 𝜂 | Λ · (2.50)

1 = Λ𝜂Λ𝑡 𝜂 | · 𝜂 (2.51)

𝜂 = Λ𝜂Λ𝑡 𝜂 2 = Λ𝜂Λ𝑡 . (2.52)

In other words, we got that if Λ is a Lorentz transformation, is a Lorentz transformation,

then the matrices Λ−1 , Λ𝑡 and Λ𝑡 −1 are matrices of Lorentz transformations. According
to definition of the group, Lorentz transformations for described group can be associated
with the usual matrix multiplication.
As can be seen, defining the relation of the Lorentz group implies the following
important statement: if the condition of preservation of the Lorentz metric is taken and
the low index of Λ is specified to be zero, then the following will be obtained:
(︀ )︀2 (︀ )︀2
𝜂𝜇𝜈 Λ𝜇0 Λ𝜈0 = Λ00 − Λ𝑖0 = 𝜂00 = 1 . (2.53)

From expression (2.53) can be seen that:

(︀ 0 )︀2 (︀ )︀2
Λ0 = 1 + Λ𝑖0 ≥ 1 . (2.54)

Expression in (2.54) is always bigger or equal to one, because a non-negative contribution

2
(Λ𝑖0 ) is added to one. Therefore there are two possibilities for the component Λ00 :

Λ00 ≥ 1 𝑜𝑟 Λ00 ≤ −1 . (2.55)

The correspondent Lorentz group is a six dimensional non-compact Lie group whose
mathematical name is 𝑂 (1, 3), where 𝑂 stands for orthogonal. This group consists of four
topologically separated spaces.
Now it is possible discuss the following statements. First of all, Lorentz transformations
may reverse the direction of time or not reverse the direction of time, e.g. keep the direction
of time. In other words, they can transform a future point in time like vector into a past
point in one. In other words, it’s related to the effect if condition (2.55) completes.
The second thing is that Lorentz transformation reverse or not reverse the four dimensional
reference frame. This related to the effect if

det Λ = 1 𝑜𝑟 det Λ = −1 . (2.56)

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We have two operations, which are also Lorentz transformations, and there are discrete
operations. One of them is parity and another is time reversal. Time reversal operation
will reverse direction of time and the parity will change the orientation of the frame
and transform the matrices with a determinant equal to one with two matrices with
determinant equal to minus one and vice versa.
Λ’s for which Λ00 ≥ 1 preserve the direction of time. For this reason they are called
orthochronous. The product of two orthochronous transformations is also an orthochronous
transformation. It can be seen easily if the following is noticed
⃒ 0⃒ ⃦ 𝑖 ⃦
⃒Λ0 ⃒ > ⃦Λ0 ⃦ . (2.57)

This implies that modulus of Λ00 is bigger than the norm of the vector Λ𝑖0 .
Now Λ𝑖0 is a 3-dimensional vector and then ‖Λ𝑖0 ‖ is the square of its norms and it’s
naturally follows from the relation (2.54).
Analogously, if in the relation (2.54) Λ → Λ𝑡 is changed, which is also Lorentz
transformation, the following will be obtained:
⃒ 0⃒ ⃦ 0⃦
⃒Λ0 ⃒ > ⃦Λ𝑖 ⃦ . (2.58)

Therefore, if the product of two Lorentz transformations Λ and Λ′ is taken, it will be:

(ΛΛ′ ) 00 = Λ00 Λ′00 + Λ0𝑖 Λ′𝑖0 . (2.59)

Now if the Cauchy-Bonyakowski-Schwarz inequality is used, which essentially tells that

for any two vectors 𝑥 and 𝑦 the following is found:

|(𝑥, 𝑦)| ≤ ‖𝑥‖ ‖𝑦‖ . (2.60)

It will be obtained that:

⃒ 0 ′𝑖 ⃒ ⃦ 0 ⃦ ⃦ ′𝑖 ⃦
⃒Λ𝑖 Λ 0 ⃒ ≤ ⃦Λ𝑖 ⃦ ⃦Λ 0 ⃦ . (2.61)

Therefore, if it is assumed that Λ00 and Λ′00 are both positive, it will be concluded that:

(ΛΛ′ ) 00 > 0 , (2.62)

which means that Λ00 must be bigger or equal to one, because any Lorentz transformation
with positive Λ00 must have the property that it’s zero-zero component is bigger or equal
to one.

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In other words, if we have two orthochronous transformations, then the product of

these two orthochronous transformations is also an orthochronous transformation, which
means that orthochronous transformations by itself form a group.
It can be also shown that the inverse of the orthochronous transformation is an
orthochronous transformation. And this group or subgroup of the Lorentz group, which
consists of orthochronous transformations is denoted as 𝑂+ (1, 3), where plus is related to
the fact that we preserve the direction of time.
Now concerning the determinant, the Lorentz transformations, which preserve orientation
are called proper. These are transformations with determinant equals to +1. The transformations
with determinant equals to -1 are called improper.
The proper Lorentz transformations form a subgroup of the Lorentz group, which is
denoted as 𝑆𝑂 (1, 3), where the letter 𝑆 as usual means special.
Now it is possible to combine these two properties of being orthochronous and being
proper and get a subgroup, which is called a restricted Lorentz group. A restricted Lorentz
group denoted as 𝑆𝑂+ (1, 3) and consists of proper orthochronous Lorentz transformations.
Actually identity matrix is a proper orthochronous Lorentz transformation.
Now it is possible to get the picture of how topologically the Lorentz group looks like:

1) A component which consists of all Lorentz transformations, which have a Λ00 ≥ 1,

and which have det Λ = 1. This is the restricted Lorentz group 𝑆𝑂+ (1, 3).

2) A component which contain Λ00 ≤ −1 and det Λ = −1. This component presents
improper transformations, inverting the direction of time. The passage from restricted
Lorentz group to this component is done by application of what is called time
reversal. Time reversal is a particular Lorentz transformation, which looks like:

𝑇 = diag (−1, 1, 1, 1) . (2.63)

3) A component which contain Λ00 ≥ 1 and det Λ = 1. This component is obtained

from the proper orthochronous subgroup by means of application, what is called
parity or space inversion denoted by 𝑃 :

𝑃 = diag (1, −1, −1, −1) . (2.64)

As can be seen from (2.64), parity changes orientation of the three spatial coordinate
axes by multiplying them by minus one.

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4) A component which contain Λ00 ≤ −1 and det Λ = 1. That is why all transformations
of the component are not orthochronous. The component is obtained from restricted
Lawrence subgroup by means of combined application of a space inversion with
the time inversion. Combined application keeps the determinant equals to one, but
since the time reversal is employed, then what will be obtained are matrices of
non-orthochronous Lawrence transformations.

Discrete operations, which relates this topologically distinguished or topologically

different components of the Lawrence group, are the following discrete operations:

{𝑒, 𝑃, 𝑇, 𝑃 𝑇 } . (2.65)

Presented discrete transformations also form what is called a discrete group or discrete
subgroup of the Lawrence group 𝑂 (3, 1), which allows the movement among these different
topologically different components.
Described list of components suggests that only component with Λ00 ≥ 1, and det Λ = 1
here is an actual subgroup of the Lawrence group. Other components, if we take them by
themselves, do not form a subgroup. There is only one subgroup and this subgroup is a
subgroup of proper orthochronous Lawrence transformations because only this component
contains an identity group. The connected subgroup of the Lorentz group of dimension
six is 𝑆𝑂+ (1, 3).
The Lorentz group is six dimensional Lie group because of the fact that Lawrence
transformations combine three rotations around three coordinate axes: 𝑥, 𝑦 and 𝑧. In
addition, there are three boosts, which involve one special direction and the time direction.
If you have a four dimensional Minkowsky space, the picture of possible transformations
will look like presented on fig. 2.3.
Then it is necessary to remind how any rotation can be parameterized. It is possible
to do it by specifying the rotation axis ⃗𝑛 and the rotation can be done by specifying a
unit vector. If ⃗𝑛 is a unit vector, it is also needed to specify a rotation angle around this
unit vector, and the rotation angle will be 𝜃.
If the following three by three matrices are defined
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 0 0 0 0 1 0 −1 0
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
𝑎1 = ⎝0 0 −1⎠ , 𝑎2 = ⎝ 0 0 0⎠ , 𝑎3 = ⎝1 0 0⎟
⎜ ⎟ ⎜ ⎟ ⎜
⎠ , (2.66)
0 1 0 −1 0 0 0 0 0

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Fig. 2.3. The picture of possible transformations of a four dimensional Minkowsky space

Fig. 2.4. The rotation axis ⃗𝑛 and the rotation angle 𝜃

then such a transformation is described by a matrix

𝑅 (⃗𝑛, 𝜃) = 𝑒𝜃⃗𝑛⃗𝑎 , (2.67)

where ⃗𝑎 comprises three matrices 𝑎1 , 𝑎2 , 𝑎3 :

⃗𝑎 = (𝑎1 , 𝑎2 , 𝑎3 ) . (2.68)

𝑅 (⃗𝑛, 𝜃) is a three by three matrix of rotations specified by the direction ⃗𝑛 and the
angle 𝜃. It is possible to compute this matrix quite explicitly. In fact, the scalar product
⃗𝑛⃗𝑎 can be written as:
(⃗𝑛⃗𝑎)𝑖𝑗 = −𝜀𝑖𝑗𝑘 𝑛𝑘 . (2.69)

Then it is practicable to exponentiate the product simplified by formula (2.69) and

this gives us an explicit three by three matrix with the following matrix elements. After
computation of this matrix elements the following expression will be obtained

𝑅𝑖𝑗 (⃗𝑛, 𝜃) = cos 𝜃 𝛿𝑖𝑗 + (1 − cos 𝜃) 𝑛𝑖 𝑛𝑗 − sin 𝜃 𝜀𝑖𝑗𝑘 𝑛𝑘 . (2.70)

It can be easily checked that the matrix 𝑅 is orthogonal. It satisfies the properties
that
𝑅𝑡 · 𝑅 = 𝑅 · 𝑅𝑡 = 1 . (2.71)

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It means that this matrix is an element of the group of rotation 𝑆𝑂 (3).

It also should be said that
−𝜋 <𝜃 ≤𝜋 , (2.72)

which covers the whole interval 2𝜋.

But in fact, it is possible to restrict the rotation angle to run from 0 to 𝜋, because
the negative values of 𝜃 from −𝜋 to 0 are equivalent to changing the direction of vector
⃗𝑛 to the opposite. The vector ⃗𝑛 can be allowed to be an identity or a unit vector, which
actually runs over 𝜃 or run over the two dimensional sphere.
Very often people use this model for a topological description of the rotation group
(fig. 2.5). They say that the angle can be identified with the lens of the vector ⃗𝑛. This
vector is not a unit one, but the wall of a radius 𝜋 is felt. Any point inside this wall of
radius 𝜋 will represent a rotation in the direction, specified by the line connecting a zero
point with a point inside the ball. The rotation angle will be given by the lens of the
interval connecting this point with the origin. A bit non-trivial in this description of the
rotations is that diametrally opposite points on a sphere should be identified because of
the fact that a rotation on angle 𝜋 and −𝜋 will give the same rotation.

Fig. 2.5. The model for a topological description of the rotation group

Now it is possible to embed the rotation group into the Lorenz group and embedding
is done in the following way:
(︃ )︃
1 0
Λ (𝑛, 𝜃) = , (2.73)
0 𝑅 (⃗𝑛, 𝜃)

where the time direction is untouched and represented by 1.

It also can be checked about the matrix 𝑅 that it satisfies the property that

𝑅 (⃗𝑛, 𝜃 + 2𝜋) = 𝑅 (⃗𝑛, 𝜃) . (2.74)

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Boosts
Now let’s come to boosts. Boosts will be represented by four by four matrices:
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 −1 0 0 0 0 −1 0 0 0 0 −1
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜−1 0 0 0⎟ ⎜ 0 0 0 0⎟ ⎜0 0 0 0⎟
⎟ ⎜ ⎟ ⎜
𝑏1 = ⎜ ⎟ , 𝑏1 = ⎜ ⎟ , 𝑏1 = ⎜ ⎟ . (2.75)
⎜ ⎟
⎜0 0 0 0⎟ ⎜−1 0 0 0⎟ ⎜0 0 0 0⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 0 0 0 0 0 0 0 −1 0 0 0

Now there will be rotation, but around what is called hyperbolic angle. The corresponding
Lorentz transformations are now specified again by the axis around which the rotation is
happened and by the rotation angle, which is called 𝜗:

⃗
Λ (⃗𝑛, 𝜗) = 𝑒𝜗⃗𝑛𝑏 , (2.76)

where ⃗𝑏 is equal to:

⃗𝑏 = (𝑏1 , 𝑏2 , 𝑏3 ) . (2.77)

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Lecture 3. Lorentz and Poincare Groups, Noether’s

Theorem

Repeating of the rotation group

At the last discussion it was considered that any rotation can represented by the
rotation 𝑅 (⃗𝑛, 𝜃) given by the formula:

𝑅 (⃗𝑛, 𝜃) = 𝑒𝜃⃗𝑛·⃗𝑎 . (3.1)

If the corresponding exponent is computed, then a rotation matrix will be obtained

𝑅𝑖𝑗 (⃗𝑛, 𝜃) = cos 𝜃 𝛿𝑖𝑗 + (1 − cos 𝜃) 𝑛𝑖 𝑛𝑗 − sin 𝜃𝜀𝑖𝑗𝑘 𝑛𝑘 , (3.2)

where 𝑖 and 𝑗 runs from 1 to 3.

It can be easily checked that the matrix 𝑅 is orthogonal, which means that

𝑅𝑡 𝑅 = 𝑅𝑅𝑡 = 1 . (3.3)

Also, the matrix 𝑅 is periodic:

𝑅 (⃗𝑛, 𝜃 + 2𝜋) = 𝑅 (⃗𝑛, 𝜃) . (3.4)

As can be seen, 𝑅 represents three dimensional rotations and embed into a six dimensional
Lawrence group in the following way:
(︃ )︃
1 0
Λ (⃗𝑛, 𝜃) = , (3.5)
0 𝑅 (⃗𝑛, 𝜃)

where Λ is a matrix of Lorentz transformation.

Lorentz boosts
When we speak about Lorentz boosts, it is necessary to consider a similar parametrization
with the rotation group, but instead of three by three matrices it is needed to consider
four by four matrices 𝑏𝑖 which have a form described in previous lection.
Exponentiating these matrices, but now multiplied with the angle 𝜗, will give the
matrix of Lawrence boosts:
⃗
Λ (⃗𝑛, 𝜗) = 𝑒𝜗⃗𝑛𝑏 . (3.6)

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If the corresponding exponentiation is performed, the following matrix will be found

(︃ )︃
cosh 𝜗 −𝑛𝑖 sinh 𝜗
Λ (⃗𝑛, 𝜗) = , (3.7)
−𝑛𝑖 sinh 𝜗 𝛿𝑖𝑗 (cos 𝜗 − 1) 𝑛𝑖 𝑛𝑗

where Λ now it is parameterized by the hyperbolic angle and the name hyperbolic is
related with the hyperbolic functions: cosh and sinh.
It is also should be noticed that the hyperbolic angle 𝜗 is not running from 0 to 2𝜋,
but it runs from −∞ to +∞. Therefore, 𝜗 is usually called rapidity because it is related to
a velocity with which we boost the Lorentz system with respect to the reference system.
It can be seen that cosh 𝜗 is always bigger or equal to one

cosh 𝜗 ≥ 1 . (3.8)

By the property, it is a variable which takes values from 1 to +∞ and, for this reason,
it is possible to parameterize it as
1
cosh 𝜗 = √︁ , (3.9)
𝑣2
1− 𝑐2

where 𝑣 is a three dimensional velocity vector and 𝑐 is the speed of light. In other words,
the rapidity is related to velocity by means of formula (3.9).
Now, the transformations, as can be seen from (3.9), are now correspond to the
orthochronous transformations. Clearly this means that the component Λ00 of the Lawrence
transformation is bigger than 0, which means that the work takes place with orthochronous
Lorentz transformations.
Equation (3.9) can be solved for the modulus of velocity and if we do that, the following
answer will be found
|⃗𝑣 | = ±𝑐 tanh 𝜗 . (3.10)

Let’s take from (3.10) the expression with the plus sign. If the plus sign is selected, it
is necessary to restrict the variable 𝜗 to run from 0 to +∞ because in this case tanh will
be positive and we will get on the right side the non negative expression.
In this case, if this choice is made, it will be seen that the variable sinh 𝜗 will be given
by
|⃗𝑣 |
sinh 𝜗 = √︁ . (3.11)
⃗𝑣 2
𝑐 1 − 𝑐2

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Now it is possible to use the parametrization of cosh and sinh in the formula for
Lorentz transformation
⎛ ⎞
+
√︁ 1 − √︁ 1 ⃗𝑣2 ⃗𝑣𝑐
2
⎜ 1− ⃗𝑣2 1− 2 ⎟
𝑐 𝑐
Λ (⃗𝑣 ) = ⎜ (︂ )︂ ⎟ , (3.12)
1+
1 ⃗𝑣 ⃗𝑣 ⊗⃗𝑣 ⎠+
− √︁ √︁ 1 −1
⎝
1− ⃗𝑣2
2 𝑐 2
1− ⃗𝑣2 ⃗𝑣 2
𝑐 𝑐

where Λ is four by four matrix and depends on ⃗𝑣 , which represented in (3.12) as a column
of size 3. It also should be noticed that ⃗𝑣 + is a row of size 3.
Variable 𝑣 in formula (3.12) is the velocity with which a moving frame is boosted with
respect to the original inertial frame.
The expression ⃗𝑣 ⊗ ⃗𝑣 + means that ⃗𝑣 is multiplied by transposed ⃗𝑣 in terms of the
tensor multiplication ⎛ ⎞
𝑣1
+
⎜ ⎟ (︁ )︁
⃗𝑣 ⊗ ⃗𝑣 = ⎜
⎝𝑣2 ⎠
⎟ ⊗ 𝑣1 𝑣2 𝑣3 , (3.13)
𝑣3
where for an element with indexes 𝑖𝑗 we will get

⃗𝑣 ⊗ ⃗𝑣 + 𝑖𝑗 = 𝑣𝑖 𝑣𝑗 .
(︀ )︀
(3.14)

Now let’s formulate the general form of Lorentz transformations. A four dimensional
vector X can be transformed to a vector X′ by a rotation matrix and also by a matrix of
Lawrence boost:
X′ = Λ (⃗𝑛, 𝜃) Λ (⃗𝑣 ) X. (3.15)
⏟ ⏞
generic Lorentz transformation

From formula (3.15) there are three parameters related to rotations and three parameters
in the velocity vector related to Lawrence boost. All together this parameters form a six
parametric group of transformations.
If expression (3.15) is written more explicitly, it will be obtained that
⎛ ⎞ ⎛ ⎞
′
𝑐𝑡 ⎛ ⎞ 𝑐𝑡
√︁ 1 √︁ 1 ⃗𝑣 +
⎜ ⎟ (︃
⎜𝑥1′ ⎟
)︃ − ⎜ ⎟
1 0 ⎜ 1− ⃗𝑣𝑐2 2
1− ⃗𝑣2
𝑐
2 𝑐 ⎟ ⎜ 𝑥1 ⎟
⎜
⎜ 2′ ⎟ = ⎟ . (3.16)
⎜ ⎟ ⎜ (︂ )︂ ⎟⎜ ⎟
1 ⃗𝑣 ⊗⃗𝑣 ⎠ ⎜𝑥2 ⎟
+
⎜𝑥 ⎟ 0 𝑅 (⃗𝑛 , 𝜃) − √︁ 1 ⃗𝑣 √︁ 1 −
⎝
+ 1 2
1− ⃗𝑣2 𝑐 ⃗𝑣
2 2
1− ⃗𝑣2
⎝ ⎠ ⎝ ⎠
𝑥3′ 𝑐 𝑐
𝑥3

If expression (3.16) is written for components by multiplication of the vector on

the right side by two matrices, it will be possible to derive the transformation loss for

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individual components of the four dimensional vector. For instance, for 𝑡′ it will be found
that 𝑡′ is related to the original coordinates by the following formula

′ 𝑡 − (⃗𝑥𝑐·⃗2𝑣)
𝑡 = √︁ . (3.17)
2
1 − ⃗𝑣𝑐2

The three dimensional vector ⃗𝑥′ is related to the original vector ⃗𝑥 and time 𝑡 by the
formula ⎡ ⎛ ⎞ ⎤
⃗𝑣 𝑡 1 ⃗𝑣 · (⃗𝑣 · ⃗𝑥) ⎦
⃗𝑥′ = 𝑅 (⃗𝑛, 𝜃) ⎣⃗𝑥 − √︁ + ⎝ √︁ − 1⎠ . (3.18)
2
1 − ⃗𝑣𝑐2
2
1 − ⃗𝑣𝑐2 ⃗𝑣 2

We may now recognize that expressions in (3.17) and (3.18) are standard formulas
of special relativity. (3.17) represents a transformation of time coordinate 𝑡, and (3.18)
represents a transformation of the spatial coordinates 𝑥 under Lawrence transformations
governed by the rotation parameters, 𝜃 and by the velocity vector. These formulas are
standard formulas which may be recognized from the course of special relativity.
As an exercise formulas can be used, for instance, to derive the law of addition of
velocities in special relativity. For that it is necessary to consider 2 successive Lawrence
transformations, which are boosts for velocities 𝑣 1 and 𝑣 2 , and use this formulas in order
to see how the velocity add in special relativity.
Now, it is possible to extend the Lorentz group to the Poincare group by including
shifts of space time coordinates of the following form

𝑥𝜇 → 𝜉 ′𝜇 = 𝑥𝜇 + 𝑎𝜇 , (3.19)

where 𝑎𝜇 is a constant shift which is a constant four dimensional vector.

Lie algebra
Lie algebra is important notion, which is related to the notion of a Lie group. And a
Lie algebra is an infinitesimal version of a Lie group.
Mathematically a Lie algebra is a linear space, which is supplied an operation called
commutator, which is bilinear, skew symmetric and satisfies a Jacobi identity.
If two elements of a vector space 𝑣 and 𝑤 are applied, the commutator of this two
vectors will have the next property:

[𝑣, 𝑤] = − [𝑤, 𝑣] . (3.20)

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The Jacobi identity means that if three elements 𝑣, 𝑤 and 𝑢 are taken, next cyclic
permutation will be obtained:

[[𝑣, 𝑤] , 𝑢] + [[𝑢, 𝑣] , 𝑤] + [[𝑤, 𝑢] , 𝑣] = 0 , (3.21)

where the operation written in (3.21) must be bilinear. Such a linear space is called a Lie
algebra.
In fact, every Lie group has a Lie algebra, and in particular a Poincare group has a Lie
algebra. And this Lie algebra is spent by generators, which we will denote by 𝑀𝜇𝜈 , where
𝜇 and 𝜈 running from 0 to 3. The 𝑀𝜇𝜈 will be Lorenz generators. In addition to Lorenz
generators, there will be also generators responsible for shifts, which will be denoted as
𝑃𝜇 and call as momentum generators.
It will be seen a little bit later that space-time shifts related to momentum of our
dynamical system and it’s very natural to call the 𝑃𝜇 momentum generators.
The Lie algebra relations between generators 𝑀𝜇𝜈 and generators 𝑃𝜇 , which make a
Lie algebra of the Poincare group, are given by the following commutation relations:
⎧
[𝑀𝜇𝜈 , 𝑀𝜌𝜎 ] = 𝑖 (−𝜂𝜈𝜌 𝑀𝜇𝜎 + 𝜂𝜇𝜌 𝑀𝜈𝜎 + 𝜂𝜈𝜎 𝑀𝜇𝜌 − 𝜂𝜇𝜎 𝑀𝜈𝜌 )
⎪
⎪
⎪
⎪
⎨
[𝑀𝜇𝜈 , 𝑃𝜎 ] = (𝜂𝜇𝜎 𝑃𝜈 − 𝜂𝜈𝜎 𝑃𝜇 ) (3.22)
⎪
⎪
⎪
⎩[𝑃𝜇 , 𝑃𝜈 ] = 0
⎪

Exponentiating of generators of the Lie algebra will give the corresponding group
elements and, in fact, this was already done , for instance, for a case of the rotation
group. It was seen that the group element is constructed by means of exponentiating of
matrices 𝑎, but the matrices 𝑎 are elements of the Lie algebra of the rotation group 𝑆𝑂 (3)
and the corresponding commutator of 𝑎𝑖 and 𝑎𝑗 of matrices is easy to compute:

[𝑎𝑖 , 𝑎𝑗 ] = 𝜀𝑖𝑗𝑘 𝑎𝑘 . (3.23)

In quantum theory, the Poincare generators become Hermitian operators. Hermitian

operators means that 𝑀𝜇𝜈 and 𝑃𝜇 are represented as operators, which satisfy the condition
(3.22).
A Hermitian operators also satisfy the following condition:
+
𝑀𝜇𝜈 = 𝑀𝜇𝜈 , 𝑃𝜇+ = 𝑃𝜇 . (3.24)

The presented generators will be explicitly realized on a certain space, called the Fock
space. We will come to the discussion of the Fock space a little bit later.

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A scheme which we want to achieve in quantum field theory connected with searching
of so called unit area and realization of unitary representations of the Poincare group.
As it can be known unitary representations play an important role in quantum mechanics
and also as a consequence in the quantum field theory because unitarity is naturally related
to the conservation of probabilities.
In quantum field theory it is also necessary to achieve unitary representations of our
symmetry groups and one of the important symmetry groups is the Poincare group, which
is a symmetry group of the space-time.
Therefore, the main goal will be to obtain the unitary representation of the Poincare
group. Finally, we can also introduce important relation between the proper orthochronous
Lorentz group, whose name is 𝑆𝑂+ (1, 3), and the special linear group of complex 2 by 2
matrices, whose mathematical name is 𝑆𝐿 (2, C).
There are three Pauli matrices 𝜎𝑖 , which are well known from quantum mechanics
(︃ )︃ (︃ )︃ (︃ )︃
0 1 0 −𝑖 1 0
𝜎1 = , 𝜎1 = , 𝜎1 = . (3.25)
1 0 𝑖 0 0 −1
Pauli matrices presented in (3.25) can be also combined with the unit matrix 𝜎 0 , which
is the unit 2 by 2 matrix (︃ )︃
1 0
𝜎0 = . (3.26)
0 1
All Pauli matrices can be combined in one definition of a matrix 𝜎 𝜇 in the following
way
𝜎𝜇 = 𝜎0, 𝜎𝑖 ,
(︀ )︀
(3.27)
where 𝑖 can be equal to 1, 2 and 3.
Simultaneously, a notion of matrices can introduced

𝜎 𝜇 = 𝜎 0 , −𝜎 𝑖 .
(︀ )︀
(3.28)

𝜎 𝜇 and 𝜎 𝜇 are collections of 2 by 2 matrices. Then, having a four dimensional vector

𝑥𝜇 , a 2 by 2 Hermitian matrix can be defined by means of the following construction:
3
∑︁ 3
∑︁
𝑋= 𝑥𝜇 𝜎 𝜇 = 𝜂𝜇𝜈 𝑥𝜇 𝜎 𝜈 . (3.29)
𝜇=0 𝜇=0

If the matrices 𝜎 in (3.29) is substituted , it will be seen that explicitly, we construct

the 2 by 2 matrices of the following type:
(︃ )︃
𝑥0 + 𝑥3 𝑥1 − 𝑖𝑥2
𝑋= . (3.30)
𝑥1 + 𝑖𝑥2 𝑥0 − 𝑥3

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It’s easy to see that since 𝑥𝜇 are real, the matrix 𝑋 is Hermitian, which means that
under commission conjugation it stays invariant:

𝑋+ = 𝑋 . (3.31)

If there is such a commission matrix, it is possible to uniquely determine the original

components 𝑥𝜇 by taking a trace of 𝑥 with a metric 𝜎 𝜇 :
1
𝑥𝜇 = Tr (𝑋 · 𝜎 𝜇 ) . (3.32)
2
It can be easily verified that computing a trace from (3.32) will lead to turning back
to the original coordinates 𝑥𝜇 .
Finally, it also can be seen that if the determinant of the commission matrix X is
computed, it will be

det 𝑋 = 𝑥0 + 𝑥3 𝑥0 − 𝑥3 − 𝑥1 − 𝑖𝑥2 𝑥1 + 𝑖𝑥2 .

(︀ )︀ (︀ )︀ (︀ )︀ (︀ )︀
(3.33)

If (3.33) is simplyfied, it will be obtained that

(︀ )︀2 (︀ )︀2 (︀ )︀2 (︀ )︀2
det 𝑋 = 𝑥0 − 𝑥1 − 𝑥2 − 𝑥3 . (3.34)

Expression (3.34) is nothing else as a four interval, which is Lawrence invariant.

Now it is possible to define on the commission matrix 𝑋 the action of the group 𝑆𝐿.
Let’s take a matrix 𝑔 from 𝑆𝐿 (2, C) and allow it to be multiplied on the commission
matrix 𝑋 in the following way:

𝑋 → 𝑋 ′ = 𝑔𝑋𝑔 + , (3.35)

It is clear that such transformation will not destroy hermiticity and will preserves it,
because due to the rules of hermitian conjugation the following expression will be obtained
)︀+ (︀ )︀+
𝑋 ′+ = 𝑔𝑋𝑔 + = 𝑔 + 𝑋 + 𝑔 + = 𝑔𝑋 + 𝑔 + .
(︀
(3.36)

It also can be seen that if we look at the determinant of 𝑋 ′ , this determinant will be
equal to
det 𝑋 ′ = det 𝑔𝑋𝑔 + = det 𝑔 det 𝑔 + det 𝑋 , (3.37)
⏟ ⏞
=det 𝑔

where the determinant of 𝑔 + is determinant of 𝑔 𝑡 but determinant does not changes with
transportation of the matrix. Since the talk is about special linear group, determinant of
𝑔 is equal to 1 by the definition of something that is called special linear group.

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Let’s remind that the special linear group is a group of 2 by 2 complex mattresses
with a unit determinant. That is why

det 𝑋 ′ = det 𝑋 . (3.38)

Let’s conclude that since these determinants are equal to one, we get determinant of 𝑋
and determinant of 𝑋 is a four interval. Therefore, completed transformations by matrices
from 𝑆𝐿 (2, C) group preserve the four interval. Once they preserve the four interval it is
possible to say that they just perform on the vector 𝑥𝜇 Lorentz transformations.
It can be seen that the conservation of the vector length is guaranteed by the transformation.
But this transformation may be written more explicitly
1 1 (︀
𝑥′𝜇 = Tr (𝑋 ′ 𝜎 𝜇 ) = Tr 𝑔𝑋𝑔 + 𝜎 𝜇 .
)︀
(3.39)
2 2
Then it is possible to simplify (3.39) using the definition of 𝑥 from (3.29):
1 (︀
𝑥′𝜇 = Tr 𝜎 𝜇 𝑔𝜎 𝜌 𝑔 + 𝜂𝜌𝜈 𝑥𝜈 .
)︀
(3.40)
2
Formula (3.40) relates the transformed coordinates to the original coordinates and
it is known how this transformation should look like in terms of matrix of Lorentz
transformation
𝑥′𝜇 = Λ𝜇𝜈 𝑥𝜈 (3.41)

and this means that the matrix Λ𝜇 is found in an explicit way and it’s given by the formula
1 (︀
Λ (𝑔)𝜇𝜈 = Tr 𝜎 𝜇 𝑔𝜎 𝜌 𝑔 + 𝜂𝜌𝜈 ,
)︀
(3.42)
2
where Λ is parameterized by an element 𝑔 of the 𝑆𝐿 (2, C).
In fact, formula (3.42) gives a map from 𝑆𝐿 (2, C) to the group 𝑆𝑂+ (1, 3).
It should be noticed why 𝑆𝑂+ (1, 3) is written. This is because, if the element Λ00 of
the gotten matrix is written, it will be obtained that
1
Λ (𝑔)00 = 𝜂00 Tr 𝜎 0 𝑔𝜎 0 𝑔 + ,
(︀ )︀
(3.43)
2
where 𝜂00 = 1, 𝜎 0 and 𝜎 0 are identity matrices.
That is why Λ (𝑔)00 is bigger than zero and therefore, it was realized by means of the
construction
1 (︀
Λ (𝑔)00 = Tr 𝑔𝑔 + > 0
)︀
(3.44)
2
the orthochronous transformation.

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It also should be pointed that all nondiagonal elements of the matrix Λ will be 0,
because the metric 𝜂 is diagonal.
Let’s prove that the transformation presented in (3.42) is proper. That means that
it is necessary to prove that determinant of the matrix is equal to one. It is possible to
compute it easily by using the definition of Λ𝜇𝜈 and it will be found that the determinant
of Λ will be reduced to the determinant of 𝑔.
As a result it will be obtained that
)︀2
det Λ = (det 𝑔)2 · det 𝑔 + = 1 .
(︀
(3.45)

What is very interesting is that the map from 𝑆𝐿 (2, C) to 𝑆𝑂+ (1, 3) is a map 2 to
1. It means that, in fact, there are two matrices of 𝑆𝐿 (2, C) which go to one matrix
of Lorentz transformation. For instance, two elements 𝑔 and −𝑔, which both belong to
𝑆𝐿 (2, C), go to one element of the Lorentz group. If 𝑔 is changed to −𝑔, the matrix Λ,
which is construct by formula (3.42) will not be changed.
And moreover, it is known that the group 𝑆𝐿 (2, C) is simply connected while 𝑆𝑂+ (1, 3)
is not simply connected. The term simply connected means that if 𝑆𝑂+ (1, 3) is considered
as a topological manifold and a closed loop such as presented at (fig. 3.1) is considered, it
will be seen that this loop cannot be contracted to a point all the time, returning inside
is 𝑆𝑂+ (1, 3).

Fig. 3.1. A topological manifold with a closed loop

In other words, in 𝑆𝑂+ (1, 3) there are non contractible loops. In a way, 𝑆𝑂+ (1, 3)
looks like a circle.
If a circle is considered, which is the simplest group 𝑈 (1) by multiplication, then, if
any loop is considered, it will be clear that it is not possible to contract it.
In fact, we discovered here that 𝑆𝐿 (2, C) represents a simply connected double cover
of 𝑆𝑂+ (1, 3) (fig. 3.2).

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Fig. 3.2. A picture how 𝑆𝐿 (2, C) double cover of 𝑆𝑂+ (1, 3)

It also can be introduced that any simply connected manifold which covers non simply
connected manifold is called the universal cover. So, 𝑆𝐿 (2, C) is the universal cover of
𝑆𝑂+ (1, 3).
As an exercise, it can be checked that indeed the matrix Λ𝜇𝜈 , which is constructed in by
formula (3.42), is a matrix of Lorentz transformation, it preserves the Minkowsky metric.
It means that you need to take Minkowsky metric 𝜂𝜇𝜈 and need to apply to it Lorentz
transformation Λ𝜇𝜌 Λ𝜈𝜆 and then you can compute this expression by using formula (3.42).
In fact, this has already been shown in a simple way by noting that transformations
of the type, which was considered, preserve the determinant and determinant is a four-
interval.
It also should be noticed that in general, if non-trivial background takes place, which
is not Minkowsky space, then this background is characterized by certain metric, which
can not be put on it. This metric might have or might not have what is called isometries.
Isometries typically would provide a symmetries of a quantum field theory on a curved
background, when the background is not flat like in Minkowsky space. It can be seen that
quantum particles not only in Minkowsky space, but, for instance, quantum particles in
the background of black hole. This means that space-time is very curved. This means that
different background requires the different metric. The possible group of isometries is a
group which preserves metric. In fact, for each manifold it is possible to define the notion
of a tangent space. Even if there is a curved manifold, there is a tangent space. Then in
this tangent space, there will be a group of transformations of tangent vectors which are
attached to this space and this space will be linear. Therefore, there will be a group of
linear transformations. If this tangent space is supplied or inherits Minkowsky metric, it
will be a Lorenz group of the tangent space. Such a way Lorenz group becomes associated
with a tangent space, but not to the whole manifold moving from point to point. There

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will be a different sort of local Lorenz groups. In this case, in general, if there is no trivial
metric, it is possible to talk about different morphisms which preserve this metric and
these different morphisms are called isometries and they form the group of isometries.

Fig. 3.3. The tangent space in very curved space-time

Noether’s theorem
Noether’s theorem has been established by me Noether in 1980. This theorem is
important, because it relates symmetries to conservation laws. If a dynamical system
are given, then a special role is played by the so called dynamical invariances or quantities,
which are invariant with respect to dynamics of the system. Very often dynamical invariances
are called as conservation laws. This means that if there is some dynamical systems, for
instance, a system of particles, which are characterized by coordinates and momenta, when
the dynamical system develops in time, the coordinates and momenta undergo changes,
but it is possible to create some complicated or simple quantities, which stay invariant
while time develops.
So, 𝑥𝑖 and 𝑝𝑖 of individual particles of the system change with time, but dynamical
invariance remains unchanged. For this reason they are called conservation laws.
If the goal is to describe a dynamical system, it is not enough to know solutions
of equations of motion. It is also important to be able to express the basic physical
characteristics of a system like, for instance, conserved energy or conserved momentum.
It is necessary to to know the energy and momentum or angular momentum of the system
as a whole and for this reason it is needed to be able to find expressions for this quantities
like energy momentum, angular momentum and so on in terms of individual dynamical
coordinates and momenta, in other words phase space variables of the system. This goal
is precisely achieved by means of Noether’s first theorem.

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Noether’s theorem tells the following: let the action of a dynamical system be invariant,
up to a boundary term with respect to a finite parametric, that is dependent on 𝑠 constant
parameters, continuous transformations of fields and space time coordinates. Then

1) there exists 𝑠 linearly independent currents 𝐽𝑛𝜇 , where n runs from 1 to 𝑠 with
divergences equal to certain linear combinations of Lagrangian derivatives.

2) on-shell, that is on solutions of equations of motion these currents are divergenceless

which means that 𝜕𝜇 𝐽𝑛𝜇 = 0, and they give rise to dynamical invariance or conservation
loss, which are conserved functionals of fields and their derivatives.

The symmetries of the action, which leaves the action invariant up to a boundary term
are called variational symmetries. The symmetries we are talking about in the Noether’s
theorem can be formulated in the following form: transformation of the coordinates is
involved as
𝑥𝜇 → 𝑥′𝜇 = 𝑥′𝜇 (𝜑, 𝑥) (3.46)

where 𝑥′𝜇 depending on fields or several fields, which might have a different nature and
can be scalars, vectors, tensors, and so on and depending on original coordinates.
Together with this, we also need to introduce transformations of fields

𝜑𝑖 (𝑥) → 𝜑′𝑖 (𝑥′ ) = 𝜑′𝑖 (𝜑, 𝑥) . (3.47)

It is assumed that the work is carried out in 𝑑-dimensional space and that is why index
𝜇 is allowded to run from 0 to 𝑑 − 1. Derivatives of fields under this transformations will
change accordingly
𝜕𝜇 𝜑𝑖 → 𝜕𝜇′ 𝜑′𝑖 . (3.48)

In general, such transformations can be non-trivial, because not only space-time coordinates
are being transformed, but also fields. Moreover, space-time coordinates may be required
to be transformed by involving fields.
Let’s now prove the Noether’s theorem. First of all, the theorem requires that the
symmetries of the action may be seen up to a boundary term. It means that if the action
functional is taken, which is an integral over 𝑑-dimensional space-time of the Lagrangian
density. Then in new coordinates, the action will look like
∫︁
′
d𝑑 𝑥′ L 𝜑′𝑖 (𝑥′ ) , 𝜕𝜇′ 𝜑′𝑖 (𝑥′ ) ,
(︀ )︀
𝑆 = (3.49)
Ω

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where the Lagrangian density depends on transformed fields 𝜑′𝑖 (𝑥′ ), and derivatives 𝜕𝜇′ 𝜑′𝑖 (𝑥′ ).
Integration will also be done over the image of the original region Ω under the coordinate
transformation.
Originally there is an integration region Ω, which can be arbitrary. Under the transformation
𝑥 → 𝑥′ it is mapped to some other region, which is called Ω′ (fig. 3.4).

Fig. 3.4. The transformation 𝑥 → 𝑥′

The condition of the theorem requires that the new action linked with the old action
by the formula
⎛ ⎞
dΛ𝜇
∫︁
⎝L (𝜑𝑖 (𝑥) , 𝜕𝜇 𝜑𝑖 (𝑥)) +
′
⎜ ⎟
𝑆 = d𝑑 𝑥 ⎜ 𝜇
⎟ . (3.50)
⏟d𝑥⏞
⎠
Ω
boundary term
dΛ𝜇
In the last lecture, it has been discussed that adding a derivative d𝑥𝜇
, where Λ𝜇 is
a function of fields 𝜑 does not influence on equations of motion. Variational symmetries
allow the change of the action up to a boundary term.
The condition (3.50) is required in the Noether’s theorem and the gotten invariance
must be evaluated on the arbitrary region Ω.
It also should be noticed, that for the Noether’s theorem it is not important that the
Λ𝜇 depends only on field 𝜑. In general, Λ𝜇 might depend on derivatives of 𝜑, but the
problem is that, in this case, this addition will lead to the problem that the physical
requirement to have a unique solution of the standard Kashi problem will not preserve,
because the new Lagrangian would involve derivatives higher than the first derivative.
But the Noether’s theorem is telling us that any boundary term is allowed.
Now it is necessary to use arbitrariness of the volume Ω and compare two integrals
for 𝑆 ′ . For that, it is possible to make a change of coordinates from 𝑥′ → 𝑥 and pass to
integration over the original variable 𝑥 over the region Ω.
Under the change of coordinates it is necessary to use a Jacobian, which is a determinant
of the Jacoby matrix. Such a way the following will be obtained
⃒ ′𝛼 ⃒
𝑑 ′
⃒ 𝜕𝑥 ⃒
d 𝑥 = ⃒⃒ 𝛽 ⃒⃒ × d𝑑 𝑥 . (3.51)
𝜕𝑥

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Then there will be two integrals over the same volume Ω and since Ω is arbitrary it
is possible to change it arbitrarily. Left hand side should be equal to the right hand side
and then this is only possible if the integrals are equal to each other.
So, the condition of the equality of the integrals can be written in the following way:
)︀ ⃒ 𝜕𝑥′𝛼 ⃒ dΛ𝜇
⃒ ⃒
L 𝜑 (𝑥 ) , 𝜕𝜇 𝜑 (𝑥 ) ⃒ 𝛽 ⃒⃒ = L (𝜑𝑖 (𝑥) , 𝜕𝜇 𝜑𝑖 (𝑥)) + 𝜇 .
(︀ ′ ′ ′ ′ ′ ⃒ (3.52)
𝜕𝑥 d𝑥

The infinitesimal version of this equation is considered. This means that infinitesimal
transformations that infinitesimal transformations are necessary, in which 𝑥𝜇 goes to 𝑥′𝜇
and a small variation of 𝑥𝜇 is infinitesimally added to 𝑥𝜇 .

𝑥𝜇 → 𝑥′𝜇 = 𝑥𝜇 + 𝛿𝑥𝜇 . (3.53)

For fields infinitesimal version 𝑝ℎ𝑖𝑖 (𝑥) goes to 𝜑′𝑖 (𝑥′ ), which is

𝜑𝑖 (𝑥) → 𝜑′𝑖 (𝑥′ ) = 𝜑𝑖 (𝑥) + 𝛿𝜑𝑖 (𝑥) , (3.54)

where again the edition to 𝑝ℎ𝑖𝑖 is an infinitesimal version of the transformation.

It is also needed to parameterize transformations with the help of individual parameters.
As it was described in the Noether’s theorem, we are talking about 𝑠 independent constant
parameters, which parameterize transformations. Transformations are linear and continuous
functions of parameters and, therefore, it is possible to write
∑︁
𝛿𝑥𝜇 = 𝑋𝑛𝜇 · 𝑛
⏟𝜖 ⏞ , (3.55)
1≤𝑛≤𝑠 parameters

where 𝜖𝑛 are parameters parameters, which are constant in a sense that they do not
depend on space-time points or fields and they just numbers, 𝑋𝑛𝜇 are quantities which
describe response of 𝑥𝜇 on the change by infinitesimal parameters 𝜖𝑛 .
Also for fields it is possible to write 𝛿𝜑𝑖 (𝑥) is given by a response of fields:
∑︁
𝛿𝜑𝑖 (𝑥) = Φ𝑖,𝑛 𝜖𝑛 , (3.56)
1≤𝑛≤𝑠

where Φ𝑖,𝑛 are responses of fields on infinitesimal transformation with parameters 𝜖𝑛 .

Then let’s look at how Lagrangian density transforms under symmetry transformations.
For this reason it is necessary to consider 𝜑′𝑖 (𝑥′ ). So infinitesimally it is written in the
following way
𝜑′𝑖 (𝑥′ ) = 𝜑′𝑖 (𝑥 + 𝛿𝑥) . (3.57)

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Now, it is possible to expand variable in (3.57) in a Taylor series in the following way

𝜑′𝑖 (𝑥′ ) = 𝜑′𝑖 (𝑥) + 𝜕𝜇 𝜑′𝑖 (𝑥) 𝛿𝑥𝜇 + . . . , (3.58)

where instead of . . . it is practicable to continue a Taylor series, but there is no interest

in that, because infinitesimal transformations are considered.
Then it is possible to simplify expression (3.58) and get

𝜑′𝑖 (𝑥′ ) = 𝜑′𝑖 (𝑥) + 𝜕𝜇 𝜑′𝑖 (𝑥) 𝑋𝑛𝜇 𝜖𝑛 + . . . . (3.59)

Now it can be seen that 𝜑′𝑖 will be different from 𝜑𝑖 also by order 𝜖, but one order
𝜖 is already in expression (3.59) and this means that there will not be a mistake at the
leading order in 𝜖 if there is a replacement in the 𝜕𝜇 𝜑′𝑖 with the original field 𝜑𝑖 . In this
case, there will not be a at a given order 𝜖 because the difference between 𝜑′ and 𝜑 is of
order 𝜖 and it will contribute to the order 𝜖2 .

𝜑′𝑖 (𝑥′ ) = 𝜑′𝑖 (𝑥) + 𝜕𝜇 𝜑𝑖 (𝑥) 𝑋𝑛𝜇 𝜖 . (3.60)

At this point, it is important to realize that the variation, which is applied to the field
𝜕
𝜑, 𝛿 does not commute with derivative 𝜕𝑥𝜇
. That is why, because variation 𝛿 an effect of
both changing coordinates and fields. This variation is due to both the change of the form
of the field, and also due to the change of the argument of the field. The variation 𝛿𝜑
incodes both variation of coordinates and the variation of the form of the field. In order to
distinguish variation of the form from the general variation, which also involves variation
of the coordinates, the variation of the form of the field is introduced, which will not be
¯ By definition this is
denoted by 𝛿.

¯ 𝑖 = 𝜑′ (𝑥) − 𝜑𝑖 (𝑥) .
𝛿𝜑 (3.61)
𝑖

In other words, it can be seen how the variation of the form of the field changes,
keeping this field at the same point. So space-time point is kept untouched. Such a way
the difference between primed and unprimed fields is considered. It is possible to find
what this variation is, because this is nothing else as the following thing:

¯ 𝑖 = (Ψ𝑖,𝑛 − 𝜕𝜇 𝜑𝑖 𝑋 𝜇 ) · 𝜖𝑛 .
𝛿𝜑 (3.62)
𝑛

Expression for (3.62) can be gotten from the next condition:

𝜑𝑖 (𝑥) + Φ𝑖,𝑛 𝜖𝑛 = 𝜑′ (𝑥) + 𝜕𝜇 𝜑𝑖 𝑋𝑛𝜇 · 𝜖𝑛 , (3.63)

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where then it is necessaru to subtract 𝜑𝑖 (𝑥) from 𝜑′𝑖 (𝑥):

¯ 𝑖 = 𝜑′ (𝑥) − 𝜑𝑖 (𝑥) = (Φ𝑖,𝑛 − 𝜕𝜇 𝜑𝑖 𝑋 𝜇 ) 𝜖𝑛 .

𝛿𝜑 (3.64)
𝑖 𝑛

From (3.64) it can be seen that variation of the form coincides with the full variation
only if 𝛿𝑥𝜇 = 0.
Now it is possible to compute the variation of the Lagrangian density and for that the
Lagrangian is considered as a function of the variable 𝑥.
The new Lagrangian density is equal to

L′ (𝑥′ ) = L 𝜑′ (𝑥′ ) , 𝜕𝜇′ 𝜑′ (𝑥′ ) .

(︀ )︀
(3.65)

As can be seen from (3.65) the new Lagrangian is an old Lagrangian density evaluated
on the new coordinates and new fields.
So now it is obtained that
dL 𝜇
L′ (𝑥′ ) = L′ (𝑥) + 𝛿𝑥 . (3.66)
d𝑥𝜇
It is possible to add the Lagrangian density and then subtract it
dL 𝜇
L′ (𝑥′ ) = L (𝑥) + L′ (𝑥) − L (𝑥) + 𝛿𝑥 (3.67)
d𝑥𝜇
which gives us variational form of the Lagrangian plus total derivative:

¯ + dL 𝛿𝑥𝜇 .
L′ (𝑥′ ) = L + 𝛿L (3.68)
d𝑥𝜇
The variation of the Lagrangian is computed in the usual way as

¯ (𝑥) = 𝜕L 𝜑¯𝑖 + 𝜕L 𝜕𝜇 𝛿𝜑
𝛿L ¯ 𝑖, (3.69)
𝜕𝜑𝑖 𝜕 (𝜕𝜇 𝜑𝑖 )
where it is possoble to use the effect that the variation of the form commutes with the
space-time derivative and, therefore, this gives us the following expression:
(︂ (︂ )︂)︂ (︂ )︂
¯ 𝜕L d 𝜕L ¯ d 𝜕L ¯
𝛿L (𝑥) = − 𝛿𝜑𝑖 + 𝜇 𝛿𝜑𝑖 . (3.70)
𝜕𝜑𝑖 d𝑥𝜇 𝜕 (𝜕𝜇 𝜑𝑖 ) d𝑥 𝜕 (𝜕𝜇 𝜑𝑖 )
⏟ ⏞
𝜀𝑖

The last thing to take into account is the infinitesimal expression for integration
measure. On the next lecture it will be shown that the integration measure is given
by the following expression
⃒ ′𝛼 ⃒ ⃒
⃒ ⃒ 𝛼 𝜕𝛿𝑥𝛼 ⃒
⃒ ⃒ ⃒
⃒ 𝜕𝑥 ⃒ ⃒ 𝜕 𝛼 𝛼
⃒ 𝜕𝑥𝛽 ⃒ = ⃒ 𝜕𝑥𝛽 (𝑥 + 𝛿𝑥 )⃒ = ⃒𝛿𝛽 + 𝜕𝑥𝛽 ⃒ . (3.71)
⃒ ⃒ ⃒ ⃒ ⃒ ⃒

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It can be seen that if there is no goal to restrict yourself to the leading order in 𝜖, then
expression (3.71) in the leading order gives us
⃒ ′𝛼 ⃒
⃒ 𝜕𝑥 ⃒ d 𝜇
⃒ 𝜕𝑥𝛽 ⃒ = 1 + d𝑥𝜇 𝛿𝑥 + . . . . (3.72)
⃒ ⃒

Now let’s return back to the original equation (3.52), which is required in the Noether’s
theorem. The infinitesimal version of this expression will be considered. That is why
Lagrangian will be replaced by Lagrangian plus its first variation. As a result on the left
side of the (3.52) it wiil be obtained that
(︂ (︂ )︂ )︂ (︂ )︂
¯ 𝑖+ d 𝜕L ¯ 𝑖 + ... + dL d
L + 𝜀𝑖 𝛿𝜑 𝛿𝜑 𝛿𝑥 𝜇 𝜇
1 + 𝜇 𝛿𝑥 + . . . . (3.73)
d𝑥𝜇 𝜕 (𝜕𝜇 𝜑𝑖 ) d𝑥𝜇 d𝑥

On the right hand side it will be obtained that

d𝛿Λ𝜇
L+ . (3.74)
d𝑥𝜇
Then it is necessary to open the brackets on the left side and collect the terms at
leading order.
It should be noticed that 𝛿Λ𝜇 is understood as Λ𝜇𝑛 · 𝜖𝑛 and the leading order is epsilon.
After collecting the terms of order 𝜖, the following expression wiil be obtained
(︂ )︂ (︂ )︂
d 𝜕L ¯ d 𝜕L 𝜕L ¯
𝛿𝜑𝑖 + L𝛿𝑥 − 𝛿Λ = 𝜇
𝜇 𝜇
− 𝛿𝜑𝑖 (3.75)
d𝑥𝜇 𝜕 (𝜕𝜇 𝜑𝑖 ) d𝑥 𝜕 (𝜕𝜇 𝜑𝑖 ) 𝜕𝜑𝑖

From expression (3.75) it is possible to read off that there is a current exist, which
will be called 𝐽 𝜇 and can be written as
𝜕L ¯
𝐽𝜇 = 𝛿𝜑𝑖 + L𝛿𝑥𝜇 − 𝛿Λ𝜇 , (3.76)
𝜕 (𝜕𝜇 𝜑𝑖 )

The current written in (3.76) has a property that:

d𝐽 𝜇 ¯ 𝑖 ,
= −𝜀𝑖 𝛿𝜑 (3.77)
d𝑥𝜇

i.e. the divergence of the current 𝐽 𝜇 is given by a linear combination of Lagrangian

derivatives 𝜀𝑖 and this linear combination involves the variation of the form of the field
𝜑𝑖 .
Index 𝑛 will be introduced in the following way: if we plug in variations, explicitly,
a variation of the form of the field, variation of the coordinates and variation of Λ𝜇 , it
will not be possible to see that for since parameters, 𝜀𝑛 is independent. For each of these

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parameters, tere will not be a conserved current. This conserved current 𝐽𝑛𝜇 , where index
𝑛 runs from 1 to 𝑠, will be given by the following expression

𝜕L
𝐽𝑛𝜇 = − (Φ𝑖,𝑛 − 𝜕_𝜈𝜑𝑖 𝑋𝑛𝜈 ) − L · 𝑋𝑛𝜇 + Λ𝜇𝑛 . (3.78)
𝜕 (𝜕𝜇 𝜑𝑖 )

Expression from (3.78) can be called as Noether’s current. On equations of motion or,
in other words, on shell all Lagrangian derivatives are equal to 0 and this is why currents
are conserved. In other words, their divergences are equal to 0
d𝐽𝑛𝜇
= 0 for ∀𝑛 = 1, . . . , 𝑠 . (3.79)
d𝑥𝜇
Expression (3.79) is basically completes the proof of the Noether’s theorem. The only
point that was not explained is the construction of the dynamical invariance, which relies
on the divergenceless currents, but this will be explained in the next lecture.

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Lecture 4. Applications of Noether’s Theorem.

Conservation Laws and Symmetries
During the previous lecture it has been found an expression for the conserved current,
which follows from the Noether’s theorem. This current has a following structure
𝜕L
𝐽𝑛𝜇 = − (Φ𝑖,𝑛 − 𝜕𝜈 𝜑𝑖 𝑋𝑛𝜈 ) − L𝑋𝑛𝜇 + Λ𝜇𝑛 , (4.1)
𝜕 (𝜕𝜇 𝜑𝑖 )

where in the gotten expression there are two indexes 𝜇 and 𝑛, where 𝜇 runs the space-time
coordinates from 0 to 𝑑 − 1. At the same time index 𝑛 represents a range of independent
parameters and runs from 1 to 𝑠. Variable 𝑠 can be characterized as a set of independent
variation parameters.
With respect to index 𝜇 there is a conservation law of the next view

𝜕𝜇 𝐽𝑛𝜇 = 0 , (4.2)

where this condition is right on shell, which means on solutions of equations of motion
(Euler-Lagrange equations).
The local conservation laws (4.2) can be used to define the integral invariance. It is
necessary to assume that all fields vanish at spatial infinity and then defines the following
integral ∫︁
𝐽𝑛 = d⃗𝑥𝐽𝑛0 , (4.3)
B
which is denoted as an integral of the zero component of the current, where zero-component
means time component. It also should be noticed that it is necessary to integrate over
𝑑 − 1-dimensional hyper surface, which is orthogonal to the time direction.
Formula (4.3) can be visualized as presented at (fig. 4.1).
Then when such a quantity 𝐽𝑛 is constructed, the statement is that this quantity is a
conserved charge, which sometimes can be called as a Noether’s charge. It is true, because
if 𝐽𝑛 is differentiated over time, there will be obtained that

𝜕𝐽𝑛0
∫︁ ∫︁
d𝐽𝑛 𝜕 0
= d⃗𝑥 𝐽𝑛 = d⃗
𝑥 . (4.4)
d𝑥0 𝜕𝑥0 B B 𝜕𝑥0

where the differentiation over 𝑥0 is moved inside the integral, because we are integrating
over spatial directions.

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Fig. 4.1. Integration over 𝑑 − 1-dimensional hyper surface orthogonal to the time direction

Now, the current which depends on time and on the spatial components becomes the
partial derivative, but then from the conservation law it follows that in terms of indexes
𝜕𝐽 0 𝜕𝐽 𝑖
+ 𝑖 =0, (4.5)
𝜕𝑥0 𝜕𝑥
where 𝑖 runs over spatial indexes from 1 to 𝑑 − 1.
The derivatives with indexes 𝑖 can be replaced by the divergence. Such a way it is
d𝐽𝑛
possible to write a new expression for d𝑥0
:

𝜕𝐽 𝑖
∫︁
d𝐽𝑛
=− d⃗𝑥 , (4.6)
d𝑥0 B 𝜕𝑥𝑖
where the total derivative in the integral was obtained and such a way it is possible to
reduce integral to the integral over the boundary of the space region (fig. 4.2).

Fig. 4.2. Integration over the boundary of the space region

A boundary of the space region will be denoted as 𝜕B and then it is possible to write
(4.6) as ∫︁
d𝐽𝑛 (︁ )︁
=− d 𝐽⃗𝑛 · ⃗𝑛 , (4.7)
d𝑥0 𝜕B

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where ⃗𝑛 is a normal vector to the boundary of the region.

The expression in the brackets of the (4.7) is a flux of the vector 𝐽𝑛 through the
boundary.
Then it is necessary to tend boundaries to infinity. If the boundary ia taken, for
instance, as a sphere, surrounding the space as a 𝑑 − 1-dimensional ball, the boundary
will be a 𝑑 − 2 dimensional sphere surrounding the ball. If our fields are assumed to have
a vanishing behavior at infinity, then in the limit where this boundary tends to infinity
the integral will tends to zero, because fields are assumed to die at special infinity.
In this case, if the boundary conditions described above is assumed, then it will be
seen that there is a next conservation law:
d𝐽𝑛
=0. (4.8)
d𝑡
Then it is possible to apply the general theorem of Noether to the discussion of concrete
examples of symmetry transformations and derive for this examples the corresponding
conserved currents.
First of all, let’s start with the case of an internal symmetry.

Internal symmetry
This is a case where the space-time variable 𝑥𝜇 is not touched and do not do anything
with 𝑥𝜇 . That is why:
𝛿𝑥𝜇 = 0 . (4.9)

From (4.9) it is clearly seen that the only variation that is possible to be done is a
variation of the field. That is why it’s called internal symmetry, because it does not involve
change of the space-time.
This variation, as it is known, is written down as
∑︁
𝛿𝜑′ (𝑥′ ) = Φ𝑖,𝑛 𝛿𝜖𝑛 , (4.10)
1≤𝑛≤𝑠

where the only non-trivial response is Φ and therefore, if we look at the general expression
(4.1), it will be seen that only the term Φ𝑖,𝑛 contributes, because 𝑋𝑛𝜈 response on variation
of space-time is absent.
Therefore, for this case the current looks rather simple:
𝜕L
𝐽𝑛𝜇 = − Φ𝑖,𝑛 . (4.11)
𝜕 (𝜕𝜇 𝜑𝑖 )

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The second very important example is an example of a conserved quantity, which is

called energy-momentum tensor.

Energy-momentum tensor
Sometimes people also can call it as stress-energy tensor. This corresponds to infinitesimal
space-time transformations, which can be firstly considered from the case of a single scalar
field.
The transformations which will be performed are translations of space time coordinates:

𝑥′𝜇 = 𝑥𝜇 + 𝛿𝑥𝜇 , (4.12)

where 𝛿𝑥𝜇 is a shift, which can be written in the following form:

𝛿𝑥𝜇 = 𝛿𝜈𝜇 𝜀𝜈 , (4.13)

where 𝑑𝑒𝑙𝑡𝑎𝜇𝜈 is a Kronecker symbol, 𝜀𝜈 are constant shifts.

From formula (4.13) it can be clearly seen that the response

𝑋𝜈𝜇 = 𝛿𝑛 𝑢𝜇 . (4.14)

For the scalar field it is known that transformation property is

𝜑′ (𝑥′ ) = 𝜑 (𝑥) + 𝛿𝜑 . (4.15)

Therefore, in this case, the 𝛿𝜑 (𝑥) is actually vanishes, and therefore response

𝛿𝜑 = 0 → Φ𝑖,𝑛 = 0 . (4.16)

That is why the formula for conserved current now looks as follows:
𝜕L
𝑇𝜈𝜇 = 𝜕𝜈 𝜑 − 𝛿𝜈𝜇 L , (4.17)
𝜕 (𝜕𝜇 𝜑)

where for the stress energy tensor people usually use a special notation for the current
𝑇𝜈𝜇 . It also should be noticed that index 𝑛 was replaced by space-time index 𝜈.
So, there is a tensor, which is directly derived by applying the general expression of
the Noether’s theorem. Expression (4.17) is called canonical stress energy tensor for a
scalar field.

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Then the next question should be asked. How this tensor will be looking, if the vector
field is placed instead of the scalar field. Let’s look at the field denoted by 𝜑𝜆 , where
lambda is a vector index, and consider the same transformations

𝑥′𝜇 = 𝑥𝜇 + 𝜖𝜇 = 𝑥𝜇 + 𝛿𝜈𝜇 𝜖𝜈 . (4.18)

The question is: what is the transformation for the field now? What is the response of
the vector field on coordinate shifts?
It turns out that this response vanishes in the same way as it was for the case of a
scalar field. As it can be seen from the following formula if the variation of vector field is
considered, then this is by definition should be response of a vector field by:

𝛿𝜑𝜆 = Φ𝜆𝜇 · 𝜖𝜇 = 0 . (4.19)

On the other hand, the 𝛿𝜑𝜆 is the transformation which comes from the general
infinitesimal version of the transformation of a vector under the coordinate shift. This
formula is known, because it was discussed in the first lecture.
𝜕𝑥′𝜆 𝜌
𝜑′𝜆 (𝑥′ ) = 𝜑 (𝑥) . (4.20)
𝜕𝑥𝜌
Expression (4.20) shows how the general formula looks like. Then the infinitesimal transformation
will be written in the following way:

𝑥′𝜆 = 𝑥𝜆 + 𝜀𝜆 . (4.21)

It can be seen that in this case, that the infinitesimal change of the derivative
𝜕𝛿𝑥′𝜆 𝜕𝜀𝜆
= =0, (4.22)
𝜕𝑥𝜌 𝜕𝑥𝜌
where 𝜀𝜆 is constant and that is why the derivative will vanish.
As a result, the variation 𝛿𝜑𝜆 vanishes. Therefore for the vector field the response Φ𝜆𝜇
vanishes, as it was for the case of a scalar field. That is why, it is seen that the stress
energy tensor for a vector field will have a similar form to what was for the scalar field:

𝜕L
𝑇𝜈𝜇 = 𝜆
𝜕𝜈 𝜑𝜆 − 𝛿𝜈𝜇 L . (4.23)
𝜕 (𝜕𝜇 𝜑 )

There is only one difference that it is necessary to sum over vector components.
Now a bit more sophisticated transformations are considered, which are related to
angular momentum and to the conservation loss related to the conservation of angular
momentum.

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Angular momentum
For this case it is necessary to consider infinitesimal rotations. For example, for
infinitesimal rotations of Minkowski space there are transformations of the following type

𝑥′𝜇 = 𝑥𝜇 + 𝑥𝜈 · 𝜀𝜇𝜈 , (4.24)

where parameters of this transformations 𝜀𝜇𝜈 satisfy the following property

𝜀𝜇𝜈 = −𝜀𝜈𝜇 . (4.25)

In other words, these parameters constitute a second rank tensor, which is antisymmetric
with respect to permutation of indices 𝜇 and 𝜈.
If our work takes place in four-dimensional space, such a tensor will have six independent
components. That is because 𝜇 runs in this case from 0 to 3 and, therefore, antisymmetric
tensor will have the number of components equal to 4 multiplied by 3 and divided by 2,
which is 6.
This six components will present six Lawrence transformations, which include three
usual rotations of space-time coordinates, plus three boosts, where there will be a rotation
between one of the spatial coordinates and the time.
Now it is necessary to find the response of coordinates on such a transformation. It is
needful to take into account that only parameters 𝜀𝜇 with 𝜇 < 𝜈 are independent because
parameters with 𝜈 > 𝜇 can be found from the condition of antisymmetry (4.25). This
means that it is possible to represent the transformation of coordinates in the following
way:
𝛿𝑥𝜆 = 𝑥𝜈 𝜀𝜆𝜈 = 𝑥𝜈 𝛿𝜇𝜆 · 𝜀𝜇𝜈 . (4.26)

As can be seen from (4.26), 𝛿𝜇𝜆 was introduced, but if there is a sum over 𝜇, then it
will be replaced by 𝜆 and we will return back to the formula:

𝛿𝑥𝜆 = 𝑥𝜈 𝜀𝜆𝜈 . (4.27)

Now, there will be a sum over all 𝜇 and 𝜈:

∑︁ ∑︁
𝛿𝑥𝜆 = 𝑥𝜈 𝛿𝜇𝜆 𝜀𝜇𝜈 + 𝑥𝜈 𝛿𝜇𝜆 𝜀𝜇𝜈 . (4.28)
𝜇<𝜈 𝜇>𝜈

Then in the second sum it will be necessary to change the index 𝜇 for 𝜈, because 𝜇
and 𝜈 are dummy summation indexes. Then it is needed to use expression (4.25) and the

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resulting formula can be written in the following way:

∑︁ (︀
𝛿𝑥𝜆 = 𝑥𝜈 𝛿𝜇𝜆 − 𝑥𝜇 𝛿𝜈𝜆 𝜀𝜇𝜈 ,
)︀
(4.29)
𝜇<𝜈

where some now goes over independent components.

Then it is possible to write the response of the coordinates on such a variation

𝜆
𝑋𝜇𝜈 = 𝑥𝜈 𝛿𝜇𝜆 − 𝑥𝜇 𝛿𝜈𝜆 . (4.30)

where 𝜇 < 𝜈.
Let’s consider again a case of a scalar field for which there is

𝜑′ (𝑥′ ) = 𝜑 (𝑥) . (4.31)

Again, from expression (4.31) we will get that 𝛿𝜑 = 0 and, again, response of variation
of fields is actually zero (𝜑𝑛 = 0).
Now the general form of the Noether’s current will be obtained and the non-trivial
response will be substituted and the following formula for the current will be obtained.
Now the current actually will have two indexes 𝜇 and 𝜈 instead of one index 𝑛, which was
before. The current for this case is usually denoted by the letter 𝑀 .
𝜕L
𝜆
(𝜕𝜇 𝜑𝑥𝜈 − 𝜕𝜈 𝜑𝑥𝜇 ) + L 𝑥𝜇 𝛿𝜈𝜆 − 𝑥𝜈 𝛿𝜇𝜆 .
(︀ )︀
𝑀𝜇𝜈 = (4.32)
𝜕 (𝜕𝜆 𝜑)
The conservation law will read now in this way:

𝜆
𝜕𝜆 𝑀𝜇𝜈 =0 (4.33)

for any 𝜇 < 𝜈.

𝜆
In arbitrary dimension the tensor 𝑀𝜇𝜈 will have

𝑑 (𝑑 − 1)
(4.34)
2
independent components.
In fact, it can be seen that it contains in addition to the usual angular momentum,
which can be associated with the spatial rotations, also conservation laws which are related
to Lorentz boosts, because as it was discussed in four dimensions the components 𝜇
and 𝜈 provide in total six independent conserved quantities. Three of these quantities
correspond to the standard angular momentum related to the rotations between spatial
directions and three other correspond to Lorentz boosts. It will be necessary to associate

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the corresponding conserved quantity with which one of the six possible transformations
according to the Noether’s theorem.
The formula for the angular momentum tensor can be rewritten in the following way:
𝜆
there will be a small rearrangement of the terms. As a result, 𝑀𝜇𝜈 will be written in the
following way:
(︂ )︂ (︂ )︂
𝜕L 𝜕L
𝜆
𝑀𝜇𝜈 = 𝑥𝜈 𝜕𝜇 𝜑 − L𝛿𝜇 − 𝑥𝜇
𝜆
𝜕𝜈 𝜑 − L𝛿𝜈 .
𝜆
(4.35)
𝜕 (𝜕𝜆 𝜑) 𝜕 (𝜕𝜆 𝜑)
Then, if a comparison of what is written in the brackets is performed, it will be obvious
that what we see is the stress energy tensor for the scalar field. Therefore, expression (4.35)
can be written as
𝜆
𝑀𝜇𝜈 = 𝑥𝜈 𝑇𝜇𝜆 − 𝑥𝜇 𝑇𝜈𝜆 , (4.36)

where 𝑇𝜇𝜆 and 𝑇𝜈𝜆 are stress tensors for the scalar field.
Further it will be needed to place the gotten expression into the conservation law and
it will be possible to see what it implies:

𝜆
= 𝜕𝜆 𝑥𝜈 𝑇𝜇𝜆 − 𝑥𝜇 𝑇𝜈𝜆 = 𝜂𝜆𝜈 𝑇𝜇𝜆 + 𝑥𝜈 𝜕𝜆 𝑇𝜇𝜆 − 𝜂𝜆𝜇 𝑇𝜈𝜆 − 𝑥𝜇 𝜕𝜆 𝑇𝜈𝜆 ,
(︀ )︀
𝜕𝜆 𝑀𝜇𝜈 (4.37)

where due to the conservation of a stress tensor two terms 𝜕𝜆 𝑇𝜇𝜆 and 𝜕𝜆 𝑇𝜈𝜆 will vanish.
Using the rule of raising of lower indexes with the help of the Minkowski metric the
following will be obtained in formula (4.37):

𝜆
𝜕𝜆 𝑀𝜇𝜈 = 𝑇𝜈𝜇 − 𝑇𝜇𝜈 . (4.38)

For the moment, expression (4.38) was derived, which actually shows us that for the
case of a Noether’s theorem, there is the divergence of angular momentum tensor vanishing
𝜆
(𝜕𝜆 𝑀𝜇𝜈 ). Then the stress energy tensor for a case of a scalar field must be automatically
symmetric:
𝑇𝜇𝜈 = 𝑇𝜈𝜇 . (4.39)
𝜆
The question is: what happens if, for instance, the energy momentum tensor 𝑀𝜇𝜈 for
a case of a vector field is considered? There is an trivial response of a vector on space
time rotations and this response can be found from the following:
∑︁
𝛿𝜑𝜆 = 𝜑𝜆𝜇𝜈 𝜀𝜇𝜈 , (4.40)
𝜇<𝜈

where 𝜇 and 𝜈 represent independent parameters of Lawrence transformations (six parameters

in four dimensions).

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On the other hand, 𝛿𝜑𝜆 should be given by the transformation law of a vector on the
coordinate transformations which are in this case
(︃ )︃
𝜆
𝜕𝛿𝑥 𝜕 ∑︁ (︀
𝛿𝜑𝜆 = 𝜑𝜌 (𝑥) = 𝑥𝜈 𝛿𝜇𝜆 − 𝑥𝜇 𝛿𝜈𝜆
)︀
. (4.41)
𝜕𝑥𝜌 𝜕𝑥𝜌 𝜇𝜈

where the underlined expression can be interpreted as a coordinate response.

The field response will have the next structure:

Φ𝜆𝜇𝜈 = 𝜂𝜌𝜈 𝛿𝜇𝜆 − 𝜂𝜌𝜇 𝛿𝜈𝜆 𝜑𝜌 = 𝜑𝜈 𝛿𝜇𝜆 − 𝜑𝜇 𝛿𝜈𝜆 .

(︀ )︀
(4.42)

Now, if the response from (4.42) is substituted together with a response of coordinates
into the general formula for the current, it will be possible to find for the tensor of angular
momentum of a vector field the following formula:
𝜕L [︀ 𝜌
𝜆 𝜌
L
]︀ (︀ 𝜆 𝜆
)︀
𝑀𝜇𝜈 =− 𝜑𝜈 𝛿 𝜇 − 𝜑𝜇 𝛿𝜈 + (𝑥 𝜇 𝜕 𝜈 𝜑 − 𝑥 𝜈 𝜕𝜇 𝜑) − 𝑥 𝜈 𝛿𝜇 − 𝑥 𝜇 𝛿 𝜈 . (4.43)
𝜕 (𝜕𝜆 𝜑𝜌 )
If the gotten expression is compared with stress tensor of a vector field, it will be
possible to see that expression (4.43) may be written in the following way:
(︂ )︂
𝜆 𝜆 𝜆 𝜕L 𝜕L
𝑀𝜇𝜈 = 𝑥𝜈 𝑇𝜇 − 𝑥𝜇 𝑇𝜈 + 𝜑𝜇 − 𝜑𝜈 . (4.44)
𝜕 (𝜕𝜆 𝜑𝜈 ) 𝜕 (𝜕𝜆 𝜑𝜇 )
If we look at expression (4.44), it will be obvious that it does not reduce to the previous
case of a scalar field, because for a scalar field we had only first two terms, but now there
is an addition term.
It should be noticed that the first two term, which was in a case of the scalar field, is
called orbital momentum. The additional piece characterizes polarization properties of the
field and relates to the notion of spin. Such a way the part in the brackets of expression
(4.44) is called a spin part.
From the discussion of the conservation law it is known that
(︂ )︂
𝜆 𝜕L 𝜕L
𝜕𝜆 𝑀𝜇𝜈 = 𝑇𝜈𝜇 − 𝑇𝜇𝜈 + 𝜕𝜆 𝜑𝜇 − 𝜑𝜈 . (4.45)
𝜕 (𝜕𝜆 𝜑𝜈 ) 𝜕 (𝜕𝜆 𝜑𝜇 )
From expression (4.45) it is not possible to conclude that the spin part and the orbital
part are separately conserved. It can be seen that this parts will be separately conserved
only in the case, while the stress tensor is symmetric. In other words, while

𝑇𝜇𝜈 = 𝑇𝜈𝜇 (4.46)

the spin part will be separately conserved.

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In general, it is unknown if the stress tensor is symmetric and it is possible to find

the non-symmetric tensor. In this case, there is no possibility to define two separate
quantities, because there is no separate conservation of the orbital part and the spin part.
In total, when the orbital part is added up to spin part, it is always conserved because
by the Noether’s theorem, it comes from a unique quantity namely the tensor of angular
momentum, which according to the Noether’s theorem should have vanishing divergence.
In particular, the case of Dirac field will be studied, where it will be found out that the
stress energy is not symmetric and therefore there is no separate conservation of spin and
orbital parts.

Lagrangians of Wess-Zumino“
”
The last example of conservation laws and the application of the Noether’s theorem
will concern Lagrangians of so called Wess-Zumino“ type. Wess and Zumino are scientists
”
who are responsible for the contribution to the discovery of super-symmetry.
In some cases, the Lagrangians have the property that they are invariant, mainly
Lagrangians which have super symmetry, only up to boundary terms. Since the appearance
of boundary terms is allowed by Noether’s theorem, there are examples where there are
conservation laws associated to the existence or appearance of the boundary terms.
The simplest Lagrangian density, which exhibit the property to be invariant up to
boundary terms, is the Lagrangian which has the next form:

L = 𝐶𝑖𝑗𝜇 𝜑𝑖 𝜕𝜇 𝜑𝑗 , (4.47)

where the Lagrangian wrote for a number of scalar fields 𝜑𝑖 , index 𝑖 is running from 1 to
the number 𝑚 and 𝜇 is a space-time index.
Alsp it should be noticed that coefficients 𝐶𝑖𝑗𝜇 are anti symmetric and this means that

𝐶𝑖𝑗𝜇 = −𝐶𝑗𝑖𝜇 . (4.48)

This is needed otherwise the Lagrangian will be a total derivative. So this anti-symmetry
property guarantees that there is no possibility to take the derivative out.
The infinitesimal transformation of fields 𝜑𝑖 , which leaves the Lagrangian invariant up
to the boundary term is a simple shift. This means that a field 𝑝ℎ𝑖𝑖 is taken and it is
shifted by constant parameter 𝜀𝑖 .
It can be seen that the variation of the Lagrangian density in this case is

𝛿L = 𝐶𝑖𝑗𝜇 𝜀𝑖 𝜕𝜇 𝜑𝑗 . (4.49)

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Because of the constant 𝐶 and 𝜀 it is possible to rewrite (4.49) as

𝛿L = 𝜕𝜇 𝐶𝑖𝑗𝜇 𝜀𝑖 𝜑𝑗 .
(︀ )︀
(4.50)

As can be seen from (4.50) the gotten variation looks like the total derivative. It is
possible to rewrite it as
𝛿L = 𝜕𝜇 Λ𝜇 , (4.51)

where Λ𝜇 is
Λ𝜇 = 𝐶𝑖𝑗𝜇 𝜀𝑖 𝜑𝑗 . (4.52)

It is also doable to write Λ𝜇 as the sum of components Λ𝜇𝑖 :

Λ𝜇 = Λ𝜇𝑖 𝜀𝑖 . (4.53)

That is why
Λ𝜇𝑖 = 𝐶𝑖𝑗𝜆 𝜑𝑗 . (4.54)

It is atteinable now to use the formula for the Noether’s current and write
𝜕L
𝐽𝑖𝜇 = − 𝑖
+ 𝐶𝑖𝑗𝜇 𝜑𝑗 , (4.55)
𝜕 (𝜕𝜇 𝜑 )

If the Lagrangian from expression (4.47) is taken and placed into (4.53), the following
will be obtained
𝐽𝑖𝜇 = −𝐶𝑗𝑖𝜇 𝜑𝑖 + 𝐶𝑖𝑗𝜇 𝜑𝑖 = 2𝐶𝑖𝑗𝜇 𝜑𝑖 . (4.56)

It is possible to immediately test the conservation law. For this case, it is necessary to
take the current from (4.56) and compute its divergence. As a result there will be received
that
𝜕𝜇 𝐽𝑖𝜇 = 2𝜕𝜇 𝐶𝑖𝑗𝜇 𝜑𝑗 ,
(︀ )︀
(4.57)

where the underlined equation is nothing else as an equation of motion for a field 𝜑𝑖 .
Solutions of equations of motion as should be according to the Noether’s theorem will
be the divergence of the corresponding current and it will vanish. Finally there will be
obtained
𝜕𝜇 𝐽𝑖𝜇 = 0 . (4.58)

The first remark, which we should to make is so-called improvement procedure.

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Improvement procedure
The point of this discussion is that, in principle, the conserved current 𝐽𝑛𝜇 is not
uniquely defined. As it has already been seen, the Noether’s theorem gives us a very
concrete expression for the current and this current is called canonical. But it turns out
that it is possible to change the current by adding to it so called improvement term:

𝐽𝑛𝜇 → 𝐽𝜇 + 𝜕𝜈 χ𝜇𝑛 , (4.59)

⏟ 𝑛⏞ ⏟ ⏞
canonical topological

where χ is a tensor which is assumed to be skew symmetric and then

χ𝜇𝜈 𝜈𝜇
𝑛 = −χ𝑛 . (4.60)

Then it can be seen that if such a term is added up, the conservation law will not be
changed or influenced, because if there is a performance with a derivative 𝜕𝜇 and compute
the divergence of this extra contribution, the following expression will be obtained:

𝜕𝜇 𝜕𝜈 χ𝜇𝜈
𝑛 = 0 , (4.61)

where there is a zero because derivatives 𝜕𝜇 and 𝜕𝜈 commute, while the tensor χ is skew
symmetric.
In other words, it is clear that adding the topological term does not break the conservation
law. Moreover, if χ is a function of fields, which vanish at infinity, it also does not break
the conservation of the Noether’s charge, because
∫︁
d⃗𝑥𝜕𝜈 χ0𝜈 . (4.62)
𝛽

It is possible to write the derivative in (4.62) as

𝜕𝜈 χ0𝜈 = 𝜕0 χ00 + 𝜕𝑘 χ0𝑘 , (4.63)

where the variable 𝜕0 χ00 simply vanishes and therefore it is possible to write (4.62) in the
following way: ∫︁
− d⃗𝑥𝜕𝑘 χ0𝑘 . (4.64)
𝛽

From (4.64) it is clearly seen that, as before, there will be a flux of the corresponding
vector χ0𝑘 over the boundary of the integration area 𝛽. Since the boundary tends to
infinity the integral (4.64) tends to zero.

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Then the next question may be asked: is it possible to make the stress tensor, which
might be not symmetrical, to become symmetric by adding a proper improvement term?
In other words, it is neccessary to take a 𝑇 𝜇𝜈 and add to it the improvement term:

𝑇 𝜇𝜈 → 𝑇 𝜇𝜈 + 𝜕𝜌 χ𝜌𝜇𝜈 , (4.65)

where according to the previous discussion

χ𝜌𝜇𝜈 = −χ𝜇𝜌𝜈 . (4.66)

If the tensor could be symmetric then, as it was discussed,the conservation of the

orbital part and the spin part will be gotten.
The answer to that question is that it turns out that there exists an improved symmetric
stress tensor J𝜇𝜈 , which is symmetric only if and only if the anti-symmetric part of the
canonical stress tensor 𝑇 𝜇𝜈 is the total derivative.
Let’s assume that the anti-symmetric part is a total derivative. This means that

𝑇 𝜇𝜈 − 𝑇 𝜈𝜇 = −𝜕𝜌 Ω𝜌𝜇𝜈 , (4.67)

where Ω satisfies the anti-symmetry property with respect to indices 𝜇 and 𝜈:

Ω𝜌𝜇𝜈 = −Ω𝜌𝜈𝜇 . (4.68)

Let’s prove that it is always possible to improve such a tensor by adding to it the
improvement term. To construct the improvement term it is needed to do the following:
the improvement term χ𝜌𝜇𝜈 is taken in the following form:
1
χ𝜌𝜇𝜈 = (Ω𝜌𝜇𝜈 + Ω𝜇𝜈𝜌 − Ω𝜈𝜌𝜇) . (4.69)
2
Then it can be seen that, in fact, the tensor χ has the desired symmetry:

χ𝜌𝜇𝜈 = −χ𝜇𝜌𝜈 , (4.70)

where it is possible to prove this statement if indexes are changed in the (4.69):
1
χ𝜌𝜇𝜈 = (Ω𝜇𝜌𝜈 + Ω𝜌𝜈𝜇 − Ω𝜈𝜇𝜌) . (4.71)
2
Then, because the Ω has a property of skew symmetry with respect to the last two
indices, it is possible to rewrite expression (4.71) as
1
χ𝜌𝜇𝜈 = (−Ω𝜇𝜈𝜌 − Ω𝜌𝜇𝜈 + Ω𝜈𝜌𝜇) . (4.72)
2
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If terms from the (4.72) are compared with terms from the (4.69), it will be clear that
the (4.70) is true.
The second statement is that if χ defined by the (4.69) is taken and χ𝜌𝜈𝜇 is subtracted,
the following will be obtained:

χ𝜌𝜇𝜈 − χ𝜌𝜈𝜇 = Ω𝜌𝜇𝜈 . (4.73)

Therefore, in the anti symmetric part (4.67) it is possible to substitute instead of the
tensor Ω𝜇𝜈 the next difference:

𝑇 𝜇𝜈 − 𝑇 𝜈𝜇 = −𝜕𝜌 (χ𝜌𝜇𝜈 − χ𝜌𝜈𝜇 ) . (4.74)

Then it is possible to open the brackets and get that:

𝑇 𝜇𝜈 − 𝑇 𝜈𝜇 = −𝜕𝜌 χ𝜌𝜇𝜈 + 𝜕𝜌 χ𝜌𝜈𝜇 . (4.75)

Let’s combine 𝑇 with derivatives in the following way:

𝑇 𝜇𝜈 + 𝜕𝜌 χ𝜌𝜇𝜈 = 𝑇 𝜈𝜇 + 𝜕𝜌 χ𝜌𝜈𝜇 , (4.76)

where it is attainable to conclude that the terms on the left and the right sides are
symmetric tensors J𝜇𝜈 and J𝜈𝜇 respectively. It is automatically means that

J𝜇𝜈 = J𝜈𝜇 . (4.77)

Indeed, it has been proved that if anti symmetric part of the stress tensor is a total
derivative, then the stress energy tensor can always be improved by adding a proper
improvement term.
In the other way, it is necessary to assume the opposite, that there exists an improved
symmetric tensor, which is the tensor J𝜇𝜈 of the next view:

J𝜇𝜈 = 𝑇 𝜇𝜈 + 𝜕𝜌 χ𝜌𝜇𝜈 . (4.78)

But then from the (4.78), it automatically follows that the anti-symmetric part of the 𝑇 𝜇𝜈
is a total divergence, because
J𝜈𝜇 = 𝑇 𝜈𝜇 + 𝜕𝜌 χ𝜌𝜈𝜇 (4.79)

and it is possible to subtract (4.79) from expression (4.78):

𝑇 𝜇𝜈 − 𝑇 𝜈𝜇 = 𝜕𝜌 (χ𝜌𝜇𝜈 − χ𝜌𝜈𝜇 ) , (4.80)

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which in (4.80) it is shown that anti symmetric part is a total divergence. Therefore, the
opposite statement is also approved.
The tensor χ𝜌𝜇𝜈 which is skew symmetric in indexes 𝜌 and 𝜇 and whose derivative can
be added to improve the stress tensor has a name of Bellinfante“ tensor.
”
The final remark concerning this improvement procedure is that even improvement is
not unique. Let’s suppose thatoer start was from a canonical tensor and then we improved
it by adding a proper Bellinfante“ tensor and made the tensor symmetric. But even if
”
the tensor is symmetric, it is still possible to add to it the following improvement term:

𝑇𝜇𝜈 → 𝑇𝜇𝜈 + 𝜕 𝜌 𝜕 𝜎 𝑊𝜇𝜌𝜈𝜎 , (4.81)

where 𝑊 is a tensor of the fourth rank, where it is required from this tensor 𝑊 the same
symmetries as the symmetries of the Riemann tensor. It’s known that the Riemann tensor
has the following symmetries:
⎧
𝑊𝜇𝜌𝜈𝜎 = −𝑊𝜌𝜇𝜈𝜎
⎪
⎪
⎪
⎪
⎨
𝑊𝜇𝜌𝜈𝜎 = −𝑊𝜇𝜌𝜎𝜈 (4.82)
⎪
⎪
⎪
⎩𝑊𝜇𝜌𝜈𝜎 = 𝑊𝜈𝜎𝜇𝜌
⎪

So if there is such a tensor as 𝑊 and it is added up with two derivatives on 𝑊 , then

there will still be a symmetric tensor.
Finally, it is possible to find the symmetric tensor directly from a gravitational theory.
There is a procedure which produce straightforwardly a symmetric stress energy tensor.
But this one has to consider the action for field, which we are interested in coupled to an
arbitrary gravitational background.
It is possible to take our "favorite"field, for instance, scalar field, vector field and so
on and couple it couple our field to arbitrary gravitational background, which essentially
means that our field is coupled to the metric. Then the action has standard form. For
instance, in four dimensions it would be derivative of the Lagrangian density:
∫︁
𝑆 = d4 𝑥L . (4.83)

The Lagrangian density in addition to the field, that is considered, also involves the
space time metric 𝑔𝜇𝜈 with the following determinant:

𝑔 = det 𝑔𝜇𝜈 (4.84)

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It’s well known that under different morphisms

𝑥′𝜇 → 𝑥𝜇 + 𝜉 𝜇 , (4.85)

where 𝜉 𝜇 is a vector field. 𝜉 𝜇 by definition is a vector which depends on the coordinates

of space-time
𝜉 𝜇 := 𝜉 𝜇 (𝑥) . (4.86)

The metric is known to transform in the following way:

𝛿𝑔 𝜇𝜈 = ∇𝜇 𝜉 𝜈 + ∇𝜈 𝜉 𝜇 . (4.87)

The transformation of the metric induces a transformation of the action. If a variation

of the action is performed, the action response in the following way:
∫︁
𝛿L
𝛿𝑆 = d4 𝑥 (∇𝜇 𝜉 𝜈 + ∇𝜈 𝜉 𝜇 ) . (4.88)
𝛿𝑔𝜇𝜈
Then the quantity can be introduced, which is stress energy tensor in the presence of
the gravitational field. By definition, this is
2 𝛿L
𝑇𝜇𝜈 = √ . (4.89)
−𝑔 𝛿𝑔 𝜇𝜈
In the gravity theory it’s proved that the following identity takes place:
(︀√ )︀ √
𝜕𝜇 −𝑔𝑇 𝜇𝜈 𝜉𝜈 = −𝑔∇𝜇 (𝑇 𝜇𝜈 𝜉𝜈 ) . (4.90)
⏟ ⏞
divergence

It is now possible to open the brackets in the (4.90) and write that
(︀√ )︀ √ √
𝜕𝜇 −𝑔𝑇 𝜇𝜈 𝜉𝜈 = −𝑔∇𝜇 𝑇 𝜇𝜈 𝜉𝜈 + −𝑔𝑇 𝜇𝜈 ∇𝜇 𝜉𝜈 . (4.91)

Formula (4.91) means that in fact, it is now possible to come back to formula (4.88)
and rewrite the variation 𝛿𝑆 as
√
∫︁
𝛿𝑆 = d4 𝑥 −𝑔𝑇 𝜇𝜈 ∇𝜇 𝜉𝜈 , (4.92)

where the integral presented in (4.92) can be evaluated by parts:

(︀√ √
∫︁ ∫︁
𝛿𝑆 = d 𝑥𝜕𝜇 −𝑔𝑇 𝜉𝜈 − d4 𝑥 −𝑔∇𝜇 𝑇 𝜇𝜈 𝜉𝜈 .
4 𝜇𝜈
)︀
(4.93)

If it is assumed that fields 𝜉𝜈 vanish at infinity, then the total derivative term does not
contribute and gets out. Since our interest is the action to be invariant with respect to

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default morphisms, then it means that the variation 𝛿𝑆 must vanish for any 𝜉𝜈 . The first
term of the (4.93) will vanish and therefore the one possible variant is

∇𝜇 𝑇 𝜇𝜈 = 0 . (4.94)

There is the expression in the (4.94), which looks like the conservation law, but it’s
not true, because the derivative in (4.94) is not the usual derivative but it’s a covariant
derivative.
If we work on the flat background and make the metric in expression (4.94) Minkowski,
then the variant derivative turns into the usual derivative and 𝑇 𝜇𝜈 will be conserved in
the standard sense.
∇𝜇 = 𝜕𝜇 → 𝜕𝜇 𝑇 𝜇𝜈 = 0 . (4.95)

On the other hand, from expression (4.89) it can be seen that since metric is symmetric
with respect to indexes 𝜇 and 𝜈, the object which will be obtained by means of the
Lagrangian variation with respect to the symmetric metric will be automatically symmetric.

𝑇 𝜇𝜈 = 𝑇 𝜈𝜇 . (4.96)

There is a way to obtain a symmetric tensor by coupling the field to the general
gravitational background with an arbitrary metric and then upon this object is derived,
it can be seen that it will have a meaning of the stress energy tensor with the usual
conservation law, if we return back in the final equation (4.94 to the flat Minkowski
metric.

Casimir operators
Tensors are objects, which transform themselves under Lorentz transformations. They
are conserved quantities with respect to the time evolution of the dynamical system.
If there is some kind of Lagrangian or Hamiltonian driving the dynamics, then these
quantities are indeed conserved. The numbers, which are computed as components of
tensors, depend on the Lorentz frame in which we compute them. If a rotation of the
Lorentz frame is performed or if our level is boosted, for instance, with respect to the
𝜆
original frame, then numbers represented by the components of, for instance, 𝑇𝜇𝜈 or 𝑀𝜇𝜈
will be changed. They will go to other numbers, which also will conserve with the time.
Because of the dynamics of our system, they always been conserved, but they will be
changed because these objects are tensors.

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From this point of view, you may ask yourself what are the quantities which actually
characterize our dynamical system in such a way that they are independent on the choice
of the Lorentz frame. In representation theory these quantities are known as Casimir’s or
Casimir operators.
The Casimir’s objects commute with all generators of a Poincare group:

[𝐶, 𝑃𝜇 ] = 0 , (4.97)

where 𝐶 is a Casimir’s object.

Expression (4.97) means that if 𝐶 is taken to be a Casimir, then it must commute
with all generators of translations 𝑃𝜇 , which are components constructed as conserved
charges related to stress tensor. So
∫︁
𝑃𝜇 = 𝑇𝜇0 d⃗𝑥 , (4.98)

where it is necessary to integrate over spatial directions. Analogously, Casimir must

commute with all generators of angular momentum

[𝐶, 𝑀𝜇𝜈 ] = 0 , (4.99)

which are obtained by ∫︁

0
𝑀𝜇𝜈 = 𝑀𝜇𝜈 d⃗𝑥 (4.100)

with respect to the space.

There may be a question: what are the Casimir elements of Casimir operators for the
case of the Poincare group? For the case of the Poincare group, there are two Casimirs.
One Casimir is simply equal to

𝐶1 = 𝑃𝜇 𝑃 𝜇 = 𝑃02 − 𝑃⃗ 2 . (4.101)

The gotten Casimir object takes the values 𝑚2 𝑐2 and therefore

𝑃02 − 𝑝⃗2 = 𝑚2 𝑐2 . (4.102)

If the terms in the (4.102) are placed to the one side, it will be possible to see that
something that was obtained is called mass-shell condition for a relativistic particle, which
is going to be studied in detail a little bit later.

𝑃02 − 𝑝⃗2 − 𝑚2 𝑐2 = 0 . (4.103)

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If the commutation relations between 𝐶1 and another 𝑃𝜎 or generator 𝑀𝜇𝜈 is written,

it will be found out that ⎧
⎨[𝑃𝜇 𝑃 𝜇 , 𝑃𝜎 ] = 0
(4.104)
⎩[𝑃 𝑃 𝜇 , 𝑀 ] = 0
𝜇 𝜇𝜈

The second Casimir is constructed by using a certain special vector 𝑊𝜇 , which is

constructed in the following way

𝑊𝜇 = 𝜀𝜇𝜈𝜌𝜆 𝑃 𝜈 𝑀 𝜌𝜆 , (4.105)

where we got a fully anti-symmetric tensor. The second Casimir object is called Pauli-
”
Lubanski“ vector. The property of this vector is that it is orthogonal with respect to the
Lorentz invariant scalar product:
𝑊𝜇 𝑃 𝜇 = 0 . (4.106)

The second property is that it may be used to build up the second Casimir of the
Poincare group, which is given by a square of the Pauli-Lubanski“ vector:
”

𝐶2 = 𝑊 𝜇 𝑊 𝜇 . (4.107)

It is possible to check by direct calculations that if 𝐶2 is taken and it is commuted

with 𝑃𝜎 , it will be found out that

[𝑊𝜇 𝑊 𝜇 𝑃𝜎 ] = 0 (4.108)

as well as it can be found with 𝑀𝜌𝜆 :

[𝑊𝜇 𝑊 𝜇 𝑀𝜌𝜆 ] = 0 . (4.109)

Values of the second Casimir in irreducible representations massive of the Poincare

group will look like as follows

𝐶2 = 𝑚2 𝑐 · 𝑠 (𝑠 + 1) , (4.110)

where 𝑠 is a half integer numbers, starting from 0, 21 , 1, 3

2
and so on, which is called the
spin.
In other words, massive irreducible representations of the Poincare group are characterized
by two numbers the mass and spin. In fact, this is in acted with characterization of
elementary particles, because every elementary particle has a mass and spin or it is

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massless, but massless is a separate case, it is a separate representations of the Poincare

group, because as can be seen for massless representation 𝐶1 and 𝐶2 will be equal to zero.
In that case, there appears a new Casimir, which is called Helicity. Helicity coincides
with a projection of a spin of a particle on the direction of motion:
𝑝⃗ · ⃗𝑠
, (4.111)
|⃗𝑝|

where (4.111) is a projection of spin on the direction of motion.

The talk about helicity will be in detail, when the discussion will be about the Dirac’s
theory, where it will be possible to realize how helicity appears, why it plays an important
role and why helicity is applied only to the case of massless particles, but not the massive
ones.
It is now necessary to introduce some definitions. First of all, let’s start with representation.
Representation means the representation of abstract generators of the Poincare algebra
by concrete operators, acting in some concrete space of states.
Basically, it is needful to realize pure abstract algebraic relations of the Poincare group.
Those commutators which have been written down in the previous lectures, by means of
concrete operators realized in a concrete Hilbert space. It is necessaty to think about this
objects as about a concrete operators, acting on states in some Hilbert space.
Then it is necessary to introduce a reducible representation. Representation is called
the reducible if there will be no invariant subspaces, where invariant subspaces means
that a vector is taken from the subspace and we act on it with generators. The result of
this action remains belong to the same subspace and it is not possible to get out of the
subspace.
If there are no subspaces except the trivial one and the whole space, then such
representation of a Lie group is called irreducible.
And on the top, a notion of unitarity is added up. Representation is called unitary if
the Lie algebra generators are realized by Hermitian operators. If the Lorentz generators
by Hermitian operators are realized, the next conditions will be true:
⎧
⎨𝑀 + = 𝑀𝜇𝜈
𝜇𝜈
(4.112)
⎩𝑃 + = 𝑃
𝜇 𝜇

The unitary representations are important, because they can tell us that probabilities
that is needed to be computed, will not depend on the particular Poincare frame in which

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it is being computed, because unitarity preserves probability as it is known from courses

on quantum mechanics.
In this way, the interest is in constructing unitary irreducible representations of Poincare
group. It is known that since Poincare group is non compact, in particular, because
it includes the Lorentz group which is non compact, then by the theorem from the
group theory it is known that such unitary representations of non compact group must
necessarily be infinity dimensional. That is why the spaces in which it is possible to
realize such unitary irreducible representations of the Poincare group must be infinity
dimensional, and that is one of the reasons why it is necessary to deal with fields, which
in court, an infinite number of degrees of freedom, on which in quantum theory unitary
irreducible representations of the Poincare group will be realized.

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Lecture 5. Conserved Charges as Symmetry

Generators. Klein-Gordon Field. Mass-Shell Condition

Conserved charges as the symmetry generators

Conserved charges that follow from Noether’s theorem play a role of infinitesimal
generators of symmetries. Symmetry generators means that they play a role of generators
of symmetry, which is a correspond to according to the Noether’s theorem.
To understand this statement it is necessary to recall the construction of the Hamiltonian
formalism for the case of field theory. Normally the start should be from the Lagrangian
description and introduced Lagrangian, Lagrangian density and the action. To pass to
the Hamiltonian formalism, it is neceassary to introduce in addition to the canonical
coordinates 𝜑𝑖 (𝑥), which in this case is played by our fields, a canonical momenta
𝛿L
𝜋𝑖 (𝑥) = , (5.1)
𝛿 𝜑˙ 𝑖
where the expression looks like an usual expression in classical mechanics. As can be
known, a momentum is a derivative of the Lagrangian with respect to velocity. In this
case, it is also possible to write (5.1) as
𝜕L
𝜋𝑖 (𝑥) = , (5.2)
𝜕 (𝜕0 𝜑𝑖 (𝑥))
where 𝜕0 is a derivative with respect to 𝑥0 , which is the same as

𝑥0 = 𝑐𝑡 . (5.3)

The relation between the momentum and the derivative of the Lagrangian density
allows to express the velocity of the field in terms of canonical momentum.

𝜑˙𝑖 (𝑥) → 𝜋𝑖 (𝑥) . (5.4)

Therefore, then it is necessary to construct the Hamiltonian using the standard prescription
of classical mechanics. It is neeeded to construct it as
∫︁ (︁ )︁
˙
𝐻 = d⃗𝑥 𝜋𝑖 𝜑𝑖 − 𝐿 , (5.5)

where it is possible to write 𝐿 as an integral of Lagrangian density and get the following
expression: ∫︁ ∫︁
[︁ ]︁
𝐻= d⃗𝑥 𝜋𝑖 𝜑˙ 𝑖 − L = d⃗𝑥H , (5.6)

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where H is an Hamiltonian density, which can be written as:

H = 𝜋𝑖 𝜑˙ 𝑖 − L . (5.7)

Then absolutely the same procedure should be implemented, but for the case of
fields. An important role in constructing the Hamiltonian formalism in the usual classical
mechanics is played by an object, which is called Poisson’s bracket, because this bracket
allows to formulate Hamilton’s equations of motion in a compact and a simple way. In
general, it is known that a Poisson’s bracket is an operation on a space of functions or a
phase space. If there is a phase space, which is parameterized by the coordinates 𝑝 and 𝑞,
and then on this phase space, there will be a space over the phase space, which will be a
space of functions, where it is possiblet to define an operation, which is called Poisson’s
bracket, which is a map cross functions into functions (fig. 5.1).

Fig. 5.1. A phase space, parameterized by the coordinates 𝑝 and 𝑞, with space of functions
over the phase space

Any two functions on a phase space it put in correspondence as an a function on the

phase space and the Poisson’s bracket satisfies the following conditions:

1) the Poisson’s bracket is skew symmetric:

{𝑓, 𝑔} = −{𝑔, 𝑓 } . (5.8)

2) the Poisson’s bracket is bilinear:

{𝜆𝑓 + 𝜇𝑔, ℎ} = 𝜆{𝑓, ℎ} + 𝜇{𝑔, ℎ} . (5.9)

3) the Poisson’s bracket must satisfy Jacobi identity. If a Poisson’s bracket of 𝑓 and
𝑔 is taken and then bracket with another function ℎ and then add up a cyclic
permutations of these functions 𝑓 , 𝑔 and ℎ, then the result will be equal to:

{{𝑓, 𝑔}, ℎ} + cyclic permutations = 0 . (5.10)

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Since our work is done with fields, for any two local in time functionals, for instance,
𝐹 [𝜋, 𝜑] and 𝐺 [𝜋, 𝜑] depending on momenta and coordinates the Poisson’s bracket will be
defined. Let’s define the Poisson’s bracket as a following functional:
∑︁ ∫︁ [︂ ]︂
𝛿𝐹 𝛿𝐺 𝛿𝐹 𝛿𝐺
{𝐹, 𝐺} = d⃗𝑥 − . (5.11)
𝑖
𝛿𝜋𝑖 (𝑥) 𝛿𝜑𝑖 (𝑥) 𝛿𝜑𝑖 (𝑥) 𝛿𝜋𝑖 (𝑥)

In the (5.11) a definition of the canonical Poisson’s bracket over the space of functionals
was written, which now parameterized to depend on canonical coordinates and momenta.
Before functions of 𝑝 and 𝑞 were studied, where 𝑝 and 𝑞 are coordinates on the classical
phase space of classical mechanics. To any point a value, which is the value of the function
at this point, is put in correspondence.
Then it is workable to go on with this definition and compute the Poisson’s brackets
between sort of coordinates on the infinite dimensional phase space, which are fields and
their momenta. A simple exercise to see that if two fields were taken at different space
points and at the same time 𝑡, then the result of evaluation of the Poisson’s bracket
between the fields will be equal to zero:

{𝜑𝑖 (𝑡, ⃗𝑥) , 𝜑𝑗 (𝑡, ⃗𝑦 )} = 0 . (5.12)

That is simply, because for the bracket to be non-zero it is necessary to vary one
functional with respect to momenta and the other with respect to the coordinate, but in
the (5.12) there are only coordinates and therefore the Poisson’s bracket would vanish.
The same expression will be right for momenta. If there is a Poisson’s bracket of two
momenta at different space points, then the result will be zero:

{𝜋𝑖 (𝑡, ⃗𝑥) , 𝜋𝑗 (𝑡, ⃗𝑦 )} = 0 . (5.13)

Finally, if the bracket between momenta and coordinates is evaluated at different space
points and at the same time 𝑡, then the result will be equal to:

{𝜋𝑖 (𝑡, ⃗𝑥) , 𝜑𝑗 (𝑡, ⃗𝑦 )} = 𝛿𝑖𝑗 𝛿 (⃗𝑥 − ⃗𝑦 ) . (5.14)

It is possible to group expressions (5.12), (5.13) and (5.14) under the name of generalization
of the canonical Poisson’s bracket in classical mechanics to the case of fields.
It is also important that in the Hamiltonian setting the Poisson’s brackets are equal-
time. It can be seen, because the right hand side of expressions (5.12), (5.13) and (5.14)
does not depend on time.

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It also should be noticed that if different times are selected, it will not be possible to
know what will be the outcome and therefore there is a way to solve equations of motion
for the fields and compute the Poisson’s bracket on the phase space.
If there is such equal time Poisson’s structure as in formulas above, it is possible to
define the time evolution of any functional. For instance, there is a way to say that F
under the Hamiltonian flow set up by our preference choice of the Hamiltonian. Some
function is selected on a phase space, which will be regarded as the Hamiltonian and then
evolution of any function on a phase space can be described by means of the following
formula:
𝐹˙ = {𝐻, 𝐹 } . (5.15)

In fact, formula (5.15) is the standard equation of classical mechanics, which goes
without any modification to the field theory.
If the time evolution of any functional is defined, it is necessary to compute the equal-
time Poisson’s bracket of 𝐻 with 𝐹 .
For the canonical Poisson’s bracket of 𝑞 and 𝑝 it is possible to write similar to (5.12)-
(5.14) expressions: ⎧
{𝑞𝑖 , 𝑞𝑗 } = 0
⎪
⎪
⎪
⎪
⎨
{𝑝𝑖 , 𝑝𝑗 } = 0 (5.16)
⎪
⎪
⎪
⎩{𝑝𝑖 , 𝑞𝑗 } = 𝛿𝑖𝑗
⎪

If, for instance, coordinate 𝑞𝑖 is taken, the 𝑞˙ with the help of the Hamiltonian 𝐻 will
be equal to:
𝜕𝐻
𝑞˙𝑖 = {𝐻, 𝑞𝑖 } = . (5.17)
𝜕𝑝𝑖
At the same time, if the time derivative of the momentum is computed, then according
to equation (5.15) in the usual mechanics the following can be found
𝜕𝐻
𝑝˙𝑖 = {𝐻, 𝑝𝑖 } = − . (5.18)
𝜕𝑞
Such a way, we got in (5.17) and (5.18) nothing else as Hamilton’s equations of motion.
Therefore (5.15) represents Hamilton’s equations in field theory.
Finally, it is practicable to generalize the procedure for evaluation of the (5.15). It is
necessary to evaluate the Poisson’s bracket in (5.15) and for this purpose it is needed to
evaluate (5.11) and pick up a functional, which will be regarded as a Hamiltonian, and
then for any functional 𝐹 Hamilton’s equations will be obtained. Such a way recalled a
standard Hamiltonian formalism.

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If we remember the general expression for the Noether’s charge, which was recieved
from the Noether’s theorem and which is given by the following formula
∫︁
𝐽𝑛 = d⃗𝑦 𝐽𝑛0 (𝑦) , (5.19)

where it is necessary to integrate over spatial directions.

Formula (5.19) can be transformed into:
∫︁ [︁ (︁ )︁ ]︁
𝐽𝑛 = d⃗𝑦 −𝜋𝑗 Φ𝑗,𝑛 (𝜑) − 𝜕𝜇 𝜑𝑖 𝑋𝑛 (𝜑) − L𝑋𝑛 (𝜑) + Λ𝑛 (𝜑) .
⃗ 𝜇 0 0
(5.20)

Then let’s try to compute a canonical Poisson’s bracket in the space of fields for
evaluating the bracket between the field the conserved Noether’s charge. For this case, it
is needed to introduce one more property of the Poisson’s bracket:

{𝑓 · 𝑔, ℎ} = {𝑓, ℎ} · 𝑔 + {𝑔, ℎ} · 𝑓 . (5.21)

Before evaluation of the Poisson’s bracket we also need to remember that:

𝜕L
𝜋𝑗 = . (5.22)
𝜕 (𝜕0 𝜑𝑗 )

Due to formulas (5.21) and (5.23) we can write that:

∫︁
{𝜑𝑖 (𝑥) , 𝐽𝑛 } = d⃗𝑦 {𝜋𝑗 (𝑦) , 𝜑𝑖 (𝑥)} (Φ𝑗,𝑛 (𝜑) − 𝜕𝜇 𝜑𝑖 𝑋𝑛𝜇 (𝜑)) + 𝜋𝑗 (𝑦) {𝜑𝑖 (𝑥) , 𝜕𝜇 𝜑𝑗 }𝑋𝑛𝜇 (𝜑) − {𝜑𝑖 (𝑥) , L (
(︀

(5.23)
where the nontrivial impact to the Poisson’s bracket was recieved only from first two
terms in the bracket of (5.20)/
Then it is needful to compute the Poisson’s bracket between 𝜑𝑖 (𝑥) and the Lagrangian
density
{𝜑𝑖 (𝑥) , L} = {𝜑𝑖 (𝑥) , L (𝜑, 𝜕𝜇 𝜑)} , (5.24)

where Lagrangian density is a function of 𝜑 and 𝜕𝜇 𝜑. Therefore, the rule presented in

(5.21) can be used and Lagrangian density can be differentiated with respect to 𝜑:
𝜕L
{𝜑𝑖 (𝑥) , L} = {𝜑𝑖 (𝑥) , 𝜕𝜇 𝜑𝑗 } . (5.25)
𝜕 (𝜕𝜇 𝜑𝑖 )

It also should be noticed that since L is a Lagrangian density and it is known that this
is a function of 𝜑𝑗 and the derivative of 𝜑𝑗 , where index 𝑗 means that, in general, there
are many fields, it can be assumed that the bracket acts on any function on the phase

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space in the same way as differentiation. For instance, we would compute the Poisson’s
bracket for the usual case of mechanics, if there are 𝑞𝑖 and a function of 𝑞𝑗 and 𝑝𝑗 , as

𝜕𝑓 𝜕𝑓
{𝑞𝑖 , 𝑓 (𝑞𝑗 , 𝑝𝑗 )} = {𝑞𝑖 , 𝑞𝑗 } + {𝑞𝑖 , 𝑝𝑗 } , (5.26)
𝜕𝑞𝑗 𝜕𝜌𝑗

where it was supposed that the bracket acts as a derivative and there is the rule that if
there is a derivative of a composite function 𝑓 (𝑔 (𝑥)), it could be simplified into
d 𝜕𝑓 𝜕𝑔
𝑓 (𝑔 (𝑥)) = . (5.27)
d𝑥 𝜕𝑔 𝜕𝑥
Due to rule (5.27), (5.26) can also be written as:
𝜕𝑓 𝜕𝑞𝑗 𝜕𝑓 𝜕𝜌𝑗
{𝑞𝑖 , 𝑓 (𝑞𝑗 , 𝑝𝑗 )} = + . (5.28)
𝜕𝑞𝑗 𝜕𝑞𝑖 𝜕𝜌𝑗 𝜕𝑞𝑖

If we return to (5.25), it will be seen that the non-trivial contribution might come only
if 𝜇 equal to zero and it is possible to get that:
𝜕L
{𝜑𝑖 (𝑥) , L} = {𝜑𝑖 (𝑥) , 𝜕0 𝜑𝑗 } . (5.29)
𝜕 (𝜕0 𝜑𝑗 (𝑥))

Let’s look at the Poisson’s bracket in (5.29). It is clear that this break will be non-
trivial only if 𝑖 equals to 𝑗.
Then it is possible to find that

{𝜑𝑖 (𝑥) , L} = 𝜋𝑗 {𝜑𝑖 (𝑥) , 𝜕0 𝜑𝑗 (𝑦)} . (5.30)

Now it is possible to substitute the gotten result back into the original expression
(5.23) and see, that there is a cancelation between two terms. As the result it will be
obtained that:
∫︁
{𝜑𝑖 (𝑥) , 𝐽𝑛 } = d⃗𝑦 {𝜋𝑗 (𝑦) , 𝜑𝑖 (𝑥)} (Φ𝑗,𝑛 (𝜑) − 𝜕𝜇 𝜑𝑖 𝑋𝑛𝜇 (𝜑)) , (5.31)

where the second term in (5.23) also non-trivial only if 𝜇 = 0.

Then (5.14) can be used to simplify formula (5.31):
∫︁
{𝜑𝑖 (𝑥) , 𝐽𝑛 } = d⃗𝑦 𝛿𝑖𝑗 𝛿 (⃗𝑥 − ⃗𝑦 ) (Φ𝑗,𝑛 (𝜑) − 𝜕𝜇 𝜑𝑖 𝑋𝑛𝜇 (𝜑)) . (5.32)

Important to mention that delta-functions 𝛿 (⃗𝑥 − ⃗𝑦 ) and 𝛿𝑖𝑗 also can be used in order to
get the final expression:
{𝜑𝑖 (𝑥) , 𝐽𝑛 } = Φ𝑖,𝑛 − 𝜕𝜇 𝜑𝑖 𝑋𝑛𝜇 . (5.33)

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What is written down in the (5.33) is nothing else as variation of the form of the field:

{𝜑𝑖 (𝑥) , 𝐽𝑛 } = 𝛿 𝑛 𝜑𝑖 . (5.34)

Sometimes the variation of the form (5.34) in classical differential geometry is called
the Lie derivative.
If generators 𝐽𝑛 are summed with parameters 𝜀𝑛 of symmetry transformations, it will
be obtained that:
{𝜑𝑖 (𝑥) , 𝐽𝑛 𝜀𝑛 } = (Φ𝑖,𝑛 − 𝜕𝜇 𝜑𝑖 𝑋𝑛𝜇 ) 𝜀𝑛 = 𝛿𝜑𝑖 , (5.35)

where we got on the right side just a variation of the form of the field 𝜑𝑖 under infinitesimal
transformations. And this is exactly what was meant by a statement that infinitesimal
variations of the form or infinitesimal symmetry transformations are generated by the
conserved Noether’s charges with respect to the canonical Poisson’s structure.

𝛿𝜑𝑖 = {𝜑𝑖 (𝑥) , 𝐽} . (5.36)

It is possible to give a concrete important example, which is a demonstration of the

property that conserved charges shows themselves up as symmetry generators.
Let’s consider, for instance, time translation. It is known that this symmetry corresponds
to the conservation of energy. Time variable gets translated by adding to 𝑥0 infinitesimal
parameter 𝜀:
𝑥0 → 𝑥′0 = 𝑥0 + 𝜀 . (5.37)

Let’s also consider for simplicity just single scalar field. It is known that there is a
conserved current corresponding to the symmetry and this current is component of a
stress energy tensor 𝑇0𝜇 . As it well known this component is conserved and the local
conservation of energy for this component is

𝜕𝜇 𝑇0𝜇 = 0 . (5.38)

The corresponding Noether’s charge for this component has the next form:
∫︁
𝐽 = d⃗𝑥𝑇00 (𝑥) , (5.39)

where there was an integration over the spatial directions.

It also should be noticed that if 𝐽 is differentiated with respect to 𝑥0 , the following
will be recieved
d
𝐽 =0. (5.40)
d𝑥0
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Let’s remember the expression for the stress-energy tensor for the case of a single
scalar field. It is found that
𝜕L
𝑇𝜈𝜇 = 𝜕𝜈 𝜑 − 𝛿𝜈𝜇 L . (5.41)
𝜕 (𝜕𝜇 𝜑)

Our takk is about 𝑇00 component and due to this fact it is possible to place 0 and 0
instead of indexes 𝜇 and 𝜈 and get that
𝜕L
𝑇00 = 𝜕0 𝜑 − L . (5.42)
𝜕 (𝜕0 𝜑)
Then the following may be done. It is known that the expression
𝜕L
(5.43)
𝜕 (𝜕0 𝜑)

is momentum of the field 𝜑 and therefore in expression (5.42) the momentum multiplied
by velocity minus Lagrangian density is used, but according to the Hamiltonian formalism
the gotten expression is nothing else as a Hamiltonian density of the field:
𝜕L
H= 𝜕0 𝜑 − L = 𝜋𝜕0 𝜑 − L . (5.44)
𝜕 (𝜕0 𝜑)
Therefore, the Hamiltonian density conserve Noether’s charge.
In this case, 𝐽 can be expressed as
∫︁ ∫︁
𝐽 = d⃗𝑥 (𝜋𝜕0 𝜑 − L) = d𝑥H . (5.45)

If the goal is to have the proper normalization constants, it is necessary to suggest that
1
𝐽= 𝐻 . (5.46)
𝑐
So, next expression is recieved
∫︁
𝐻=𝑐 d𝑥H . (5.47)

So, Hamiltonian appears to coincide with a Noether’s charge corresponding to time

translations. This means that in fact,

{𝜑 (𝑥) , 𝐽} = −𝜕0 𝜑 (𝑥) , (5.48)

where the fact that for a scalar field the response Φ on translations of coordinates and
time is equal to zero and 𝑋𝑛𝜇 is proportional to 𝛿0𝜇 was used.

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It also should be noticed that other components responsible for the conservation of
the total momentum are related to the
∫︁
𝑃𝑖 = d⃗𝑥𝑇𝑖0 , (5.49)

where 𝑖 is running from 1 to 3. This would be the next example which will be studied,
but before this it is necessary to use formula (5.46) to get that
𝜕
{𝜑 (𝑥) , 𝐻} = − 𝜑 (𝑥) . (5.50)
𝜕𝑡
If the place of 𝜑 and 𝐻 is changed, according to the rule of replacement of the Poisson’s
bracket the next formula will be obtained:

{𝐻, 𝜑 (𝑥)} = 𝜑˙ (𝑥) . (5.51)

The exact manifestation of Hamilton’s equations of motion was received in the (5.51),
which was declared at the beginning of lecture. It can be seen that the infinitesimal
generator of time translations according to this formalism is nothing else as the Hamiltonian
for the scalar field. It is possible to make a conclusion that the Hamiltonian generates
time translation. On the other hand, translation of time is the symmetry, such that the
corresponding conserved Noether’s charge coincides with the Hamiltonian according to
the Noether’s theorem and the Hamiltonian generates infinitesimal translations in time.
It is possible to do more general setting without split into spatial components. Corresponding
to momentum conservation and energy conservation it is possible to construct the generator
𝑃𝜇 , which is a generator of space-time shifts, where index 𝜇 running from 1 to 3. According
to this discussion it will be found out that the 𝑃𝜇 will be equal to
∫︁
𝑃𝜇 = d⃗𝑥 𝜋 𝜌 𝜕𝜇 𝜑𝜌 − L𝛿𝜇0 .
(︀ )︀
(5.52)

It is also attainable to ask the same question about space-time rotations. It is known
that they are generated by the components of the angular momentum 𝑀𝜇𝜈 :
∫︁
0
𝑀𝜇𝜈 = d⃗𝑥𝑀𝜇𝜈 (5.53)

A very similar computation should be completed and it can be completed as an

additional exercise. As a result the following expression should be found:
∫︁
𝑀𝜇𝜈 = d⃗𝑥 𝜑𝜇 𝜋𝜈 − 𝜑𝜈 𝜋𝜇 + 𝑥𝜈 𝜋 𝜌 𝜕𝜇 𝜑𝜌 − L𝛿𝜇0 − 𝑥𝜇 𝜋 𝜌 𝜕𝜈 𝜑𝜌 − L𝛿𝜈0 .
(︀ (︀ )︀ (︀ )︀)︀
(5.54)

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Such a way we got expressions for two generators of Poincare group (5.52) and (5.54)
as functionals on a phase space generated by fields 𝜑𝜇 and momentums 𝜋𝜇 . Even more to
that, it is possible to compute Poisson’s brackets between the gotten functionals by using
fundamental Poisson’s brackets. And it will be obtained that with respect to Poisson’s
brackets generators will reproduce the standard relations between the generators of the
Poincare algebra:
⎧
⎨{𝑀𝜇𝜈 , 𝑃𝜎 } = 𝜂𝜈𝜎 𝑃𝜇 − 𝜂𝜇𝜎 𝑃𝜈
(5.55)
⎩{𝑀 , 𝑀 } = 𝜂 𝑀 − 𝜂 𝑀 − 𝜂 𝑀 + 𝜂 𝑀
𝜇𝜈 𝜌𝜎 𝜈𝜌 𝜇𝜎 𝜇𝜌 𝜈𝜎 𝜈𝜎 𝜇𝜌 𝜇𝜎 𝜈𝜌

It should be noticed that in quantum theory all Poisson’s brackets will be replaced by
commutators. Such a way it will be possible to get relations between Poincare generators
as operators, and then this will be a relations of the Lie algebra of the Poincare group.
Now let’s come to the next important topic which is called Klein-Gordon field.

Klein-Gordon field
First of all, let’s write the action for this field:
∫︁ [︂ ]︂
1 4 1 𝜇 1 (︁ 𝑚𝑐 )︁2 2
𝑆 [𝜑] = d 𝑥 𝜕𝜇 𝜑 (𝑥) 𝜕 𝜑 (𝑥) − 𝜑 (𝑥) , (5.56)
𝑐 2 2 ~

where it was continued to denote the field as 𝜑 (𝑥). It was recieved that the Klein-Gordon
field is a massive relativistic scalar field. The first term in brackets of the (5.56) is a
term, which contains kinetic energy and it has already been derived something similar in
the first lecture. This term was derived from the discrete approach, when the model of
masses connected by springs in the limiting case to infinity case and the length of springs
tended to zero was considered. And on the top of that one may add something, which is
proportional to 𝜑2 multiplied with the coefficient, which is designed in such a way that
the physical dimension of the kinetic energy and the term which contains 𝜑2 is the same.
Four-dimensional integration measure d4 𝑥 is understood as

d4 𝑥 = d𝑥0 d𝑥1 d𝑥2 d𝑥3 , (5.57)

where as it is known
d𝑥0 = 𝑐d𝑡 (5.58)

and, therefore, the speed of light in front of the action stands to cancel.

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The action is relativistic because derivatives 𝜕𝜇 and 𝜕 𝜇 are paired in such a way that
the corresponding term will be invariant under Lawrence transformations.
Field 𝜑 itself is a scalar and it is known how it transforms under coordinate transformations.
From the action 𝑆 it is attainable to straightforwardly derive equations of motion and
they have the following form:
(︂ (︁ 𝑚𝑐 )︁2 )︂
𝜇
𝜕𝜇 𝜕 + 𝜑 (𝑥) = 0 . (5.59)
~
Equation (5.59) is the Euler-Lagrange equation.
It is also possible to rewrite (5.59) more explicitly in the following way:
1 𝜕2 𝜕2
(︂ )︂
2
− + () 𝜑 (𝑥) = 0 . (5.60)
𝑐2 𝜕𝑡2 𝜕𝑥2𝑖
Equation (5.60) has a name of Klein-Gordon equation. Sometimes equation (5.59) also
written in a way where the operator 𝜕𝜇 𝜕 𝜇 is replaced by

 = 𝜕𝜇 𝜕 𝜇 . (5.61)
𝑚𝑐
It also can be noticed that in fact, the combination ~
in an inverted way has a
meaning of
~
, 𝜆= (5.62)
𝑚𝑐
where 𝜆 has a dimension of length and is called Compton wavelengths associated with
the particle of mass 𝑚.
Usually people prefer to work in the natural units, where

[~] = [𝑐] = 1 (5.63)

𝑚𝑐
and therefore the term ~
is simply becomes 𝑚. That is why the action for the field 𝜑
can be written as ∫︁ (︂ )︂
1 𝜇4 2 2
𝑆 [𝜑] = d 𝑥 𝜕𝜇 𝜑𝜕 𝜑 − 𝑚 𝜑 . (5.64)
2
It can be seen that if standard units are selected, then it is possible to write the action
in the following form:
∫︁ [︂ ]︂
1 ˙ 2 1 (︁ ⃗ )︁2 1 (︁ 𝑚𝑐 )︁2 2
𝑆 [𝜑] = d𝑡d𝑥 𝜑 − ∇𝜑 − 𝜑 . (5.65)
2𝑐2 2 2 ~
Now it is available to straightly develop the Hamiltonian formalism and this is something
that is needed for quantization. So, it is possible to derive canonical momentum for the
field 𝜑:
𝛿𝐿
𝜋 (𝑥) = . (5.66)
𝜕 𝜑˙ (𝑥)

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By the way, what is written down in the (5.65) is an integral over time of Lagrangian:
∫︁
𝑆 [𝜑] = d𝑡𝐿 . (5.67)

Also it can be seen that if the Lagrangian is differentiated or the variation of the
˙ it will be found that:
Lagrangian is taken with respect to 𝜑,
𝜑˙ (𝑥) 1
𝜋 (𝑥) = 2
= 𝜕0 𝜑 (𝑥) . (5.68)
𝑐 𝑐
If the canonical procedure of passing from Lagrangian to the Hamiltonian is used, the
Hamiltonian for the Klein-Gordon field will be obtained, which has the following form:
∫︁ [︂ (︁ 𝑚𝑐 )︁2 ]︂
1 2 2
𝐻= d⃗𝑥 𝑐 𝜋 + 𝜕𝑖 𝜑𝜕𝑖 𝜑 + 𝜑2 . (5.69)
2 ~
It is also interesting to know what is the physical dimension of the field 𝜑. So this 𝜑
will be denoted in the square brackets [𝜑] and this will be physical notion for the physical
dimension. Such a notation is used for dimension of any physical quantity. It is just taken
in square brackets and transferred to the physical dimension.
It is known that the physical dimension of the Hamiltonian is a dimension of energy:

[𝐻] = 𝜀 . (5.70)

It is possible to apply this approach to the (5.69) and get that

1
[𝜀] = 𝑙3 × 2
× [𝜑]2 = 𝑙 [𝜑]2 , (5.71)
𝑙
(︀ 𝑚𝑐 )︀2
where the fast that d⃗𝑥 has a dimension of 𝑙3 and the inverse Compton length ~
has
1
a dimension of 𝑙2
was used. Such a way it will be recieved that
√︂
𝜀
[𝜑] = . (5.72)
𝑙
Such a quantity in classical physics 𝜀𝑙 , which has a meaning of energy per unit length, is
called tension. So field 𝜑 has a physical dimension of a square root of tension.
It is possible to use the same procedure to find the physical dimension of the momentum.
By doing analogous computation it will be found that
√︂
1 𝜀
[𝜋] = . (5.73)
𝑐𝑙 𝑙
It is also attainable to rewrite (5.73) as
√︂ √︂ √︂
𝜀 𝑚𝑐2 𝑚
[𝜋] = = = . (5.74)
𝑐2 𝑙 3 𝑐2 𝑙 3 𝑙3

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It was obtained that the momentum has a physical dimension of square root of mass
divided by volume or square root of mass density. If the action of the field 𝜑 is considered,
it will be seen that it is possible to compute the physical dimension of the action:
1 1 𝑙2 𝜀 𝑙×𝜀
[𝑆] = × 𝑙4 × 2 × [𝜑]2 = = (5.75)
𝑐 𝑙 𝑐 𝑙 𝑐
The length divided by velocity was recieved in the (5.75), which gives us time and it
can be finally computed that:
[𝑆] = 𝑡 × 𝜀 . (5.76)

Also it should be noticed that time multiplied by energy is nothing else as momentum
multiplied by length, which is the same as as angular momentum:

[𝑆] = 𝑝 × 𝑙 = [~] . (5.77)

This is a very important fact that the physical dimension of action is always the same
as a dimension of the Planck constant or a dimension of the angular momentum. It does
not only concern a scalar field it concerns any field.
Another important fact is that if there are some two functionals of field momentum
and the field itself, for instance, 𝐹 [𝜋, 𝜑] and 𝐺 [𝜋, 𝜑].
Then it will be possible to find the physical dimension of the equal time Poisson’s
bracket given by the formula that was discussed at the beginning of the lecture. The
physical dimension of this bracket, which will be denoted as square bracket of {𝐹, 𝐺}, will
be offset from the physical dimension of the product:

[𝐹 ] · [𝐺] (5.78)

by ~. In other words, it is possible to write that:

[𝐹 ] · [𝐺]
[{𝐹, 𝐺}] = . (5.79)
~
This can be easily understood from the definition of the Poisson’s bracket. It is known
that the Poisson’s bracket is a differential operation, which explicitly given by the following
formula: ∫︁ [︂ ]︂
𝛿𝐹 𝛿𝐺 𝛿𝐹 𝛿𝐺
{𝐹, 𝐺} = d⃗𝑥 − , (5.80)
𝛿𝜋 (𝑥) 𝛿𝜑 (𝑥) 𝛿𝜑 (𝑥) 𝛿𝜋 (𝑥)
It has been already computed that dimensions of 𝜋 (𝑥) and 𝜑 (𝑥) are:
⎧
⎨[𝜋 (𝑥)] = 1 √︀ 𝜀
𝑐𝑙 𝑙
(5.81)
⎩[𝜑 (𝑥)] = √︀ 𝜀
𝑙

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Therefore, in order to compute the dimension of the right hand side of the (5.80)
first of all it is necessary to find what is the dimension of the variational derivative. To
complete this it is necessary to consider what is the variation 𝛿𝐹 :
∫︁
𝛿𝐹
𝛿𝐹 = d⃗𝑥 𝛿𝜑 (𝑥) . (5.82)
𝛿𝜑 (𝑥)

Thus, the result is in (5.82) is what is called variation of functional 𝐹 . From this formula
it can be seen that the dimension of the variation is the same as a dimension of 𝐹 and
equal to: ]︂ [︂
𝛿𝐹 3
[𝛿𝐹 ] = [𝐹 ] = 𝑙 × × [𝜑] , (5.83)
𝛿𝜑
where the fact that the integral over spatial coordinates equal to 𝑙3 and that dimension
of 𝛿𝜑 is the same as a dimension of 𝜑 was used.
It is also possible to simplify (5.83) and get that:
[︂ ]︂ √︂
𝛿𝐹 3 𝜀
[𝛿𝐹 ] = ×𝑙 × . (5.84)
𝛿𝜑 𝑙
[︁ ]︁
It is practicable to express 𝛿𝐹 𝛿𝜑
from the (5.84) and get that:
[︂ ]︂
𝛿𝐹 [𝐹 ]
= . (5.85)
𝛿𝜑 𝑙5/2 × 𝜀1/2

Before evaluation of the dimension of (5.80) it is also necessary to compute variational

derivative with respect to the momentum 𝜋 (𝑥). To make this we need to find variation
of 𝐺: ]︂ [︂ [︂ ]︂ √︂
3𝛿𝐺 𝛿𝐺 3 𝜀
[𝛿𝐺] = [𝐺] = 𝑙 × × [𝜋] = 𝑙 . (5.86)
𝛿𝜋 𝛿𝜋 𝑐 𝑙3
2

The final expression for the (5.86) will have a next view:
[︂ ]︂
𝛿𝐺 [𝐺] × 𝑐
= 3/2 . (5.87)
𝛿𝜋 𝑙 × 𝜀1/2

Now let’s compute the dimension of the Poisson bracket. It can be seen that according
to the definition (5.80), it can be obtained that

[𝐹 ] [𝐺] × 𝑐
[{𝐹, 𝐺}] = 𝑙3 × × 3/2 . (5.88)
𝑙5/2 ×𝜀 1/2 𝑙 × 𝜀1/2
If (5.88) is simplified, it will be found that:

[𝐹 ] × [𝐺] × 𝑐 [𝐹 ] × [𝐺] [𝐹 ] × [𝐺]

[{𝐹, 𝐺}] = = = . (5.89)
𝑙×𝜀 𝑡×𝜀 ~

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It has been declared that the dimension of the Poisson’s bracket of two functionals is
not equal to the product of the dimensions of this functionals but it’s offset from them
by 1/~:
[𝐹 ] × [𝐺]
[{𝐹, 𝐺}] = . (5.90)
~
Then let’s return back to the Klein-Gordon equation and will write it down in the
Hamiltonian form.
Let’s write the evolution of the field and the momentum, which will be given by the
Poisson’s bracket of the field and the momentum with the Hamiltonian:
⎧
⎨𝜑˙ (𝑥) = {𝐻, 𝜑 (𝑥)}
(5.91)
⎩𝜋˙ (𝑥) = {𝐻, 𝜋 (𝑥)}

Expressions in the (5.91) are Hamiltonian equations of motion. If this Poisson’s brackets
are evaluated , it will be obtained that:
⎧
⎨𝜑˙ = 𝑐2 𝜋
(5.92)
⎩𝜋˙ = 𝜕 2 𝜑 − (︀ 𝑚𝑐 )︀2 𝜑
𝑖 ~

It is possible to exclude momentum of the field in the (5.92) by taking a second

derivative of 𝜑:
(︁ 𝑚𝑐 )︁2
𝜑¨ = 𝑐2 𝜋˙ = 𝑐2 𝜕𝑖2 𝜑 − 𝑐2 𝜑 (5.93)
~
If both sides of the (5.93) are sevided by 𝑐2 and everything is transferred to the left
hand side, thw dollowing will be obtained
1¨ 2
(︁ 𝑚𝑐 )︁2
𝜑 − 𝜕𝑖 𝜑 + 𝜑=0, (5.94)
𝑐2 ~
which is exactly the Klein-Gordon equation.
Also it is possible to rewrite (5.94) in the following form:

1 𝜕2 𝜕2
(︂ (︁ 𝑚𝑐 )︁2 )︂
− + 𝜑=0, (5.95)
𝑐2 𝜕𝑡2 𝜕𝑥2𝑖 ~

The next important point to discuss is so called mass-shell condition and in a way this
scalar field on a mass-shell amounts just to solving the Klein-Gordon equation. So there
is the Klein-Gordon equation and now our goal is to solve it. How to do it?

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How to solve Klein-Gordon equation?

The most efficient way to look at the solutions and to understand how solutions look
like is to go to the Fourier space. So it is just necessary to make a Fourier transform by
using the Fourier integral:
∫︁
1
𝜑 (𝑥) = d4 𝑘𝑒𝑖𝑘𝑥 𝜑˜ (𝑘) , (5.96)
(2𝜋)3/2
where 𝜑˜ (𝑘) is a Fourier image of the field 𝜑.
The expression 𝑘𝑥 in the exponent of the (5.96) is a Lawrence invariant scalar product,
which is by definition:
𝑘 · 𝑥 = 𝑘𝜇 𝑥𝜇 = 𝑘 0 𝑥0 − ⃗𝑘 · ⃗𝑥 . (5.97)

Very often the value ⃗𝑘 from the product (5.97) has a name of wave vector. It is also
possible to place 𝑥0 into this equation and get that:

𝑘 · 𝑥 = 𝑐𝑘 0 · 𝑡 − ⃗𝑘 · ⃗𝑥 = 𝜔𝑡 − ⃗𝑘 · ⃗𝑥 , (5.98)

where 𝜔 is called frequency and from this equation:

𝜔
𝑘0 = . (5.99)
𝑐
It also should be noticed that if (5.98) is placed into the exponent, it will be obtained
that
⃗
𝑒𝑖𝑘𝑥 = 𝑒𝑖(𝜔𝑡−𝑘·⃗𝑥) (5.100)

and in quantum mechanics this combination is usually called plane wave.

So Fourier transform for the field 𝜑 (𝑥) is in fact the decomposition of a plane waves.
And it can be said that the integral in the (5.96) is an integral of a plane waves which is
taken.
Then it is necessary to take this Fourier representation of field 𝜑 (𝑥) and plug it in
Klein-Gordon equation and see what will be recieved for the corresponding Fourier image.
One more thing that it should be noticed that very often people use in the books
slightly different representation for the Fourier transform, namely, they use so-called
energy type variables. It may be found very popular in quantum mechanics, where Einstein
and de-Broil formulas are used, which relate energy, frequency and the momentum with a
wave vector. It’s like a transition between wave characteristics and the particle characteristics.
This is done with the help of the Einstein formula where energy is equal to

𝐸 = ~𝜔 . (5.101)

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Formula (5.101) was discovered by Einstein, when he studied the photo effect.
The second formula, which is due to de-Broil is that momentum of a particle is
related to the wave vector of the corresponding wave process, represented by the particle.
According to the formula
𝑝⃗ = ~⃗𝑘 . (5.102)

It is possible to express 𝜔 and ⃗𝑘 from formulas (5.101) and (5.102) and get that

𝐸 𝑝⃗
𝜔= and ⃗𝑘 = . (5.103)
~ ~

When 𝜔𝑡 − ⃗𝑘 · ⃗𝑥 is replaced with the new variables, then the exponent will have the
next form:
⃗
𝑒𝑖(𝜔𝑡−𝑘·⃗𝑥) → 𝑒𝑖(𝐸·𝑡−⃗𝑝 ·⃗𝑥 )/~ . (5.104)

The same plane-wave was recieved in the (5.104), but written by using energy type
variables. In this case also, when the expression for the field 𝜑 is written, the next
expression will be obtained:
∫︁
d𝐸d⃗𝑝
𝜑 (𝑥) = 3/2
𝑒𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ 𝜑˜ (𝑝) . (5.105)
(2𝜋) 𝑐~4

The final thing is that if one of representations is taken, for instance, the representation
presented in the (5.105) and plug it into the Klein-Gordon equation, it will be discovered
that Fourier image satisfies the following equation:
[︃(︂ )︂ ]︃
2
𝐸
− 𝑝⃗ 2 − 𝑚2 𝑐2 𝜑˜ (𝑝) = 0 , (5.106)
𝑐

where 𝜑 depends on 𝑝𝜇 : (︂ )︂
𝜇
(︀ 0
)︀ 𝐸
𝑝 = 𝑝 , 𝑝⃗ = , 𝑝⃗ . (5.107)
𝑐
According to (5.107), it is possiblet to rewrite (5.106) in the following form:

𝑝𝜇 𝑝 − 𝑚2 𝑐2 𝜑˜ (𝑝) = 0 .
[︀ 𝜇 ]︀
(5.108)

There may be a thought that equation (5.108) has only trivial solution 𝜑˜ (𝑝) = 0,
because it is possible to think that the only thing to do is to just cancel the bracket before
𝜑˜ (𝑝). This would be so if the usual functions are used, but, in fact, one of the important
things, which happen in quantum field theory and also in classical field theory, is that,
in general, there should not be a thought about the field as about usual function on

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space-time. The right way is to think about the field as a distribution. So, if there is 𝜑 (𝑥)
or 𝜑 (𝑝), they will be distributions on space-time. Another name for this distributions is
generalized functions.
A very characteristic example of a distribution is, for instance, Dirac’s delta function.
This is an example of a function, which is should not be understood as a usual function,
but rather it is a distribution. Such a way distributions are linear continuous functionals
on a space of basic functions. In any case, if the field 𝜑 is understood in the distributional
sense, then equation (5.108) has the following solution:

𝜑˜ (𝑝) = 𝛿 𝑝2 − 𝑚2 𝑐2 𝜙 (𝑝) ,
(︀ )︀
(5.109)

where 𝜙 (𝑝) is a good continuous function without having any zero, where 𝑝2 − 𝑚2 𝑐2 = 0.
The condition
𝑝𝜇 𝑝𝜇 = 𝑚2 𝑐2 (5.110)

is called mass-shell condition, because it gives a relationship between the energy and the
particle momentum. According to this statement (5.110) has next alternative form:

𝐸 2 = 𝑝⃗ 2 𝑐2 + 𝑚2 𝑐4 . (5.111)

The condition (5.111) is exactly the relation between energy and momentum of a single
relativistic particle from the special relativity:
√︀
𝐸 = ± 𝑝⃗ 2 𝑐2 + 𝑚2 𝑐4 . (5.112)

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Lecture 6. More on Klein-Gordon equation. Canonical

Quantization

How to solve Klein-Gordon equation?

Let’s continue with solving the Klein-Gordon equation. It has been already introduced
the Fourier transform of the classical field 𝜑 (𝑥). Such a way, the next transform is made:

𝜑 (𝑥) → 𝜑˜ (𝑝) (6.1)

and it was found out that for the Fourier image of 𝜑 solution exists in the sense of
generalized functions:
𝜑˜ (𝑝) = 𝛿 𝑝2 − 𝑚2 𝑐2 𝜙 (𝑝) ,
(︀ )︀
(6.2)

where 𝜑˜ (𝑝) is a solution of the Klein-Gordon equation in the momentum space.

And the condition
𝑝2 − 𝑚2 𝑐2 = 0 (6.3)

is called mass-shell condition for relativistic particle and it gives a standard relativistic
relationship, because between the energy and the momentum of a particle.
If (6.3) is written with the help of a 3-dimensional vector of momentum and energy,
then it will have the following form:
(︂ )︂2
𝐸
− 𝑝⃗ 2 − 𝑚2 𝑐2 = 0 . (6.4)
𝑐

The solution of the (6.4) is given by

√︀
𝐸 = ±𝑐 𝑝⃗ 2 + 𝑚2 𝑐2 . (6.5)

There are two solutions with a positive and a negative energy. If the goal is to depict the
function 𝐸 as a function of the three dimensional momentum, a two sheeted hyperboloid
as it shown on the (fig. 6.1) will be obtained.
Then let’s take the solution (6.5) and return it back into the Fourier transform formula
and write the field 𝜑 (𝑥) as:
∫︁
d𝐸d⃗𝑝
𝜑 (𝑥) = 3/2
𝛿𝑝2 − 𝑚2 𝑐2 𝑒𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ 𝜙 (𝑝) , (6.6)
(2𝜋) 𝑐~4

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Fig. 6.1. The function 𝐸 as a function of the three dimensional momentum

where it is also possible to write the (6.6) in the natural units and take

𝑐~4 = 1 . (6.7)

In the Fourier transform there is a delta-function and this delta-function is useful,

because it is possible to use it in order to make an explicit integration. At least, it is
possible to integrate over energy variable 𝐸 and for that it is necessary to apply the
formula of a delta-function of a composite variable:
∑︁ 𝛿 (𝑥 − 𝑎𝑖 )
𝛿 (𝑓 (𝑥)) = , (6.8)
𝑖
|𝑓 ′ (𝑎𝑖 )|

where it is apologized that: ⎧

⎨𝑓 (𝑎𝑖 ) = 0
(6.9)
⎩𝑓 ′ (𝑎 ) ̸= 0
𝑖

Formula (6.8) can be called as a formula for change of variables in the delta-function.
Let’s apply (6.8) for the next delta-function
(︂ 2 )︂
𝐸 2 2 2
𝛿 − 𝑝⃗ − 𝑚 𝑐 , (6.10)
𝑐2

where (6.10) can be gotten from the 𝛿 (𝑝2 − 𝑚2 𝑐2 ) using (6.5).

Then expression (6.8) will be used for the delta-function 𝛿 (𝑝2 − 𝑚2 𝑐2 ) :
(︁ √︀ )︁ (︁ √︀ )︁
2 2 2 2 2 2
)︀ 𝑐𝛿 𝐸 − 𝑐 𝑝⃗ + 𝑚 𝑐 𝑐𝛿 𝐸 + 𝑐 𝑝⃗ + 𝑚 𝑐
𝛿 𝑝2 − 𝑚2 𝑐2 =
(︀
√︀ + √︀ . (6.11)
2 𝑝⃗ 2 + 𝑚2 𝑐2 2 𝑝⃗ 2 + 𝑚2 𝑐2

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The (6.11) will be easy to understand, if the delta-function is represented as:

𝛿 (𝑓 (𝐸)) , (6.12)

where
𝐸2
𝑓 (𝐸) = − 𝑝⃗ 2 − 𝑚2 𝑐2 . (6.13)
𝑐2
Such a way 𝑓 ′ (𝐸) will be equal to:
2𝐸
𝑓 ′ (𝐸) = . (6.14)
𝑐2
The derivative 𝑓 ′ (𝐸) should be evaluated at zeros of the function 𝑓 (𝐸). In one case,
there is ⃒ √︀
⃒ 2 𝑝⃗ 2 + 𝑚2 𝑐2
𝑓 ′ (𝐸)⃒ = . (6.15)
⃒
⃒ √ 𝑐
𝐸=𝑐 ⃗ 2 +𝑚2 𝑐2
𝑝

The same should be done with a negative root and then the formula will be obtained:
⃒ √︀
⃒ 2 𝑝⃗ 2 + 𝑚2 𝑐2
𝑓 ′ (𝐸)⃒ =− . (6.16)
⃒
⃒ √ 𝑐
2 2 2
𝐸=−𝑐 𝑝
⃗ +𝑚 𝑐

The answer, which will be obtained, if delta-function 𝛿 (𝑝2 − 𝑚2 𝑐2 ) is integrated in

(6.6) is:
[︂∫︁ ]︂ ∫︁
𝑐 d⃗𝑝 𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ d⃗𝑝 −𝑖(𝐸𝑡+⃗𝑝 ⃗𝑥 )/~
𝜑 (𝑥) = 𝑒 𝜙 (𝐸, 𝑝⃗ ) + 𝑒 𝜙 (−𝐸, 𝑝⃗ ) , (6.17)
(2𝜋)3/2 ~4 2𝐸 2𝐸

where 𝐸 here is not an independent variable anymore, but represents a positive branch
of the dispersion relation (6.5):
√︀
𝐸=𝑐 𝑝⃗ 2 + 𝑚2 𝑐2 . (6.18)

It is also possible to make a change 𝑝⃗ → −⃗𝑝 in the second integral of the (6.17) and
get the formula which people usually use:
[︂∫︁ ]︂ ∫︁
𝑐 d⃗𝑝 𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ d⃗𝑝 −𝑖(𝐸𝑡+⃗𝑝 ⃗𝑥 )/~
𝜑 (𝑥) = 3/2 4
𝑒 𝜙 (𝐸, 𝑝⃗ ) + 𝑒 𝜙 (−𝐸, −⃗𝑝 ) , (6.19)
(2𝜋) ~ 2𝐸 2𝐸

where the first term in the bracket is called positive frequency part of 𝜑 (𝑥) and the second
integral is called negative frequency of 𝜑 (𝑥).
It also should be noticed that d⃗𝑝 does not change the sign, when 𝑝⃗ → −⃗𝑝 is
transformed, because Jacobian modulus is always positive.

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The standard way to proceed further is to introduce the complex amplitudes. One of
them is called 𝑎* (𝑝) and this is taken to be

𝜙 (𝐸, 𝑝⃗ )
𝑎* (⃗𝑝 ) = √ . (6.20)
~3 2𝐸
Analogously, one defines an amplitude 𝑎 (𝑝), which is by definition is:

𝜙 (−𝐸, −⃗𝑝 )
𝑎 (⃗𝑝 ) = √ . (6.21)
~3 2𝐸
The gotten functions or amplitudes are only functions of momentum, because 𝐸 is also
a function of momentum and it comes from this function which was originally a function
of 4-momentum 𝑝 in the case of 4-dimensional Minkowski space. In other words, function
of 𝐸 and the function of 3-dimensional momentum 𝑝⃗ . It is known that because the delta-
function has been integrated, 𝐸 in the (6.19) is not an independent variable, but it is a
positive solution (6.18).
Then it is also possible to check that for a real scalar field amplitudes 𝑎 and 𝑎* right
the following
𝜙* (𝐸, 𝑝⃗ ) = 𝜙 (−𝐸, −⃗𝑝 ) . (6.22)

Expression (6.22) comes from considering the Fourier transform and using the fact
that our talk is about the real scalar field. Therefore 𝜑 (𝑥) is a real function. Then on
Fourier amplitude there will be the relation (6.22) and this tells us that amplitudes 𝑎
and 𝑎* are simply complex conjugate of each other. That’s why the star (*) has a simple
meaning of complex conjugation.
The Fourier transform, which is used to solve the Klein-Gordon equation, in terms of
amplitudes 𝑎 and 𝑎* takes the following form:
∫︁
d⃗𝑝 1 [︀ *
1/2
𝑎 (⃗𝑝 ) 𝑒𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ + 𝑎 (⃗𝑝 ) 𝑒−𝑖(𝐸𝑡+⃗𝑝 ⃗𝑥 )/~ .
]︀
𝜑 (𝑥) = 𝑐~ 3/2
√ (6.23)
(2𝜋~) 2𝐸

Then obviously it is also possible to compute the expression for canonical momentum.
It is known that canonical momentum is equal to
d𝜑 (𝑥)
𝜋 (𝑥) = . (6.24)
d𝑡
Computing the time derivative from (6.24), the following will be obtained:

d⃗𝑝 √ [︀ *
∫︁
𝑖 𝑖(𝐸𝑡−⃗
𝑝⃗𝑥 )/~ −𝑖(𝐸𝑡+⃗
𝑝⃗𝑥 )/~
]︀
𝜋 (𝑥) = 2𝐸 𝑎 (⃗
𝑝 ) 𝑒 − 𝑎 (⃗
𝑝 ) 𝑒 . (6.25)
2𝑐~1/2 (2𝜋~)3/2

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The important point is the following: it will be seen that the Hamiltonian, which will
be obtained, will be well defined and corresponds only to solutions with positive energy
and this can be seen in the following way. The idea now is to take our original Hamiltonian
𝐻 and rewrite it in terms of complex amplitudes 𝑎 and 𝑎* . What the thing that should
be done is to take just the Hamiltonian 𝐻, which is equal to:
∫︁ [︂ (︁ 𝑚𝑐 )︁2 ]︂
1 2 2
𝐻= d⃗𝑥 𝑐 𝜋 + 𝜕𝑖 𝜑𝜕𝑖 𝜑 + 𝜑2 . (6.26)
2 ~

So what is necessary to make next is to take expressions for 𝜑 (𝑥) (6.23) and for 𝜋 (𝑥)
and place it instead of 𝜑 (𝑥) and for 𝜋 (𝑥) in the (6.26).
When this substitution is done, one integration over ⃗𝑥 and two integration over
momentum 𝑝⃗ will be recieved. It is possible to exchange the order of integration and
it is practicable to integrate over ⃗𝑥 firstly. This will affect only on exponentials, because
only they contain variable ⃗𝑥 . So integration over ⃗𝑥 will give us a delta-function. Delta-
function will depend on the difference of momentum related to one copy of 𝜑 and another
will be related to another copy of 𝜑. This means that it is possible to use the delta-function,
which is obtained after integration over ⃗𝑥 to integrate over one momentum variable. At
the end we will be left over with just one integration over momentum. In the process of
evaluating 𝐻 it is needed to use the expression for delta-function, which has the following
form: ∫︁
d⃗𝑥 𝑖⃗𝑝 ⃗𝑥 /~
𝛿 (⃗𝑝 ) = 𝑒 . (6.27)
(2𝜋~)3
Then, if the integration over 𝑥 is done, the result for the Hamiltonian will look as
follows: ∫︁
1
𝐻= d⃗𝑝 𝐸 (⃗𝑝 ) 𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 ) , (6.28)
~
where 𝐸 (⃗𝑝 ) has the form of (6.18).
The expression for the Hamiltonian (6.28) has been written down in terms of complex
amplitudes and the Hamiltonian is real and expression 𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 ) is positive and equal
to:
𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 ) = |𝑎 (⃗𝑝 )|2 . (6.29)

It is also possible to rewrite the Hamiltonian in terms of frequency 𝜔 (𝑝), which is

related to the energy by Einstein formula:

𝐸 (⃗𝑝 ) = ~𝜔 (⃗𝑝 ) . (6.30)

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Expression (6.30) will remove Plank’s constant from expression (6.28). If (6.29) is
rewritten with the help of (6.30), then the Hamiltonian takes the form:
∫︁
𝐻 = d⃗𝑝 𝜔 (⃗𝑝 ) 𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 ) . (6.31)

Further, it is possible to rewrite also an expression for 𝜑 (𝑥) in terms of frequency:

d⃗𝑘
∫︁
1 (︁ (︁ )︁
⃗ −𝑖(𝜔𝑡−⃗𝑘 ⃗
𝑥)
(︁ )︁
* ⃗ 𝑖(𝜔𝑡−⃗𝑘 ⃗
𝑥)
)︁
𝜑 (𝑥) = 𝑐 √︂ (︁ )︁ 𝑎 𝑘 𝑒 +𝑎 𝑘 𝑒 . (6.32)
(2𝜋)3/2 ⃗
2𝜔 𝑘

(︁ )︁
The piece in the (6.32), which is related to 𝑎* ⃗𝑘 is called positive frequency and
𝑎 is called negative frequency. The positive and negative is a convention. Sometimes in
the books, people use an opposite convention, they call what we call positive frequency
negative and what we call negative, they call positive. The definition depends on which
literature, you take. This is related to the fact that an opposite way to cause this amplitude
is related to ideas that if it is possible to see the Schrodinger equation:
𝜕 −𝑖𝜔𝑡
𝑖~ 𝑒 = ⏟~𝜔 𝑒−𝑖𝜔𝑡 , (6.33)
𝜕𝑡 ⏞
𝐸

then the wave with a negative frequency −𝑖𝜔𝑡 corresponds to the positive solution.
If the sign in exponent of (6.33) is changed, then the following expression will be
obtained:
𝜕 𝑖𝜔𝑡
𝑒 = −𝐸𝑒𝑖𝜔𝑡 .
𝑖~ (6.34)
𝜕𝑡
That’s why in the literature sometimes exponent of −𝑖𝜔𝑡 associated oscillator to it is
called positive frequency. But this convention will not be used since Schrodinger equation
is not important in our case and Klein-Gordon equation is used. Therefore, positive
frequency will be associated with positive exponent of 𝑖𝜔𝑡 correspondent in quantum
theory to creation operators. The exponent of −𝑖𝜔𝑡 will correspond negative frequency
and annihilation operators.
Let’s look at the physical meaning of amplitudes 𝑎* and 𝑎. It can be seen that from the
solution for field 𝜑 (𝑥) it is useful to define the time-dependent amplitudes. So, 𝑎* (⃗𝑝 , 𝑡)
we can associate with 𝑎* (⃗𝑝 ):
⎧
⎨𝑎* (⃗𝑝 , 𝑡) = 𝑎* (⃗𝑝 ) 𝑒𝑖𝜔𝑡
(6.35)
⎩𝑎 (⃗𝑝 , 𝑡) = 𝑎 (⃗𝑝 ) 𝑒−𝑖𝜔𝑡

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It can be seen that this time-dependent amplitudes can be obtained as a solution of

the Hamilton’s equations of motion for 𝑎 and 𝑎* and Hamilton’s equations means that
⎧
⎨ d𝑎 = {𝐻, 𝑎}
d𝑡
(6.36)
⎩ d𝑎* = {𝐻, 𝑎* }
d𝑡

where 𝐻 is a Hamiltonian, given by formula (6.31). The Poisson bracket is the usual
oscillator bracket, which is equal to

{𝑎 (⃗𝑝 ) , 𝑎* (⃗𝑝 ′ )} = 𝑖𝛿 (⃗𝑝 − 𝑝⃗ ′ ) , (6.37)

where now there is an infinite collection of oscillators parameterized by the continuous

parameter 𝑝. It is also possible to write following Poisson’s brackets:

{𝑎 (⃗𝑝 ) , 𝑎 (⃗𝑝 ′ )} = 0 = {𝑎* (⃗𝑝 ) , 𝑎* (⃗𝑝 ′ )} . (6.38)

Such a way a system in terms of complex amplitudes and its equations of motion
according to the Hamiltonian formalism were described, where Hamiltonian is given by
formula (6.31).
If the classical massive Klein-Gordon field is considered, it is possible to say that this
is nothing is just an infinite collection of harmonic oscillators, which oscillate with the
relativistic frequency 𝜔 (⃗𝑝 ).
Such a representation of a scalar Klein-Gordon field 𝜑 (𝑥) in terms of complex amplitudes
𝑎 and 𝑎* is called holomorphic representation.
If the Poisson’s bracket is used between the oscillators and use this expression is used
for fields 𝜑 (𝑥) and 𝜋 (𝑥), which we have showed before, it is possible to compute the
Poisson’s bracket between 𝜑 (𝑥) and 𝜑 (𝑥′ ), as well as between 𝜑 (𝑥) and 𝜋 (𝑥′ ) and also
between 𝜋 (𝑥) and 𝜋 (𝑥′ ):
⎧
{𝜑 (𝑥) , 𝜑 (𝑥′ )} = 0
⎪
⎪
⎪
⎪
⎨
{𝜑 (𝑥) , 𝜋 (𝑥′ )} = 𝑖𝛿 (𝑥 − 𝑥′ ) (6.39)
⎪
⎪
⎪
⎩{𝜋 (𝑥) , 𝜋 (𝑥′ )} = 0
⎪

Such a way the standard formulas were restored for canonical Poisson’s brackets
between field and its momentum, that has been already discussed.
It is also available to write down expressions for other generators of the Poincare group.
For instance, Hamiltonian is one of the generators of the Poincare group, but there are

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other generators: spatial shifts, spatial rotations and Lorenz boosts. Let’s now use this
holomorphic representation to obtain expression for other generators of Poincare group
in terms of amplitudes 𝑎 and 𝑎* .
In particular, let’s write down the expression for generators of spatial shifts
∫︁
𝑃𝑖 = d⃗𝑥 𝜋𝜕𝑖 𝜑 (6.40)

in terms of oscillators, which takes the following form:

∫︁
1
𝑃𝑖 = d⃗𝑝 𝑝𝑖 𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 ) . (6.41)
~
Generators of spatial rotations 𝑀𝑖𝑗 is given by:
∫︁
𝑀𝑖𝑗 = d⃗𝑥 (𝑥𝑗 𝜕𝑖 𝜑 − 𝑥𝑖 𝜕𝑗 𝜑) , (6.42)

where this rotation generators were obtained from Noether’s theorem. If solution for the
field 𝜑 is substituted in terms of complex amplitudes, the following expression will be
found: ∫︁
𝑀𝑖𝑗 = 𝑖 d⃗𝑝 𝑎* (⃗𝑝 ) (𝑝𝑖 𝜕𝑗 − 𝑝𝑗 𝜕𝑖 ) 𝑎 (⃗𝑝 ) . (6.43)

Then it is necessary to find expression for Lorenz boosts. It has the following form:
∫︁
1
𝑀0𝑖 = d⃗𝑥 𝑥𝑖 H − 𝑐𝑡𝑃𝑖 . (6.44)
𝑐
So Lorenz boosts are the most complicated generators and they have explicit dependence
on time that one of the features of this generators.
If (6.31) is rewritten in terms of oscillators, the following expression will be obtained:
∫︁
𝑖
𝑀0𝑖 = d⃗𝑝 𝐸 (⃗𝑝 ) (𝑎* (⃗𝑝 ) 𝜕𝑖 𝑎 (⃗𝑝 ) − 𝜕𝑖 𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 )) − 𝑐𝑡𝑃𝑖 , (6.45)
2𝑐
where 𝑃𝑖 is the same thing as in the (6.41), which is given in terms of oscillators.
So Lorenz boost have an explicit time-dependence and this is a manifestation of the fact
that in the Hamiltonian formulation the boost symmetries broken. This is not surprising,
because in order to develop the Hamiltonian formalism it is necessary to fix the direction
of time.
In the Hamiltonian formalism time plays always a distinguished role, while on the
other hand, it is known that when Lorenz boost is done, the time direction is mixed with
spatial directions.

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If a boost is considered, then it is possible to get an explicit time dependence in the

expression for this generator, but the conservation law provided by the Noether’s theorem
is of course the law, which holds. So, if a total derivative of boost generator over time is
taken, it is known that in the Hamiltonian mechanics, if our generator has an explicit time
dependence and it is needed to compute the total derivative accurately, first the partial
derivative is computed, keeping all the other variables fixed and the boost is differentiated
over its explicit time dependence and then a Poisson bracket of the Hamiltonian is added
up with 𝑀0𝑖 :
d𝑀0𝑖 𝜕𝑀0𝑖
= + {𝐻, 𝑀0𝑖 } . (6.46)
d𝑡 𝜕𝑡
That is how time evolution equation should be written for any function, which has an
explicit time dependence, and therefore, if this expression for the boost is differentiated
with respect to the time variable, it will be obtained that:
d𝑀0𝑖
= {𝐻, 𝑀0𝑖 } − 𝑐𝑃𝑖 . (6.47)
d𝑡
d𝑀0𝑖
In other words, if 𝑀0𝑖 is conserved in time, then the total derivative d𝑡
is equal to
zero and from here it is available to get an expression that the Poisson bracket

{𝐻, 𝑀0𝑖 } = 𝑐𝑃𝑖 . (6.48)

Expression (6.48) is nothing else as a part of the Poincare algebra. So 𝐻 generates

shifts in time and shifts in time commuting with boosts with respect to the Poisson bracket
produce the generators of shifts in spatial directions. This is one way how relations of the
Poincare algebra may be reproduced. But if the Poisson bracket between the oscillators
is used and the Poisson relations of all these generators is computed between themselves,
it will be discovered that they form nothing else as Poincare algebra with respect to this
Poisson brackets for oscillator variables 𝑎 and 𝑎* .
The classical part is finished and the work with classical fields is also finished. The
Klein-Gordon equation was solved and rewrote all physical quantities such as Hamiltonian
and generators of the Poincare group were rewritten. Then the canonical quantization will
be proceeded.

Canonical quantization
Canonical quantization consists in replacing the equal time Poisson bracket of classical
fields with quantum Poisson brackets with what is called quantum Poisson bracket denoted

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by the same expression as a classical Poisson bracket with a subscript ~:

{, } → {, }~ . (6.49)

The realization of this quantum Poisson bracket is known and according to Dirac, this
quantum Poisson bracket must be equal to:
𝑖
[, ] . (6.50)
~
Expression (6.50) represents a Dirac way of quantizing fields and it’s called canonical
quantization.
On the one hand, one uses for the quantum Poisson bracket expression (6.49), while
for the right hand side of this quantum Poisson bracket one uses the same expression
as for the classical Poisson bracket. This means that when we quantize, we promote our
classical field 𝜑 (𝑥) becomes an operator. The same with momentum 𝜋 (𝑥): it becomes an
operator, acting in some space, which will be described a bit later.
The essential point is that this operators 𝜑 (𝑥) and 𝜋 (𝑥) have the following quantum
Poisson brackets:
𝑖
{𝜑 (𝑡, ⃗𝑥 ) , 𝜑 (𝑡, ⃗𝑦 )}~ =
[𝜑 (𝑡, ⃗𝑥 ) , 𝜑 (𝑡, ⃗𝑦 )] . (6.51)
~
In expression (6.51) the equal time Poisson bracket is replaced with an equal time
commutator. Since this canonical quantization is considered the quantum Poisson bracket
is replaced by the value of the classical Poisson bracket which is zero, therefore:
𝑖
[𝜑 (𝑡, ⃗𝑥 ) , 𝜑 (𝑡, ⃗𝑦 )] = 0 . (6.52)
~
This means that fields commutes at different space points, but at the same moment
of time. The same should be done for canonical momenta. Also there is zero on the right
hand side:
𝑖
{𝜋 (𝑡, ⃗𝑥 ) , 𝜋 (𝑡, ⃗𝑦 )}~ =
[𝜋 (𝑡, ⃗𝑥 ) , 𝜋 (𝑡, ⃗𝑦 )] = 0 , (6.53)
~
while for the quantum Poisson bracket between momentum and field the following is
obtained:
𝑖
{𝜋 (𝑡, ⃗𝑥 ) , 𝜑 (𝑡, ⃗𝑦 )}~ =
[𝜋 (𝑡, ⃗𝑥 ) , 𝜑 (𝑡, ⃗𝑦 )] = 𝛿 (⃗𝑥 − ⃗𝑦 ) . (6.54)
~
Expression (6.54) can be rewritten as a equal time commutator of canonical momentum
with the field is given by:

[𝜋 (𝑡, ⃗𝑥 ) , 𝜑 (𝑡, ⃗𝑦 )] = −𝑖~𝛿 (⃗𝑥 − ⃗𝑦 ) . (6.55)

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It’s more convenient to write it as a following commutator:

[𝜑 (𝑡, ⃗𝑥 ) , 𝜋 (𝑡, ⃗𝑦 )] = 𝑖~𝛿 (⃗𝑥 − ⃗𝑦 ) . (6.56)

Relations between quantum fields are called canonical commutation relations. Similarly,
upon quantization the classical amplitudes 𝑎* and 𝑎 are replaced by operators 𝑎† and 𝑎:
⎧
⎨𝑎* (⃗𝑝 ) → 𝑎† (⃗𝑝 )
(6.57)
⎩𝑎 (⃗𝑝 ) → 𝑎 (⃗𝑝 )

They are understood as creation and annihilation operators. Meaning of creation and
annihilation will be clarified a little bit later. So the creation and annihilation operators for
harmonic oscillator labeled by this momentum variable 𝑝⃗ and the commutation relation
between 𝑎 and 𝑎† are given by the formula:
⎧
[𝑎 (⃗𝑝 ) , 𝑎 (⃗𝑝 ′ )] = 0
⎪
⎪
⎪
⎪
⎨[︀
𝑎† (⃗𝑝 ) , 𝑎† (⃗𝑝 ′ ) = 0
]︀
(6.58)
⎪
⎪
⎪
⎩ 𝑎 (⃗𝑝 ) , 𝑎† (⃗𝑝 ′ ) = ~𝛿 (⃗𝑝 − 𝑝⃗ ′ )
⎪ [︀ ]︀

In a way, it can be said that now an abstract algebra generated by generators 𝑎 (⃗𝑝 )
and 𝑎† (⃗𝑝 ) is studied. This algebra satisfies the commutation relations (6.58).
The commutativity of fields at different space-time points, but at the same moment of
time is important. In fact, from the quantum mechanical point of view commutativity of
operators means that these operators can be measured simultaneously. The eigenvalues
for this operators can be simultaneously measured. This fields then are independent
observable. It is like the same field, but taken as the one and the same moment of time,
but at different space points represent different observables, because the values of this
field at the same time, but at different space time points commute.
Then, an important representation will be constructed for this moment abstract variables
𝑎 and 𝑎† . Construction of representation means that the 𝑎 and 𝑎† will be identified as
concrete operators acting in some space. It is needed that this space to be also the Hilbert
space. The norm in this space will be introduced and the 𝑎 and 𝑎† will be realized as
explicit operators acting on this space. The construction of this particularly important
representation of quantum field theory is carried out in the following way: first of all, it
is needed to introduce a state in the direct notation bra and ket to introduce a particular
state, which can be called as vacuum state

|0⟩ , (6.59)

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which is a state without particles from physical point of view. Then it is assumed that this
vacuum state is specified by the condition that, when any of the annihilation operators
act on it, it gives zero:
𝑎 (⃗𝑝 ) |0⟩ = 0 . (6.60)

That’s why these operators have a name of annihilation operators. So, annihilation operators
annihilate the vacuum state. Acting on this state with creation operators, which are 𝑎† (⃗𝑝 )
will create a one particle state with momentum 𝑝⃗ , which is denoted in this way as:
√︀
2𝜔 (⃗𝑝 )𝑎† (⃗𝑝 ) |0⟩ = |⃗𝑝 ⟩ , (6.61)
√︀
where a normalization 2𝜔 (⃗𝑝 ) is indroduced.
In expression (6.61) it is recieved that 𝑎† acting on the vacuum gives one particle
states, which are labeled by the momentum 𝑝⃗ .
So, our start was from the vacuum |0⟩, then one particle state with momentum |⃗𝑝 1 ⟩ was
constructed. Then it is possiblet to construct a state of two particles with momentums 𝑝⃗ 1
and 𝑝⃗ 2 - |⃗𝑝 1 , 𝑝⃗ 2 ⟩ and so on. More and more creation operators with different momentum
are applied and then some number of particles are created, which in physical interpretation
carries this momentum from 𝑝⃗ 1 up to 𝑝⃗ 𝑛 - |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩. The space of all these states from
vacuum up to 𝑛’s particle state is called Fock space. Sometimes the representation of field
operators in the Fock space is called representation of second quantization. This term will
be explained a little bit later. Where does it come from? Why second? What does it mean
second quantization? Why people talk about second quantization, when they talk about
representation of field operators in the infinity dimensional Fock space?
So, the question is why this is a representation. In the Fock space operators 𝑎 and 𝑎†
act in the following way: operator 𝑎† simply adds a new particle with momentum 𝑝⃗ . If
the creation operator is taken and applied to a state, which has already 𝑛 particles in it
like, for instance, state |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩, then what this operator does, it’s just adds a new
particle:
1
𝑎† (⃗𝑝 ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = √︀ |⃗𝑝 , 𝑝⃗ 1 , . . . , 𝑝⃗ 𝑛 ⟩ , (6.62)
2𝜔 (⃗𝑝 )
where to keep the normalization it is necessary to divide by the square root of 2𝜔 (⃗𝑝 ).
The action operator or annihilation operator 𝑎 (⃗𝑝 ) is a bit more complicated. When
this operator acts on a state with 𝑛 particles, it’s supposed to annihilate:
𝑛
∑︁ ⃒ ⟩
𝛿 (⃗𝑝 − 𝑝⃗ 𝑖 ) ⃒𝑝⃗ 1 , . . . , ⃗ˆ𝑝𝑖 , . . . , 𝑝⃗ 𝑛 ,
√︀
𝑎 (⃗𝑝 ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = ~ 2𝜔 (⃗𝑝 ) (6.63)
⃒
𝑖=1

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where ~ 2𝜔 (⃗𝑝 ) is normalization and ⃗ˆ𝑝𝑖 means that this particle is absent. And also it
√︀

is needed to show distinguish action on the vacuum:

𝑎 (⃗𝑝 ) |0⟩ = 0 . (6.64)

So, all the 𝑎† operators commute between themselves, all 𝑎 operators commute between
themselves, that it is possible to verify that this actions are compatible with the non trivial
property is that what happens when 𝑎 (⃗𝑝 ) and 𝑎† (⃗𝑝 ) meets. Let’s write:

𝑎 (⃗𝑝 ) 𝑎† (⃗𝑝 ′ ) (6.65)

and then let’s act with this operators on the 𝑛 particle state and use formulas (6.62) and
(6.63):
1
𝑎 (⃗𝑝 ) 𝑎† (⃗𝑝 ′ ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = √︀ 𝑎 (⃗𝑝 ) |⃗𝑝 ′ , 𝑝⃗ 1 , . . . , 𝑝⃗ 𝑛 ⟩ . (6.66)
2𝜔 (⃗𝑝 ′ )
Thenit is neessary to use 𝑎 (⃗𝑝 ) operator:

𝑎 (⃗𝑝 ) 𝑎† (⃗𝑝 ′ ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = ~𝛿 (⃗𝑝 − 𝑝⃗ ′ ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ +

√︁ ⃒ ⟩ (6.67)
+~ 𝜔(⃗ 𝜔(⃗𝑝 ) ∑︀𝑛
𝛿 (⃗
𝑝 − 𝑝
⃗ 𝑖 )
⃒ ′
𝑝
⃗ , 𝑝
⃗ 1 , . . . , ˆ
⃗
𝑝 𝑖 , . . . , 𝑝
⃗ 𝑛 .
𝑝 ′) 𝑖=1 ⃒

Now the goal is to act with an opposite order of operators:

𝑛
∑︁ ⃒ ⟩
𝛿 (⃗𝑝 − 𝑝⃗ 𝑖 ) 𝑎 (⃗𝑝 ) ⃒𝑝⃗ 1 , . . . , ⃗ˆ𝑝𝑖 , . . . , 𝑝⃗ 𝑛 . (6.68)
† ′
√︀ † ′ ⃒
𝑎 (⃗𝑝 ) 𝑎 (⃗𝑝 ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = ~ 2𝜔 (⃗𝑝 )
𝑖=1

Finally, the operator 𝑎† (⃗𝑝 ′ ) will add one more particle and we will get:
√︃
𝑛
† ′ 𝜔 (⃗𝑝 ) ∑︁ ⃒
⃒ ′ ˆ
⟩
𝑎 (⃗𝑝 ) 𝑎 (⃗𝑝 ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = ~ 𝛿 (⃗𝑝 − 𝑝⃗ 𝑖 ) ⃒𝑝⃗ , 𝑝⃗ 1 , . . . , ⃗𝑝𝑖 , . . . , 𝑝⃗ 𝑛 . (6.69)
𝜔 (⃗𝑝 ′ ) 𝑖=1

Then it is necessary to compare the results of two gotten expressions (6.67) and (6.69)
for actions of operators 𝑎 and 𝑎† , but in different order. If we subtract from the first
actions, the second one, an expression for the commutator will be recieved. If a state with
𝑛 particles is studied or, in other words, a more or less arbitrary state with any number
of particles is considered, it will be clear that under the commutator the second term of
(6.67) cancels the first term of (6.69). Such a way, we will get the following formula:

𝑎 (⃗𝑝 ) , 𝑎† (⃗𝑝 ′ ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ = ~𝛿 (⃗𝑝 − 𝑝⃗ ′ ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ .

[︀ ]︀
(6.70)

The gotten expression is valid for any state in the Fock space. The number of particles
plays no role. Therefore, it can be said that the representation of algebra of operators 𝑎
and 𝑎† has been realized.

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This way shows how to prove that representation of the oscillator algebra of creation
and annihilation operators in the Fock space has been constructed.
An arbitrary state in the Fock space is a superposition of 𝑛 particle states. An arbitrary
state, which can be denoted by 𝜒, will be given by:
∞ 𝑛
∫︁ ∏︁
∑︁ 1 d⃗𝑝
|𝜒⟩ = √ √︀ 𝑖 𝜒𝑛 (⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ) |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩ . (6.71)
𝑛=0 𝑛! 𝑖=1 2𝜔 (⃗𝑝 𝑖 )
Expression (6.71) represents a superposition, where states with a different number of
particles are superposed and this is reflected by the term with sum from zero to infinity
and also particles with different momentum are superposed.
The momentum could be different, but since momentum continues, summing over
continuous variable is the same as to integrate over this variable and arbitrary coefficients
in front of different momentum can be implemented by putting functions 𝜒𝑛 (⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ).
Of course, it is possible to compute the norm of such a state. First of all, there is an
assumption about the vacuum state that this is a state with a well defined norm equal to
one. The norm of this state is scalar product of the state with itself and by definition is
taken to be equal to one:
⟨0|0⟩ = 1 . (6.72)

It is possible to use the definition of the 𝑛 particle states to compute the scalar product
of states with different particles:

⟨⃗𝑞 1 , . . . , ⃗𝑞 𝑛 |⃗𝑝 1 , . . . , 𝑝⃗ 𝑚 ⟩ . (6.73)

What is meant by definition is that it is understood that if a state is created by acting

with a creation operator on the vacuum, then conjugation means that the conjugate state
will be a state with bra vacuum acted by 𝑎 (⃗𝑝 ):

𝑎† (⃗𝑝 ) |0⟩ → ⟨0| 𝑎 (⃗𝑝 ) . (6.74)

In other words, operators 𝑎 and 𝑎† are considered to be Hermitian conjugate to each other.
Then, if then this assumption or this convention is used for how the conjugate states are
understood, then it is possible to write in terms of annihilation operators the bra state
⟨⃗𝑞 1 , . . . , ⃗𝑞 𝑛 | and with creation operators the ket state |⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ⟩. Then it is needed to
move operators of annihilation through the creation operators to reach the right vacuum
and produce zero:
𝑛
∏︁ 𝑚
∏︁
√︀ √︀
2𝜔 (⃗𝑝 𝑙 ) 0⃒𝑎 (⃗𝑞 1 ) . . . 𝑎 (⃗𝑞 𝑛 ) 𝑎† (⃗𝑝 1 ) . . . 𝑎† (⃗𝑝 𝑚 )⃒0
⟨︀ ⃒ ⃒ ⟩︀
(6.73) = 2𝜔 (⃗𝑞 𝑘 ) (6.75)
𝑘=1 𝑙=1

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Then it is necessary to use commutation relations between 𝑎 and 𝑎† to move 𝑎 through

𝑎† to the right. Such a way 𝑎 will reach the right vacuum and annihilate it and then
analogously 𝑎† move to the left and then annihilate the left vacuum. Every time 𝑎 is
commuted through 𝑎† a delta-function will be recieved, due to the commutation relations
between 𝑎 and 𝑎† . If this computations are done, the following result will be obtained:
𝑛
∑︁ ∏︁
(6.73) = 𝛿𝑚𝑛 (2~𝜔 (⃗𝑝 𝑖 )) 𝛿 (⃗𝑝 𝑖 − ⃗𝑞 P𝑖 ) . (6.76)
P 𝑖=1

where symbol P means a sum of all permutations. Permutation is understood as a map

(︃ )︃
1 2 ... 𝑛
P= (6.77)
𝛼1 𝛼2 . . . 𝛼𝑛

and permutation is a bijective map from {1, 2, . . . , 𝑛} to itself. That is what is called a
symmetric group. So, such bijective maps from 1 to 𝑛 form a symmetric group, which
mathematical name is 𝑆𝑛 . And permutation is an element of the group 𝑆𝑛 .
Expression (6.77) shows how the scalar product between particles with definite momentum
looks like. If now the scalar product is used to compute the scalar product of an arbitrary
state ⟨𝜓|𝜒⟩, which are built on the functions 𝜓 and 𝜒, the following result will be found:
∞ ∫︁
∑︁ 𝑛
∏︁
⟨𝜓|𝜒⟩ = 𝜓𝑛* (⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ) 𝜒𝑛 (⃗𝑝 1 , . . . , 𝑝⃗ 𝑛 ) d⃗𝑝 𝑖 . (6.78)
𝑛=0 𝑖=1

This is, in a way the standard scalar product, which will be used use in quantum mechanics
to make the space of square integrable functions on a space 𝑅𝑛 or simply on 𝑅 to become
a Hilbert space. The (6.78) represents a structure of the Hilbert space is introduced by
means of the scalar product and in a sense this is the quantum mechanical generalization
or simply generalization to the case of infinite number of particles of the standard quantum
mechanical scalar product, which supplies the space of square integrable functions with
the structure of the Hilbert space. So, Fock space is a Hilbert space. This also partially
√︀
explains why the 2𝜔 (⃗𝑝 ) is introduced in the definition of states obtained by creation
operator, it’s explained by the fact that, if a scalar product of two states is considered,
for instance, of two one particle states ⟨⃗𝑝 |⃗𝑞 ⟩, then it gives us:

⟨⃗𝑝 |⃗𝑞 ⟩ = 2~𝜔 (⃗𝑝 ) 𝛿 (⃗𝑝 − ⃗𝑞 ) , (6.79)

which is the same as

⟨⃗𝑝 |⃗𝑞 ⟩ = 2𝑐𝑝0 𝛿 (⃗𝑝 − ⃗𝑞 ) , (6.80)

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if it is written in terms of zero component of the momentum, then the scalar product,
because of the factor 2~𝜔 (⃗𝑝 ), is relativistic invariant.
What does it mean? It means that if the Lorentz transformation with 𝑝 and 𝑞 or is
done, in other words, go to another frame by applying Lorentz transformation on both 𝑝
and 𝑞:
𝑝′𝜇 = Λ𝜇𝜈 𝑝𝜈 , 𝑞 ′𝜇 = Λ𝜇𝜈 𝑞 𝜈 , (6.81)

then the product ⟨⃗𝑝 |⃗𝑞 ⟩ will be relativistic invariant.

So, if ⟨Λ⃗𝑝 |Λ⃗𝑞 ⟩ is done, where Λ is an arbitrary Lorentz transformation, then the scalar
product remains invariant:
⟨Λ⃗𝑝 |Λ⃗𝑞 ⟩ = ⟨⃗𝑝 |⃗𝑞 ⟩ . (6.82)

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Lecture 7. Second Quantization. Commutation and

Green’s Functions, Pauli-Jordan Function
Last lecture the procedure of canonical quantization was discussed, where the Poisson
brackets for fields and the momenta of the fields have been essentially replaced with
quantum Poisson brackets that are commutators and the construction of the Fock space
has been also discussed, which is an infinitive dimensional Hilbert space constructed by a
successive application of creation operators to the unique vacuum state.
In this lecture our goal is to develop these operator concepts.
Now it is possible to come to the question about the Hamiltonian. In the classical
theory it is known that the Hamiltonian can be written in terms of creation and annihilation
operators in the following way:
∫︁
𝐻= d⃗𝑝 𝜔 (⃗𝑝 ) 𝑎* (⃗𝑝 ) 𝑎 (⃗𝑝 ) . (7.1)

The (7.1) was a classical expression for the Hamiltonian. If a similar computation for
the Hamiltonian 𝐻 in quantum field theory is done, of course, it is known that 𝑎* (⃗𝑝 )
and 𝑎 (⃗𝑝 ) will be replaced by 𝑎† (⃗𝑝 ) and 𝑎 (⃗𝑝 ), but it should be done carefully, because
these two operators: 𝑎† (⃗𝑝 ) and 𝑎 (⃗𝑝 ), do not commute with each other and, therefore,
the order of these operators in any operator expression, which was constructed with the
help of the 𝑎 and 𝑎† , matters. For the case of the Hamiltonian one accepts to write the
following expression for the quantum Hamiltonian:
∫︁
𝐻 = d⃗𝑝 𝜔 (⃗𝑝 ) 𝑎† (⃗𝑝 ) 𝑎 (⃗𝑝 ) (7.2)

meaning that operator 𝑎† stands on the left from an operator 𝑎. This way of ordering of
operators 𝑎 and 𝑎† is called normal ordering. In other words, normal ordering is a rule of
ordering of operators 𝑎 and 𝑎† in such a way that operators 𝑎† always go to the left or
always stand on the left from the operator 𝑎.
In other words, if there is a certain expression from operators 𝑎 and 𝑎†

𝑎𝑎† 𝑎𝑎† 𝑎 (7.3)

and then the procedure of normal ordering is applied to this expression, then this expression
should go to
𝑎† 𝑎† 𝑎𝑎𝑎 . (7.4)

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It should be recalled that the order of operators 𝑎 between themselves and 𝑎† -s between
themselves does not matter, because operators 𝑎 commute between themselves and 𝑎† also
commute. Their relative order does not matter, but what matters is that all 𝑎† -s stand
on the left from operators 𝑎. To symbolize this way of ordering people use the special
notation, which is called normal ordering and they write double dots on the left from the
expression, which will be put in normal order and to the right of it. So, expression (7.3)
in the normal order form is given by the expression:

: 𝑎𝑎† 𝑎𝑎† 𝑎 : . (7.5)

The same idea of normal ordering is also applied to fields. If there is a set of fields
{𝜑1 (⃗𝑥 1 ) , . . . 𝜑𝑛 (𝑥𝑛 )} and our goal is to put their product in normal order, that means
that in the resulting expression first of all all the fields will be written in terms of creation
and annihilation operators and then all creation operators will be placed to the left from
all annihilation operators.
The idea of adopting such ordering prescription goes to the fact that the energy of any
state in Fock space becomes a well-defined quantity. In particular, vacuum carries zero
energy, because if 𝐻 is applied to the vacuum state then that would be
∫︁
𝐻 |0⟩ = d⃗𝑝 𝜔 (⃗𝑝 ) 𝑎† (⃗𝑝 ) 𝑎 (⃗𝑝 ) |0⟩ . (7.6)

Since vacuum is annihilated by operator 𝑎 (⃗𝑝 ), then the result of application of 𝐻 to

the vacuum state will give us zero. On the other hand, if the Hamiltonian is written in
the opposite way or if anti-normal ordering is assumed, for instance, where 𝑎† is placed
to the right, then, in fact, our Hamiltonian will create first a one particle state out of the
vacuum and then this one particle state will be annihilated by means of operator 𝑎:

𝑎 (⃗𝑝 ) 𝑎† (⃗𝑝 ) |0⟩ . (7.7)

As can be seen, in order to compute the result of this action it is necessary to compute
𝑎 (⃗𝑝 ) through 𝑎† (⃗𝑝 ), which will result into the 𝛿 (0), because it would be 𝛿 (⃗𝑝 − 𝑝⃗ ). So, the
𝛿 (0) is actually infinite and in this case it can be seen that the action of the Hamiltonian
on the vacuum state will give us a non-sensic equation, because the result of application
will give us just zero. In order to prevent the problem with defining the Hamiltonian
of any state in the Fock space and just to make it well defined, it is possible to apply
the normal ordering prescription. It’s very important to remember about this, that in
quantum field theory for operators acting in the Fock space one usually uses the normal

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order and prescription. In particular, this is done for the Hamiltonian in order to render
the energy of Fock states well defined.
Now if our field 𝜑 is considered again, then in terms of creation and annihilation
operators the field 𝜑, which now is an operator has the following expansion. First of all
there is the measure factor. If it is written with all ~ and 𝑐 constants it will look like:
∫︁
d⃗𝑝 1 (︀ † 𝑖(𝐸𝑡−⃗
𝑝⃗𝑥 )/~ −𝑖(𝐸𝑡−⃗
𝑝⃗𝑥 )/~
)︀
𝜑 (𝑥) = 𝑐 𝑎 (⃗
𝑝 ) 𝑒 + 𝑎 (⃗
𝑝 ) 𝑒 . (7.8)
(2𝜋~)3/2 2𝜔 (⃗𝑝 )
√︀

Let’s now take the positive frequency part of the field 𝜑 (⃗𝑥 ), which contains 𝑎† . This
piece will be denoted as 𝜑+ :
∫︁
d⃗𝑝
+
𝜑 (⃗𝑥 , 0) = 𝑐 3/2
𝑎+ (⃗𝑝 ) 𝑒−𝑖⃗𝑝 ⃗𝑥 /~ . (7.9)
(2𝜋~)
Let’s take the vacuum state and act on this vacuum state by operator 𝜑+ . Such a way
shows that: ∫︁
1d⃗𝑝
+
𝜑 (⃗𝑥 , 0) |0⟩ = 𝑐 √︀ 𝑒−𝑖⃗𝑝 ⃗𝑥 /~ |⃗𝑝 ⟩ .
3/2
(7.10)
(2𝜋~) 2𝜔 (⃗𝑝 )
In particular, if another one particle state with a non-trivial momentum 𝑝⃗ is taken
and a bra state ⟨⃗𝑝 | is taken, then it will be seen that
∫︁
⟨︀ ⃒ + ⃒ ⟩︀ 𝑐 𝛿⃗𝑞 ⃒𝑎 (⃗𝑞 )⃒0 𝑒−𝑖⃗𝑞 ⃗𝑥 /~ ,
⟨︀ ⃒ + ⃒ ⟩︀
𝑝⃗ 𝜑 (⃗𝑥 , 0) 0 =
⃒ ⃒ 𝑝
⃗ (7.11)
(2𝜋~)3/2
√︀
2𝜔 (⃗𝑞 )
where the variable, over which the integration is, is denoted by ⃗𝑞 and 𝑝⃗ is a certain fixed
momentum, which defines the one particle state ⟨⃗𝑝 |. It is possible to compute the quantity
described by (7.11) and it is necessary, first of all, to write down an expression for a state
⟨⃗𝑝 |. It is known that it can be represented as:
√︀
⟨⃗𝑝 | = ⟨0| 𝑎 (⃗𝑝 ) 2𝜔 (⃗𝑝 ) . (7.12)

When (7.12)is plugged in the scalar product from (7.11), 𝑎 (⃗𝑝 ) will be commuted with
𝑎† (⃗𝑞 ) and 𝑎 (⃗𝑝 ) will be moved to the right vacuum to annihilate it and as a result of this
evaluation the delta function on 𝑝⃗ − ⃗𝑞 will be obtained:
⟨︀ ⃒ ⃒ ⟩︀
0⃒𝑎 (⃗𝑝 ) 𝑎+ (⃗𝑞 )⃒0 ∼ 𝛿 (⃗𝑝 − ⃗𝑞 ) ⟨0|0⟩ ∼ 𝛿 (⃗𝑝 − ⃗𝑞 ) . (7.13)

Then it is necessary to substitute expression (7.13) into the integral (7.11). Such a
way, it will be seen that the result of the evaluation will become simply an integral of a
delta function and when the delta function is integrated it will be found that:
⟨︀ ⃒ + ⃒ ⟩︀ 𝑒−𝑖⃗𝑝 ⃗𝑥 /~
𝑝⃗ ⃒𝜑 (⃗𝑥 , 0)⃒0 = . (7.14)
(2𝜋~)3/2

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Expression (7.14) is recognized from quantum mechanics. It is known that this is a wave
function of a free particle. Every particle moving with momentum 𝑝⃗ . In other words, the
procedure of applying the positive frequency part to the vacuum positive frequency part
of our scalar field to the vacuum creates a particle with a standard quantum mechanical
wave function, which corresponds to free particle moving with momentum 𝑝⃗ .
Let’s also make a comment about the procedure of constructing the Fock space, which
also is known under the name second quantization.

Second quantization
The Klein-Gordon equation, which has been solved and already studied, was introduced
by Schrodinger himself. His motivation for introducing the Klein-Gordon equation was
rather simple. Basically it is known that there is the standard relativistic dispersion
relation for relativistic particle:

𝐸2
− 𝑝⃗ 2 − 𝑚2 𝑐2 = 0 . (7.15)
𝑐2
The relation (7.15) is called as Mass-Shell condition, but in fact this is a dispersion
relation for relativistic particle, which relates energy of this particle with its momentum
and Schrodinger’s idea was simply to pass to quantum mechanics by replacing a three-
momentum 𝑝⃗ with an operator of momentum, which equals to
𝜕
𝑝⃗ = −𝑖~ (7.16)
𝜕⃗𝑥
and replacing energy with
𝜕
𝐸 → 𝑖~
. (7.17)
𝜕𝑡
Energy is, in fact, an eigenstate of the hamiltonian and the hamiltonian generates
time evolution. So, shifts in time, while momentum is responsible for the space shifts and
it’s realized as a derivative from (7.16). After this replacements on the left hand side of
the dispersion relation the Klein-Gordon operator is discovered. If this little calculation
is done, it will be seen that the operator will have the next form:

1 𝜕2 𝜕 𝑚2 𝑐2
− + . (7.18)
𝑐2 𝜕𝑡2 𝜕𝑥2𝑖 ~2

Instead of the usual mass-shell condition, which is formulated in terms of 𝐸 and 𝑝⃗

now there is a second order differential operator and in a way the natural substitution

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of the dispersion relation would be to take this differential operator and apply it to the
wave function 𝜑:
1 𝜕2 𝑚2 𝑐2
(︂ )︂
𝜕
− + 𝜑=0. (7.19)
𝑐2 𝜕𝑡2 𝜕𝑥2𝑖 ~2
In fact, then it is possible to say that the expression, which is in the (7.18), on the
one hand mathematically is the kernel of the wave operator. On the other hand, from
a physical point of view, written in the (7.19), is a Schrodinger equation for the wave
function 𝜑. And the fact that this equation contains second order derivative in time rather
than first derivative in time is due to the fact that relativism is considered, which puts
time and space variables. Since this variables are on equal footing and they are rotated
into each other by means of Lorentz transformations, the second derivatives in the space
directions are accompanied by second time derivatives in the time direction. In a way this
is a relativistic version of the usual non-relativistic Schrodinger equation and this is an
equation which defines the wave function 𝜑. When the wave function is defined by means
of equation (7.19) it is necessaryt to continue with the usual quantum mechanical analogy
for the wave function 𝜑. Then a number of problems arise immediately. Especially, in the
case, when non-trivial interactions or self-interactions are considered for this theory. To
clarify this let’s introduce instead of a real field actually complex field, because complex
field is more suitable to regard it as a wave function, because wave function in the standard
quantum mechanics is a complex function. So, if a real scalar field is passed to complex
scalar field, then the action for this field is written in the following way:
∫︁ [︂ ]︂
1 4 𝜇 *
(︁ 𝑚𝑐 )︁2
*
𝑆 [𝜑] = d 𝑥 𝜕𝜇 𝜑𝜕 𝜑 − 𝜑𝜑 . (7.20)
𝑐 ~

Similar to what happens in the non-relativistic case, it is possible to introduce two

important quantities. One of them is a probability density 𝜌 and another is the probability
current denoted as ⃗𝑗 . These quantities are defined explicitly as
𝑖~
𝜌= (𝜑* 𝜕𝑡 𝜑 − 𝜑𝜕𝑡 𝜑* ) (7.21)
2𝑚𝑐2
and
⃗𝑗 = − 𝑖~ 𝜑* ∇
(︁ )︁
⃗ 𝜑 − 𝜑∇⃗ 𝜑* . (7.22)
2𝑚
These quantities are introduced, because in the non-relativistic quantum mechanics it
can be seen that these quantities satisfy the standard continuity equation
𝜕𝜌 ⃗ ⃗
+ ∇𝑗 = 0 . (7.23)
𝜕𝑡
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The continuity equation is satisfied due to the Klein-Gordon equation, especially, it

reduces to the Klein-Gordon equation for a field 𝜑, when 𝜌 and ⃗𝑗 are plugged into formula
(7.23) will be seen that there will be zero, because the Klein-Gordon equation is satisfied.
Now it looks like the standard paradigm of usual quantum mechanics is followed. 𝜌
and ⃗𝑗 were introduced, which satisfy the continuity equation as our goal is to have this
quantity in the non-relativistic case. In fact, it is also possible to understand that what
is written down in the (7.23) is nothing else as they’re written down the covariant form
of the current conservation
𝜕𝜇 𝐽 𝜇 = 0 , (7.24)

where 𝐽 𝜇 is a 4 current with components

(︁ )︁
𝜇 ⃗
𝐽 = 𝑐𝑝, 𝑗 . (7.25)

The current 𝐽 𝜇 is conserved in equations of motions and is simply a direct consequence

of Noether’s theorem, because the existence of this conserved quantity is due to the fact
that the action for relativistic complex scalar field has the following symmetry

𝜑 → 𝑒𝑖𝛼 𝜑 , (7.26)

where 𝛼 is a constant parameter. From expression (7.26) it can be seen that if the field 𝜑 is
changed by multiplying it with the phase and 𝜑* is simultaneously changed by multiplying
it with a conjugate phase, then this phase disappears and the action is simply invariant
with respect to these transformations. By Noether’s theorem it is known that symmetry
must correspond to a conserved current and this current exactly or the probability for
current was written in expression (7.25). On equations of motion, which are a Klein-
Gordon equations for 𝜑 and 𝜑* the current 𝐽 𝜇 must be conserved and this is exactly what
can be verified by direct means.
The problem is however that probability density is not positive. So, 𝜌 is not positive
in contrast to the expression for 𝜌 which is in non-relativistic quantum mechanics. In
non-relativistic quantum mechanics 𝜌 is defined as

𝜌 = 𝜑* 𝜑 = |𝜑|2 . (7.27)

This is an explicitly positive density. It can be given an interpretation of probability

density: probability is is always non-negative.
For our case it can be seen that 𝜌 is not positive from an explicit expression for 𝜌
(7.21). That is because in the relativistic case the second order differential equation is

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considered. Since we deal with the second order differential equation supplying it and
solving it it is necessary to supply it with initial conditions and since it is of the second
order, it is needed to supply as initial conditions not only the value of 𝜑, but also the
value of its time derivative at an initial moment of time. This values of 𝜑 and 𝜑˙ can be
chosen in an arbitrary way and if they can be chosen in an arbitrary way, so, 𝜑 and 𝜑˙
might also be taken negative values and therefore 𝜌 is not any more positive defined. In
general actually 𝜌 goes in the non-relativistic limit: when a speed of light tends to infinity,
it goes to the density of non-relativistic quantum mechanics. This can be explicitly traced
by using the expression for 𝜑 in the so-called WKB approximation and this expression for
𝜑 then starts from
2
𝜑 ∼ 𝑒𝑖𝑚𝑐 𝑡 𝜓 + . . . , (7.28)

where . . . are corrections to the expression and 𝜓 is taken as a wave function from non-
relativistic theory. Then it can be seen that, when expression (7.28) will be differentiated
with respect to time, the first time derivative will be not acting on 𝜓. It will just
2
𝜑𝑡 ∼ 𝑖𝑚𝑐2 𝑒𝑖𝑚𝑐 𝑡 𝜓 . (7.29)

Then from expressions (7.28) and (7.29) it can be seen that at leading order in the
limit 𝑐 → ∞, it will actually be obtained that:

𝜑* 𝜕𝑡 𝜑 ∼ 𝜓 * 𝜓 (7.30)

and time derivative will disappear. So, 𝜌 is nice in all respects that it’s a part of the
conserved for current. It is a quantity, which has a correct non-relativistic limit, but it loses
one important property, which allows to interpret it really as a probability density. Namely
it’s not positive anymore and, therefore, the whole quantum mechanical interpretation
of the quantity 𝜑 as, in particular, a quantum mechanical wave function breaks. If the
interaction is considered, thenit is possible to show that 𝜌 cannot be served as a quantum
mechanical probability density.
The correct interpretation of field 𝜑 has been found later by Pauli and Weiskopf.
According to Pauli and Weiskopf, the field 𝜑 must be treated as an operator rather than
the wave function. So, for us 𝜑 is not a wave function, but rather 𝜑 is an operator, which
then should be used to define with its help the amplitudes and a consequence creation
and annihilation operators, which are used to build the Fock space of the scalar field. We
naturally come to the Fock space representation treating 𝜑 as an operator, because this
operator should act in some space and the natural space, where it acts is a Fock space.

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Such a treatment of 𝜑 as an operator acting in the Fock space is exactly what is called
second quantization. This term is historical, because when second quantization is said,
in a way it is suggested that under the first quantization act the classical scalar field is
replaced by a wave function 𝜑 just like Schrodinger did, but in fact it’s not a wave function
and it is needed to quantize it again to promote it to an operator acting in some Hilbert
space, which in this case is a Fock space.
Today, perhaps, it is not a good thing to to talk about first or second quantization, it is
just known that it is necessary to treat 𝜑 from the very beginning as an operator acting in
the Fock space and in this respect it is possible to refer to this as quantization of the scalar
field, as a process by means of which we an operator is directly put in correspondence to a
classical field and then for this operator a representation is constructed. It is represented
in terms of creation and annihilation operators.
The next topic that is necessary to introduce is commutation and greens functions.

Commutation and Greens functions

In the theory of interacting fields even in the classical theory, it is known that solutions
of inhomogeneous field equations with point-like sources play a special role. These solutions
of such inhomogeneous equations with point-like sources have a special name, they are
called Greens functions.
As it is known from the classical electrodynamics, greens functions can be of different
nature. For instance, there could be retarded Greens functions, advanced Greens functions
and, in particular, in quantum field theory the interest is in the so-called causal Greens
functions, which are also known under the name of Feynman propagator.
Now let’s to discuss how these functions are defined and explicitly constructed for
the case of the Klein-Gordon field. Therefore, what is needed to consider here is the
Klein-Gordon field.
Let’s start from solutions of the Klein-Gordon equation without sources. This is related
to such objects as commutators of quantum fields at different space-time points. Since
our talk is about commutators, these functions sometimes are referred to as commutation
functions.
It is necessary to answer a question: what is the commutator of quantum fields at
different space-time points is? Not at the same time, where the commutation relations are
canonical, because we have already discussed it, but at different space and time points.

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The first very important commutation function, which will be discussed is the so-called
Pauli-Jordan function. This function is defined in the following way

[𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] = 𝑖~𝑐D (𝑥 − 𝑥′ ) , (7.31)

where D is a Pauli-Jordan function and 𝑥 and 𝑥′ are points in the four-dimensional space-
time. By definition a Pauli-Jordan function is a commutator of two quantum fields at
different space-time points. As it will be seen D is a function and it is not an operator.
Although the operator is being calculated, the result is a number 𝑐, which is a number or
a function multiplied by the identification operator.
Before the derivation is sketched, an explicit form of what is going to found will be
written. Since D (𝑥 − 𝑥′ ) depends on one argument, it can be written as follows
⎡ (︁ √ )︁ ⎤
𝑚𝑐 2
𝜖 (𝑡) ⎣ (︀ 2 )︀ 1 𝑚𝑐 2 (︀ 2 )︀ 1 ~ 𝑥 ⎦
(︁ )︁ 𝐽
D (𝑥) = − 𝛿 𝑥 − 𝜃 𝑥 𝑚𝑐
√ , (7.32)
2𝜋 2 ~ ~
𝑥2

where 𝜃 (𝑥2 ) is Heaviside function, 𝐽1 is a Bessel function. It should be also noticed that
in our case instead of 𝑥2 the following is used

2 2 2
(𝑥 − 𝑥′ ) = 𝑐2 (𝑡 − 𝑡′ ) − (⃗𝑥 − ⃗𝑥 ′ ) . (7.33)

One more variable that it is needed to introduce is 𝜖:

⎧
+1 , 𝑡 > 0
⎪
⎪
⎪
⎪
⎨
𝜖 (𝑡) = 𝜃 (𝑡) − 𝜃 (−𝑡) = 0 , 𝑡 = 0 . (7.34)
⎪
⎪
⎪
⎩−1 , 𝑡 < 0
⎪

Let’s also remind what is 𝜃 (𝑡):

⎧
⎨+1 , 𝑡 > 0
𝜃 (𝑡) = (7.35)
⎩0 , 𝑡 ≤ 0

It can be seen from formula (7.32) that, because of the presence of the prefactor 𝜖 (𝑡),
D (𝑥) turns out to be zero if 𝑡 = 𝑡′ . This means that the equal time commutator vanishes,
which is good, because it shows that it’s compatible with the fact that for the same time
argument fields commute. So,

[𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡)] = 0 (7.36)

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and this is the result of canonical quantization and this is how canonical Poisson bracket
requires to have when to the quantum theory is passed under the procedure of canonical
quantization.
The second feature is that it also can be seen that commutator of two local fields
vanishes if their space time points are separated by the space-like interval. So, 𝑥2 is
space-like, if 𝑥2 < 0. In our case our talk is about condition that if (𝑥 − 𝑥′ )2 < 0, then
Pauli-Jordan function vanishes. Therefore, fields, which are separated by such an interval
are commute and this is in a way if our goal is implementation of causality principle
in quantum field theory. This is an important fact, which is manifestation of causality
principle in quantum field theory. Why it is so? That’s because there is no signal known
that can propagate faster than the speed of light and this means that if there are two
events or fields which will be measured at points, which are separated by the space-like
interval, then measurement of field in one point and the measurement of the field in the
other point do not correlate, because these measurements are completely independent.
Signal is not enough to propagate from the point 𝑥, where a measurement of the field 𝜑 is
made to the point 𝑥′ , where it is implied that a measurement of the field at the point 𝑥′
is made. In other words, field values separated by the space-like interval are independent
observables and it is known that if there are two independent observables in order to be
able to measure them according to the basic principles of quantum mechanics, they must
commute. The last comment about the explicit expression, which is going to be found by
means of computation is that expanding the Pauli-Jordan function around the light cone
the following expression will be found:
[︂ ]︂
𝜖 (𝑡) (︀ 2 )︀ 1 (︁ 𝑚𝑐 )︁2 (︀ 2 )︀
D (𝑥) = − 𝛿 𝑥 − 𝜃 𝑥 + ... . (7.37)
2𝜋 2 ~
Essentially what we are expanding in the (7.37) is the piece of the (7.32) with 𝐽1 , where
√ √
around light cone means that this function is expanded around 𝑥2 ≈ 0. So, when 𝑥2
√
tends to 0 the ratio 𝐽1 to the 𝑥2 goes to 1. Therefore, it can be seen that on the light cone
there are severe singularities for the Pauli-Jordan function. Passing through the light cone
D (𝑥) experience singularities of two types. One of them is just delta-function singularity
and the second type of singularity is finite discontinuity is given by the function 𝜃 (𝑥2 ). So,
finite discontinuity is something which is usually understood as function jump, because
of the 𝜃-function.
Now the question may be asked: how to derive the gotten result? In particular, how
to construct the function D?

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Let’s explain how it can be done and let’s construct a scheme.

1) It is possible to use an expression for fields in terms of oscillators. Let’s start from
the expression for the commutator

[𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] (7.38)

given in terms of creation and annihilation operators. Up to a prefactor, which

contains 𝑐, 2𝜋 and ~, the following will be obtained:
∫︀ d⃗𝑘 d⃗𝑘 ′ (︁ [︁ + (︁ )︁ (︁ ′ )︁]︁ 𝑖(𝜔𝑡−𝜔′ 𝑡′ )−𝑖(⃗𝑘 ⃗𝑥 −⃗𝑘 ′ ⃗𝑥 ′ )
7.38 ∼ √ 𝑎 ⃗𝑘 , 𝑎 ⃗𝑘 𝑒 +
(︁ ′)︁
[︁ 4𝜔𝜔 (︁ )︁]︁ ⃗ ⃗′ ′
)︁ (7.39)
+ 𝑎 ⃗𝑘 , 𝑎+ ⃗𝑘 ′ 𝑒−𝑖(𝜔𝑡−𝜔 𝑡 )+𝑖(𝑘 ⃗𝑥 −𝑘 ⃗𝑥 ) .
′ ′

So, the Fourier decomposition of 𝜑 was used and then formula (7.39) was recieved.

2) Commutation relations between quantum oscillators are used and it is replaced by

delta-functions. Then it is possible to integrate the corresponding delta-function.
So, the first and the second commutators in the (7.39) will be replaced by
(︁ )︁ (︁ )︁
𝛿 ⃗𝑘 − ⃗𝑘 ′ and − 𝛿 ⃗𝑘 − ⃗𝑘 ′ (7.40)

respectively. Then it is possible to integrate over ⃗𝑘 ′ by using these delta functions,

simplify a bit and as a result the following expression will be obtained:
∫︁ ⃗
−2𝑐 d𝑘 𝑖⃗𝑘 (⃗𝑥 −⃗𝑥 ′ )
7.38 = 𝑖~𝑐 3 𝑒 sin 𝜔 (𝑡 − 𝑡′ ) . (7.41)
(2𝜋) 2𝜔

where the coefficient was added before expression (7.38). If the gotten expression is
compared with the definition of the Pauli-Jordan function, it will be seen that this
prefactor 𝑖~𝑐 is also in the (7.31). So, it is recieved that

d⃗𝑘 𝑖⃗𝑘 ⃗𝑥
∫︁
2𝑐
D (𝑥) = − 𝑒 sin 𝜔𝑡 . (7.42)
(2𝜋)3 2𝜔

So, a subject of our further computation is obtained.

3) Then it is necessary to continue the computation and it is needed to evaluate

the three-dimensional integral or evaluate it in terms of known functions like, for
instance, a Bessel function. It is convenient here to come to spherical coordinates.
So, it is necessary to introduce a quantity 𝑟 which is
√
𝑟= ⃗𝑥 2 (7.43)

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Fig. 7.1. Change of coordinates to 𝑟 and 𝑘 variables

and 𝑘, which is √︀
𝑘 = ⃗𝑘 2 . (7.44)

Our coordinate system is oriented along direction of ⃗𝑟 (fig. 7.1).

Therefore, the D (𝑥) will be obtained in spherical coordinates in the following form:
∫︁ ∞ 2 ∫︁ 2𝜋 ∫︁ 𝜋
2𝑐 𝑘 d𝑘
D (𝑥) = − sin 𝜔𝑡 d𝜙 sin 𝜃d𝜃𝑒𝑖𝑘𝑟 cos 𝜃 , (7.45)
(2𝜋)3 0 2𝜔 0 0

where it was used that

⃗
𝑒𝑖𝑘 ⃗𝑥 = 𝑒𝑖𝑟𝑘 cos 𝜃 . (7.46)

The integral in the (7.45) can be elementary computed. It can be seen that the
measure sin 𝜃d𝜃 can be replaced by

sin 𝜃d𝜃 = −d (cos 𝜃) . (7.47)

Such a way, integral with exponent can be elementary integrated and then integral
over d𝜑 is also elementary taken and, therefore, D (𝑥) upon this integration becomes
∫︁ ∞
𝑐 𝑘d𝑘
D (𝑥) = − 2 sin 𝜔𝑡 sin 𝑘𝑟 . (7.48)
2𝜋 𝑟 0 𝜔

In the (7.48) evaluation was reduced to the one-dimensional integral and this can
be further written conveniently in the following way:
∫︁ ∞
𝑐 1 𝜕 d𝑘
2
sin 𝜔𝑡 cos 𝑘𝑟 . (7.49)
2𝜋 𝑟 𝜕𝑟 0 𝜔 (𝑘)

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where 𝜔 is the function of 𝑘. When we will be integrating back over 𝑟, the cosine
will turn into a sine, but it also produce 𝑘 that was in the numerator of the (7.48).

Let’s remember what 𝜔 is. Now it can be seen that the integral (7.49) is a bit non-
trivial, it’s not easy to take, because 𝜔 (𝑘) is an explicit function of 𝑘, which is not
pleasant function, it’s equals to
√︁
𝜔 (𝑘) = 𝑐 ⃗𝑘 2 + 𝜇
√︀
⃗ 2 = 𝑐 𝑘 2 + 𝜇2 , (7.50)

where 𝜇 is
𝑚𝑐
𝜇= . (7.51)
~
So, explicitly D (𝑥) then if we substitute the expression for 𝜔 take the following
form: ∫︁ ∞
1 1 𝜕 d𝑘 √︀
D (𝑥) = √︀ cos 𝑘𝑟 sin 𝑥0 𝑘 2 + 𝜇2 , (7.52)
4𝜋 𝑟 𝜕𝑟 −∞ 𝑘 2 + 𝜇2
where integral limits were changed to from −∞ to +∞. The integration was extended
for negative values of 𝑘, because the function, that is integrated is even. Then the
integral is divided by two and that’s why 2𝜋 2 turns into 4𝜋 2 in denominator. An
argument of sine was also changed according to the formula
𝑥0
𝑡= . (7.53)
𝑐

Such a way an one dimensional integral (7.52) is obtained, which will be taken. It
can be seen that it is quite non-trivial, because it involves the parameter 𝑘 under
the square roots.

It is convenient to introduce the function 𝐹 (𝑟, 𝑡) and simply define it as

1 +∞
∫︁
d𝑘 √︀
𝐹 (𝑟, 𝑡) = √︀ cos 𝑘𝑟 sin 𝑥0 𝑘 2 + 𝜇2 . (7.54)
𝜋 −∞ 𝑘 2 + 𝜇2
To proceed, of course, it is needed to find a clever choice of variables, which somehow
√︀
should efficiently deal with the 𝑘 2 + 𝜇2 . The way to do it is to introduce the
following change of variables:

𝑘 = 𝜇 sinh 𝜙 , −∞ ≤ 𝜙 ≤ +∞ , (7.55)

where 𝜙 is not an angle, but just a variable. That’s efficient change, because if
√︀
𝑘 2 + 𝜇2 is now computed, it will be seen that this is
√︀ √︁
𝑘 + 𝜇 = 𝜇2 sinh2 𝜙 + 𝜇2 = 𝜇 cosh 𝜙 > 0 .
2 2 (7.56)

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The gotten in the (7.56) expression is always positive. Moreover, it can be seen that
if measure decay is taken it will give us

d𝑘 = 𝜇 cosh 𝜙d𝜙 (7.57)

and cancels exactly with 𝜇 cosh 𝜑, which is in the denominator .

Under such change of variables function 𝐹 will take the following form

1 +∞
∫︁
d𝜙 cos (𝜇𝑟 sinh 𝜙) × sin 𝜇𝑥0 cosh 𝜙 .
(︀ )︀
𝐹 (𝑟, 𝑡) = (7.58)
𝜋 −∞

The integral (7.58) is obtained, but the formula can be used further for the product
of two trigonometric functions cosine multiplied by sine. So let’s proceed and write
(7.58) as [︁
1
∫︀ +∞
𝐹 (𝑟, 𝑡) = 2𝜋 −∞
d𝜙 sin (𝜇𝑟 sinh 𝜙 + 𝜇𝑥0 cosh 𝜙) −
]︁ (7.59)
− sin (𝜇𝑟 sinh 𝜙 − 𝜇𝑥0 cosh 𝜙) .

The first step in our next considerations is to start distinguishing three different cases
depending on inequalities between 𝑥0 and 𝑟. For definiteness in further computation
the case where 𝑥0 , which is the same as 𝑐𝑡 is bigger than 𝑟 and 𝑟 is bigger than 0, will
be considered. That is needed, because every time a certain inequality is picked up
between the 𝑐𝑡 and 𝑟, it is necessary to make its own change of variables in order to
proceed. Thus, for the case that is being considered here, it is necessary to perform
appropriate further variable modification, namely

𝑥0 𝑟
√︁ = cosh 𝜙0 √︁ = sinh 𝜙 . (7.60)
0 2 2 0 2 2
(𝑥 ) − 𝑟 (𝑥 ) − 𝑟

Once again, a new change of variables is needed and this is needed in order to turn
the sum of sinh and cosh into something simple. Indeed, if a change of variables is
done, then the expression for 𝐹 (𝑟, 𝑡) takes the following form
[︂ √︁ ]︂
1
∫︀ +∞ 2
𝐹 (𝑟, 𝑡) = 2𝜋 −∞ d𝜙 sin 𝜇 (𝑥0 ) − 𝑟2 cosh (𝜙 + 𝜙0 ) +
[︂ √︁ ]︂ (7.61)
2
+ sin 𝜇 (𝑥0 ) − 𝑟2 cosh (𝜙 − 𝜙0 ) .

An interesting point is that since the integration is over all 𝜑 in the first integral it
is possible to make a shift of variables 𝜑 − 𝜑0 to remove the 𝜑0 from integration and

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in the second integral it is possible to make a change 𝜑 + 𝜑0 to remove this shift.

Then integrals are equal to each other and the answer doubles and gives us
1 +∞
∫︁ [︂ √︁ ]︂
0 2 2
𝐹 (𝑟, 𝑡) = d𝜙 sin 𝜇 (𝑥 ) − 𝑟 cosh (𝜙) . (7.62)
𝜋 −∞
Now finally if we look at the books looking for an explicit integral representation
for the Bessel function, then it will be seen that what is written down in the (7.62)
is exactly the integral representation for the Bessel function:
(︂ √︁ )︂
2
𝐹 (𝑟, 𝑡) = 𝐽0 𝜇 (𝑥0 ) − 𝑟2 . (7.63)

So our goal was achieved and the function 𝐹 (𝑟, 𝑡) was reduced to just a Bessel
function of index zero. If this calculation is niw done for other inequalities between
𝑐𝑡 and 𝑟, the following will be found:
⎧ (︂ √︁ )︂
0 2 2
⎪
⎪
⎪
⎪ +𝐽0 𝜇 (𝑥 ) − 𝑟 , 𝑐𝑡 > 𝑟
⎪
⎨
𝐹 (𝑟, 𝑡) = 0 − 𝑟 < 𝑐𝑡 < 𝑟 (7.64)
⎪
⎪ (︂ √︁ )︂
⎪ 2
⎩−𝐽0 𝜇 (𝑥0 ) − 𝑟2 , 𝑐𝑡 < −𝑟
⎪
⎪

The three gotten results for 𝐹 (𝑟, 𝑡) can be combined in one expression by saying
that 𝐹 (𝑟, 𝑡) is given by
(︂ )︂
𝑚𝑐
√︁
𝐹 (𝑟, 𝑡) = 𝜖 (𝑡) 𝜃 (𝑐𝑡)2 − 𝑟2 𝐽0 2
(︀ )︀
(𝑐𝑡) − 𝑟2 . (7.65)
~

4) The final step consists in evaluating D (𝑥), which is derivative of the expression for
𝐹 (𝑟, 𝑡) with respect to one over 𝑟. So, the following is recieved
[︂ (︂ √︁ )︂]︂
𝜖 (𝑡) 𝜕 𝑚𝑐
D (𝑥) = 2 2 2
(︀ )︀
𝜃 (𝑐𝑡) − 𝑟 𝐽0 (𝑐𝑡) − 𝑟2 . (7.66)
4𝜋𝑟 𝜕𝑟 ~
When the (7.66) is differentiated over 𝜕𝑟, differentiating of terms in the brakets
will give us delta-function, which has been seen in the expression for 𝛿 (𝑥) and then
differentiating the Bessel function with zero index will produce the Bessel function
with index one. So, finally the expression for the Pauli-Jordan function will be found:
⎡ (︁ √ )︁ ⎤
𝑚𝑐 2
𝜖 (𝑡) ⎣ (︀ 2 )︀ 1 𝑚𝑐 2 (︀ 2 )︀ 1 ~ 𝑥 ⎦
(︁ )︁ 𝐽
D (𝑥) = − 𝛿 𝑥 − 𝜃 𝑥 𝑚𝑐
√ . (7.67)
2𝜋 2 ~ ~
𝑥 2

That’s what concerns important commutation function, which is called Pauli-Jordan

function and which is a commutator between quantum fields at different space-time
points.

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In fact, what can be shown further is that it is clear that the commutator has been
calculated
[𝜑 (𝑥, 𝑡) , 𝜑 (𝑥′ , 𝑡′ )] = 𝑖~𝑐D (𝑥 − 𝑥′ ) . (7.68)

It is possible then to show that since 𝑥′ and 𝑡′ are different from 𝑥 and 𝑡, it is possible to
act on both sides of the relation with the Klein-Gordon’s wave operator, which is

1 𝜕2 𝜕2 (︁ 𝑚𝑐 )︁2
𝑥,𝑡 = 2 2− 2+ . (7.69)
𝑐 𝜕𝑡 𝜕𝑥𝑖 ~

It is possible to act with (7.69) on both sides of the (7.68). So, then the next expression
will be obtained:
[𝑥,𝑡 𝜑 (𝑥, 𝑡) , 𝜑 (𝑥′ , 𝑡)] = 𝑖~𝑐𝑥,𝑡 D (𝑥 − 𝑥′ ) . (7.70)

Since 𝜑 is a solution of the Klein-Gordon equation, the first term in the commutator of the
(7.70) is zero and, therefore, on the right hand side it is necessary to find that the result
of applying box with respect to coordinates 𝑥 and 𝑡 to Pauli-Jordan function also gives
us 0. Thus, in fact, what was obtained also means that D (𝑥 − 𝑥′ ) is an explicit solution
of the Klein-Gordon equation:
𝑥,𝑡 D (𝑥 − 𝑥′ ) . (7.71)

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Lecture 8. Retarded and Advanced Green’s Functions.

Feynman Propagator. Yukawa Force
In the last lecture we have introduced the Pauli-Jordan function. The Pauli-Jordan
function is a commutator of two field operators taken at different space-time points.
Now we will define two more interesting and important functions, namely the retarded
Green’s function and the advanced Green’s function. We will look how they are explicitly
constructed and then we will come to the very important notion in quantum field theory,
the notion of Feynman propagator.
Let’s start with the definition of the retarded Green’s function.

Retarded Green’s function

So, the retarded Green’s function is a ∆𝑟𝑒𝑡 , which is defined in the following way

∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) = 𝜃 (𝑡 − 𝑡′ ) [𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] . (8.1)

We see that if we compare this to the definition of the Pauli-Jordan function, then
what is written down in the (8.1) is simply

∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) = 𝑖~𝑐𝜃 (𝑡 − 𝑡′ ) D (𝑥 − 𝑥′ ) . (8.2)

Advanced Green’s function

Analogously, the advanced Green’s function is defined as

∆𝑎𝑑𝑣 (𝑥 − 𝑥′ ) = −𝜃 (𝑡′ − 𝑡) [𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] . (8.3)

Again, we need to compare the (8.3) with the definition of the Pauli-Jordan function:

∆𝑎𝑑𝑣 (𝑥 − 𝑥′ ) = −𝑖~𝑐𝜃 (𝑡′ − 𝑡) D (𝑥 − 𝑥′ ) . (8.4)

We have to show that the functions introduced in such a way as in (8.2) and (8.4)
are really Green’s functions. In other words, they solve the Klein-Gordon equation with
a delta-function source on the right hand side. This showing can be done in different
ways. The simplest way, probably, to take a ∆𝑟𝑒𝑡 and act on it with the Klein-Gordon
propagator.

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Of course, we need also to use then the fact that the field 𝜑 (⃗𝑥 , 𝑡) solves the Klein-
Gordon equation. Let’s see how it works.
So, we take the Klein-Gordon operator, which is essentially
(︀ −2 2
𝑐 𝜕𝑡 − 𝜕𝑖2 + 𝜇2 ,
)︀
(8.5)

where we denoted 𝜇 as
𝑚𝑐
. 𝜇= (8.6)
~
And then we apply (8.5) to retarded Green function:

−𝑐2 𝜕𝑡2 − 𝜕𝑖2 + 𝜇2 (𝜃 (𝑡 − 𝑡′ ) [𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )]) .

(︀ )︀
(8.7)

We need first to apply derivative with respect to time and applying it, we will get the
following:

− 𝑐2 𝜕𝑡2 : 𝜕𝑡 −𝑐2 𝛿 (𝑡 − 𝑡′ ) [𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] + 𝜃 (𝑡 − 𝑡′ ) [𝜋 (𝑥, 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] .

(︀ )︀
(8.8)

Then we have the part of the Klein-Gordon operator, which straightforwardly gets
inside the bracket and acts on the field 𝜑. So, these are derivatives with respect to spatial
directions. What we get here is

− 𝜕𝑖2 + 𝜇2 : 𝜃 (𝑡 − 𝑡′ ) −𝜕𝑖2 + 𝜇2 𝜑 (⃗𝑥 , 𝑡) , 𝜑 (𝑥′ , 𝑡′ ) .

[︀(︀ )︀ ]︀
(8.9)

Now we have to continue and act with the remaining time derivative on the result
inside the brackets of the (8.8), but before doing that, we know that, because of the delta-
function we would have a non-trivial result only when 𝑡 = 𝑡′ . So, we go to the support of
the delta-function, but in this case, when 𝑡 = 𝑡′ , we have an equal time commutator of
two quantum fields 𝜑 and we know that from their canonical commutation relations for
𝑡 = 𝑡′ , this commutator vanishes. So, in fact, the whole term with delta-function produce
a vanishing contribution and we can just drop it.
Then we need to apply the time derivative for the second term in the brackets of the
(8.7) and the resulting expression for the (8.7) will have the next form:

8.7 = 𝛿 (𝑡 − 𝑡′ ) [𝜋 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )]+𝜃 (𝑡 − 𝑡′ ) 𝑐2 𝜕𝑡2 − 𝜕𝑖2 + 𝜇2 𝜑 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ ) . (8.10)

[︀(︀ )︀ ]︀

What do we see from formula (8.10)? The first thing that we see is that since 𝜑 (⃗𝑥 , 𝑡) is a
solution of the Klein-Gordon equation, then the first term of the commutator (𝑐2 𝜕𝑡2 − 𝜕𝑖2 + 𝜇2 ) 𝜑 (⃗𝑥 , 𝑡)
in the (8.10) vanishes, because 𝜑 is a solution.

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Such a way the whole result turns to

8.7 = 𝛿 (𝑡 − 𝑡′ ) [𝜋 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] . (8.11)

We got something, which we can easily compute, because due to the delta-function
in the (8.11), the time argument 𝑡′ can be replaced for 𝑡. As a result we got an equal
time commutator between the momentum of the field and the field itself. Because of the
canonical commutation relation, we know that what is written down in the commutator
brackets is in fact

[𝜋 (⃗𝑥 , 𝑡) , 𝜑 (⃗𝑥 ′ , 𝑡′ )] = −𝑖~𝛿 (𝑡 − 𝑡′ ) 𝛿 (⃗𝑥 − ⃗𝑥 ′ ) . (8.12)

In other words, we restore on the right hand side of the (8.12) the four dimensional
delta-function and the final expression can be written as:

8.7 = −𝑖~𝛿 (4) (𝑥 − 𝑥′ ) , (8.13)

where
𝑥0 = 𝑐𝑡 . (8.14)

Let’s act the bracket ( + 𝜇2 ) on the retarded Green’s function ∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ). Then
the result of this will be

 + 𝜇2 ∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) = −𝑖~𝑐𝛿 (4) (𝑥 − 𝑥′ ) ,

(︀ )︀
(8.15)

where ∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) is a Green’s function for the Klein-Gordon operator.

Why the ∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) is called as retarded? It’s called retarded, because due to the
presence of 𝜃 function in the definition, it can be seen that the Green’s function has a
distinguished property that ∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) vanishes for 𝑡 < 𝑡′ . The Green’s function with
such a property is called retarded.
For ∆𝑎𝑑𝑣 (𝑥 − 𝑥′ ) computation is absolutely the same. It can be found that ∆𝑎𝑑𝑣 (𝑥 − 𝑥′ )
is also a fundamental solution of the Klein-Gordon equation. The fundamental means that
it’s a solution with a right hand side represented in the form of a point-like source.
There is another way to show how the ∆𝑟𝑒𝑡 (𝑥 − 𝑥′ ) allows to obtain a useful representation
for the Green’s function.
Let’s look at the following representation, which in fact uses the integral representation
for the Pauli-Jordan function
d⃗𝑘 [︀ 𝑖𝜔𝑡
∫︁
𝜃 (𝑡) ]︀ ⃗
∆𝑟𝑒𝑡 (𝑥) = −~𝑐 𝑒 − 𝑒−𝑖𝜔𝑡 𝑒+𝑖𝑘 ⃗𝑥 . (8.16)
(2𝜋)3 2𝑘 0

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Expression (8.16) is integral representation for the function ∆𝑟𝑒𝑡 (𝑥), where 𝑘0 is a
component of the wave vector
√︁
𝜔
𝑘0 = ⃗𝑘 2 + 𝜇2 = . (8.17)
𝑐
According to the (8.17), the integral representation can be modified. Let’s start from
some auxiliary or additional integral, which apparently at first sight has nothing to do with
the ∆𝑟𝑒𝑡 . Then the complex analysis should be used, in particular, as a residue theorem
to compute the following integral:
∫︁ +∞ 0
𝑒−𝑖𝑐𝑘 𝑡 d𝑘 0
. (8.18)
−∞ (𝑘 0 + 𝑖𝜖)2 − ⃗𝑘 2 − 𝜇2
The integral (8.18) is taken along the real line, but it is more convenient to use a complex
analysis. In the (8.18) the complex plane of the variable 𝑘 0 was considered for computation
of the integral over variable 𝑡0 .
Let’s look at the denominator of the (8.18) and forget about the little piece 𝑖𝜖, which
was added to 𝑘 0 . An 𝜖 is a positive and small and, therefore, it is a little shift in the
imaginary direction. This little term should be used, because without it the denominator
in formula (8.18) looks like
(︀ 0 )︀2
𝑘 − ⃗𝑘 2 − 𝜇2 (8.19)

and then, when 𝑘 0 will be running through the whole real line, eventually it will meet the
problem, because at certain values of 𝑘 0 the expression in the denominator may become
equal to zero. This happens, when
√︁
𝑘 = ± ⃗𝑘 2 + 𝜇2 .
0
(8.20)

Therefore, without 𝑖𝜖 term singularities along the real line will exist and then the
integral will not be well defined and will diverge.
Exactly, to overcome this problem and make sense of the integral one adds up a little
term 𝑖𝜖 for shift the pole in the lower half plane. It can be seen that now poles are shifted
from the real line a little bit down (fig. 8.1).
That’s because a different from the (8.19) equation should be solved:
)︀2
𝑘 + 𝑖𝜖 − ⃗𝑘 2 − 𝜇2 = 0 .
(︀ 0
(8.21)

The square root of the (8.21) will give:

√︁
𝑘 + 𝑖𝜖 = ± ⃗𝑘 2 + 𝜇2 .
0
(8.22)

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Fig. 8.1. The poles of the integral (8.18) at 𝑘 0 plane

Therefore, √︁
𝑘 = −𝑖𝜖 ± ⃗𝑘 2 + 𝜇2 .
0
(8.23)

Because of the −𝑖𝜖 poles are shifted in the lower half plane. This means there are no
problems, if integral over the real line will be taken. It also should be noticed that the
Cauchy’s theorem can be applied to compute the integral by closing the correspondent
contour in the lower half plane. Integral goes through infinity and close the contour in the
lower half plane like presented at the (fig. 8.1).
Then the integral can be computed by Cauchy’s residue theorem in the following way:
the two poles are exist and residues at these poles should be taken. So, the integral will
be transformed to
⎡ ⎤
∫︁ +∞ 0 −𝑖𝑐𝑘0 𝑡 −𝑖𝑐𝑘0 𝑡
d𝑘 𝑒 𝑒
8.18 = √︁ ⎣ √︁ − √︁ ⎦ , (8.24)
−∞
2 ⃗𝑘 2 + 𝜇2 𝑘 0 − ⃗𝑘 2 + 𝜇2 + 𝑖𝜖 𝑘 0 + ⃗𝑘 2 + 𝜇2 + 𝑖𝜖

where the expression was transformed into 2 decomposed terms, which are residues at the
first and the second pole. According to the Cauchy’s residue theorem, the direction of the
integration should be taken into account and the integration contour should be enclosed
in such a way, that the region enclosed by this contour must remain on the left. The way,
which was presented in the (fig. 8.1) stays the integration contour on the right and that
is why the sign should be changed.

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One more point to discuss is why the contour was enclosed below the real line. That
is because for 𝑡 > 0 in the exponential
0𝑡
𝑒−𝑖𝑐𝑘 (8.25)

𝑘 0 should have a negative imaginary part. It should behave like −𝑖 |𝑘 0 |, in order to produce
minus in front of the argument of the exponential. It should be so to provide the dumping
in the spatial infinity, which can be achieved by increasing of the radius of a circle. For 𝑡
bigger than zero the integral was closed below the real line.
Then, of course, two poles should be encountered. One pole is at
√︁
𝑘 = ⃗𝑘 2 + 𝜇2 − 𝑖𝜖
0
(8.26)

and the other pole is at √︁

𝑘 = − ⃗𝑘 2 + 𝜇2 − 𝑖𝜖 .
0
(8.27)

By Cauchy’s theorem, the result will equal to

[︂ √ √ ]︂
2𝜋𝑖 𝑖𝑐 ⃗𝑘 2 +𝜇2 𝑡 −𝑖𝑐 ⃗𝑘 2 +𝜇2 𝑡
8.18 = √︁ 𝑒 −𝑒 . (8.28)
2 ⃗𝑘 2 + 𝜇2

Expression (8.28) is the answer for this auxiliary integral. In the original integral 𝑡
can be anything. It can be bigger than zero and it can be less than zero. For 𝑡 > 0 the
contour should be closed in the lower half plane. For 𝑡 < 0 the situation is actually the
opposite in order to guarantee the exponential damping of the function at large values of
𝑘 0 . Therefore, for 𝑡 < 0 the contour should be closed in the upper half plane, but in the
upper half plane there are no poles and the function has its poles only in the lower half
plane. That is why the integral will be equal to zero.
It can be obtained that that the integral has interesting property: for positive 𝑡, it
gives a non-trivial answer given by expression (8.28), for negative 𝑡 it actually vanishes.
On the other hand, it can be obtained that the integral, in fact, is the next quantity from
(8.16):
1 [︀ 𝑖𝜔𝑡
𝑒 − 𝑒−𝑖𝜔𝑡 ,
]︀
0
(8.29)
2𝑘
√︁
where 𝜔 and 𝑘 should be replaced by 𝑐𝑘 and ⃗𝑘 2 + 𝜇2 respectively.
0 0

Instead of expression (8.28), which was gotten before, the original integral representation
can be substituted for the retarded Green’s function:
d4 𝑘
∫︁
𝑖
∆𝑟𝑒𝑡 (𝑘) = ~𝑐 4 𝑒−𝑖𝑘𝑥 , (8.30)
(2𝜋) (𝑘 0 + 𝑖𝜖) − ⃗𝑘 2 − 𝜇2
2

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where integration was performed not only for spacial directions of 𝑘, but also for the 𝑘 0
variable. The scalar product 𝑘𝑥 in the (8.30) is a relativistic scalar product:

𝑘𝑥 = 𝑘𝜇 𝑥𝜇 . (8.31)

Let’s write one important property of the ∆𝑟𝑒𝑡 . From the courses on mathematical
physics, it can be known that the Green’s function is not unique, because of the fact that
the following equation should be solved:

 + 𝜇2 𝐺 = 𝛿 (4) (𝑥 − 𝑥′ ) .
(︀ )︀
(8.32)

It can be clearly seen that if one particular solution of the equation was found, then a
solution of the homogeneous equation can always be added to this solution and the gotten
expression will be again a Green’s function:

𝐺 = 𝐺0 + 𝐺ℎ𝑜𝑚 (8.33)

Solution of the homogeneous equation means that

 + 𝜇2 𝐺ℎ𝑜𝑚 = 0 .
(︀ )︀
(8.34)

Playing with the solution of the homogeneous equation, the properties of the Green’s
function can be adjusted. For instance, we can have retarded solution or advanced solution.
Another type of solution is also exist, which is more interesting to observe and called
Feynman propagator. But before that let’s denote what is retarded means. Retarded
means just a specific property that Green’s function appears to be developed after the
acting of the impulse. For instance, if a delta-function source is exist, which switches at
𝑡 = 𝑡′ , then the signal propagates at later times and that is what retarded Green’s function
is taken care of. In a more convenient way, a retarded Green’s function has a retarded
property, which means that at the later values of time, we expect to get consequences of
the signal. This consequences is observed, because

∆𝑟𝑒𝑡 (𝑡 − 𝑡′ > 0) ̸= 0 . (8.35)

On the other hand, before the source switched on

∆𝑟𝑒𝑡 (𝑡 − 𝑡′ < 0) = 0 . (8.36)

An advanced Green’s function has an opposite property.

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With the integral representation (8.30) it can be also elementary checked that Klein-
Gordon operator acting on ∆ doesn’t produce a delta-function. And if the integral representation
is used, then it will be seen that:
2
d4 𝑘 (𝑘 0 ) − ⃗𝑘 2 − 𝜇2
∫︁
 + 𝜇2 ∆𝑟𝑒𝑡 = 𝑖~𝑐 𝑒−𝑖𝑘𝑥 .
(︀ )︀
4 (8.37)
0 2
(2𝜋) (𝑘 + 𝑖𝜖) − 𝑘 − 𝜇⃗ 2 2

As can be seen from the (8.37), the numerator and the denominator have the same
expression and there is only different that a little 𝑖𝜖 in the denominator exist. Then the
limit 𝜖 → 0 can be taken and the resulting expression will be:
d4 𝑘 −𝑖𝑘𝑥
∫︁
2
(︀ )︀
lim  + 𝜇 ∆𝑟𝑒𝑡 = −𝑖~𝑐 𝑒 . (8.38)
𝜖→0 (2𝜋)4
The integral in the (8.38) is a Fourier image of the four dimensional delta-function:

lim  + 𝜇2 ∆𝑟𝑒𝑡 = −𝑖~𝑐𝛿 (4) (𝑥) .

(︀ )︀
(8.39)
𝜖→0

Finally, according to (8.39), Klein-Gordon operator acting on delta retarded produce the
delta-function source.
The most interesting function to consider, which is essentially used as an element of
quantum field theory, is a so-called causal Green’s function. This is the same as Feynman
propagator.

Feynman propagator
Feynman propagator can be denoted simply by symbol ∆. It also depends on 𝑥 − 𝑥′
and the motivation for its introduction can be given in the following way. Let’s look at
vacuum expectation value of two quantum fields:

⟨0|𝜑 (𝑥) 𝜑 (𝑥′ )|0⟩ , (8.40)

where 𝜑 (𝑥) 𝜑 (𝑥′ ) is a two point function. Two point means that it depends on 𝑥 and on
𝑥′ . It is also important to notice that 𝜑 is a quantum field.
Expression (8.40) can be written in terms of creation and annihilation operators. After
that the product of operators should be put between two vacuum states. Often this is
called as vacuum expectation value for the product of two operators. The result of this
evaluation will have the following form:

⟨0|𝜑 (𝑥) 𝜑 (𝑥′ )|0⟩ = 0⃒𝜑− (𝑥) 𝜑+ (𝑥′ )⃒0 = 0⃒ 𝜑− (𝑥) 𝜑+ (𝑥′ ) ⃒0 .
⟨︀ ⃒ ⃒ ⟩︀ ⟨︀ ⃒[︀ ]︀⃒ ⟩︀
(8.41)

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where the fact that field 𝜑 (𝑥) can be split into positive and negative frequency parts was
used. It should be noticed that the contribution of the 𝜑− (𝑥′ ) will produce zero, because
this part depends on the annihilation operator and the annihilation operator annihilates
vacuum. Analogously, if 𝜑+ (𝑥) is used, than due to the fact that 𝜑+ (𝑥) depends on the
creation operator, zero will be produced after acting on the left vacuum.
The last expression in the (8.41) with commutator was written, because
⟨︀ ⃒ + ′ − ⃒ ⟩︀
0⃒𝜑 (𝑥 ) 𝜑 (𝑥)⃒0 = 0 . (8.42)

Then the commutator from the (8.41) can be evaluated and, therefore, the result will
have the following form:

⟨0|𝜑 (𝑥) 𝜑 (𝑥′ )|0⟩ = 𝑖~𝑐D− (𝑥 − 𝑥′ ) . (8.43)

where D− is the negative frequency part of the Pauli-Jordan function.

On the other hand, the expression on the left hand side of the (8.43) can be physically
interpreted as an amplitude for the process, when at the moment 𝑡 > 𝑡′ a particle at a
point 𝑥′ is created and a particle at a point 𝑥 is destroyed (fig. 8.2).

Fig. 8.2. Creation of a particle at a point 𝑥′ and destroying a particle at a point 𝑥

The process presented at (fig. 8.2) can be described in a bit different way. A similar
amplitude may be considered, when a particle is created at 𝑥, but then it destroyed at 𝑥′ .
To combine this processes in one go, one introduces the notion of the Feynman
propagator ∆ (𝑥 − 𝑥′ ), which is constructed in the following way:

∆ (𝑥 − 𝑥′ ) = 𝜃 (𝑡 − 𝑡′ ) ⟨0|𝜑 (𝑥) 𝜑 (𝑥′ )|0⟩ + 𝜃 (𝑡′ − 𝑡) ⟨0|𝜑 (𝑥′ ) 𝜑 (𝑥)|0⟩ , (8.44)

where 𝜃 function prescribes the way how creation and annihilation happens in time. For
instance, 𝜃 (𝑡 − 𝑡′ ) means that particle is created at a point 𝑥′ and a time 𝑡′ and then
annihilated at a point 𝑥 and a time 𝑡.
The Feynman propagator can be written in a different way. The two terms presented
in the (8.44) can be encoded into one formula:

∆ (𝑥 − 𝑥′ ) = ⟨0|𝑇 (𝜑 (𝑥) 𝜑 (𝑥′ ))|0⟩ , (8.45)

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where the operation of T ordering explicitly means that the 𝑇 checks the time arguments
of fields 𝜑 (𝑥) and 𝜑 (𝑥′ ) and then it orders them according to the time dependence of fields.
The field with a less value of 𝑇 always go to the right and this rule can be remembered
by slogan that Youth is always right“.
”
The Feynman propagator also can be expressed in terms of a Pauli-Jordan function
and, in fact, half the work to obtain this representation has already been done. Additionally,
a Pauli-Jordan function should be supplied with 𝜃-function:

∆ (𝑥 − 𝑥′ ) = 𝑖~𝑐 𝜃 (𝑡 − 𝑡′ ) D− (𝑥 − 𝑥′ ) + 𝜃 (𝑡′ − 𝑡) D− (𝑥′ − 𝑥) .

[︀ ]︀
(8.46)

The Pauli-Jordan function have a positive and negative frequency parts, which are
connected with each other according to the formula:

D+ (𝑥) = −D− (−𝑥) . (8.47)

Let’s consider the Feynman propagator in more detail. It is useful to repeat the same
exercise as was done for the retarded Green’s function. First of all, the integral expression
for D− should be written:

d⃗𝑘 −𝑖𝑐𝑘0 𝑡+𝑖⃗𝑘 ⃗𝑥

∫︁
𝑖
D =−
−
𝑒 . (8.48)
(2𝜋)3 2𝑘 0

Then, expression (8.48) need to be put in the Feynman propagator. The result will
have the following form:

d⃗𝑘 𝑖⃗𝑘 ⃗𝑥
∫︁ [︂ ]︂
′ 𝜃 (𝑡) −𝑖𝑐𝑘0 𝑡 𝜃 (−𝑡) 𝑖𝑐𝑘0 𝑡
∆ (𝑥 − 𝑥 ) = ~𝑐 𝑒 × 𝑒 + 𝑒 , (8.49)
(2𝜋)3 2𝑘0 2𝑘0

where 𝑘0 as it was before equal to

√︁
𝑘0 = ⃗𝑘 2 + 𝜇2 . (8.50)

The integral can be evaluated by the same trick as before. So, the integral taken along
the real line in the complex 𝑘0 plane should be found or, in other words, an auxiliary
integral should be found:
d𝑘0 𝑒−𝑖𝑐𝑘0 𝑡
∫︁
, (8.51)
𝑘 2 − 𝜇2 + 𝑖𝜖
where the term 𝑖𝜖 was used in a different way in comparison to what was done before,
when 𝑖𝜖 was put close to 𝑘0 :
(𝑘0 + 𝑖𝜖) . (8.52)

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It also should be noticed that

𝑘 2 = 𝑘02 − ⃗𝑘 2 (8.53)

and, therefore (8.51) can be written as

d𝑘0 𝑒−𝑖𝑐𝑘0 𝑡
∫︁
. (8.54)
𝑘 2 − ⃗𝑘 2 − 𝜇2 + 𝑖𝜖
0

Poles of the integral are now located as shown at the (fig. 8.3). Due to the (8.54), the
left pole will be shifted up and the right pole a little bit down.

Fig. 8.3. The poles of the integral (8.54) at 𝑘 0 plane

What is the purpose of such a different shift? The purpose of the shift is to actually
realize the properties of the 𝜃-function. One function should be equal to 1, when 𝑡 is
positive and another should be equal to one when 𝑡 is negative, where 𝑡 is a difference
between 𝑡 and 𝑡′ .
Similarly to the previous contour integral, a contour should be closed in the lower half
plane. The difference is that now only one pole in this integration contour exists. And so
the non-trivial contribution in this case will be found.
The denominator of the integral should be split into simple poles and the expression
with simple poles will be equal to:
⎡ ⎤
∫︁ −𝑖𝑐𝑘0 𝑡 −𝑖𝑐𝑘0 𝑡
d𝑘 𝑒 𝑒
8.54 = √︁ 0 ×⎣ √︁ − √︁ ⎦ . (8.55)
⃗ 2
2 𝑘 +𝜇 2 ⃗ 2 2 ⃗ 2
𝑘0 − 𝑘 + 𝜇 + 𝑖𝜖 𝑘0 + 𝑘 + 𝜇 − 𝑖𝜖2

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Condition for 𝑘0 , which defines the poles, should be written in the following way:
√︁
𝑘0 = ± ⃗𝑘 2 + 𝜇2 − 𝑖𝜖 , (8.56)

where the term 𝑖𝜖 is under the square root. But 𝜖 is small and for small 𝜖 the following
replacement is right:
√︁ √︁
± 𝑘 + 𝜇 − 𝑖𝜖 → ± ⃗𝑘 2 + 𝜇2 ∓ 𝑖𝜖 .
⃗ 2 2 (8.57)

According to the (8.55) the next statement can be formed: for 𝑡 bigger than 0, the
contour should be closed in the lower half plane and only one pole gives the contribution;
for 𝑡 less than 0 the contour should be closed in the upper half plane. For instance, for
the retarded Green’s function, the integral’s result was 0, because there was no pole in
the upper half plane.
Now the integral can be evaluated by using the Cauchy residue theorem:
1
[︂ √ √ ]︂
−𝑖𝑐 ⃗𝑘 2 +𝜇2 𝑡 𝑖𝑐 ⃗𝑘 2 +𝜇2 𝑡
8.55 = 2𝜋𝑖 √︁ −𝜃 (𝑡) 𝑒 − 𝜃 (−𝑡) 𝑒 . (8.58)
2 ⃗𝑘 2 + 𝜇2
The result of evaluation of the integral is that now the integral representation for the
Feynman propagator can be written:
d4 𝑘
∫︁
𝑖
∆ (𝑥) = ~𝑐 4 2 2 + 𝑖𝜖
𝑒−𝑖𝑘𝑥 . (8.59)
(2𝜋) 𝑘 − 𝜇
The causal properties are now encoded exactly in the little shift 𝑖𝜖 in the denominator.
From expression (8.59) it also can be seen that the propagator in the Fourier space will
be simply:
𝑖
∆ (𝑘) = . (8.60)
𝑘2
− 𝜇2 + 𝑖𝜖
It also can be explained where the term 𝑖𝜖 comes from. It comes from encoding
the causality acts of creation and annihilation particles with the help of the T-ordering
product. So, T-ordering seats in the 𝑖𝜖 shift in the denominator of the Feynman propagator.
The integral can be actually computed explicitly and as the result ∆ in the 𝑥-space
will be gotten in terms of known special functions:
(︀ √
~𝑐 (︁ 𝑚𝑐 )︁2 𝐾1 𝑚𝑐
)︀
−𝑥2 + 𝑖𝜖
∆ (𝑥) = 2 𝑚𝑐
√
~
, (8.61)
4𝜋 ~ ~
−𝑥 2 + 𝑖𝜖

where 𝐾1 is the Macdonald function, 𝑥2 is a relativistic interval, which looks like:

𝑥2 = 𝑥20 − ⃗𝑥 2 . (8.62)

The last thing that needed to be discussed concerning the scalar field is what is called
Yukawa force.

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Yukawa force
The Yukawa force topic is linked with the action of a scalar field. The expression for
the action was already introduced and can be simply written as:
𝜇2 2
∫︁ [︂ ]︂
1 4 1 𝜇
𝑆 [𝜑] = d 𝑥 𝜕𝜇 𝜕 𝜑 − 𝜑 + 𝐽 (𝑥) 𝜑 (𝑥) , (8.63)
𝑐 2 2
where 𝐽 (𝑥) is called as source term. This term can be considered as a source for a
field 𝜑 and the simultaneously what it does, it changes the Hamiltonian of the field 𝜑
itself. In other words, the Hamiltonian and the energy receives an extra contribution from
interaction of the field 𝜑 with the source. The source here is supposed to be in a way
external.
It is interesting here to study the response of the field 𝜑 on the presence of the source
𝐽 (𝑥). In fact, the term with 𝐽 is a part of the theory, which is called linear response
theory.
It is important that in the presence of the source the Hamiltonian 𝐻 receives an extra
contribution, which can be denoted as 𝐻𝐽 . This is given simply by
∫︁
𝐻𝐽 = − d⃗𝑥 𝐽 (𝑥) 𝜑 (𝑥) . (8.64)

𝐻𝐽 comes with a minus sign, because if the Hamiltonian will be derived from the
action 𝑆, the canonical procedure should be applied, where the Hamiltonian is given by:

𝐻 = 𝜋𝜑 − 𝐿 . (8.65)

When taking the minus Lagrangian, then the term comes with a minus sign and this
will produce an extra contribution to the Hamiltonian of the field.
On the other hand, equations of motion for the field will be changed. It will not just
the Klein-Gordon equation, but it will be inhomogeneous Klein-Gordon equation, where
on the right hand side a source term will exist:

𝜕𝜇 𝜕 𝜇 + 𝜇2 𝜑 (𝑥) = 𝐽 (𝑥) .
(︀ )︀
(8.66)

When the action will be varied to find equations of motion, the field 𝜑 will be also
varied and a contribution from the source term will be gotten.
Then all the knowledge about the Green’s function can be used. So, Green’s function
becomes important because it is known that if an homogeneous equation exists, it can
always been solved by using the method of Green’s functions. The right solution is:
∫︁
𝑖
𝜑 (𝑥) = d4 𝑦𝐺 (𝑥 − 𝑦) 𝐽 (𝑦) . (8.67)
~𝑐

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If a Klein-Gordon operator is applied to the field 𝜑 (𝑥), then the Klein-Gordon operator
will hit the Green’s function under the integral:
∫︁
𝜇 2 𝑖
d4 𝑦 𝜕𝜇 𝜕 𝜇 + 𝜇2 𝐺 (𝑥 − 𝑦) 𝐽 (𝑦) .
(︀ )︀ (︀ )︀
𝜕𝜇 𝜕 + 𝜇 𝜑 (𝑥) = (8.68)
~𝑐
When Klein-Gordon operator hits Green’s function, it will produce by definition of Green’s
function a term:
𝜕𝜇 𝜕 𝜇 + 𝜇2 𝐺 (𝑥 − 𝑦) = −𝑖~𝑐𝛿 (4) (𝑥 − 𝑦) .
(︀ )︀
(8.69)

Then the integral should be taken and due to the delta-function presence, the result
will be:
𝜕𝜇 𝜕 𝜇 + 𝜇2 𝜑 (𝑥) = 𝐽 (𝑥) .
(︀ )︀
(8.70)

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Lecture 9. Yukawa Force. Dirac Equation

Yukawa force
At the previous lecture the action for a scalar field was introduced in the presence of
the linear source and this action is given by the following expression:
∫︁ [︂ ]︂
1 4 1 𝜇 1 2 2
𝑆 [𝜑] = d 𝑥 𝜕𝜇 𝜑𝜕 𝜑 − 𝜇 𝜑 + 𝐽𝜑 , (9.1)
𝑐 2 2
where 𝐽 is a linear source.
If the dimension of the field 𝜑 is recalled, the dimension of the source will be easily
deduced. Recalling that the dimension of the action is the same as the dimension of ~ (the
Plank constant) and dimension of ~ is the same as a dimension of angular momentum, it
can be found out that the linear source has the following physical dimension:
√
~𝑐
[𝐽] = 3 . (9.2)
𝑙
√
An interesting thing is that the quantity ~𝑐 has the same dimension as an electric
charge. Therefore, the source 𝐽 has a dimension of the density of electric charge.
As soon as the source is added up, the energy is changed and the Hamiltonian receives
an extra contribution from the source term. This extra contribution to the Hamiltonian
can be denoted as 𝐻𝐽 and it can be given by
∫︁
𝐻𝐽 = − d⃗𝑥 𝐽 (𝑥) 𝜑 (𝑥) , (9.3)

where 𝜑 (𝑥) is a real scalar field.

It is also can be seen that equations of motion for 𝜑 in the presence of the source gets
change and new equations of motion will have the following form:

𝜕𝜇 𝜕 𝜇 + 𝜇2 𝜑 (𝑥) = 𝐽 (𝑥) .
(︀ )︀
(9.4)

The solution of equation (9.4) can be written as

∫︁
𝑖
𝜑 (𝑥) = d4 𝑦𝐺 (𝑥 − 𝑦) 𝐽 (𝑦) , (9.5)
~𝑐
where 𝐺 (𝑥 − 𝑦) is the Green’s function, which is a solution of the fundamental Klein-
Gordon equation, where on the right hand side of this equation a delta-source or point
like source:
𝜕𝜇 𝜕 𝜇 + 𝜇2 𝐺 (𝑥 − 𝑥′ ) = −𝑖~𝑐𝛿 (4) (𝑥 − 𝑥′ ) .
(︀ )︀
(9.6)

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It is well known that there are many solutions of equation (9.6), because to any
solution of the inhomogeneous equation the homogeneous part can be added up. That
is why, there are such Green’s functions as retarded Green’s function, advanced Green’s
function and Feynman propagator exist. But for the present discussion it does not matter
which Green’s function will be picked up. The (9.4) can be easily gotten if 𝜑 (𝑥) in the
(9.4) will be replaced by the (9.5) and if Klein-Gordon operator will be moved into the
integral for acting on the Green’s function:
∫︁
𝜇 2 𝑖
d4 𝑦 𝜕𝜇 𝜕 𝜇 + 𝜇2 𝐺 (𝑥 − 𝑦) 𝐽 (𝑦) .
(︀ )︀ (︀ )︀
𝜕𝜇 𝜕 + 𝜇 𝜑 (𝑥) = (9.7)
~𝑐
According to the (9.6), expression (9.7) can be modified:
∫︁
9.7 = d4 𝑦𝛿 (4) (𝑥 − 𝑦) 𝐽 (𝑦) = 𝐽 (𝑦) . (9.8)

The goal now is to calculate the interaction energy. This is quantity, which is mediated
by a scalar field between two equal charge static sources, represented by point-like particles,
sitting at positions 𝑥1 and 𝑥2 (fig. 9.1). So, these particles are static and this means
that their positions are fixed and they are not changing with the time. But due to the
interaction mediated by field 𝜑 around, these particles will interact with each other and
there will be a force, which this particles will exert on each other, in particular, because
a particle at 𝑥2 will be interpreted as a source of the field. Another particle will find itself
under the influence of the scalar field and will be considered as a test particle.

Fig. 9.1. Interaction of the point-like particles, sitting at positions 𝑥1 and 𝑥2

In terms of mathematical formulas this means that:

√
𝐽2 (𝑥) = ~𝑐𝛿 (⃗𝑥 − ⃗𝑥 2 ) , (9.9)

where 𝐽2 is a source of a scalar field 𝜑 (𝑥).

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It can be seen that, when 𝑥 is equal to 𝑥2 , then the delta function will be different from
zero and the magnitude of the source is proportional to the quantity with a dimension of
electric charge.
The source function 𝐽2 can be used in the right hand side of equation (9.4) and then the
response to the presence of the source will be a scalar field 𝜑 (𝑥) arising in the surrounding
space due to the fact that the source was put at a position 𝑥2 .
The first particle, on the other hand, will be represented in a similar manner. A
function 𝐽1 (𝑥) will be associated for this source with the same properties:
√
𝐽1 (𝑥) = ~𝑐𝛿 (⃗𝑥 − ⃗𝑥 1 ) , (9.10)

where 𝐽1 is a function of the test particle.

The energy 𝐻𝐽 can be interpreted as an additional contribution 𝑉 :

𝑉 := 𝐻𝐽 . (9.11)

The energy of field 𝜑 due to the interaction of the test particle with a field 𝜑 will be
given by: ∫︁
𝑉 =− d⃗𝑥 𝐽1 (𝑥) 𝜑 (𝑥) . (9.12)

Formula (9.12) is similar to formula (9.3), except the fact that now a particle concentrated
at the position 𝑥1 .
On the other hand, since 𝜑 (𝑥) is a scalar field produced by the source 𝐽2 , an explicit
solution for 𝜑 can be written in terms of the Green’s function:
∫︁
𝑖
𝑉 =− d⃗𝑥 d4 𝑦𝐽1 (𝑥) 𝐺 (𝑥 − 𝑦) 𝐽2 (𝑦) , (9.13)
~𝑐
where in the (9.13) 𝜑 (𝑥) was replaced by the integral with the Green’s function according
to the (9.5). 𝐽2 (𝑦) is the source, which generates a field and this a field interacts with a
test particle, which can be also considered as another source. Therefore, the source of the
field is interchangeable. It is also can be seen actually that formula (9.13) is completely
symmetric with respect to 𝑥 and 𝑦. There is only little difference for the moment that d⃗𝑥
is a three dimensional integration and d𝑦 is the four dimensional integration, but it can
be reduced also to three dimensional integration.
Integrals in the (9.13) can be computed by using the knowledge of the Green’s function.
First thing to do is substitute the expression for the source. For simplicity a source with
a unit value of charge will be introduced, but if an arbitrary value is needed, then the

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expression may be multiplied with a number. So, let’s multiply each of these sources with
a number 𝑞, which is just number, and then
√
𝑞 ~𝑐 (9.14)

will have the dimension of electric charge.

And the 𝑞 value will be chosen in the same way and, therefore, particles will have the
same charge. Upon substituting the point like sources 𝐽1 and 𝐽2 , the following expression
will be gotten:
∫︁
2
𝑉 = −𝑖𝑞 d⃗𝑥 d4 𝑦𝐺 (𝑥 − 𝑦) 𝛿 (⃗𝑥 − ⃗𝑥 1 ) 𝛿 (⃗𝑦 − ⃗𝑥 2 ) . (9.15)

Then, the transformation to the Fourier’s representation of the Green’s function can be
done. As a Green’s function the Feynman propagator will be used, but it does not really
matter which one to take. The result of the transformation will have the following form:
−𝑖𝑘0 (𝑥0 −𝑦 0 ) 𝑖⃗𝑘 (⃗
𝑞 2 ~𝑐 𝑒 𝑥 −⃗𝑦 )
∫︁
𝑒
𝑉 = d⃗𝑥 d 𝑦d⃗𝑘 d𝑘
4 0
𝛿 (⃗𝑥 − ⃗𝑥 1 ) 𝛿 (⃗𝑦 − ⃗𝑥 2 ) . (9.16)
(2𝜋)4 (𝑘 0 )2 − ⃗𝑘 2 − 𝜇2 + 𝑖𝜖
First of all, expression (9.16) can be integrated over ⃗𝑥 and ⃗𝑦 . This is trivial, because
⃗𝑥 and ⃗𝑦 in arguments of delta-functions will be replaced: ⃗𝑥 by ⃗𝑥 1 and ⃗𝑦 by ⃗𝑥 2 . Another
thing, which can be done immediately is the integration over 𝑦 0 , because

d4 𝑦 = d⃗𝑦 d𝑦 0 (9.17)

and then the integration over 𝑦 0 can be performed, because 𝑦 0 comes only with the
exponential
𝑒−𝑖𝑘 (𝑥 −𝑡 ) = 𝑒−𝑖𝑘 𝑥 𝑒𝑖𝑘 𝑦 .
0 0 0 0 0 0 0
(9.18)

When the integration over 𝑦 0 will be completed, the proportionality coefficient will be a
delta-function on the variable 𝑘 0 , because an integral with exponential is just proportional
to the delta-function: ∫︁
1 0 0
d𝑦 0 𝑒𝑖𝑘 𝑦 = 𝛿 𝑘 0 .
(︀ )︀
(9.19)
2𝜋
According to the facts, which was discussed above, the simplified expression will have
the following form:
⃗
𝑞 2 ~𝑐 𝑒𝑖𝑘 (⃗𝑥 1 −⃗𝑥 2 )
∫︁
𝑉 =− d⃗𝑘 , (9.20)
(2𝜋)3 ⃗𝑘 2 + 𝜇2 − 𝑖𝜖
where in the denominator since the integration over 𝑘 0 was completed, 𝑘 0 can be replaced
by 0.

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The term 𝑖𝜖 in the denominator of the (9.20) can be removed by taking a limit 𝜖 → 0,
because the integral over ⃗𝑘 has no problems with the denominator, because ⃗𝑘 2 + 𝜇2 is
non negative. That’s actually shows that any Green’s function can be chosen, because all
Green’s functions differ by the 𝜖 prescription.
The simplified integral for the 𝑉 value will have the next form:
⃗
𝑞 2 ~𝑐 𝑒𝑖𝑘 (⃗𝑥 1 −⃗𝑥 2 )
∫︁
𝑉 =− ⃗
d𝑘 . (9.21)
(2𝜋)3 ⃗𝑘 2 + 𝜇2

The integral (9.21) can be computed explicitly. This can be done by passing to spherical
coordinates. Let’s introduce the modulus of vector ⃗𝑘 , which can be denoted by 𝑘 without
vector symbol on it ⃒ ⃒
𝑘 = ⃒⃗𝑘 ⃒ (9.22)
⃒ ⃒

and the distance between points 𝑥1 and 𝑥2 should be also introduced, which is:

𝑟 = |⃗𝑥 1 − ⃗𝑥 2 | . (9.23)

Then the spherical coordinate system can be chosen by directing the axis 𝑧 along
⃗𝑥 1 − ⃗𝑥 2 and then the integral will be written in the following way:
𝑞 2 ~𝑐 ∞ 2
∫︁ 𝜋 ∫︁ 2𝜋
𝑒𝑖𝑘𝑟 cos 𝜃
∫︁
𝑉 =− 𝑘 d𝑘 sin 𝜃d𝜃 d𝜙 . (9.24)
(2𝜋)3 0 0 0 𝑘 2 + 𝜇2
The spherical coordinate system can be described in the following way: the direction
of the 𝑧 axis is the same as the direction of the vector ⃗𝑥 1 − ⃗𝑥 2 and 𝑥 and 𝑦 axes are also
existed. Then the vector ⃗𝑘 is also defined, which is a running vector in the integral, and
then angles 𝜃 and 𝜙 should be introduced as it shown on the (fig. 9.2).
The standard measure of the d⃗𝑘 is:

d⃗𝑘 = 𝑘 2 sin 𝜃d𝑘 . (9.25)

Nothing depends on angle 𝜙 in the integral (9.24) and it can be immediately integrated.
The result will give just 2𝜋.
Then the integral over angle 𝜃 can also be taken, because:

sin 𝜃d𝜃 = −d (cos 𝜃) , (9.26)

which is very convenient, because cos 𝜃 is standing in the exponential 𝑒𝑖𝑘𝑟 cos 𝜃 . Then, the
integral will have the following form:
∞
𝑞 2 ~𝑐
∫︁
4𝜋 sin (𝑘𝑟)
𝑉 =− 𝑘 2 d𝑘 · , (9.27)
(2𝜋)3 0 𝑘𝑟 𝑘 2 + 𝜇2

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Fig. 9.2. The spherical coordinate system

where the expression

sin (𝑘𝑟)
(9.28)
𝑘𝑟
come from integrating of the exponential in the (9.24).
Expression (9.27) can be simplified to the next form:

𝑞 2 ~𝑐 ∞ 𝑘 sin (𝑘𝑟)
∫︁
𝑉 =− 2 d𝑘 2 . (9.29)
2𝜋 𝑟 0 𝑘 + 𝜇2

The integral from the (9.29) will be denoted as 𝐼 (𝑟), where:

∫︁ ∞
𝑘 sin (𝑘𝑟)
𝐼 (𝑟) = d𝑘 2 . (9.30)
0 𝑘 + 𝜇2
This integral can be computed by using the Cauchy residue theorem in the theory of
complex variables.
The 𝐼 (𝑟) can be also rewritten as:
∫︁ ∞
𝜕 cos (𝑘𝑟)
𝐼 (𝑟) = − d𝑘 . (9.31)
𝜕𝑟 0 𝑘 2 + 𝜇2
Then the cosine can be represented as a difference of two exponentials and, in fact, the
denominator should be also expanded into simple fractions. One more step to complete
is to make integration also from −∞ to +∞ using the fact that the function under the
integral is even. Therefore,
+∞
𝑒𝑖𝑘𝑟 𝑒−𝑖𝑘𝑟
∫︁ (︂ )︂
1 𝜕
𝐼 (𝑟) = − + , (9.32)
4 𝜕𝑟 −∞ (𝑘 + 𝑖𝑎) (𝑘 − 𝑖𝑎) (𝑘 + 𝑖𝑎) (𝑘 − 𝑖𝑎)

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where, the variable 𝑎 is equal to:

𝑚𝑐 1
𝑎=𝜇= = . (9.33)
~ 𝜆
𝑚𝑐 1
In fact, in physical dimensions the quantity ~
is known as an inverse lens 𝜆
and this is
something which is called as Compton wavelengths of a particle with mass 𝑚.
As can be seen the integral (9.32) is always real line, but the contour can be closed
in the complex plane and, in particular, for the first integral exponential 𝑒𝑖𝑘𝑟 it should
be closed in the upper half plane. For the second integral the situation is opposite and
it should be closed in the lower half plane. Then it can be seen, which poles ends up in
the lower half plane and the residue theorem should be used to compute the integral. The
corresponding pole in the up half plane will be equal to 𝑘 = −𝑖𝑎. In the lower half plane
it will be equal to 𝑘 = −𝑖𝑎.
So, the integral reduces to:

𝑒−𝑎𝑟 𝑒−𝑎𝑟
(︂ )︂
1 𝜕
𝐼 (𝑟) = − 2𝜋𝑖 − 2𝜋𝑖 . (9.34)
4 𝜕𝑟 2𝑖𝑎 −2𝑖𝑎

Then it can be simplified to the following expression:

𝜋 𝜕 −𝑎𝑟
𝐼 (𝑟) = − 𝑒 , (9.35)
2𝑎 𝜕𝑟
where after differentiating the final result will be equal to:
𝜋 −𝑎𝑟
𝐼 (𝑟) = 𝑒 . (9.36)
2
In terms of Compton wavelengths expression (9.36) can be written as:
𝜋 −𝑟/𝜆
𝑒 . (9.37)
2
Now expression (9.29) can be simplified and the result will be equal to:

1 𝑞 2 ~𝑐 −𝑟/𝜆
𝑉 =− 𝑒 . (9.38)
4𝜋 𝑟

Thanks to the (9.38) the potential between two equally charged static sources was gotten
and this potential is called as Yukawa potential.
In particular, from expression (9.38), several facts can be highlighted. First of all, a
very interesting thing is that this potential appears to be negative and changing 𝑞 → −𝑞
will produce the same expression, because of the existence of 𝑞 2 . And this potential with

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a minus sign actually means that the force is an attractive, because having a smaller
distance between the particles will decrease the potential and it will take more and more
negative values.
It should be also noticed that it’s preferable to approach each other to diminish
the energy. Diminishing the energy corresponds, therefore, in this case, to the desire
of particles to be closer to each other.
The second thing which can be seen is that there is the exponential, which dumps the
potential, when 𝑟 becomes sufficiently large, in particular, when 𝑟 becomes bigger than
the wavelengths of the particles potential shows exponential decay. So, potential goes to
0 very fast, actually exponentially with a distance increasing.
It is also should be noticed that the interaction range set up by the exponential does
depend on the Compton wavelengths. It’s governed by the effective length, which is turns
out to be Compton wavelengths.
This additionally shows, why people call the Greens function as a propagator. The
propagator propagates interaction from the one source to the other source or from a
source to a test particle. And it propagates interaction, represented by the field 𝜑.
Generally, a very physical calculation was performed, which showed explicitly what
happens if two particles are existed inside the space field with a scalar field 𝜑. And because
of interacting with this field they start to attract.
One more important fact is that when 𝜆 → ∞, which essentially means that 𝑚 → 0
according to
1 𝑚𝑐
= , (9.39)
𝜆 ~
the field will be massless scalar field, instead of massive scalar field and then the potential
between two static sources actually becomes a Coulomb potential:
1 𝑞 2 ~𝑐
𝑉 (𝑟) → − . (9.40)
4𝜋 𝑟
When mass is non trivial, the Coulomb potential gets screened and becomes short range
because of presence of the exponential.
In the mid of 1930s Yukawa tried to understand the origin of the strong force, which
holds together protons and neutrons in the nucleus. So, the nucleus consists of protons
and neutrons. It was also known that the force acting between these protons and neutrons
inside nuclear is a very short range.
Yukawa made a conjecture that the strong force was mediated by particle similar to
photon, which like photon in electrodynamics mediates interactions between electrons and

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positrons. He assumed that something similar creates place inside nucleus and there is a
hypothetical particle, which mediates interactions between this protons and neutrons.
Assuming that this mediating particle is massive and following absolutely the same
analogy with electromagnetism, one sees that one needs to add mass to the wave equation
and this produces as a result, the short range mediating force between two static sources.
Yukawa actually predicted the existence of a novel type of particle and from the known
data on the range of nuclear forces he was actually able to predict with quite a good
accuracy, the mass of this particle. So, if the range of the strong force is known, which
can be actually done experimentally, then 𝜆 can be deduced or equivalently the mass of
the particle can be found. And indeed later, this particle called 𝜋 meson or sometimes
people call it simply pion, pion was discovered in cosmic rays. It also appeared later in
this cyclotron experiments, where it was predicted by Yukawa.
Although, today, there are several types of mesons are known and it is also known that
mesons are not elementary themselves, but rather they are composite particles made of
the quark and it’s anti quark. Nevertheless, Yukawa theory made a very right qualitative
picture and it also, in a way, paved the development of what is called QCD or quantum
chroma dynamics. Nevertheless, that was very important step towards development of
QCD by assuming the nature of strong force as a rising from a massive particle with zero
spin.
As a general computation the field of any spin can be applied to a mediator and
similar computation takes place also for electromagnetism, where instead of minus sign
plus sign can be seen in the potential, and, therefore, there will be repulsive force instead
of attraction. Then for particles, it is more preferable to take a large distance between
them, in order to decrease the energy, because then energy is positive and it will be
decreasing when the distance will tend to infinity. So, sign is very important and sign
eventually comes from the equations of motion that induced by the type of action. For
instance, if a particle of a given speed was chosen to describe, as a result it leads to the
physical consequences of attraction or repulsion.
Then, a new chapter in the quantum field theory, namely the Dirac equation should
be introduced.

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Dirac equation
It’s a new and very interesting story called the Dirac equation. In 1928, the Dirac
discovered his relativistic equation trying to overcome difficulties with negative probability
densities of the Klein-Gordon equation.
As it was already shown Schrodinger made this attempt before to write down relativistic
quantum mechanics by taking dispersion relation between energy and momentum of
relativistic particle and transformed this condition in the language of operators by replacing
momentum and energy with the corresponding operators, shifts in a space and shifts in
time. And also he wrote down the Klein-Gordon equation, which was later rediscovered
many times by Klein, Gordon, Fock and many others.
But it was the problem to treat this equation as a relativistic equation for a wave
function 𝜑, because it has difficulties with probabilistic interpretation, because equation
have the second order. That’s allows to fix initial conditions and the initial velocities and
leads to the appearance of negative probability densities.
The reason, which led Dirac create his equation was essentially the following. First of
all, the target is to prevent the occurrence of negative probability densities. This means
that time derivatives in the formula for the probability density should be avoided. So,
the equation must, therefore, not contain time derivatives higher than the first order. For
instance, the Schrodinger equation in the usual quantum mechanics has the first order of
time:
𝜕
𝑖~ 𝜓 = 𝐻𝜓 . (9.41)
𝜕𝑡
To achieve a good probabilistic interpretation, a wave function must satisfy the equation,
which contains only first derivative.
On the other hand, if relativistic covariance is wanted, then it requires that the
spacial and time components, in particular, spacial derivatives and the time derivative
must be treated on equal footing. For example, in the standard Schrodinger equation of
quantum mechanics, this is not the case, because, typically, the Hamiltonian contains
second spacial derivative. That’s immediately shows that the equation is non-relativistic,
because relativism requires that 𝑡 can be converted to 𝑥 and 𝑥 can be converted to 𝑡 by
means of linear Lawrence transformations. It is impossible to have this, if derivatives in
the equation have a different order.
Now also, one needs linearity to have a superposition principle like in quantum mechanics.
So, equation must be linear.

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Like in the usual quantum mechanics a superposition principle exists, because linearity
means that if there are two solutions of the equation, then they sum will be also a solution.
The last important point is, that since the relativistic particle is described, it must
still satisfy the standard on-shell condition for relativistic particles. So,

𝐸 2 = 𝑝2 𝑐2 + 𝑚2 𝑐4 . (9.42)

For 𝜓 this would mean that 𝜓 must satisfy the second order Klein-Gordon equation:

1 𝜕2 𝜕2
(︂ (︁ 𝑚𝑐 )︁2 )︂
− + 𝜓=0, (9.43)
𝑐2 𝜕 2 𝑡 𝜕𝑥2𝑖 ~

because this equation will produce the relativistic dispersion relation.

Dirac said the following: let’s try to write down something, which would be linear for
𝜓 and contains derivative of the first order:
(︂ )︂
𝜕𝜓 ~𝑐 𝜕𝜓 𝜕𝜓 𝜕𝜓
𝑖~ = 𝛼1 1 + 𝛼2 2 + 𝛼3 3 + 𝛽𝑚𝑐2 𝜓 , (9.44)
𝜕𝑡 𝑖 𝜕𝑥 𝜕𝑥 𝜕𝑥

where the left hand side looks like non-stationary Schrodinger equation, but on the right
hand side, there are terms, which are the linear combination of first order derivatives
function 𝜓.
Coefficients 𝛼𝑖 cannot be just usual numbers, because if they’re usual numbers, which
have a certain values, then the equation is not even invariant with respect to usual three
dimensional rotations, because three dimensional rotations will act on coordinates 𝑥1 , 𝑥2
and 𝑥3 and they will be just then produce changes of coefficients 𝛼1 , 𝛼2 , 𝛼3 and will
immediately destroy the equation.
So, Dirac said that it’s not possible for these 𝛼1 , 𝛼2 , 𝛼3 to be numbers and supposed
them to be matrices. If they are matrices, then 𝜓 cannot be just scalar and it should
be some multi component object. 𝜓 should be something, which would contain many
components: 𝜓1 and so on: ⎛ ⎞
𝜓1
⎜ . ⎟
𝜓=⎜ . ⎟
⎝ . ⎠ . (9.45)
𝜓𝑛
𝜓 also cannot be a scalar also for another reason. From relativistic covariance it is
known that probability density, which would be

𝜌 = 𝜓*𝜓 , (9.46)

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it must be relativistic invariant or to be a zero component of a vector, because for 𝜌 the

continuity equation exists, which is written as
𝜕𝜌
+ ÷⃗𝑗 = 0 . (9.47)
𝜕𝑡
Expression (9.47) describing a vector probability, where a vector, probability density
(︁ )︁
⃗ ⃗
and probability current are defined. And 𝜌 with 𝑗 forms a four vector 𝜌, 𝑗 . That is
why, 𝜌 must be a component of a 4-vector. It means that 𝜓 cannot be just scalar in this
case.
Indeed, the suggestion is to treat 𝛼𝑖 as matrices and 𝜓 as a column with a number of
components 𝜓1 , . . . , 𝜓𝑛 . Dirac called the object 𝜓, presented in the (9.45) as spinor.
By the way, if 𝛼 are some matrices, then 𝛽 is also a matrix and 𝜓 is a column.
Then, equation (9.44) can be written in the following form:
(︂ )︂
𝜕 𝑖 𝜕 2
𝑖~ + 𝑖~𝑐𝛼 − 𝛽𝑚𝑐 𝜓 = 0 . (9.48)
𝜕𝑡 𝜕𝑥𝑖
Let’s get the relativistic dispersion relation for 𝜓. This can be seen as follows. If such
an equation for a 𝑛-component object exists, it can be hit by the following operator:
𝜕 𝜕
𝑖~ − 𝑖~𝑐𝛼𝑖 𝑖 + 𝛽𝑚𝑐2 , (9.49)
𝜕𝑡 𝜕𝑥
which is a matrix differential operator. Firstly, it’s differential operator, because it contains
derivatives. Secondly, it’s a matrix.
When the (9.48) will be hit by the operator (9.49) from the left, the expression will
have the following form:
𝜕2 𝑖 𝑗 𝑗 𝑖
2 2𝛼 𝛼 + 𝛼 𝛼 𝜕 2𝜓 𝑖 𝜕𝜓
(︀ 2 )︀2 2
− ~2 3
(︀ 𝑖 )︀
+ ~ 𝑐 + 𝑖~𝑚𝑐 𝛼 𝛽 + 𝛽𝛼 − 𝑚𝑐 𝛽 𝜓 = 0 , (9.50)
𝜕𝑡2 2 𝜕𝑥𝑖 𝜕𝑥𝑗 𝜕𝑥𝑖
where initially there are nine terms to proceed, but four terms will cancel out. The second
derivative of 𝜓 with respect to 𝑥𝑖 and 𝑥𝑗 . This two derivatives acting on 𝜓 commute and
the derivative can be written in different ways, because it’s symmetric.
Obtained expression (9.50) is a consequence of equation (9.48). It was assumed that 𝜓
satisfy the linear equation and as a consequence it might satisfy the second order equation
(9.50), because it contains second order derivatives. It also contains a first derivative and
the term without derivative.
If the equation is multiplied by − ~21𝑐2 , then the equation will take the following form:

1 𝜕 2 𝜓 𝛼𝑖 𝛼𝑗 + 𝛼𝑗 𝛼𝑖 𝜕 2 𝑚𝑐 (︀ 𝑖 𝑖 𝜕𝜓
(︁ 𝑚𝑐 )︁2
𝛽 2𝜓 = 0 .
)︀
− − 𝑖 𝛼 𝛽 + 𝛽𝛼 + (9.51)
𝑐2 𝜕𝑡2 2 𝜕𝑥𝑖 𝜕𝑥𝑗 ~ 𝜕𝑥𝑖 ℎ
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Now it’s pretty clear what conditions on matrices 𝛼𝑖 and 𝛽 should be imposed to turn
equation (9.51) into the Klein-Gordon equation.
It is wanted to make exactly the Klein-Gordon equation for each component of 𝜓 and
each component of 𝜓 must satisfy the Klein-Gordon equation, because 𝜓 is supposed to
describe relativistic particle with a standard dispersion relation. It can be seen that:

𝛼𝑖 𝛼𝑗 + 𝛼𝑗 𝛼𝑖 = 2𝛿 𝑖𝑗 , (9.52)

where 𝑖 and 𝑗 run from 1 to 3. Then the second term of (9.51) will be turned simply into
𝜕2𝜓
second derivative 𝜕𝑥𝑖2
.
Then the following should be required:

𝛼𝑖 𝛽 + 𝛽𝛼𝑖 = 0 . (9.53)

Such a way the term with the first derivative will be removed.
Finally, to get the standard Klein-Gordon equation for a particle of mass 𝑚 the next
condition should be completed:
𝛽2 = 1 . (9.54)

So, the equation, which will be gotten for 𝜓 is the following:

1 𝜕 2𝜓 𝜕 2 𝜓 (︁ 𝑚𝑐 )︁2
− + 𝜓=0. (9.55)
𝑐2 𝜕𝑡2 𝜕𝑥𝑖2 ~
This equation is valid for each component of 𝜓, because there are no matrices. They all
now proportional to identity matrix. It also can be written in the next form:
⎛ ⎞
(𝜕𝜇 𝜕 𝜇 + 𝜇2 ) 𝜓1
⎜ .. ⎟
𝜕𝜇 𝜕 𝜇 + 𝜇2 𝜓 = ⎜
(︀ )︀
. ⎟=0. (9.56)
⎝ ⎠
𝜇 2
(𝜕𝜇 𝜕 + 𝜇 ) 𝜓𝑛

From relations (9.52) - (9.54) it is immediately can be seen that if indexes 𝑖 and 𝑗 will
be equal to each other, than expression (9.52) will have the following form:

𝛼𝑖 𝛼𝑖 + 𝛼𝑖 𝛼𝑖 = 2 (9.57)

and then
(︀ 𝑖 )︀2
𝛼 =1. (9.58)

Including the condition (9.54), it was gotten that all squares of matrices equal to 1, where
1 here is the identity matrix, because 𝛼 and 𝛽 are matrices.

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Conditions (9.54) and (9.58) means all values of this matrices can be only equal to +1
or -1.
Second important observation is that all these matrices have zero trace and it can be
seen from the condition (9.53). 𝛽 is invertible, because of the (9.54) and, therefore, it can
be written that:
𝛼𝑖 = −𝛽𝛼𝑖 𝛽 . (9.59)

Now a trace of 𝛼𝑖 can be taken and due to cyclic property of a trace it will be equal
to:
Tr𝛼𝑖 = −Tr 𝛽𝛼𝑖 𝛽 = −Tr 𝛼𝑖 𝛽 2 = −Tr𝛼𝑖 = 0 .
(︀ )︀ (︀ )︀
(9.60)

Analogously trace for matrix 𝛽 will be equal to:

(︁(︀ )︀ )︁
2
𝛼𝑖 𝛽 = Tr 𝛼𝑖 𝛽𝛼𝑖 = −Tr𝛽 ,
(︀ )︀
Tr𝛽 = Tr (9.61)

where the following property was used:

𝛼𝑖 𝛽 + 𝛽𝛼𝑖 = 0 → 𝛼𝑖 𝛽𝛼𝑖 = −𝛽 . (9.62)

Therefore, ⎧
⎨Tr𝛼𝑖 = 0
(9.63)
⎩Tr𝛽 = 0

It means that the number of positive eigenvalues must be the same as the number of
negative eigenvalues. And as it was gotten eigenvalues can be equal only to +1 or -1.
One more important conclusion is that the dimension of matrices 𝛼𝑖 and 𝛽 should be
even or, in other words, 𝑛 must be even.

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Lecture 10. The Dirac Equation and Lorentz

Transformations
In the last lecture the Dirac’s equation was introduced. The physical principles, which
are standing behind the derivation of this equation was basically discussed and an object
which is called spinor was introduced.
The main consequence of Dirac equation as it was computed is that this equation also
satisfies the Klein-Gordon equation. In other words, every component of our spinor satisfies
the Klein-Gordon equation, which means that every component describes a relativistic
particle, that means a dispersion relation for this particle is relativistic.
The equation also involves a set of matrices which satisfy the following properties:
there are three matrices 𝛼𝑖 , which satisfy the following algebraic relation:

𝛼𝑖 𝛼𝑗 + 𝛼𝑗 𝛼𝑖 = 2𝛿 𝑖𝑗 . (10.1)

And there is one matrix 𝛽, for which the following conditions are right:
⎧
⎨𝛼𝑖 𝛽 + 𝛽𝛼𝑖 = 0
(10.2)
⎩𝛽 2 = 1

It also was shown that traces of all these matrices are equal to 0. So,
⎧
⎨Tr𝛼𝑖 = 0
(10.3)
⎩Tr𝛽 = 0

According to (10.1) and (10.2) eigenvalues of matrices 𝛼𝑖 and 𝛽 are equal to +1 and
-1. Due to the condition (10.3) matrices can be realized only in even dimension.
Then the concrete realization of these matrices should be gotten. The first idea is that
since matrices must exist only in even dimensions, let’s try the minimal even dimension,
where such matrices have a chance to exist. The dimension, which satisfies this condition
is a dimension with 𝑛 = 2. Therefore, matrices 𝛼𝑖 and 𝛽 will be 2x2 matrices.

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In two dimensions there are three distinguished matrices, which are Pauli matrices
⎧ ⎛ ⎞
⎪
⎪ 0 1
𝜎1 = ⎝
⎪
⎪ ⎠
⎪
⎪
⎪
⎪
⎪
⎪ 1 0
⎪
⎪ ⎛ ⎞
0 −𝑖
⎪
⎨
𝜎2 = ⎝ ⎠ (10.4)
⎪
⎪
⎪
⎪ 𝑖 0
⎪
⎪ ⎛ ⎞
1 0
⎪
⎪
⎪
𝜎3 = ⎝
⎪
⎪ ⎠
⎪
⎪
⎩ 0 −1

They actually obey the desired relation, namely the anti-commutator of these matrices

{𝜎 𝑖 , 𝜎 𝑗 } = 2𝛿 𝑖𝑗 , (10.5)

which is similar to the anti-commutator of matrices 𝛼𝑖 and 𝑎𝑙𝑝ℎ𝑎𝑗 . However in two

dimensions the fourth independent matrix would be just the identity matrix, but, unfortunately,
this cannot be identified with 𝛽, because identity matrix would commute with Pauli
matrices, while we should have a non-trivial relation between 𝜎-s and matrix 𝛽, which is
not just a simple commutation relation.
This means that in dimension with 𝑛 = 2 such realization of Dirac matrices cannot be
found. The next trial would be to ask for such matrices in dimension with 𝑛 = 4. Indeed,
it appears that the minimal dimension, where one can construct the four matrices 𝛼𝑖 and
𝛽 with all properties above is 4. Concretely, the following construction for 𝛼𝑖 can be taken:

0
(︃ )︃
𝑖
𝜎
𝛼𝑖 = . (10.6)
−𝜎 𝑖 0

At the same time, for 𝛽 the next matrix can be taken

1 0
(︃ )︃
𝛽= , (10.7)
0 −1
It can be checked that matrices, indeed, satisfy all the relations above and they are
traceless. So, that’s concrete and explicit realization of the matrices, which should feature
in the Dirac equation.
In the last lecture it was found that the Dirac Hamiltonian and the Dirac equation in
the Hamiltonian form can be written in the following way:
𝜕𝜓
𝑖~ = H𝜓 , (10.8)
𝜕𝑡
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where H is the Dirac Hamiltonian, which can be called as first quantized Hamiltonian
meaning, that here the pedestrian approach to the theory was used. 𝜓 can be treated as a
multi component wave function for the Dirac equation, which has a form of the standard
Schrodinger equation.
The Dirac Hamiltonian is a matrix, which can be written in the following form:

H = 𝑐⃗𝑎 𝑝⃗ + 𝛽𝑚𝑐2 . (10.9)

In a more explicit form, expression (10.9) can be presented as:

(︂ )︂
𝜕
𝑖
H = 𝑐𝑎 −𝑖~ 𝑖 + 𝛽𝑚𝑐2 , (10.10)
𝜕𝑥
where 𝑝⃗ was replaced according to the definition of the momentum operator. H is a
differential matrix acting on four component wave function 𝜓 or simply on column
⎛ ⎞
𝜓
⎜ 1⎟
⎜𝜓 ⎟
⎜ 2⎟
𝜓=⎜ ⎟ . (10.11)
⎜𝜓3 ⎟
⎝ ⎠
𝜓4

Now, the algebra of the matrices, which enter the Dirac’s equation, should be investigated
and this is needed in order to discuss covariant properties of the Dirac’s equation, in
particular, to understand how this equation transforms under Lawrence group, because
the Lawrence group is very important and it’s a symmetry group of relativistic quantum
field theory.
In order to to do this the structure of the matrices should be better understood. And
𝛽
to complete it the Dirac’s equation should be multiplied by the ratio 𝑐
. Also, the new
notation can be introduced:
𝛾 0 = 𝛽 , 𝛾 𝑖 = 𝛽𝛼𝑖 , (10.12)

where 𝑖 is running from 1 to 3.

Let’s also identify 𝑥0 with:
𝑥0 = 𝑐𝑡 (10.13)

and, therefore,
𝜕 𝜕
0
= . (10.14)
𝜕𝑥 𝑐𝜕𝑡
Then the Dirac’s equation can be written in a more symmetric form:
[︂ (︂ )︂ ]︂
0 𝜕 𝑖 𝜕
𝑖~ 𝛾 +𝛾 − 𝑚𝑐 𝜓 = 0 , (10.15)
𝜕𝑥0 𝜕𝑥𝑖

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where in the last term the variable 𝛽 is disappeared due to the property (10.2).
Now, the relativistic definition of a relativistic scalar product can be used
(︁
𝜇 𝑚𝑐 )︁
𝑖𝛾 𝜕𝜇 − 𝜓=0. (10.16)
~
With a natural choice of units
~=𝑐=1 (10.17)

equation (10.17) takes the form:

(𝑖𝛾 𝜇 𝜕𝜇 − 𝑚) 𝜓 = 0 . (10.18)

And the (10.18) is the final form of the Dirac’s equation.

Is also can be seen that with the new matrices 𝛾 𝜇 the algebraic relations, which was
introduced for 𝛼 and 𝛽 can be also uniformized in one relation for matrices 𝛾 𝜇 , namely

𝛾 𝜇 𝛾 𝜈 + 𝛾 𝜈 𝛾 𝜇 = 2𝜂 𝜇𝜈 · 1 . (10.19)

The free algebra generated by symbols 𝛾 𝜇 modulo the relation (10.19) or subject to this
relation is called Clifford algebra.
The name goes back to the Clifford and this algebra has been known to mathematicians
before the discovery of the Dirac’s equation, but miraculously turns out that this mathematical
background, on which the Dirac’s equation is based on actually already existed in mathematics
in the form of the Clifford algebra.
From the definition of matrices 𝛾 𝜇 it is clear that matrix 𝛾 0 is Hermitian. So, it satisfies
the following relation:
(︀ 0 )︀+
𝛾 = 𝛾0 , (10.20)

while the matrices 𝛾 𝑖 are anti-Hermitian:

(︀ 𝑖 )︀+
𝛾 = −𝛾 𝑖 . (10.21)

(10.21) are right, because 𝛾 𝑖 are given by the product of 𝛽 with 𝛼𝑖 . 𝛽 and 𝛼𝑖 are
Hermitian, but under Hermitian conjugation the order of matrices should be changed.
Therefore,
(︀ 𝑖 )︀+ (︀ 𝑖 )︀+ +
𝛾 = 𝛼 𝛽 = −𝛽𝛼𝑖 = −𝛾 𝑖 . (10.22)

The presented in (10.20) and (10.21) properties can be combined in one relation, which
can be written as
(𝛾 𝜇 )+ = 𝛾 0 𝛾 𝜇 𝛾 0 . (10.23)

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For instance, when 𝜇 is specified to zero,

(︀ 0 )︀+
𝛾 = 𝛾 0𝛾 0𝛾 0 = 𝛾 0 (10.24)
2
according to the fact that (𝛾 0 ) = 1.
One more important fact is that 𝜂 𝜇𝜈 in the (10.19) is Minkowski metric. Another
important observation is that actually one concrete realization of 𝛾 matrices, which
satisfies the Clifford algebra relations, is found. In fact, there exist an infinite number
of possible realizations and this can be recognized by noting that actually if the following
similarity transformation is performed, 𝛾 𝜇 can be taken and transformed as:

𝛾 𝜇 → 𝑈 𝛾 𝜇 𝑈 −1 , (10.25)

where 𝑈 is an arbitrary unitary matrix. Then 𝛾 matrices obtained after this transformation
will satisfy the same Clifford algebra relations and also the same Hermiticity properties.
The fact that Clifford algebra relations are satisfied is trivial, because

𝑈 (𝛾 𝜇 𝛾 𝜈 + 𝛾 𝜈 𝛾 𝜇 ) 𝑈 −1 = 𝑈 𝛾 𝜇 𝛾 𝜈 𝑈 −1 + 𝑈 𝛾 𝜈 𝛾 𝜇 𝑈 −1 . (10.26)

Then the following property can be used

𝑈 −1 𝑈 = 𝑈 𝑈 −1 = 1 . (10.27)

to simplify the (10.26):

10.26 = 𝑈 𝛾 𝜇 𝑈 −1 𝑈 𝛾 𝜈 𝑈 −1 + 𝑈 𝛾 𝜈 𝑈 −1 𝑈 𝛾 𝜇 𝑈 −1 . (10.28)

The (10.28) can be written as:

10.28 = 𝛾 ′𝜇 𝛾 ′𝜈 + 𝛾 ′𝜈 𝛾 ′𝜇 . (10.29)

As a result
10.29 = 2𝜂 𝜇𝜈 1 . (10.30)

So the new 𝛾 matrices obtained by the unitary transformation satisfy the same Clifford
algebra relations and it can be also checked in an analogous way that due to unitarity of
𝑈 , hermeticity property of newly introduced 𝛾 matrices are the same as for the old ones.
The concrete representation, which was introduced for 𝛼𝑖 and 𝛽 in terms of 𝛾 matrices,
can be read in the following way:

1 0 0
(︃ )︃ (︃ )︃
𝜎𝑖
𝛾0 = , 𝛾𝑖 = . (10.31)
0 −1 −𝜎 𝑖 0

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The concrete representation of 𝛾 matrices introduced above is called Dirac representation.

In this representation 𝛾 0 is diagonal and has the form presented in the (10.31).
It also should be noticed, that other representations can be obtained by applying, for
instance, to the Dirac representation the unitary transformation with arbitrary 𝑈 .
Let’s investigate properties of 𝛾 matrices in more detail. From matrices 𝛾 𝜇 16 linearly
independent Hermitian matrices was constructed in the following way: first of all, an
identity matrix can be built if
1 = 𝜂 𝜇𝜇 𝛾 𝜇 𝛾 𝜇 (10.32)

will be constructed, where there is no summation over index 𝜇 and it should be either 0
or 1, 2 and 3. It can be seen that from Clifford algebra the product of 𝛾 matrices with
the same index and multiplied with Minkowski component of the Minkowski metric will
obtain identity.
Then four Hermitian matrices 𝛾 0 exist and then 𝛾 𝑖 , which will be also Hermitian
matrices if they will be multiplied by 𝑖.
Six matrices, which can be denoted as 𝜎 𝜇𝜈 , also can be formed and they will be given
by the following formulas:
⎧
⎨𝜎 𝑖𝑗 = 𝑖 𝛾 𝑖 𝛾 𝑗 −𝛾 𝑗 𝛾 𝑖 = 𝑖𝛾 𝑖 𝛾 𝑗 𝑖 < 𝑗 𝑖, 𝑗 = 1, 2, 3
𝜇𝜈 2
𝜎 = (10.33)
⎩𝜎 0𝑗 = 𝛾 0 𝛾 𝑗 −𝛾 𝑗 𝛾 0 = 𝛾 0 𝛾 𝑗 , 𝑗 = 1, 2, 3
2

The number of matrices in the (10.33) is equal to 6. It also should be noticed that
these matrices are skew symmetric.
Then, there is one important matrix, which has a special name of 𝛾 5 , where 5 stands
for historical reasons. These matrix can be introduced as:

𝛾 5 = 𝑖𝛾 0 𝛾 1 𝛾 2 𝛾 3 , (10.34)

where 𝑖 is added to provide uh the hermiticity property for 𝛾 5 .

Finally, there are also 4 matrices 𝜏 𝜇 introduced in the following way:
⎧
⎨𝜏 𝑖 = 𝛾 𝑖 𝛾 5 , 𝑖 = 1, 2, 3
𝜏𝜇 = (10.35)
⎩𝜏 0 = 𝑖𝛾 0 𝛾 5

So, in total 16 hermitian 𝛾 matrices was constructed with the help of 𝛾-s.
It is also should be noticed that 𝛾 5 matrix introduced in the (10.34) satisfies the
following properties:
(︀ 5 )︀2 (︀ )︀+
𝛾 = 1 , 𝛾5 = 𝛾5 . (10.36)

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In the Dirac representation due to the definition of 𝛾 5 (10.34) the explicit form of this
matrix is:
0 1
(︃ )︃
𝛾5 = . (10.37)
1 0
Then 𝛾 5 is off diagonal in Dirac representation as soon as 𝛾 0 is diagonal.
It can be checked that all other possible products of 𝛾 matrices and their linear
combinations can be expressed by 16 hermitian matrices that was introduced. This means
that these 16 matrices provide the basis in the space of 4x4 hermitian matrices.
Indeed, they form a basis it is needed to be shown that they are actually linearly
independent. This can be done in the following way. To show linear independence it
may be noticed that all these matrices except identity matrix have vanishing trace. For
instance, let’s do it for 𝛾 𝜇 and use the fact that

𝛾𝜈 𝛾 𝜈 = 1 , (10.38)

where one matrix 𝛾 have low index 𝜈 and another is upper index 𝜈 and there is no
summation over index 𝜈. The result is gotten due to the Clifford algebra.
Indeed, if the upper index 𝜇 in the (10.19) is replaced by the lower index 𝜇, then the
expression will have the following form:

𝛾𝜇 𝛾 𝜈 + 𝛾 𝜈 𝛾𝑚 𝑢 = 2𝛿𝜇𝜈 · 1 , (10.39)

where Minkowski metric was replaced by the Kronecker’s delta 𝛿𝜇𝑛 𝑢. Now, if index 𝜇 is
equal to index 𝜈, then formula (10.39) will be transformed into:

2𝛾𝜈 𝛾 𝜈 = 2𝛿𝜈𝜈 · 1 = 2 · 1 . (10.40)

That is why, (10.38) is right.

Coming back to the trace of 𝛾 𝜇 , the following standard trick to prove that trace
vanishes can be used:
⃒
⃒ 1
Tr𝛾 𝜇 = Tr1𝛾 𝜇 = Tr (𝛾𝜈 𝛾 𝜈 𝛾 𝜇 )⃒ = Tr (𝛾𝜈 (𝛾 𝜈 𝛾 𝜇 + 𝛾 𝜇 𝛾 𝜈 )) , (10.41)
⃒
⃒ 2
𝜈̸=𝜇

where the property (10.38) and cyclic property was used. This is the way how to manipulate
the product 𝛾𝜈 𝛾 𝜈 under the trace and then due to the Clifford algebra relation since index
𝜈 was chosen not to coincide with index 𝜇, the sum in the brackets of the (10.41) is

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actually 0. So, the result of the (10.41) is 0, which shows that all traces of 𝛾 𝜇 is equal to
0:
Tr𝛾 𝜇 = 0 . (10.42)

Analogously it can be shown that traces of all other matrices are 0. In particular, trace
of 𝛾 5 is 0 and in the Dirac representation it can be explicitly seen from the form of the
matrix (10.37):
Tr𝛾 5 = 0 . (10.43)

The fact that traces of all matrices excepting the identity matrix vanish can be used
to show that these 16 matrices are linearly independent. This can be shown by assuming
the opposite: it should be supposed that this matrices are linearly dependent and then if
a linear combination of all of them is created, it will be equal to

𝐹 = 𝑎 · 1 + 𝑏𝜇 𝛾 𝜇 + 𝑐𝜇𝜈 𝜎 𝜇𝜈 + 𝑑𝜇 𝜏 𝜇 + 𝑒𝛾 5 , (10.44)

where 𝑎, 𝑏𝜇 , 𝑐𝜇𝜈 , 𝑑𝜇 and 𝑒 are arbitrary coefficients. The statement that this matrices are
linearly dependent means that 𝐹 can be equal to zero with some non-zero coefficients. In
other words, the goal is to show that the situation when 𝐹 = 0 and some coefficients of
the linear combination are not equal to zero is not possible.
The proof can be done in the following way. First of all, a trace of the left and the
right hand side should be taken and since on the left hand side 𝐹 is zero, trace of 𝐹 is
also zero. So, the expression will have the following form:

0 = 𝑎Tr1 = 4 · 𝑎 → 𝑎 = 0 , (10.45)

where the fact that traces of all matrices excepting the identity matrix are vanished was
used.
Since on the left hand side of the (10.45) stands zero, 𝑎 must be equal to zero. So, the
term with 𝑎 is absent. Then the expression for 𝐹 should be taken again without identity
matrix and it should be multiplied with some 𝛾 matrix 𝛾 𝜆 . So,

Tr 𝛾 𝜆 𝐹 = 𝑏𝜇 Tr 𝛾 𝜆 𝛾 𝜇 + 𝑐𝜇𝜈 Tr 𝜎 𝜇𝜈 𝛾 𝜆 + 𝑑𝜇 Tr 𝛾 𝜆 𝜏 𝜇 + 𝑒Tr 𝛾 𝜆 𝛾 5 .
(︀ )︀ (︀ )︀ (︀ )︀ (︀ )︀ (︀ )︀
(10.46)

Now, it can be shown by analyzing all traces, that all of them are equal to zero except
the trace with the argument 𝛾 𝜆 𝛾 𝜇 . For 𝜇 equal to 𝜆 due to the Clifford algebra the identity
matrix will be gained and, therefore,

Tr 𝛾 𝜆 𝐹 = 4𝑏𝜇 → 𝑏𝜇 = 0 .
(︀ )︀
(10.47)

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In this case, only coefficients 𝑐𝜇𝜈 , 𝑑𝜇 and 𝑒 remain. Then multiplying the rest by 𝛾 𝜆 𝛾 𝛽 it
can be shown that coefficients 𝑐𝜇𝜈 are equal to zero, then the same is for 𝑑𝜇 and, finally,
for 𝑒. In this way proceeding similarly to what was explained before, it can be shown that
all coefficients in the linear combination must be equal to zero. Therefore, these matrices
are linearly independent. This means that all found matrices form a basis in the space of
all complex 4x4 matrices.
Linear combinations of 16 matrices with complex coefficients allow to reconstruct any
complex 4x4 complex matrix. This statement is very similar to two dimensional case with
the Pauli matrices. If there are three Pauli matrices and an identity matrix, then it can
be seen that linear combinations of these matrices with complex coefficients can form any
2x2 complex matrix.
It also should be noticed that transforming 𝛾 matrices to another set of 𝛾 matrices
would be equivalent to transforming the wave function 𝜓 by unitary transformation
and unitary transformations are always allowed in quantum field theory and quantum
mechanics. This is the same statement as that changing the basis of 𝛾 matrices can be
compensated by the change of the wave function. And it can be seen from the Dirac
equation
(𝑖𝛾 𝜇 𝜕𝜇 − 𝑚) 𝜓 = 0 . (10.48)

Now, this equation should be passed to another basis. So, 𝛾 matrices will be changed
to the new basis according to:

𝛾 𝜇 → 𝑈 𝛾 𝜇 𝑈 −1 = 𝑈 𝛾 𝜇 𝑈 + , (10.49)

where
𝑈 −1 = 𝑈 + . (10.50)

Then in the new basis of 𝛾-s expression (10.48) will be changed in the following way:

𝑖𝑈 𝛾 𝜇 𝑈 + 𝜕𝜇 − 𝑚 𝜓 = 0 .
(︀ )︀
(10.51)

Since the fact that 𝑈 𝑈 + = 1, (10.51) can be transformed to the following expression:

𝑖𝑈 𝛾 𝜇 𝑈 + 𝜕𝜇 − 𝑚𝑈 𝑈 + 𝜓 = 0 .
(︀ )︀
(10.52)

Equation (10.52) then should be multiplied from the left by 𝑈 −1 and after this manipulation
the expression will look like
(𝑖𝛾 𝜇 𝜕𝜇 − 𝑚) 𝑈 + 𝜓 = 0 , (10.53)

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(10.53) returned to the original form of the Dirac equation, where wave function 𝜓
was modified by a unitary transformation to wave function 𝜓 ′ :

𝜓′ = 𝑈 +𝜓 . (10.54)

The change of the wave function by unitary transformation should not result in the
change of the probability.
Now, the discussion of transformation properties of the Dirac equation under Lorentz
transformations can be performed.
Let’s remember what is spinor 𝜓. 𝜓 is a function on space-time, which represents a
4-component object of the following form:
⎛ ⎞
𝜓1 (𝑥)
⎜ ⎟
⎜𝜓 (𝑥)⎟
2
𝜓 (𝑥) = ⎜ ⎟ . (10.55)
⎜ ⎟
⎜𝜓3 (𝑥)⎟
⎝ ⎠
𝜓4 (𝑥)

So, it consists of 4 functions on space-time. The question, which should be understood is

what happens to 𝜓 under performing the Lorentz transformation.

Lorentz transformations of spinors

It is already known that all Lorentz transformations form a Lorentz group and the
Lorentz group has four different components and one component, which contains identity
is a subgroup of the Lorentz group. It’s a subgroup, which contains proper orthochronous
transformations. Other components of the Lorentz group are obtained from this one by
means of parity or time reversal operations or combined parity with time reversal.
Under proper Lorentz transformations, it is known how the variable 𝑥 transforms. So,
𝑥 is a space-time coordinate and under Lorentz transformations it undergoes the following
change:
𝑥′𝜇 = Λ𝜇𝜈 𝑥𝜈 , (10.56)

where Λ𝜇𝜈 is a matrix of Lorentz transformations with the defining property that

𝜂𝜇𝜈 Λ𝜇𝛼 Λ𝜈𝛽 = 𝜂𝛼𝛽 . (10.57)

The (10.57) is the defining property of the Lorentz transformation, which actually tells
what Lorentz transformation is. Lorentz transformation is a linear transformation of space-
time coordinates, which preserves the Minkowski metric.

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Fig. 10.1. Four components of a Lorentz group

It is very natural to assume that what happens under Lorentz transformation is that
wave function 𝜓 (𝑥) goes to 𝜓 ′ (𝑥′ ), which is related to the old function 𝜓 (𝑥) by means of
application of some matrix, which is unknown, but which is needed to be found, and act
on 𝜓 (𝑥):
𝜓 (𝑥) → 𝜓 ′ (𝑥′ ) = 𝑆𝜓 (𝑥) , (10.58)
where 𝑆 is some matrix and this matrix must be a function of the transformation Λ, which
is applied to transform space-time points:

𝑆 = 𝑆 (Λ) . (10.59)

This matrix 𝑆 should be found. This can be done by applying the Einstein relativity
principle. This is a general principle, which can be applied to all equations of physics
saying that in the new Lorentz frame the Dirac equation must look the same as in the
original frame.
Namely, after the Lorentz transformation new function 𝜓 ′ (𝑥′ ) must still satisfy the
same Dirac equation: (︂ )︂
𝜇 𝜕 𝑚𝑐
𝑖𝛾 − 𝜓 ′ (𝑥′ ) = 0 . (10.60)
𝜕𝑥′𝜇 ~
Then, the reverse procedure should be completed. It should be turned back to the
original form of the Dirac equation, which would allow to relate 𝜓 ′ (𝑥′ ) with 𝜓 (𝑥). To do
𝜕
this, first of all, the derivative 𝜕𝑥𝜇
should be investigated:
𝜕 𝜕𝑥′𝜈 𝜕
= , (10.61)
𝜕𝑥𝜇 𝜕𝑥𝜇 𝜕𝑥′𝜈
where the transformation of derivative under the change of coordinates was written and
then using the (10.56) the expression can be simplified:
𝜕 𝜕
𝜇
= Λ𝜈𝜇 ′𝜈 . (10.62)
𝜕𝑥 𝜕𝑥
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The relation (10.62) can be inverted and written in the following form:
)︀𝜈 𝜕 𝜕
Λ−1
(︀
= . (10.63)
𝜇 𝜕𝑥𝜈 𝜕𝑥′𝜈
Expression (10.63) can be gotten by the (10.62) by multiplying both sides of the
equation with an inverse matrix Λ−1 .
Then, the Dirac equation in the new frame (10.60) should be multiplied from the left
with the matrix 𝑆 −1 : (︂ )︂
−1 𝜇 𝜕 𝑚𝑐
𝑖𝑆 𝛾 𝑆 ′𝜇 − 𝜓 (𝑥) = 0 , (10.64)
𝜕𝑥 ~
where 𝜓 ′ (𝑥′ ) was replaced according to the (10.58) and in the second term the expression
𝑆 −1 𝑆 was gotten, which is actually equal to 1.
Let’s also substitute the change of the derivative. So,
(︂ )︂
−1 𝜈
(︀ −1 )︀𝜇 𝜕 𝑚𝑐
𝑖𝑆 𝛾 𝑆 Λ 𝜈 − 𝜓 (𝑥) = 0 , (10.65)
𝜕𝑥𝜇 ~

The old form of the Dirac equation in the original frame will be reproduced if the
condition
)︀𝜇
𝑆 −1 𝛾 𝜈 𝑆 Λ−1 𝜈 = 𝛾 𝜇
(︀
(10.66)

is right.
Expression (10.66) means that matrix 𝑆 is a solution of this expression. So, it is needed
to understand how to solve this equation and find how 𝑆 is related to Λ.
The (10.66) can be multiplied by matrix Λ from the right:

𝑆 −1 𝛾 𝜇 𝑆 = Λ𝜇𝜈 · 𝛾 𝜈 . (10.67)

Another way to write it is to multiply the original equation by 𝑆 from the left and by
𝑆 −1 from the right and also then it will lead to another form:
)︀𝜇
𝑆𝛾 𝜇 𝑆 −1 = Λ−1 𝜈 𝛾 𝜈 .
(︀
(10.68)

It can be seen that the index 𝜈 as the index of a vector is transforming with matrix
Λ the same as index of a vector should transform under Lorentz transformation, but this
should be compensated by acting of on 𝛾 𝜇 with the matrix 𝑆 by similarity transformation.
And one needs to find the matrix 𝑆, which does the Lorentz rotation on the indices of 𝛾
matrices by means of similarity transformation. So, it’s very convenient to come to the
infinitesimal form of the equation by assuming that a Lorentz transformation is close to

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identity. In other words, there exists a small parameter, which can be called as 𝜖 and Λ𝜇𝜈
can be considered as a function of this small parameter 𝜖 and then we expand it in powers
of 𝜖. Terms of this operation can be written in the form:

Λ𝜇𝜈 (𝜖) = 𝛿𝜈𝜇 + 𝜖𝜂 𝜇𝜌 𝜔𝜌𝜈 + . . . , (10.69)

where 𝜔𝜌𝜈 is a infinitesimal generator of Lorentz transformations. 𝜂 𝜇𝜌 is a Minkowski

metric, which raises the index of 𝜌. It was introduced just for convenience, because indices
of Λ are located on a different level: 𝜇 is up, 𝜈 is down. Then, for Lorentz generator one
of lower indices up with the help of the Minkowski metric.
Then, it can be actually found what are 𝜔-s in the equation by taking the defining
relation for Lorentz transformation:

𝜂𝜇𝜈 Λ𝜇𝛼 (𝜖) Λ𝜈𝛽 (𝜖) = 𝜂𝛼𝛽 . (10.70)

Λ-s in the (10.70) can be expanded and then according to the (10.69) only terms of
linear order in 𝜖 will be kept:

𝜂𝜇𝜈 (𝛿𝛼𝜇 + 𝜖𝜂 𝜇𝜌 𝜔𝜌𝛼 + . . .) 𝛿𝛽𝜈 + 𝜖𝜂 𝜈𝜏 𝜔𝜏 𝛽 + . . . = 𝜂𝛼𝛽 .

(︀ )︀
(10.71)

Then brackets in the (10.71) should be opened and only the leading term in 𝜖 should
be kept. It will be seen that in the equation on 𝜔 the leading term on the left hand side
will be equal to the leading term on the right hand side and this is 𝜂𝛼𝛽 . So, there will be
cancelation of terms and at linear order in 𝜖 it will be found that

𝜔𝜇𝜈 = −𝜔𝜈𝜇 , (10.72)

which is simply means that the generator of Lorentz transformation with lower indices is
simply a skew symmetric, where 𝜇 and 𝜈 run values from 0 to 3. Matrix has six components
from the expression
4×3
=6. (10.73)
2
And these are exactly 6 Lorentz transformations making all 6 parameters of Lorentz
transformations: 3 rotations and 3 Lorentz boosts. So, 𝜔𝜇𝜈 are infinitesimal parameters of
arbitrary Lorentz transformation.
Then, a relationship between the Lie algebra and its Lie group should be used. The
representation of the one parametric subgroup of the Lorentz group can be written down
by simply exponentiating their Lie algebra element representing the Lorentz transformation:

Λ (𝜖) = exp (𝜖𝜂𝜔) . (10.74)

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The exponential (10.74) will give a one parametric subgroup of the Lorentz group,
where the parameter of one parametric subgroup is exactly the 𝜖. 𝜂𝜔 is a matrix, which
by definition equal to
(𝜂𝜔)𝜇𝜈 = 𝜂 𝜇𝜌 𝜔𝜌𝜈 . (10.75)

Two matrices 𝜂 and 𝜔 was already introduced and 𝜂𝜔 is one parametric subgroup of
a Lorentz group. This means that the original equation, which is wanted to be solved can
be now written in the following way:

𝑆 −1 (𝜖) 𝛾 𝜇 𝑆 (𝜖) = Λ𝜇𝜈 (𝜖) 𝛾 𝜈 , (10.76)

where the equation is written by picking up as a Lorentz transformation elements Λ𝜇𝜈 (𝜖)
of the one parametric subgroup of the Lorentz group.
When 𝜖 is equal to zero, the identity matrix will be gotten and, therefore, on the right
hand side 𝛾 𝜇 will be gotten and on the left hand side a matrix 𝑆 must satisfy the property
that when 𝜖 = 0, it’s just an identity matrix:

𝑆 (0) = 1 . (10.77)

The first strategy will be to find a differential equation for 𝑆 with respect to the
parameter 𝜖 and then solve this differential equation. In this way, to find what exactly the
matrix 𝑆 is. Therefore, the (10.76) should be differentiated with respect to 𝜖:
d𝑆 d𝑆 dΛ𝜇𝜈 (𝜖) 𝜈
− 𝑆 −1 𝑆 −1 𝛾 𝜇 𝑆 + 𝑆 −1 𝛾 𝜇 = 𝛾 . (10.78)
⏟ d𝜖⏞ d𝜖 d𝜖
d(𝑆 −1 (𝜖))
d𝜖

This relation further can be written in the following way. If the auxiliary variable 𝑋 is
introduced, which is by definition
d𝑆 −1
𝑋 := 𝑆 , (10.79)
d𝜖
then equation (10.78) can take the following form
dΛ𝜇𝜈 𝜈 −1
[𝛾 𝜇 , 𝑋] = 𝑆𝛾 𝑆 . (10.80)
d𝜖
In other words, in the (10.80) equation (10.78) was multiplied by 𝑆 from the right and by
𝑠−1 from the left. Then, the left hand side reduces to a commutator of 𝛾 𝜇 with 𝑋. The
expression 𝑆𝛾 𝜈 𝑆 −1 comes from multiplications and then this is actually nothing as:
)︀𝜈
𝑆𝛾 𝜈 𝑆 −1 = Λ−1 𝛽 𝛾 𝛽 .
(︀
(10.81)

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Therefore, (︂ )︂𝜇
𝜇 dΛ −1
[𝛾 , 𝑋] = Λ 𝛾𝜈 . (10.82)
d𝜖 𝜈

If the solution for the one parametric subgroup is substituted into the (10.82), than
the result will be equal to:
[𝛾 𝜇 , 𝑋] = 𝜂 𝜇𝜌 𝜔𝜌𝜈 𝛾 𝜈 . (10.83)

Equation (10.83) for 𝑋, which is needed to be solved, is purely matrix equation. So,
all elements here are matrices. This equation tells that the commutator of two matrices
𝛾 𝜇 and 𝑋 is another matrix, where 𝜔𝜌𝜈 are parameters of Lorentz transformations.

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Lecture 11. Lorentz Boosts. Parity

Last time, the matrix 𝑆 was introduced, which performs the transformation of wave
function 𝜓 in such a way that the transformed Dirac equation has the same form as an
original one. And it was found out that if for a Lorentz transformation a one parametric
subgroup given by
Λ (𝜖) = exp (𝜖𝜂𝜔) , (11.1)

where 𝜂𝜔 has components (𝜂𝜔)𝜇𝜈 given by the formula:

(𝜂𝜔)𝜇𝜈 = 𝜂 𝜇𝜌 𝜔𝜌𝜈 . (11.2)

Then for matrix 𝑆 the corresponding formula for Lorentz transformation turns into
the following equation:
𝑆 −1 (𝜖) 𝛾 𝜇 𝑆 (𝜖) = Λ𝜇𝜈 (𝜖) 𝛾 𝜈 . (11.3)

As the next step equation (11.3) should be differentiated with respect to 𝜖 and the
quantity 𝑋 given by
d𝑆 −1
𝑋= 𝑆 (11.4)
d𝜖
should be introduced.
Then the equation for 𝑋 takes the form:

[𝛾 𝜇 , 𝑋] = 𝜂 𝜇𝜌 𝜔𝜌𝜈 𝛾 𝜈 . (11.5)

And this is an equation, which should be solved for the matrix 𝑋.

The convenient way to solve it is to look for a solution in the following form:

𝑋 = 𝜆𝜌𝜆 𝜔𝜌𝜆 . (11.6)

So, a solution, which is linear in generators 𝜔 and 𝜔 are generators of Lorentz transformations,
carries indices 𝜌 and 𝜆 and this is the way how the form, in which the expression for matrix
𝑋 contains, looks.
Substituting the (11.6) into formula (11.5) will lead to an equation of the following
form:
1 (︀ 𝜇𝜌 𝜆
𝑔 𝜇 𝜆𝜌𝜆 − 𝜆𝜌𝜆 𝛾 𝜇 = 𝜂 𝛾 − 𝜂 𝜇𝜆 𝛾 𝜌 ,
)︀
(11.7)
2
where on both sides 𝜔 is canceled. Also the fact that 𝜔 is skew symmetric was used:

𝜔𝜌𝜆 = −𝜔𝜆𝜌 . (11.8)

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𝜆 is also can be taken to be skew symmetric

𝜆𝜌𝜆 = −𝜆𝜆𝜌 . (11.9)

to satisfy the condition of this type.

With this assumption equation for 𝜆𝜌𝜆 can be solved and, directly, the solution will
have the following form:
1 𝜇 𝜈 1
𝜆𝜇𝜈 = [𝛾 , 𝛾 ] = (𝛾 𝜇 𝛾 𝜈 − 𝛾 𝜈 𝛾 𝜇 ) . (11.10)
8 8
Due to the Clifford algebra expression (11.10) can also be written as
1
𝜆𝜇𝜈 = 𝛾 𝜇 𝛾 𝜈 , (11.11)
4
where index 𝜇 is not equal to 𝜈. This is explicitly taken into account by the anti-
symmetrization procedure and, therefore, it can be checked that this is solution by
substituting matrix 𝜆 into equation (11.7) and using the Clifford algebra relation for
𝛾 matrices.
Then, on the other hand, since 𝜆 was found, the equation for 𝑆 can be now specified
as follows:
d𝑆 1
= 𝑋𝑆 = 𝛾 𝜇 𝛾 𝜈 𝜔𝜇𝜈 · 𝑆 . (11.12)
d𝜖 4
The (11.12) is an explicit differential equation, on the right hand side of which is the
product of a four by four matrix, because 𝛾 𝜇 and 𝛾 𝜈 are 4x4 matrices multiplied with
matrix 𝑆. This equation must be supplemented with the initial condition. If 𝜖 is equal to
zero, the corresponding matrix 𝑆 must be simply equal to the identity matrix:

𝑆 (𝜖 = 0) = 1 . (11.13)

The differential equation together with the initial condition has a unique solution and
this unique solution is simply given by:
(︂ )︂
1 𝜇 𝜈
𝑆 (𝜖) = exp 𝜖𝛾 𝛾 𝜔𝜇𝜈 . (11.14)
4

If expression (11.14) is differentiated with respect to epsilon, then the matrix

1 𝜇 𝜈
𝛾 𝛾 𝜔𝜇𝜈 (11.15)
4
comes down and exponential remains and, therefore, expression (11.12) will be right.

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So, what is written down in the (11.14) is a solution, where 𝜖 is a one parametric
subgroup and it is equal to any real number. In the described case, the variable 𝜖 is small
in order to consider expansion around 𝜖 is equal to zero. It can be fixed, for instance, to
be equal to one and, in this way, 𝑆 is taken as:
(︂ )︂
1 𝜇 𝜈
𝑆 = exp 𝛾 𝛾 𝜔𝜇𝜈 , (11.16)
4

which can be also written due to the skew symmetry of the matrix 𝜔 in the following
form: (︃ )︃
1 ∑︁ 𝜇 𝜈
𝑆 = exp 𝛾 𝛾 𝜔𝜇𝜈 . (11.17)
2 𝜇<𝜈

The (11.17) is the final formula for the Lorentz transformation of a spinor. This means
that under the Lorenz transformation 𝜆 from the proper Lorenz group the Dirac spinner
transforms in the following way:
(︃ 3
)︃
1 ∑︁ 𝜇 𝜈
𝜓 ′ (𝑥′ ) = exp 𝛾 𝛾 𝜔𝜇𝜈 𝜓 (𝑥) , (11.18)
4 𝜇,𝜈=0

where 𝑥′ is obtained from 𝑥 by means of Lorenz transformation with the element Λ:

𝑥′ = Λ𝑥 . (11.19)

The answer that has been found is rather interesting. The first thing that can be
noticed is that the matrices Λ𝜇𝜈 , which feature in Lorenz transformations, satisfy the
following algebraic relation, which is the commutator of two matrices:

[𝜆𝜇𝜈 , 𝜆𝜌𝜎 ] = 𝜂 𝜈𝜌 𝜆𝜇𝜎 − 𝜂 𝜇𝜌 𝜆𝜈𝜎 − 𝜂 𝜈𝜎 𝜆𝜇𝜌 + 𝜂 𝜇𝜎 𝜆𝜈𝜌 . (11.20)

In fact, it can be recognized that the (11.20) is nothing else as the commutation
relation between generators of the Lie algebra of the Lorenz group. This means that the
Lorenz transformations, which were constructed, are transformations realizing the spinor
representation of the Lorenz group.
It’s important to realize that this representation is different from the vector representation
acting on coordinates 𝑥𝜇 with the help of Λ𝜇𝜈 .
Let’s investigate this Lorenz transformation that was found before in more details.
First of all, the transformation can be specified for the case of spatial rotations.

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Spatial rotations
This means that just rotations are considered. As it well known the usual three-
dimensional rotations can be specified by given a rotation axis and this can be done by
specifying the unit vector ⃗𝑛 with components 𝑛𝑖 :
⃒
⃒
⃗𝑛 = {𝑛𝑖 }⃒ , (11.21)
⃒
⃒
𝑖=1,2,3

where unit means that if the length of this vector is computed, then the length is equal
to one:
|⃗𝑛 | = 1 . (11.22)

So, ⃗𝑛 is a rotation axis e.g. the axis around which rotations will be performed. A
rotation angle should also be specified and it can be denoted by 𝜃. In other words, there is
axis, which is specified by the unit vector and an angle theta by means of each rotations
will be completed (fig. 11.1).

Fig. 11.1. Rotational axis formed by the unit vector ⃗𝑛

Then there are also elements 𝜔𝑖𝑗 , which should be expressed and which are equal to

𝜔𝑖𝑗 = −𝜃 · 𝜖𝑖𝑗𝑘 𝑛𝑘 , (11.23)

where 𝑛𝑘 is the infinitesimal matrix of a transformation.

Then, the following expression can be computed

𝛾 𝑖 𝛾 𝑗 𝜔𝑖𝑗 = −𝜃𝛾 𝑖 𝛾 𝑗 𝜖𝑖𝑗𝑘 𝑛𝑘 . (11.24)

Expression (11.24) sometimes can be written in the following way:

𝛾 𝑖 𝛾 𝑗 𝜔𝑖𝑗 = 2𝑖𝜃Σ𝑘 𝑛𝑘 , (11.25)

where the following property was used:

𝛾 𝑖 𝛾 𝑗 𝜖𝑖𝑗𝑘 = −2𝑖Σ𝑘 . (11.26)

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where Σ with an upper index due to the Clifford algebra relations of 𝛾-s can be written
in the next form:
Σ𝑖 = 𝛾 5 𝛾 0 𝛾 𝑖 . (11.27)

It can be shown that due to the Clifford algebra relations (11.26) is the same as an
expression (11.27). It can be also seen that the object, which was introduced is the matrix
and this matrix has 3 components. Σ is the collection of 3 matrices labeled by the index
𝑖:
Σ𝑖 = {Σ1 , Σ2 , Σ3 } . (11.28)

In this way, Σ can be identified with 3 vector:

⃗ .
Σ𝑖 = Σ (11.29)

In fact, if a variable 𝐽𝑖 is equal to

1 1
𝐽𝑖 = Σ𝑖 = − Σ𝑖 , (11.30)
2 2
then it satisfies the following algebraic relation:

[𝐽𝑖 , 𝐽𝑗 ] = −𝑖𝜖𝑖𝑗𝑘 𝐽𝑘 . (11.31)

From expression (11.31) it can be recognized that this algebraic relation is nothing
else as the algebraic relations of the generators of the Lie algebra of the rotation group.
On the other hand, it is known that the (11.31) is a commutation relation for angular
momentum. In this way, operators 𝐽𝑖 can naturally interpret as spin operators. So Σ and
𝐽 can be called as spin operators. These objects are related to the notion of spin and the
notion of spin comes from characterizing the transformation properties of an object with
respect to three-dimensional rotations. Spin is a property of a wave function to transform
under three-dimensional rotations of space.
If not rotation by infinitesimal, but by finite angle 𝜃 is performed, then the general
formula should be used and for such a transformation 𝑆 will be equal to
(︂ )︂
𝑖⃗
𝑆 (⃗𝑛 , 𝜃) = exp Σ ⃗𝑛 𝜃 . (11.32)
2
where indices 𝜈 and 𝜇 are taken only values 1, 2 and 3.
Expression (11.32) can be gotten explicitly starting from the formula
(︂ )︂
1 𝑖 𝑗
𝑆 (⃗𝑛 , 𝜃) = exp 𝛾 𝛾 𝜔𝑖𝑗 𝜃 , (11.33)
2

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where the realization of 𝜔𝑖𝑗 (11.23) can be used, which involves the Levi-Civita tensor.
Then the definition of Σ should be used, which are 4x4 matrices. Such a way, the
exponential can be computed explicitly in the usual way by expanding in the Taylor
series:
𝜃 ⃗ ⃗𝑛 sin 𝜃 .
𝑆 (⃗𝑛 , 𝜃) = 1 · cos
+ 𝑖Σ (11.34)
2 2
Formula (11.34) realizes a transformation by finite angle 𝜃. What is interesting about this
formula is that if a rotation by angle 2𝜋 is performed, then because in the expression the
arguments of half of the angles are involved, this will be nothing else as:

𝑆 (⃗𝑛 , 𝜃 + 2𝜋) = −𝑆 (⃗𝑛 , 𝜃) . (11.35)

In other words, when a full rotation is performed, the spinor does not come to itself, but
it changes the sign. If the transformation of a spin is considered, then

𝜓 ′ (𝑥′ ) → −𝜓 (𝑥) , (11.36)

but it does not return to itself. The point 𝑥′ in the opposite actually returns to itself,
because 𝑥′ transforms with a matrix Λ and matrix Λ is a matrix, which depend on angle 𝜃
in such a way that it is periodic, up to upon rotation by 2𝜋 and, therefore, after rotation
by 2𝜋
𝜓 ′ (𝑥) = −𝜓 (𝑥) . (11.37)

Such a representation, which has the property described above is called double valued
representation of the rotation group. This is a specific of the spinor representation, because
the property occurs at the one and the same space-time point. Under performing a loop
in space and returning back to the original position a spinor can change the sign. From
this transformation it can be seen that such representations have a specific name in the
representation theory, they are called double valued representations of the rotation group.
So, the rotation group is 𝑆𝑂(3) has a topological property of being not simply connected.
If one try to draw it from a topological point of view, it is looks like a circle and it is
connected, but it’s not simply connected, which means that there are loops in 𝑆𝑂(3) that
are not contractable to an identity. Such loops, which are not contractable to the to the
identity, mean that the corresponding space is not simply connected.
At each point of 𝑆𝑂(3) two values of another group exists, which is called a double
cover of 𝑆𝑂(3) and it turns out that this double cover of 𝑆𝑂(3) is is know as 𝑆𝑈 (2) group.
And 𝑆𝑈 (2) group is a double cover of 𝑆𝑂(3), which is simply connected (fig. 11.2).

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Fig. 11.2. 𝑆𝑂(3) and 𝑆𝑈 (2) groups

This leads to the existence of a special class of representations, which are called as
spinor representations or double valued representations of the rotational group. In fact,
this also means something else. This means that the spinor cannot be observed itself,
because if one is an observer and a transformation of the Lorentz frame by 2𝜋 is performed,
in fact, one returns back to the original point by making a loop in space. This means that
𝜓 (𝑥) cannot be observed, but what can be observed is the quadratic combinations, for
instance, quadratic combinations of spinor components or some quantities like 𝜓𝜓.
So, these quantities are quadratic in 𝜓 and these quantities remain invariant upon
making the rotation in space. Such quantities would be invariant and such quantities then
provide observables, but not the components of the spinor itself.
It is also should be noticed that every representation is characterized by a spin. Spin
can take half integer values, which are 0, 21 , 1, 23 , . . . and any representation of a given spin
has 2𝑗 + 1 components. For instance, for the case of spin zero, only one component exists
and this is a scalar. For the case of one half there are two components and although the
representation is realized on 4-dimensional spinors, the following is can be seen: if one
compute explicitly what Σ is in the Dirac representation, then it will be found out that
this is given by: (︃ )︃
𝜎𝑖 0
Σ𝑖 = . (11.38)
0 𝜎𝑖
Expression (11.38) is a diagonal in terms of Pauli matrices. Then with this explicit
form of Σ𝑖 the formula for 𝑆 can be written in the following way:

1 · cos 2𝜃 + 𝑖⃗𝜎 ⃗𝑛 sin 2𝜃 0

(︃ )︃
𝑆 (⃗𝑛 , 𝜃) = (11.39)
0 1 · cos 2𝜃 + 𝑖⃗𝜎 ⃗𝑛 sin 2𝜃

If a four component spinor is represented in terms of two component spinors, the

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expression for 𝜓 can be written as follows:

(︃ )︃
𝜑
𝜓= , (11.40)
𝜒

where two components 𝜑 and 𝜒 transform independently with respect to three-dimensional

rotations and, moreover, they transform in the same way. That is exactly the representation
of spin 21 , which realized by means of the following 2x2 matrices

𝜃 𝜃
𝑅 (⃗𝑛 , 𝜃) = 1 · cos + 𝑖⃗𝜎 ⃗𝑛 sin . (11.41)
2 2
In fact, one can build up representation corresponding to any half integer spin. In
particular, vector representation is a three-dimensional representation.
1
It should be noticed that for the representation of spin 2
the matrix 𝑆 factorizes
into diagonal blocks and a spinor can be reduced on two component spinor, which do
not mix with each other. So, they transform completely independent for the reason that
𝜑 and 𝜒 do not mix at all. That means that the four-dimensional representation just
breaks into two representations for two component spinors and each of them is just a
1
representation of spin 2
realized by 2x2 matrices. This classification of representations
characterized by spin is connected with the term of irreducible representations. In other
words, representations, which cannot be decomposed further and do not have inside
themselves invariant subspaces, are so-called irreducible representations. The classification
of irreducible representations of the rotation group tells that any such representation is
uniquely characterized up to unitary equivalence by a half integer number, which is called
spin. And if this number is fixed, then the dimension in which the representation is realized
is related to spin by the formula 2𝑗 + 1.
Once again, every reducible representation in representation theory sometimes are
called as D𝑗 and parameterized by half integer number, where 𝑗 is half integer number
called spin. The dimension of this representation D𝑗 for a given spin 𝑗 is equal to

dimD𝑗 = 2𝑗 + 1 . (11.42)

Lorentz boosts
For Lorentz boosts there is 𝜔0𝑖 exists, which is given by by the formula:

𝜔0𝑖 = 𝑛𝑖 𝜗 , (11.43)

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where 𝜗 ia an hyperbolic angle. Then the matrix of Lawrence boosts 𝑆 (⃗𝑛 , 𝜗) is given by:
(︂ )︂
1 0 𝑖 𝑖
𝑆 (⃗𝑛 , 𝜗) = exp 𝛾 𝛾𝑛𝜗 , (11.44)
2

which is the same as (︂ )︂

1
𝑆 (⃗𝑛 , 𝜗) = exp 𝛼
⃗ ⃗𝑛 𝜗 , (11.45)
2
where the following replacement was applied:

𝛾 0 𝛾 𝑖 = 𝛼𝑖 . (11.46)

The (11.45) again can be explicitly computed and the following expression can be
found:
𝜗 𝜗
𝑆 (⃗𝑛 , 𝜗) = 1 cosh
+𝛼⃗ ⃗𝑛 sinh . (11.47)
2 2
The fact that we are dealing with Lorentz boosts now shows itself in the appearance of
the hyperbolic functions. Hyperbolic functions signaling that now transformations occurs
due to Lawrence boosts.
Since in the Dirac representation matrices 𝛼𝑖 are block off-diagonal, the spinors 𝜑 and
𝜒, which was introduced above mix under Lawrence boosts, because explicitly matrices
𝛼𝑖 are given by:
0 𝜎𝑖
(︃ )︃
𝛼
⃗ = . (11.48)
𝜎𝑖 0
As a result, when the matrix 𝛼
⃗ will be computed explicitly, it will be seen that this
matrix will mix the 𝜑 and 𝜒. Therefore, this matrices are not relativistic covariant.
Let’s also mention the Hermeticity properties of Lorentz transformations that were
found. For space rotations it can be seen that if one take 𝑆 + e.g. compute the Hermitian
conjugate of matrix 𝑆, then this will be equal to:
(︃ )︃
1 ∑︁
𝑆 + = exp 𝛾 𝑗 + 𝛾 𝑖+ 𝜔𝑖𝑗 , (11.49)
2 𝑖<𝑗

where 𝜔𝑖𝑗 are real numbers. Using the fact that matrices 𝛾 𝑖 are anti-Hermitian, then the
(11.49) can be written in the following form:
(︃ )︃
+ 1 ∑︁ 𝑗 𝑖
𝑆 = exp 𝛾 𝛾 𝜔𝑖𝑗 . (11.50)
2 𝑖<𝑗

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Then if the matrices 𝛾 is put in the old order by anti-commuting them, it will be gotten
that: (︃ )︃
1 ∑︁
𝑆 + = exp − 𝛾 𝑖 𝛾 𝑗 𝜔𝑖𝑗 . (11.51)
2 𝑖<𝑗

This, finally, it was gotten that

𝑆 + = 𝑆 −1 . (11.52)

In other words, what was found is that

𝑆 +𝑆 = 1 . (11.53)

Obviously, such matrices was called as unitary. It is seen that rotations are realized
by unitary matrices or, in other words, a unitary representation of the rotation group was
realized.
For Lorenz boosts the following is right:
(︂ )︂
+ 1 𝑖+ 0+
𝑆 = exp 𝛾 𝛾 𝜔0𝑖 . (11.54)
2

It is known that 𝛾 𝑖 is anti-hermitian and 𝛾 0 is hermitian and, therefore

(︂ )︂
+ 1 𝑖 0
𝑆 = exp − 𝛾 𝛾 𝜔0𝑖 . (11.55)
2

Then the anti-commutation relation can be used for matrices 𝛾 𝑖 and 𝛾 0 by using the
Clifford algebra and the following expression will be gotten:
(︂ )︂
+ 1 0 𝑖
𝑆 = exp 𝛾 𝛾 𝜔0𝑖 = 𝑆 . (11.56)
2

This is different from the unitarity condition, rather the condition (11.56) shows that
matrix 𝑆 realizing Lorenz boosts is Hermitian. In fact, both properties under Hermitian
conjugation e.g. behavior of rotation matrices and behavior of Lorenz boosts can be
combined in one formula, which looks as follows:

𝑆 + = 𝛾 0 𝑆 −1 𝛾 0 , (11.57)

2
when it can be seen that actually 𝛾 0 commutes through matrix 𝑆 −1 and (𝛾 0 ) is equal to
1. On the other hand, if 𝑆 is a Lorenz boosts, then 𝛾 0 anti-commutes with 𝛾 𝑖 standing in
the exponential and for this reason the equality of 𝑆 −1 and 𝑆 will be gotten.

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Both together Hermiticity properties are encoded in one formula, which looks like
(11.57). In fact, this formula motivates to introduce the notion of a Dirac conjugate
spinor, which is defined as follows:

𝜓 = 𝜓 *𝑡 𝛾 0 . (11.58)

So the 𝜓 is not a column, but this is a row consisting of 4 numbers, which can be also
written as
𝜓 = 𝜓+𝛾 0 , (11.59)

which means conjugation and transposition multiplied with 𝛾 0 . This is because it’s spinor
𝜓 has a simple transformation property under Lorenz transformation:

𝜓 ′ (𝑥′ ) = (𝑆𝜓)* 𝑡 𝛾 0 , (11.60)

which is according to the discussion the same as

𝜓 ′ (𝑥′ ) = 𝜓 + 𝑆 + 𝛾 0 . (11.61)

In all cases, 𝑆 + is given by formula (11.57) and then

𝜓 ′ (𝑥′ ) = 𝜓 + 𝛾 0 𝑆 −1 𝛾 0 𝛾 0 . (11.62)

The (11.62) can be simplified and then the final expression for the 𝜓 ′ (𝑥′ ) will look as
follows:
𝜓 ′ (𝑥′ ) = 𝜓 (𝑥)𝑆 −1 . (11.63)

In other words, under Lorenz transformations the Dirac conjugate spinor transforms
simply by multiplying it from the right by 𝑆 −1 . The Dirac equation now can be written
in the following form:
(︁ 𝑚𝑐 )︁
𝜓 𝑖𝛾 𝜇 𝜕𝜇 + =0, (11.64)
~
where now the derivative acting on the 𝜓, which is on the left hand side. The (11.64)
represents the Dirac equation for a Dirac conjugate spinner. It is of course follows from
the original Dirac equation by taking conjugation of the original Dirac equation and
multiplying it with 𝛾 0 . Then it can be independently checked that this equation for the
Dirac conjugate spinor remains to be Lorentz covariant exactly due to the properties that
under Lorentz transformations a Dirac conjugate spinor transforms by multiplying with
matrix 𝑆 −1 from the right.

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In general, the transformation properties of spinors with respect to the rotation group
and with respect to Lorentz boosts, which all together provide the realization of the
proper orthochronous Lorentz transformations were introduced and it was done in the
Dirac representation of 𝛾 matrices. And in this Dirac representation for 𝛾 matrices it
was shown that a rotation group is realized in the form of a reducible representation,
which splits into two irreducible two-dimensional representations, while Lorentz boosts
mix these two-dimensional representations between themselves.
Now discrete transformations should be introduced, which are left over. They have to
be taken into account if we want to talk about the full Lorentz group. Therefore, it should
be realized how spinors transform under parity and under time reversal. Let’s start from
parity.

Parity
Parity is an operation, which takes three-dimensional ⃗𝑥 and sends it to minus −⃗𝑥 :

P : ⃗𝑥 → −⃗𝑥 , (11.65)

where time remains unchanged.

While coordinates 𝑥𝑖 are reflected this means that
⎧
⎨⃗𝑥 ′ = −⃗𝑥
(11.66)
⎩𝑡′ = 𝑡

Then, again for 𝜓 ′ (𝑥′ ) in the new Lorentz frame obtained from the old one by means
of the parity transformation as follows:

𝜓 ′ (𝑥′ ) = P𝜓 (𝑥) , (11.67)

where P is some matrix is needed to be found and which leaves the Dirac equation to be
a covariant. It is very simple to check, because if for the Dirac equation in the old frame

(𝑖𝛾 𝜇 𝜕𝜇 − 𝑚) 𝜓 (𝑥) = 0 , (11.68)

then this must be the same in the new frame

𝑖𝛾 𝜇 𝜕𝜇′ − 𝑚 𝜓 ′ (𝑥′ ) = 0 .
(︀ )︀
(11.69)

Then, from the (11.69) it can be seen that the term

𝑖𝛾 0 𝜕0 (11.70)

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will not be changed, because time is not changing under the parity transformation, while
all spatial derivative must change a sign, because of the (11.66):
(︂ )︂
𝑖 𝜕
0
𝑖𝛾 𝜕0 − 𝑖𝛾 − 𝑚 P𝜓 (𝑥) = 0 . (11.71)
𝜕𝑥𝑖

From the (11.71) it is seen that if this equation is multiplied by P−1 from the left,
then to get the original Dirac equation following conditions should be to satisfied:

P−1 𝛾 0 P = 𝛾 0 , (11.72)

while
P−1 𝛾 𝑖 P = −𝛾 𝑖 . (11.73)

If the matrix P, which satisfies conditions (11.72) and (11.73) is found, then the
transformation of spinors under parity transformation will be gotten.
Equations (11.72) and (11.73) can be elementary solved if matrix P will be proportional
to 𝛾 0 with some proportionality coefficient, which can be denoted by 𝜂P :

P = 𝜂P 𝛾 0 . (11.74)

From the (11.74), P will commute with 𝛾 0 and the (11.72) is trivially satisfied and
then due to the fact that 𝛾 0 is anti-commuting with 𝛾 𝑖 the second equation (11.73) will
be satisfied.
It is also should be noticed that a number 𝜂P is called as internal parity. P can be
required to be a unitary operator
P+ P = 1 , (11.75)

which means that the number 𝜂P must be a phase and it’s modulus should be equal to

|𝜂P | = 1 . (11.76)

Then, if parity is performed twice, the original point must be achieved. Therefore,

P2 = 1 . (11.77)

On the other hand, it is known that we are dealing with a spinor and actually in this
case we can say that after application of parity twice ones returning to itself up to a minus
sign. And, therefore, in the second situation the following can be required:

P2 = −1 . (11.78)

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So, in the first case (11.77) for internal parity

𝜂P = ±1 , (11.79)

while in the second case

𝜂P = ±𝑖 . (11.80)

It can also be found that P anti-commutes with 𝛾 5 :

P−1 𝛾 5 P = P−1 𝑖𝛾 0 𝛾 1 𝛾 2 𝛾 3 P .
(︀ )︀
(11.81)

Then P is proportional to 𝛾 0 and 𝛾 0 should be moved through all 𝛾 matrices. As a result

it will be found that
11.81 = −𝛾 5 . (11.82)

If a four component spinor written in terms of two components 𝜑 and 𝜒 is taken, then
the parity operation act on it as follows:
(︃ )︃ (︃ )︃
𝜑 𝜑
P = 𝜂P 𝛾 0 . (11.83)
𝜒 𝜒

In the Dirac representation this is nothing else as

(︃ )︃
𝜂P 𝜑
11.83 = , (11.84)
−𝜂P 𝜒

where parity preserves two component spinors and they do not mix, but from this computation
it is seen that components have a different parity.
If the internal parity is chosen, then with respect to the parity it can be seen that the
bilinear combinations of spinors can be classified in the following way:

1) 𝜓𝜓, which is the scalar product of the Dirac conjugate spinor with the spinor itself
and, therefore 𝜓𝜓 is a number. And under Lorentz transformations this object here
transforms as a scalar.

2) 𝜓𝛾 5 𝜓. Under Lorentz transformations, because Lorentz transformations anti-commute

with matrix 𝛾 5 , then this object appears to be a pseudo scalar. It also can be seen
that this object goes to minus itself under parity transformation due to the property
that under parity
P𝛾 5 = −𝛾 5 . (11.85)

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3) 𝜓𝛾 𝜇 𝜓, which transforms under Lorentz transformation as a vector. It is very important

combination, because it’s a current, which has the dependence on the index 𝜇.

4) 𝜓𝛾 𝜇 𝛾 5 𝜓, which is a pseudo vector. It’s transforms as a usual vector with respect to

Lorentz transformations from proper orthochronous Lorentz group, but it does not
change a sign under parity.

5) 𝜓𝜎 𝜇𝜈 𝜓 is bilinear combination of spinors, where 𝜎 𝜇𝜈 is anti-symmetric tensor, which

was introduced earlier.

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Lecture 12. Time Reversal. Weyl Spinors and Weyl

Equations
At the previous lecture the transformations of the Dirac equation with respect to
Lorenz group was discussed and it was finished by understanding how a spinor transforms
under parity transformations.
Now, it is time to discuss the remaining type of transformations, which wasn’t discussed,
namely, time reversal.

Time reversal
At this topic it will be observed how a Dirac spinor transforms under transformations
of the type
𝑡 → −𝑡 . (12.1)

Classically, having such a transformation or having such a symmetry means that all
solutions or all trajectories 𝑞 (𝑡), which are solutions of equations of motion, are time
reversible.
In other words, if a solution is passed to another solution or to another function 𝑞𝑟𝑒𝑣 (𝑡),
which is
𝑞𝑟𝑒𝑣 (𝑡) = 𝑞 (−𝑡) , (12.2)

then this 𝑞𝑟𝑒𝑣 (𝑡) also appears to be a solution of equations of motion.

In this respect, it can be said that this transformation is in fact the symmetry of
equations. In a quantum mechanical situation of the Dirac equation, the procedure is
absolutely the same as was done, for instance, for parity.
First of all, one assume that time reversal symmetry is some operation, which can be
denoted as T. This transformation is assumed to act on a spinor 𝜓 in the following way:

T : 𝜓 ′ (𝑥′ ) = T𝜓 (𝑥) , (12.3)

where an operation of time reversal does not do anything to spatial coordinates, but
transforms the time coordinate. So,

𝑥′0 = −𝑥0 , (12.4)

while
𝑥′𝑖 = 𝑥𝑖 . (12.5)

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Let’s look at what actually happens in quantum mechanics under time reversal without
going even to a field theory to the Dirac equation. First it should be understood how
this operator actually acts in usual quantum mechanics. And in quantum mechanics,
transformations of operators under symmetries is done in the following way. If some
operator 𝐴 exists, then under time reversal it transforms as:

T𝐴T −1 = 𝐴′ . (12.6)

In quantum mechanics, it should be assumed that if an operator of coordinate exists,

then this operator does not change under the T transformation:

T⃗𝑥 T −1 = ⃗𝑥 . (12.7)

So, coordinates of a particle do not change.

In addition to an operator of coordinate, there is also an operator of momentum and
since momentum through its definition involves the time:
d⃗𝑥
𝑝⃗ = ·𝑚 , (12.8)
d𝑡
where the (12.8) is a classical definition of the momentum. Therefore, in non relativistic
quantum mechanics under such an operation operator of momentum must change the
sign. So,
𝑝⃗ → −⃗𝑝 . (12.9)

On the other hand, if the fundamental commutation relation of quantum mechanics

𝑝𝑖 , 𝑥𝑗 = 𝑖~𝛿𝑖𝑗
[︀ ]︀
(12.10)

is introduced, which is Heisenberg commutation relation, then this commutation relation

will be invariant under such a transformation with T only if this operation T acts also on
imaginary unit in the following way

T𝑖T −1 = −𝑖 . (12.11)

If the property (12.11) is applied to the (12.10), then

T 𝑝𝑖 , 𝑥𝑗 T −1 = 𝑝′𝑖 , 𝑥′𝑗 = − 𝑝𝑖 , 𝑥𝑗 ,
[︀ ]︀ [︀ ]︀ [︀ ]︀
(12.12)

where the fact that 𝑝⃗ is change the sign and ⃗𝑥 is not is used. On the right hand side of
the (12.10) the following expression will be gotten:

− 𝑝𝑖 , 𝑥𝑗 = T𝑖T −1 ~𝛿𝑖𝑗 = −𝑖~𝛿𝑖𝑗 .

[︀ ]︀
(12.13)

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It is clearly seen that the (12.13) is the same as the (12.10). This means that in
the quantum mechanics Hamiltonian is invariant and, therefore, Heisenberg equations
are invariant with respect to time reversal operation. In general, such a transformation
is realized by the so called anti-linear anti-unitary operators. Anti-linear transformation
is clear from the definition of what linear transformation is. If a wave function exists
and it multiplied with a complex number 𝛼 and then for this construction T is applied,
an usual linear transformation would mean that 𝛼 can be taken out, but for anti-linear
transformation the complex number 𝛼 will be conjugated:

T (𝛼𝜓) = 𝛼* T𝜓 . (12.14)

Anti-unitarity means something else. The notion of unitarity is defined with respect
to a chosen scalar product in a Hilbert space and anti-unitarity means that if there is a
scalar product of two elements of an space 𝑥 and 𝑦 is existed and the operator T is applied
to this scalar product, then the result of this transformation will be equal to

⟨T𝑥|T𝑦⟩ = ⟨𝑥|𝑦⟩ = ⟨𝑦|𝑥⟩ , (12.15)

where the conjugated expression was gotten.

Time reversal quantum mechanics is different from what was normal for transformations
and symmetries in a sense that symmetries were realized by unitary operators. And the
time reversal is realized by anti-linear anti-unitary transformations. Of course, this feature
of quantum mechanics goes to the quantum field theory and the operator which will be
found in the Dirac theory, which corresponds to this time reversal, has a property of
similar types.
Therefore, the following should be assumed: transformation from 𝜓 (𝑥) → 𝜓 ′ (𝑥′ ),
which is transformation
𝜓 ′ (𝑥′ ) = T𝜓 (𝑥) (12.16)

can be explicitly realized in the next way. First of all the complex conjugation should be
involved. From a spinor 𝜓 we come to a complex conjugate spinor and then this is not the
end of the story, because the equation is a matrix equation and, therefore, the possible
multiplication of the conjugate spinor 𝜓 * with some matrix 𝑇 should be admitted, where
𝑇 is a normal unitary matrix:
12.16 = 𝑇 𝜓 * (𝑥) . (12.17)

Now, to find what is actually the matrix 𝑇 should be, the Dirac equation should be
taken and, as before, we assume that in the transformed Lawrence frame by means of

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time reversal, the Dirac equation should look like in the original frame:
(︂ )︂
𝜇 𝜕 𝑚𝑐
𝑖𝛾 − 𝜓 ′ (𝑥′ ) = 0 . (12.18)
𝜕𝑥′𝜇 ~
In other words, the (12.18) is the usual Dirac equation, but with 𝑥 replaced by 𝑥′ . That’s
the standard way of implementing the Lawrence covariance of the Dirac equation.
First of all, the usual Dirac equation can be taken and then transformed into the
(12.18) by means of complex conjugation and multiplication with the matrix 𝑇 . The
usual Dirac equation has the next form:
(︂ )︂
𝜇 𝜕 𝑚𝑐
𝑖𝛾 − 𝜓 (𝑥) = 0 , (12.19)
𝜕𝑥𝜇 ~
Then the (12.19) should be conjugated:
(︂ )︂
*0 𝜕 *𝑖 𝜕 𝑚𝑐
−𝑖𝛾 0
− 𝑖𝛾 𝑖
− 𝜓 * (𝑥) = 0 . (12.20)
𝜕𝑥 𝜕𝑥 ~
The (12.20) can be simplified and the result will be equal to
(︂ )︂
*0 𝜕 *𝑖 𝜕 𝑚𝑐
𝑖𝛾 ′0
− 𝑖𝛾 ′𝑖
− 𝜓 * (𝑥) = 0 . (12.21)
𝜕𝑥 𝜕𝑥 ~
The final thing which can be done is a multiplication of the equation from the left by
𝑇 . Then the equation will have the following form:
(︂ )︂
*0 −1 𝜕 *𝑖 𝜕 𝑚𝑐
𝑖𝑇 𝛾 𝑇 − 𝑖𝑇 𝛾 − 𝑇 𝜓 * (𝑥) = 0 . (12.22)
𝜕𝑥′0 𝜕𝑥′𝑖 ~
Now, expression (12.22) should be compared with the (12.18). From this comparison,
it is clearly seen that ⎧
𝜓 ′ (𝑥′ ) = 𝑇 𝜓 * (𝑥)
⎪
⎪
⎪
⎪
⎨
𝑇 𝛾 *0 𝑇 −1 = 𝛾 0 (12.23)
⎪
⎪
⎪
⎩𝑇 𝛾 *𝑖 𝑇 −1 = 𝛾 𝑖
⎪

To solve relations for a matrix 𝑇 the representation for 𝛾 0 and 𝛾 𝑖 should be specified,
because we need to explicitly realize how the conjugation acts on this matrices.
For instance, in the Dirac representation for 𝛾 matrices

𝑔 *0 = 𝛾 0 (12.24)

and
𝛾 *2 = −𝛾 2 . (12.25)

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That’s because 𝛾 2 is constructed with the help of the Pauli matrix 𝜎 2 and this is a diagonal
matrix containing 𝑖 and −𝑖.
And matrices 𝛾 1 and 𝛾 3 are real, because they are constructed with the help of Pauli
matrices 𝜎 1 and 𝜎 3 , which are real matrices. Therefore, eventually for this matrices the
following is right: ⎧
⎨𝛾 *1 = 𝛾 1
(12.26)
⎩𝛾 *3 = 𝛾 3

Finally, for expressions in the (12.23) can be rewritten as follows:

⎧
⎪
⎪
⎪
⎪ 𝑇 𝛾 0 𝑇 −1 = 𝛾 0
⎪
⎪
⎨𝑇 𝛾 2 𝑇 −1 = 𝛾 2
⎪
(12.27)
⎪
⎪
⎪
⎪ 𝑇 𝛾 1 𝑇 −1 = −𝛾 1
⎪
⎪
⎩𝑇 𝛾 3 𝑇 −1 = −𝛾 3
⎪

It can be easily seen that the solution for 𝑇 is equal to:

𝑇 = 𝛾 1𝛾 3 , (12.28)

because 𝛾 0 anti-commutes with 𝛾 1 and anti-commutes with 𝛾 3 . Therefore, when 𝛾 1 𝛾 3 will

be passed through 𝛾 0 one will get 𝛾 0 . In other words, matrix 𝑇 must commute with 𝛾 0
and 𝛾 2 , but anti-commute with 𝛾 1 and 𝛾 3 .
In principle 𝑇 is defined up to an arbitrary phase and very often people just fix this
phase, take a convenient choice and usually they pick up 𝑇 to be equal to

𝑇 = 𝑖𝛾 1 𝛾 3 . (12.29)

This means that the action of the time reversal operation on the spinor is built up
and it looks like the 𝑇 operation apply to 𝜓 (𝑥):

𝜓 ′ (𝑥′ ) = T𝜓 (𝑥) = 𝑖𝛾 1 𝛾 3 𝜓 * (𝑥) . (12.30)

So, this operation is known and in the presented case it is called as a Wigner time reversal
operation and it is represented by anti-linear anti-unitary operator, which can explicitly
realized as
T =𝑇 ·𝐾 , (12.31)

where 𝐾 is an operator of complex conjugation.

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The same result about the form of time reversal operator can be alternatively derived
just by requiring that Dirac Hamiltonian is invariant with respect to this operation of
time reversal.
Now, an another important class of spinors can be introduced and the question about
reducibility of the proper orthochronous Lawrence transformations can be clarified. And
the subject, which will be discussed is the subject about Weyl spinners.

Weyl spinoes
In the last two lectures a representation of the proper orthochronous Lawrence group
was built on 4 component spinors. This representation is realized in terms of 4x4 matrices
and this matrices were denoted by letter 𝑆 (𝑅). They depend on the Lawrence transformation,
which can be performed on the space-time points or vectors. And it was shown that 𝑆 is
a spinor representation of the Lawrence group. Very interestingly that it turns out that
the representation, which was constructed is reducible.
There are two definitions of what can be called a reducible representation and this
definitions are equivalent. The definition number one says that representation of a Lie
group, which is Lawrence group is irreducible if there are no proper invariant subspaces,
where proper means subspaces, which are different from zero and the space representation
space itself. If zero subspace is excluded, then the definition of irreducibility means that
the representation is irreducible if there are no other invariant subspaces exist.
In other words, if start from any vector is performed in a representation space, then by
group transformations any other vector in a space will be reached and it does not happen
such a situation that there will be a subspace in a representation space, whose elements
transform via themselves and never get out of this subspace.
The second definition is telling that representations of a Lie group is irreducible if only
operators, which commute with all group elements are only those, which are proportional
to the identity operator (𝑐 · 1). In other words, these operators, which commute with any
group elements are operators of the form constant multiplied with identity operator and
no any other non-trivial operators except this one exists.
If the action of the group element in the representation space is exist and depend on
the group element 𝑔, it’s like a matrix 𝜋 (𝑔) acting on the concrete representation space.
Then, irreducibility means that there is non-trivial operator, like 𝑂 such that it commutes

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with the 𝜋 (𝑔) for any group element 𝑔:

𝜋 (𝑔) 𝑂 = 𝑂𝜋 (𝑔) , (12.32)

for ∀𝑔 ∈ 𝐺 where
𝑂 ̸= 𝑐 · 1 , (12.33)
because 𝑐 · 1 always commute with any 𝜋 (𝑔), but if this case is excluded, then there any
such 𝑂 should not exist.
In fact, in representation theory, a theorem that the first and the second definitions
are equivalent exist.
If there is an invariant subspace, then there is also a projector on this invariant
subspace and an nontrivial projection operator, which projects on this invariant subspace,
can be constructed. But then, because it is invariant, it can be seen that this projection
operator, which does appear, is an operator which breaks the irreducibility from the point
of view of the second definition.
So, it is then clear how the proof should go. One should rely on the fact that actually
the projection of the representation space can be built and it gives a nontrivial operator
which breaks irreducibility.
If we look from the point of view of the second definition on the representation, which
was constructed, and ask ourselves why a representation that was constructed by 𝑆 (𝑅)
acting on spinor representation of the proper orthochronous Lorentz group is reducible.
It will be reducible if the nontrivial operator, which does commute with all group
elements or, in other words, with all 𝑆 (𝑅) representing elements of the proper orthochronous
Lorentz group can be found. The statement is that such an nontrivial operator does exist
and it’s very easy to see that this operator is simply equal to

𝛾 5 · 𝑆 (𝑅) , (12.34)

because
[︀ 5 ]︀
𝛾 , 𝑆 (𝑅) = 0 . (12.35)
The reason for that is that 𝑆 (𝑅) realized explicitly as
(︂ )︂
1 𝜇 𝜈
𝑆 (𝑅) = exp 𝛾 𝛾 𝜔𝜇𝜈 , (12.36)
4
where 𝛾 𝜇 is Dirac matrices and an index 𝜇 is from 0 to 3. Then, 𝛾 5 is a matrix, which has
a property that the anticommutator of 𝛾 5 and 𝛾 𝜇 is equal to

{𝛾 5 , 𝛾 𝜇 } = 0 . (12.37)

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But if 𝛾 5 anticommutes with 𝛾 𝜇 , then, in fact, it commutes with the product of any two
𝛾 matrices. More explicitly, it can be written that
𝜇𝛾𝜈 𝜔
𝛾5 𝑆 (𝑅) = 𝛾5 𝑒𝛾 𝜇𝜈 /4
𝛾5 𝛾5 , (12.38)

where the fact that (𝛾5 )2 = 1 was used. Then the (12.38) can be rewritten as
𝜇𝛾𝜈 𝛾 𝜔
𝛾5 𝑆 (𝑅) = 𝑒𝛾5 𝛾 5 𝜇𝜈 /4
𝛾5 . (12.39)

𝛾5 can be moved through 𝛾 𝜇 𝛾 𝜈 and 𝛾5 matrices will cancel each other and, as a result, the
following expression will be gotten:

𝛾5 𝑆 (𝑅) = 𝑆 (𝑅) 𝛾5 . (12.40)

According to the logic of the definition of irreducibility the representation of the

proper orthochronous Lorentz group is reducible and the matrix 𝛾 5 helps use them to
construct invariant subspaces, which remain invariant under proper orthochronous Lorentz
transformations. This is done by means of constructing two projectors 𝑃± , which are given
by
1 (︀
1 ± 𝛾5 .
)︀
𝑃± = (12.41)
2
𝑃± have properties of being a projector. This means that

𝑃±2 = 𝑃± (12.42)

and that 𝑃± are orthogonal:

𝑃+ 𝑃− = 𝑃− 𝑃+ = 0 . (12.43)

The third property is that these projectors realize the decomposition of identity:

𝑃+ + 𝑃− = 1 . (12.44)

Other properties follow immediately from the properties of 𝛾 5 . For instance,

1 (︀ )︀ 1 (︀ )︀ 1 (︀
1 + 𝛾 5 1 + 𝛾 5 = 1 + 2𝛾 5 + 1 = 1 + 𝛾 5 = 𝑃+ .
)︀ (︀ )︀
𝑃+ 𝑃+ = (12.45)
4 4 2
At the same time, 𝑃+ 𝑃− is equal to:
⎛ ⎞
1 1 2
1 + 𝛾 5 1 − 𝛾 5 = ⎝1 − 𝛾 5 ⎠ = 0 .
(︀ )︀ (︀ )︀ (︀ )︀
𝑃+ 𝑃− = (12.46)
4 4 ⏟ ⏞
=1

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Projectors provide a decomposition of the identity and this means also that 𝑃±
commute with all proper orthochronous Lorentz transformations. This means that objects,
which realize the invariant subspaces, can be defined and these objects are special spinors,
which are called Weyl spinors.
A Weyl spinor is defined is defined as a two component complex spinor, which transforms
irreducibly under proper orthochronous Lorentz transformations, where this transformations
form a group with the name 𝑆𝑂+ (1, 3). Complex means that two components of the Weyl
spinor are complex numbers. In other words, a four component Dirac spinor realizes the
reducible representation of the proper orthochronous Lorentz group, but it can be split into
two component spinors and each of these two component spinors will realize irreducible
representation of a proper orthochronous Lorentz group.
Explicitly, these two component spinors are constructed from a four component Dirac
spinor by applying the projection 𝑃+ 𝜓 and the projection 𝑃− to 𝜓:
⎧
⎨𝜓+ = 𝑃+ 𝜓
(12.47)
⎩𝜓 = 𝑃 𝜓
− −

where 𝜓 is a four component Dirac spinor and 𝜓± are two component Weyl spinors.
In fact, Weyl spinors 𝜓± are called chiral and people say that 𝜓± have different chirality.
To understand more how two component Weyl spinors look like and why they are two
component a transformation from the Dirac representation of the Clifford algebra, which
explicitly is given by 𝛾 matrices of the next form

1 0 0
(︃ )︃ (︃ )︃
𝜎𝑖
𝛾0 = , 𝛾𝑖 = , (12.48)
0 1 −𝜎 𝑖 0
to another representation, which is called as Weyl representation, can be done, where in
the Weyl representation matrices are realized in the following way:

0 1 0
(︃ )︃ (︃ )︃
𝑖
𝜎
𝛾𝑐0 = , 𝛾𝑐𝑖 = , (12.49)
1 0 −𝜎 𝑖 0

where 𝛾𝑐0 is taken to be off-diagonal and 𝛾𝑐𝑖 remains to be the same as 𝛾 𝑖 . Lower index 𝑐
means chiral.
It can be seen that 𝛾𝑐0 is nothing else as 𝛾 5 in the Dirac representation and 𝛾𝑐5 , which
can be construct through chiral matrices in the chiral representation is equal to

𝛾𝑐5 = −𝛾 0 . (12.50)

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As it was pointed earlier all representations of the Clifford algebra in 4 dimensions

are related to each other by means of unitary transformation and, therefore, there should
be a unitary transformation, which relates these two explicit realizations of the Clifford
algebra. Indeed, it can be seen that there is a matrix 𝑈 , which does this job. When it is
applied to all matrices 𝛾 of the Dirac representation, matrices of the Weyl representation
will be gotten:
𝑈 𝛾 𝜇 𝑈 −1 = 𝛾𝑐𝜇 . (12.51)

Explicitly, this matrix 𝑈 is given by

1 −1
(︃ )︃
1
𝑈=√ (12.52)
2 1 1
and it has the following property:
𝑈 +𝑈 = 1 . (12.53)

What is good about new 𝛾 matrices is that if spinor is written

(︃ )︃
𝜑
𝜓= , (12.54)
𝜒

where each of this 𝜑 and 𝜒 is two component, then the following can be seen.
In the Dirac representation the matrix 𝑆 has the form

1 cos 2𝜃 + 𝑖⃗𝜎 ⃗𝑛 sin 2𝜃 0

(︃ )︃
𝑆 (⃗𝑛 , 𝜃) = (12.55)
0 1 cos 2𝜃 + 𝑖⃗𝜎 ⃗𝑛 sin 2𝜃
and shows that under rotations 𝜑 and 𝜒 transforms independently. This matrix was
constructed from the general formula:
1 𝑖𝛾𝑗 𝜔
𝑆 (⃗𝑛 , 𝜃) = 𝑒 4 𝛾 𝑖𝑗
, (12.56)

where the indices of 𝛾 matrices are 1, 2 and 3. Therefore, the construction (12.55) does
not involve 𝛾 0 , but since passing from Dirac basis to Weyl basis of 𝛾 matrices there is no
change of 𝛾 𝑖 and then this result for 𝑆 (⃗𝑛 , 𝜃) for Lorentz transformations corresponding
to rotations will hold also in the Weyl representation and the result will be the same.
For Lorentz boosts in the case of the Dirac representation 𝜑 and 𝜒 mix, but it is
interesting to see what happens in the new basis, because here the matrix 𝛾 0 in the Weyl
basis is off-diagonal and Lorentz transformations corresponding to Lorentz boosts have
the form:
0𝛾𝑖𝜔
𝑒𝛾 0𝑖 /2
. (12.57)

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Now, the following can be seen: 𝛾 0 is off-diagonal and 𝑔𝑎𝑚𝑚𝑎𝑖 are off-diagonal and
if two such off-diagonal matrices is multiplied, a diagonal matrix will be gotten. If it is
diagonal, then it can be seen that the matrix corresponding to Lorentz boosts will be
diagonal and it can be computed explicitly:

1 cosh 𝜗2 − ⃗𝑠 ⃗𝑛 sinh 𝜗2 0
(︃ )︃
𝑆 (⃗𝑛 , 𝜗) = . (12.58)
0 1 cosh 𝜗2 + ⃗𝑠 ⃗𝑛 sinh 𝜗2

The meaning of the (12.58) and (12.55) is that 𝜑 and 𝜒 do not mixed under any proper
Lorentz transformations neither under rotations nor under Lorentz boosts, because the
corresponding matrices in the Weyl representation are block diagonal. That’s, of course,
shows that, indeed, 𝜑 and 𝜒 are Weyl spinors, they transform independently under proper
orthochronous Lorentz transformations.
The other important remark is the following. It can be seen that the difference between
𝜑 and 𝜒 is also visible from formulas (12.58) and (12.55). Under rotations 𝜑 and 𝜒
transform completely in the same manner, but for Lorentz boosts 𝜑 and 𝜒 looks different
and the difference is in the sign of the elements of 𝑆 (⃗𝑛 , 𝜗).
Once again, a Lorentz group topologically looks like some set set of transformations.
Within this set there is a smaller subset, which has a name of proper orthochronous
Lorentz transformations and this subset has a name 𝑆𝑂+ (1, 3). Now we established a
transformations of the spinor 𝜓 with respect to 𝑆𝑂+ (1, 3) and when it was realized that
this representation of 𝑆𝑂+ (1, 3) on 4 dimensional spinors turns out to be reducible and
it splits into two invariant subspaces.

Fig. 12.1. Topological representation of the Lorentz group

The following question can be asked: if components of 𝜓 transform with different

matrices realizing Lorenz boosts, then representations can be still equivalent. Two representations
are called equivalent if there exists a unitary transformation such as 𝑈 and matrices of the
first representation 𝜋 (𝑔) transformed with this unitary transformation have the following

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form:
𝑈 𝜋 (𝑔) 𝑈 −1 = 𝜌 (𝑔) . (12.59)

In this case equivalence is just a statement that one representation is gotten from the
other by changing the basis with the help of the unitary transformation. In fact, in
representation theory people agree to consider unitary representations of Lie groups up
to unitary equivalence.
The question can be formulated as follows: in spite of the fact that in the (12.58) there
are different signs, representations may be still equivalent and changing the basis in the
representation space say of spinor 𝜒 will lead to a representation of spinor 𝜑 and in this
case these two representations will be equivalent. The answer to this question is negative
and it can be proofed in the following way. Let’s suppose that they are equivalent. It
means that there exists a matrix 𝑈 such that it takes a matrix of the first representation
(︂ )︂
𝜗 𝜗
𝑈 1 cosh − 𝜎 𝑛 sinh
𝑖 𝑖
𝑈 −1 (12.60)
2 2
and takes out of it the matrix of the second representation
𝜗 𝜗
11.61 = 1 cosh + 𝜎 𝑖 𝑛𝑖 sinh . (12.61)
2 2
The (12.61) be valid for any 𝜗 and any 𝑛. If this should be valid for any 𝜗 and any 𝑛,
then 𝑛 can be picked up, for instance, in the form (1, 0, 0) or (0, 1, 0) or (0, 0, 1). Then
three equations will be gotten
(︂ )︂
𝜗 𝜗 𝜗 𝜗
𝑈 1 cosh − 𝜎 sinh
𝑖
𝑈 −1 = 1 cosh + 𝜎 𝑖 sinh , (12.62)
2 2 2 2

where now instead of summed up 𝜎 𝑖 , there are individual 𝜎 𝑖 standing. In the (12.62) 𝑈
can be passed through the first term and then the gotten expression can be divided by
sinh 𝜗2 . As a result the following formula will be exist:

𝑈 𝜎 𝑖 𝑈 −1 = −𝜎 𝑖 , (12.63)

which can be rewritten as follows:

𝑈 𝜎 𝑖 = −𝜎 𝑖 𝑈 . (12.64)

The (12.64) is an equation, which should be satisfied. From this expression it can be
seen that if a matrix 𝑈 exists, it must be 2x2 matrix, which anti-commutes with all three
Pauli matrices, but such a matrix does not exist. This means that these two representations

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of the proper orthochronous group acting on spinors 𝜑 and 𝜒 are inequivalent. Therefore,
these two representations of spin one half people agreed to denote differently. So, representation
for 𝜑 is denoted as 12 , 0 and the representation realized by 𝜒 is called 0, 21 . The spinors
(︀ )︀ (︀ )︀

realizing this representations for 𝜑 and 𝜒 can be denoted as 𝜓𝛼 , which is a two component
spinor with index 𝛼, where 𝛼 takes values 1 and 2. But to distinguish them on writing we
will denote spinor of the type 12 , 0 by putting dot on 𝛼.
(︀ )︀

The two spinors 𝜓𝛼 and 𝜓𝛼˙ are two Weyl spinors, which realize different representations
of the Lorenz group. Earlier, when the construction of the Lorenz group was discussed, it
was mentioned that, in fact, the proper orthochronous Lorenz group, which is 𝑆𝑂+ (1, 3) is
not simply connected and it looks like a circle topologically, but it has a simply connected
cover. So it can be covered by another group, which is simply connected, and the name
of this group is 𝑆𝐿 (2, C). 𝑆𝑂+ (1, 3) and 𝑆𝐿 (2, C) have the same Lee algebra, where the
same means that Lee algebras are isomorphic, but these are not isomorphic as groups and
one is the universal cover of the other. In fact 𝑆𝐿 (2, C) is a double cover, which covers
twice the group 𝑆𝑂+ (1, 3) and, in fact, 𝑆𝐿 (2, C) are 2x2 complex matrices. And spinors
𝜓𝛼 and 𝜓𝛼˙ realize two inequivalent two-dimensional representations of 𝑆𝐿 (2, C). In fact,
this is a kind of a general statement. So, in general irreducible representations of a proper
orthochronous Lorentz are characterized by two half-integer numbers (𝑗1 , 𝑗2 ). In one case,
representation 𝜓𝛼˙ is a representation, which corresponds to the choice 𝑗1 to be 1/2 and
𝑗2 to be 0, while the undotted spinor is characterized by choice of 0 and 1/2.
From the point of view of rotation group the group 𝑆𝑂+ (1, 3) looks like a multi-valued
representation and it’s related to the fact that when the rotation by 2𝜋 is performed a
spinor is not returning to itself, but it changes the sign. From the point of view of universal
covers 𝑆𝐿 (2, C) there is nothing strange happens and we deal with two-dimensional
spinors, which are usual two-dimensional objects transforming by multiplying with the
2x2 complex matrices.
Another remark is the following. Let’s return to the (fig. 12.1), which was drawn earlier.
It is understood that the representation of 𝑆𝑂+ (1, 3) is reducible and it can be reduced
to Weyl spinors, which form irreducible representation of 𝑆𝑂+ (1, 3). But what about
full Lorentz group, which includes all the other elements, which can be obtained from a
𝑆𝑂+ (1, 3) by applying parity and time reversal operations? For instance, under parity
operation components transforms not independently and if one look at parity operation,
which was discussed at the previous lecture, it will found that the parity operation is

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realized by means of
P𝜓 = 𝜂P 𝛾 0 𝜓 . (12.65)

In the Dirac representation 𝛾 0 was diagonal and thanks to that (12.65) can be written as

1 0
(︃ )︃
P𝜓 = 𝜂P 𝜓. (12.66)
0 −1

If 𝜓 is written in the matrix form, than the expression for the (12.66) will be simplified
to: (︃ )︃
𝜂P 𝜑
P𝜓 = . (12.67)
−𝜂P 𝜒
Therefore, in the Dirac representation components of 𝜓 do not mix. However, when pass to
the Weyl representation will be completed, parity will be still given by the same formula,
but 𝛾𝑐0 is off-diagonal and
0 1
(︃ )︃ (︃ )︃
𝜑
P𝜓 = 𝜂P . (12.68)
1 0 𝜒
Therefore, (︃ )︃ (︃ )︃
𝜑 𝜂P 𝜒
P = . (12.69)
𝜒 𝜂P 𝜑
According to the (12.69) to describe the full Lorentz group both spinors are needed,
because with respect to the full Lorentz group the parity operation should be realized.
If the Dirac equation is observed from the point of view of 𝜑 and 𝜒, it will be seen the
following. The Dirac equation in the Weyl representation will take the following form
⎧
⎨ 1 𝜕𝜑 − 𝜎 𝑖 𝜕𝜑𝑖 + 𝑖 𝑚𝑐 𝜒 = 0
𝑐 𝜕𝑡 𝑥 ~
(12.70)
⎩ 1 𝜕𝜒 + 𝜎 𝑖 𝜕𝜒 + 𝑖 𝑚𝑐 𝜑 = 0
𝑐 𝜕𝑡 𝑥𝑖 ~

In fact equations (12.70) are coupled and equation for 𝜑 contains 𝜒 and equation for 𝜒
contains 𝜑. But there is one case, when they actually decoupled from each other and
become independent and this is the case, when mass of a particle is equal to 0, because
when mass is 0 the last term disappears. The equation becomes independent and this
equation for massless particle are called Weyl equations
⎧
⎨ 1 𝜕𝜑 − 𝜎 𝑖 𝜕𝜑𝑖 = 0
𝑐 𝜕𝑡 𝑥
(12.71)
⎩ 1 𝜕𝜒 + 𝜎 𝑖 𝜕𝜒 = 0
𝑐 𝜕𝑡 𝑥𝑖

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Weyl equation is covariant with respect to the proper orthochronous Lorentz transformation,
but it’s is not invariant under parity.
The time reversal operation can be also included. So, time reversal operation in the
Wiley representation can be written as

T𝜓 (𝑥) = 𝑇 𝜓 * = 𝑖𝛾𝑐1 𝛾𝑐3 𝜓 * = 𝑖𝛾 1 𝛾 3 𝜓 * , (12.72)

where 𝛾 1 and 𝛾 3 are the same in Weyl and in Dirac representations. They are both off-
diagonal and their product is diagonal. So, under time reversal nothing bad happens and
spinors 𝜑 and 𝜒 continue to transform independently.

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Lecture 13. Solution of the Dirac Equation

In the last lecture the concept of Weyl spinors, the concept of irreducibility were
discussed and it was found that representation of the proper orthochronous Lorentz
group on four-dimensional spinors is reducible and it is decomposable into two irreducible
representations, which are realized by means of two-dimensional Weyl spinors, which are
called as dotted and undotted spinors. Also it was found that under parity transformation
one species of Weyl spinors, say, dotted spinors go to undotted ones and vice versa.
From this point of view these four-component spinors realize irreducible representation
of the full Lawrence group, which includes in addition to proper orthochronous Lorentz
transformations also parity and time reversal. Especially, parity acts non-trivial by transforming
into each other Weyl spinors.
Let’s write the Dirac equation again. This can be written as
(︁ 𝑚𝑐 )︁
𝑖𝛾 𝜇 𝜕𝜇 − 𝜓=0. (13.1)
~
It is interesting to look for solutions of equation (13.1), which are given by plane waves or
which are described by plane waves. An answer for 𝜓 (⃗𝑥 , 𝑡) given by the following formula

𝜓 (⃗𝑥 , 𝑡) = 𝑢 (𝑝) 𝑒−𝑖(𝐸𝑡−⃗𝑝 ·⃗𝑥 )/~ , (13.2)

where 𝑢 (⃗𝑝 ) is an amplitude, which depends on components of the momentum 𝑝 only.

So, the (13.2) is a usual quantum mechanical plane wave and this wave is called plane,
because the value of the amplitude 𝑢 (𝑝) depends only on momentum, but it does not
depend on 𝑥 and the phase of this wave is one and the same follow points on the same
plane. This is a plane for which the scalar product

𝑝⃗ · ⃗𝑥 = 𝑐𝑜𝑛𝑠𝑡 . (13.3)

In fact, this plane is perpendicular to vector 𝑝⃗ and 𝑝⃗ points to the direction of propagation
of this wave. In other words, if there is momentum 𝑝⃗ pointing in a certain direction, then
a plane, which is orthogonal to 𝑝⃗ to every point on this plane the phase 𝑝⃗ multiplied by
𝑥 is one and the same. That is why the front of this wave is just a plane. It can be seen
geometrically (fig. 13.1), because every plane is described by the following equation

⃗𝑥 = ⃗𝑥 0 + 𝑠⃗𝑎 + 𝑡⃗𝑏 , (13.4)

where 𝑠 and 𝑡 belongs to R.

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Fig. 13.1. Plane, for which (13.3) is satisfied

⃗𝑎 and ⃗𝑏 will be two independent vectors on the plane. Then (13.4) is how any point
on the plane is described. Then it can be seen that product (13.3) will be equal according
to the (13.4) to
𝑝⃗ · ⃗𝑥 = 𝑝⃗ · ⃗𝑥 0 = 𝑐𝑜𝑛𝑠𝑡 , (13.5)

where 𝑝⃗ is a vector orthogonal to the plane and, therefore, it’s orthogonal to vectors ⃗𝑎
and ⃗𝑏 .
Now the ansatz (13.2) should be taken and pluged into the Dirac equation and in this
way an equation for the amplitude 𝑢 (𝑝) takes the following form:
(︀ 0
𝐸𝛾 − 𝑐𝛾 𝑖 𝑝𝑖 − 𝑚𝑐2 𝑢 (𝑝) = 0 .
)︀
(13.6)

Now, it should be noticed that what is written down in the (13.6) is nothing else as
a eigenvalue problem for the Dirac Hamiltonian 𝐻, for which the following expression is
right:
𝐻𝜓 = 𝑐𝛼𝑖 𝑝𝑖 + 𝑚𝑐2 𝛽 𝑢 = 𝐸𝑢 ,
(︀ )︀
(13.7)
⏟ ⏞
Dirac Hamiltonian

where 𝐸 is an eigenvalue of the (13.2) which coincides with the energy and to come
from equation (13.7). And to transform the (13.6) into (13.7) the first equation should be
multiplied with 𝛾 0 from the left and 𝛾 0 is the same as:

𝛾0 = 𝛽 . (13.8)

In this way, 𝛾 𝑖 will turn into

𝛾 𝑖 = 𝛼𝑖 . (13.9)

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And then we should get the Dirac Hamiltonian and put the energy to the other side.
Then, it will be seen that what is gotten is simply eigenvalue for the energy. This is very
similar to the way of solving Schrodinger equation by means of the plane wave. Plane
wave should be plug into the Schrodinger equation and then a eigenvalue problem for the
Schrodinger Hamiltonian will be absolutely the same.
It is convenient to take 𝑢 and represent it in terms of two component spinors. So, the
𝑢 is a four-component, but it can be split into 𝜑 and 𝜒:
(︃ )︃
𝜑
𝑢= (13.10)
𝜒

and plug into equation (13.6) and then the expression for 𝛼 and 𝛽 in terms of Pauli
matrices can be used. Therefore, the equation will take the following form:
⎧
⎨(𝐸 − 𝑚𝑐2 ) 𝜑 − 𝑐⃗𝜎 · 𝑝⃗ · 𝜒 = 0
(13.11)
⎩−𝑐⃗𝜎 · 𝑝⃗ · 𝜑 + (𝐸 + 𝑚𝑐2 ) 𝜒 = 0

The (13.11) is simply a system of 4 homogeneous equations for the components of

spinors 𝜑 and 𝜒. In terms of the 2x2 blocks the corresponding matrix, which acting on 𝜑
and 𝜒 can be written as follows:
(︃ )︃ (︃ )︃
𝐸 − 𝑚𝑐2 −𝑐⃗𝑠 𝑝⃗ 𝜑
=0. (13.12)
−𝑐⃗𝜎 𝑝⃗ 𝐸 + 𝑚𝑐2 𝜒

This is a homogeneous system of of four algebraic equations for components of this spinors
𝜑 and 𝜒 and it is known that in order for this system to have a non-trivial solution it
is necessary that the determinant of the matrix, which defines this homogeneous system
must be equal to zero. Otherwise there is no non-trivial solution of this system at all.
This means that the condition to have a non-trivial solution should be imposed and
determinant of the matrix from the (13.12) must be equal to zero:
⃒ ⃒
⃒𝐸 − 𝑚𝑐2 −𝑐⃗𝑠 𝑝⃗ ⃒
⃒=0. (13.13)
⃒ ⃒
⃒
⃒ −𝑐⃗𝜎 𝑝⃗ 𝐸 + 𝑚𝑐2 ⃒

The determinant (13.13) is a

)︀2
13.13 = 𝐸 2 − 𝑚𝑐2 − 𝑐2 𝑝⃗ 2 = 0 ,
(︀
(13.14)

From the (13.14), it can be seen that the condition of vanishing of determinant of
the matrix is nothing else as a relativistic dispersion relation or dispersion relation for

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relativistic particle, which tells that the energy of the plane wave solution is related to
the momentum of the plane wave by the following condition:
√︀
𝐸 = ± 𝑝⃗ 2 𝑐2 + 𝑚2 𝑐4 . (13.15)

The relation (13.15) can be also called in the context of Klein-Gordon equation as the
on-shell condition.
It is interesting that the classical Dirac theory contains solutions with both signs of
energy. It has positive solutions and it has negative solutions. Therefore, from classical
point of view, the Dirac theory is not well defined, because solutions with negative energy
should not exist. Energy typically in usual dynamical systems is quantity which must be
restricted from below. But it can be seen that solutions with any value of negative energy
are possible. It will be shown that this kind of unwanted feature of the classical Dirac
theory will turn into advantage of this theory, when a transition to the quantum case will
be completed.
Meanwhile, the sign in front of the expression for the energy can be chosen and an
explicit solution can be produced. If the positive sign is fixed, then the following solution
can be gotten: (︃ )︃
𝜑
𝑢+ (𝑝) = 𝑐⃗
𝜎𝑝⃗
, (13.16)
𝑝 )+𝑚𝑐2
𝐸(⃗
𝜑
where 𝑢+ means the solution with positive sign of the energy. So, solution is given in
terms of two components spinor 𝜑, which remains to be unrestricted. Once forever 𝐸 (⃗𝑝 )
will be fixed to be positive:
√︀
𝐸 (⃗𝑝 ) = 𝑝⃗ 2 𝑐2 + 𝑚2 𝑐4 . (13.17)

If the negative sign is fixed in front of the square root, then the following solution will
be gotten: (︃ )︃
𝜑
𝑢− (𝑝) = 𝑐⃗
𝜎𝑝⃗
, (13.18)
𝑝 )+𝑚𝑐2
−𝐸(⃗
𝜑
where 𝐸 (⃗𝑝 ) again given by the same positive solution (13.17).
Since 𝜑 remains arbitrary there are two solutions with positive energy, where two
solutions means two components of 𝜑, and two solutions with negative energy. The fact
that we have two solutions hints that there exists another operator, which commutes with
the Dirac Hamiltonian and which distinguishes between themselves two components of
𝜑. So, the fact that 𝜑 remains to be arbitrary and unspecified and has two components

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means that there is another operator, which commutes with the Dirac Hamiltonian and
which distinguishes between these two components of 𝜑 and this operator indeed can be
constructed in an explicit way and it has a name Helicity operator.

Helicity operator
The Helicity operator is constructed as follows:
1 5 0 𝑖 𝑖
J= 𝛾 𝛾 𝛾𝑝 . (13.19)
|⃗𝑝 |
If 𝛾 matrices are presented in the Dirac representation, then the following answer will be
found: (︃ )︃
1 ⃗𝜎 𝑝⃗ 0
J= . (13.20)
|⃗𝑝 | 0 ⃗𝜎 𝑝⃗
Then from the discussion of the law of matrices transformation, which realize a representation
of the Lorenz group, the (13.20) can be written as:
(︁ )︁
J= Σ ⃗ · ⃗𝑛 , (13.21)

where ⃗𝑛 is a unit vector in the direction of momentum and it’s equal to

𝑝⃗
⃗𝑛 = . (13.22)
|⃗𝑝 |
⃗ is a vector
Σ is understood as a 4x4 matrix on the one hand. On the other hand, Σ
built up from three Pauli matrices

0
(︃ )︃
⃗ = ⃗𝜎
Σ . (13.23)
0 ⃗𝜎
⃗ satisfy the
and the operator Σ can be called as the spin operator, because 3 matrices Σ
Lie algebra relations for the angular momentum and this is an operator, which can be
identified with the spin of a particle.
It can be seen that the operator, which was constructed in (12.19)-(12.21) is nothing
else as a projection of spin on the direction of propagation or the direction of motion.
Now, it can be checked that an operator J commutes with the Dirac Hamiltonian:

[𝐻, J] = 0 . (13.24)

From the course on linear algebra it is known that if there are two commuting operators,
then they can be simultaneously diagonalized and this means that these operators will

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have the same basis in the Hilbert space made from common eigenstates of these two
commuting operators.
Properties of this operator are the following. The square of the Helicity operator is
equal to
J2 = 1 , (13.25)

which means that the eigenvalues of this operator are simply ±1. Thus a state with positive
energy can also be an eigenstate of the helicity operator with either positive or negative
helicity and the same, of course, true for states with negative energy. If we now take the
general solution for states with positive energy and the same general solution for states
with negative energy and further use these states to diagonalize the Helicity operator
on these states, which are already made to be eigenstates of the Hamiltonian, then the
following explicit solution for this the amplitude 𝑢 will be found, which is simultaneously
an eigenstate of both operators of 𝐻 and of J. Somehow to distinguish states with positive
and negative Helicity indices 1 and 2 will be used. For instance, for 𝑢+ (𝑝) a solution with
positive energy and positive Helicity can be written in the following way:
⎛ ⎞
𝑝3 + |⃗𝑝 |
⎜ 1 ⎟
⎜ 𝑝 + 𝑖𝑝2 ⎟
𝑢1+ (⃗𝑝 ) = 𝜂 (𝑝) ⎜ 𝑐|⃗𝑝 |(𝑝3 +|⃗𝑝 |) ⎟ , (13.26)
⎜ ⎟
⎜ 2 ⎟
⎝ 𝑚𝑐 +𝐸(⃗𝑝 ) ⎠
𝑝 |(𝑝 +𝑖𝑝 )
𝑐|⃗ 1 2

𝑚𝑐2 +𝐸(⃗
𝑝)

where 𝜂 (𝑝) is a normalization, which can be used for normalization of the spinor in a
convenient way. This is how the solution with positive Helicity and the positive energy
looks like.
Then, it will also be an eigenstate of the Helicity operator with negative sign and the
solution for positive energy and this eigenstate will look as follows:
⎛ ⎞
− (𝑝1 − 𝑖𝑝2 )
⎜ 3 ⎟
⎜ 𝑝 + |⃗𝑝 | ⎟
𝑢2+ (𝑝) = 𝜂 (𝑝) ⎜ 𝑐|⃗𝑝 |(𝑝1 −𝑖𝑝2 ) ⎟ . (13.27)
⎜ ⎟
⎜ ⎟
⎝ 𝑚𝑐2 +𝐸(⃗𝑝 ) ⎠
𝑐|⃗𝑝 |(𝑝3 +|⃗
𝑝 |)
− 𝑚𝑐2 +𝐸(⃗𝑝 )

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Analogously, solutions with negative energy and different helicities can be written as
( )
⎛ 𝑐|⃗𝑝 | 𝑝3 +|⃗𝑝 | ⎞
−
⎜ 𝑚𝑐2 +𝐸(⃗ 𝑝) ⎟
⎜ 𝑐|⃗𝑝 |(𝑝1 +𝑖𝑝2 ) ⎟
1 ⎜−
𝑢− (⃗𝑝 ) = 𝜂 (𝑝) ⎜ 𝑚𝑐2 +𝐸(⃗𝑝 ) ⎟ (13.28)
⎟
⎜ 𝑝3 + |⃗𝑝 | ⎟
⎝ ⎠
𝑝1 + 𝑖𝑝2

and
( )
⎛ 𝑐|⃗𝑝 | 𝑝1 −𝑖𝑝2 ⎞
− 𝑚𝑐2 +𝐸(⃗𝑝 )
⎜ 𝑐|⃗𝑝 |(𝑝3 +|⃗𝑝 |) ⎟
⎜ ⎟
𝑢2− (𝑝) = 𝜂 (𝑝) ⎜ 𝑚𝑐2 +𝐸(⃗𝑝 ) ⎟ . (13.29)
⎜ ⎟
⎜ − (𝑝1 − 𝑖𝑝2 ) ⎟
⎝ ⎠
3
𝑝 + |⃗𝑝 |
Overall, normalization can be chosen in a convenient way to provide a convenient
orthogonality relations for the four different solutions presented above. It turns out then
if the normalization is chosen in the following way:
√︃
1 𝑚𝑐2 + 𝐸 (⃗𝑝 )
𝜂 (𝑝) = , (13.30)
2 𝑚𝑐2 (⃗𝑝 2 + 𝑝3 |⃗𝑝 |)

then solutions are conveniently normalized in the following way:

𝑢𝑟+ (⃗𝑝 ) 𝑢𝑠+ (⃗𝑝 ) = 𝛿 𝑟𝑠 , (13.31)

where 𝑢¯ is Dirac conjugate spinor. In other words, the scalar product of Dirac conjugate
spinor with positive energy spinor of different helicity gives 0. For negative energy solutions
with different helicity the story is similar with a little exception that on the right hand
side it will be -1:
𝑢𝑟− (⃗𝑝 ) 𝑢𝑠− (⃗𝑝 ) = −𝛿 𝑟𝑠 . (13.32)

In fact, these relations also can be found by using the projector operations or projectors
on solutions with a definite helicity simply by using the fact (13.25). So helicity plus and
minus projection operators can be introduced, which are simply equal to
1±J
J± = . (13.33)
2
Because of the fact that J squared is equal to 1, J± have all the properties of projectors
and, in particular, it can be also checked that they satisfy the following properties. If J±
is an operator, which depends on momentum, then it can be seen that

J± (⃗𝑝 ) = J∓ (−⃗𝑝 ) . (13.34)

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This is one of the properties that can be verified explicitly from the form of the operator.
In fact, solutions that was found for an amplitude 𝑢 look pretty complicated, because
there is a certain amount of symmetry broken, because of the fact that components of
momentum entering differently in different elements of the corresponding spinors. But, in
fact, these formulas immensely simplify if we go to the rest frame of a massive particle.
So, if particle is massive, a transition to the rest frame can be done, where

𝑝⃗ = 0 . (13.35)

Then it can be checked that components with positive and negative energy and definite
helicity are simply turn into
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 0 0 0
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜0⎟ ⎜1⎟ ⎜0⎟ ⎜0⎟
1 2 1 2
𝑢+ = ⎜ ⎟ , 𝑢+ = ⎜ ⎟ , 𝑢− = ⎜ ⎟ , 𝑢− = ⎜ ⎟ , (13.36)
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜0⎟ ⎜0⎟ ⎜1⎟ ⎜0⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 0 0 1

where matrices are turned to the standard basis in the space C4 .

In fact, this simple expression for components of the spinors can be used to perform a
Lorentz transformation in the arbitrary Lorentz frame. And then the result, which already
was found, will be obtained. The only point is that solutions, which will be obtained in this
way do not have a definite helicity and an extra rotation of the spinors should be performed
to diagonalize the helicity operator by taking linear combinations of two positive energy
solutions and then separately two negative energy solutions.
If solutions with positive energy are considered, then solutions for positive energy,
but for different helicities are orthogonal. The same is true for solutions with negative
energies.
It is interesting to discuss what happens if now solutions with positive and negative
energy are taken and a scalar product is needed to be evaluated. There is a little problem,
because helicity operator commutes with the matrix 𝛾 0 that can be understood from the
explicit form of this operator.
J, 𝛾 0 = 0
[︀ ]︀
(13.37)

Indeed helicity operator does commute with 𝛾 0 and 𝛾 0 is important, because it is used
to define the Dirac conjugate spinor, which can be recalled as

𝑢 (⃗𝑝 ) = 𝑢+ (⃗𝑝 ) 𝛾 0 . (13.38)

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It should be noticed that the Dirac conjugate spinors, but not Hermitian conjugate
spinors, are interested, because Dirac conjugate spinors transform under Lorentz transformation
with the inverse matrix of Lorentz transformation, in comparison to the matrix of under
which the spinor 𝑢 transforms. That is useful, because then the Lorentz covariant combinations
can be easily built up. For instance,

𝑢 (𝑝) 𝑢 (𝑝) (13.39)

is a scalar,
𝑢 (𝑝) 𝛾 𝜇 𝑢 (𝑝) (13.40)

is a vector and so on. So, that is very useful for constructing objects with well-defined
transformation properties under Lorentz transformations.
With respect to Helicity everything is fine, Helicity goes through 𝛾 0 and act on the 𝑢
itself, because Helicity commutes with 𝛾 0 , but this is not the case for the Hamiltonian 𝐻,
for which such a nice property does not exist. It does not commute with 𝛾 0

𝐻, 𝛾 0 =
[︀ ]︀
̸ 0. (13.41)

This leads to the fact that if two spinners of different energy are taken, then the scalar
product will not be equal to zero. 𝛾 0 enters into the definition of the Dirac conjugate
spinner rather the exact relation between 𝐻 and 𝛾 0 is as follows:

𝛾 0 𝐻 (⃗𝑝 ) 𝛾 0 = 𝐻 (−⃗𝑝 ) . (13.42)

That is clear from the definition of 𝐻:

𝐻 (⃗𝑝 ) = 𝑐𝛼𝑖 𝑝𝑖 + 𝑚𝑐2 𝛽 , (13.43)

where 𝛼𝑖 are simply given by

𝛼𝑖 = 𝛾 0 𝛾 𝑖 (13.44)

and 𝛽 is again equal to:

𝛽 = 𝛾0 . (13.45)

In fact, the property (13.42) can be used in the following way. First of all, it can be
seen that if a solution with a positive energy is taken and the Hamiltonian is applied to
it, then a positive eigenvalue will be gotten:

𝐻 (⃗𝑝 ) 𝑢+ (𝑝) = 𝐸 (⃗𝑝 ) 𝑢+ (𝑝) . (13.46)

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Then from the (13.46) the next expression follows. Since the Hamiltonian is Hermitian,
then:
𝑢†+ (𝑝) 𝐻 (⃗𝑝 ) = 𝐸 (⃗𝑝 ) 𝑢†+ (𝑝) . (13.47)

𝑢†+ (𝑝) contains 𝛾 0 and now the 𝛾 0 can be compensated by:

𝑢+ (𝑝) 𝛾 0 𝐻 (⃗𝑝 ) 𝛾 0 . (13.48)

Then the (13.48) is simply the following thing:

13.48 = 𝐸 (⃗𝑝 ) 𝑢+ (𝑝) , (13.49)

because what is written down here is simply the same relation as (13.47), but with plug
between 𝑢†+ (𝑝) and 𝐻 (⃗𝑝 ) 𝛾 0 squared matrix and multiplied by 𝛾 0 from the right.
On the other hand, it is known that

𝛾 0 𝐻 (⃗𝑝 ) 𝛾 0 = 𝐻 (−⃗𝑝 ) (13.50)

and, therefore, the (13.49) can be written as:

𝑢+ (𝑝) 𝐻 (−⃗𝑝 ) = 𝐸 (⃗𝑝 ) 𝑢+ (𝑝) . (13.51)

The relation (13.51) can be multiplied by a spinor 𝑢− (−𝑝) from the right:

𝑢+ (𝑝) 𝐻 (−⃗𝑝 ) 𝑢− (−𝑝) = 𝐸 (⃗𝑝 ) 𝑢+ (𝑝) 𝑢− (−𝑝) . (13.52)

The (13.52) can be simplified further if the following property will be used:

𝐻 (−⃗𝑝 ) 𝑢− (−𝑝) = −𝐸 (⃗𝑝 ) 𝑢− (−𝑝) , (13.53)

where minus sign appeared due to the fact that −𝐸 (⃗𝑝 ) is eigenvalue for the negative
energy solution.
According to the (13.53), the (13.52) can be simplified and the following equation can
be gotten:
− 𝐸 (⃗𝑝 ) 𝑢+ (𝑝) 𝑢− (−𝑝) = 𝐸 (⃗𝑝 ) 𝑢+ (𝑝) 𝑢− (−𝑝) . (13.54)

From the (13.54) it can be concluded that

𝑢+ (𝑝) 𝑢− (−𝑝) = 0 . (13.55)

This means that solutions 𝑢+ (𝑝) and 𝑢− (−𝑝) carry a different sign of energy: one energy
is positive, another energy is negative. Moreover they are orthogonal to each other and

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this orthogonality is proved by using the consideration that is done. This is what should
be taken as an orthogonality condition of positive-negative solution, because if the pairing
of solutions with the same momentum is done, then it will be found that it is not-zero
for the reason, which was mentioned that 𝛾 0 does not commute with the Hamiltonian.
Although, if the sign of solution with negative energy is changed, then the result is zero.
𝑠
This consideration suggests that it is convenient to introduce a spinor 𝑣− , which should
be defined or as
𝑠
𝑣− (⃗𝑝 ) = 𝑢𝑠− (−⃗𝑝 ) . (13.56)

Then the full set of the orthogonality conditions, which can be deduced from relations
above is the following:
𝑢𝑟+ (⃗𝑝 ) 𝑢𝑠+ (⃗𝑝 ) = 𝛿 𝑟𝑠 . (13.57)

Then it also can be checked that with minuses or with 𝑣-s, which correspond to negative
energy solutions, these solutions are orthogonal for different helicities:

𝑣 𝑟− (⃗𝑝 ) 𝑣−
𝑠
(⃗𝑝 ) = −𝛿 𝑟𝑠 . (13.58)

Finally, the orthogonality condition for different energy solutions is

𝑣 𝑟− (⃗𝑝 ) 𝑢𝑠+ (⃗𝑝 ) = 0 (13.59)

for any 𝑟 and 𝑠. Analogously,

𝑢𝑟+ (⃗𝑝 ) 𝑣−
𝑠
(⃗𝑝 ) = 0 . (13.60)

For helicity two projections, which project on states with positive and negative helicity,
were introduced, where in fact the same also can be done for projections on solutions
with positive and negative energy. This is convenient if an arbitrary spinor exist, it can be
decomposed into solutions of positive and negative energy by using the projectors. One
projector projects on solution with positive energy and the other will project on solutions
with negative energy and these projections are explicitly constructed as follows:
⎧
⎨Λ+ = 𝑚𝑐·1+
𝑝
2𝑚𝑐
(13.61)
⎩Λ = 𝑚𝑐·1−
𝑝
− 2𝑚𝑐

This is how projectors on positive and negative energy are constructed and if 𝑝
is written
down explicitly, then the (13.61) will look as follows:
⎧
⎨Λ+ = 𝛾 0 𝐸(⃗𝑝 )−𝑐𝛾 𝑖 𝑝𝑖 +𝑚𝑐2 ·1
2𝑚𝑐
(13.62)
⎩Λ = − 𝛾 0 𝐸(⃗𝑝 )−𝑐𝛾 𝑖 𝑝𝑖 −𝑚𝑐2 ·1
− 2𝑚𝑐

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Projectors Λ+ and Λ− are projectors, which means that the following relations are
satisfied: ⎧
⎨Λ2 = Λ+
+
(13.63)
⎩Λ2 = Λ
− −

Then, these projectors are orthogonal, which means that

Λ+ Λ− = 0 . (13.64)

Finally, it can be seen that

Λ+ + Λ − = 1 / (13.65)

It cab be also checked that Λ+ applied to spinors will give the following expression

Λ+ 𝑢1,2 1,2
+ = 𝑢+ . (13.66)

Analogously, for 𝐿− the following expression can be gotten:

1,2 1,2
Λ− 𝑣− = 𝑢− . (13.67)

So, 𝑢+ and 𝑣− form a basis in the full space of solutions and these are separately
eigenstates in the subspaces, which are defined by means of projectors Λ+ and Λ− . There
are also important formulas, which people often use, namely, Λ+ and Λ− can be expressed
in the components of spinors 𝑢+ and 𝑣− .
In particular, Λ+ is a 4x4 matrix and its elements 𝑖 and 𝑗 can be written in the
following way:
2
∑︁ 2
∑︁
(︀ 𝑟 )︀ (︀ 𝑟 )︀ (︀ 𝑟 )︀ (︀ 𝑟 + 0 )︀
(Λ+ )𝑖𝑗 = 𝑢+ 𝑖 𝑢+ 𝑗 = 𝑢+ 𝑖 𝑢+ 𝛾 𝑗 . (13.68)
𝑟=1 𝑟=1

Also the similar expression can be written for Λ− :

2
∑︁ 2
∑︁
(︀ 𝑟 )︀ (︀ 𝑟 )︀ (︀ 𝑟 )︀ (︀ 𝑟 + 0 )︀
(Λ− )𝑖𝑗 = − 𝑣− 𝑖 𝑣 − 𝑗 = − 𝑣− 𝑖 𝑣− 𝛾 𝑗 . (13.69)
𝑟=1 𝑟=1

That is basically all, which allows now to write down the general solution of the Dirac
equation or basically a decomposition of the Dirac spinor via the basis of plane waves. If
a superposition of solutions is taken, then it will be gotten a general solution of the Dirac
equation. So, plane waves actually provide the basis over which an arbitrary solution can
be expanded and this means that a superposition principle can be implemented, because
Dirac equation is the linear equation. If there are two solutions, then the sum of these

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two solutions is another solution of this equation. If we want to construct an arbitrary

solution, then a sum of all solutions should be taken, which means that an integral with
a certain integration measure implied by the normalization of spinors should be taken:
)︂1/2 2
𝑚𝑐2
∫︁ (︂
1 d⃗𝑝 ∑︁ 𝜇 𝜇
𝑏𝑟 (⃗𝑝 ) 𝑢𝑟+ (⃗𝑝 ) 𝑒−𝑖𝑝𝜇 𝑥 /~ + 𝑑*𝑟 (⃗𝑝 ) 𝑣−
𝑟
(⃗𝑝 ) 𝑒𝑖𝑝𝜇 𝑥 /~ ,
(︀ )︀
𝜓 (𝑥) = 𝑐 2 ×
(2𝜋~)3/2 𝐸 (⃗𝑝 ) 𝑟=1
(13.70)
where 𝑏𝑟 (⃗𝑝 ) and 𝑑*𝑟 (⃗𝑝 ) are amplitudes. The written above integral is a general solution
of the Dirac equation, which is written as a superposition of plane wave solutions that
𝑟
was found. And the fact that the 𝑢𝑟+ (⃗𝑝 ) and 𝑣− (⃗𝑝 ) form a basis in the space of solutions
was used. Then for each momentum 𝑝⃗ the corresponding spinors are exist.
By direct calculation it can be checked that the (13.70) is a solution of the Dirac
equation. All what is needed to do is just to take the Dirac operator
(︂ )︂
0 𝜕 𝑖 𝜕 𝑚𝑐
𝑖𝛾 + 𝑖𝛾 − (13.71)
𝑐𝜕𝑡 𝜕𝑥𝑖 ~
and act with it on 𝜓 (𝑥) and when the Dirac operator acts on 𝜓 (𝑥) this essentially means
that it can be passed under the integral and act on the exponent, because it’s an exponent,
which depends on axis over which differentiation is performed. Therefore,
𝛾0
(︂ )︂
0 𝜕 𝑖 𝜕 𝑚𝑐 −𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ 𝑟
𝑖𝛾 + 𝑖𝛾 − 𝑒 𝑢+ = (𝐸 (⃗𝑝 ) − 𝐻 (⃗𝑝 )) 𝑢𝑟+ 𝑒−𝑖(𝐸𝑡−⃗𝑝 ⃗𝑥 )/~ . (13.72)
𝑐𝜕𝑡 𝜕𝑥𝑖 ~ 𝑐~
𝑢𝑟+ is a positive energy solution of the Dirac equation with eigenvalue 𝐸 (⃗𝑝 ) and,
therefore
𝐸 (⃗𝑝 ) − 𝐻 (⃗𝑝 ) = 0 . (13.73)

and
13.72 = 0 . (13.74)

Analogously, it can be shown for the negative energy solutions that they satisfy Dirac
equation.
In the quantum theory the operator 𝑏𝑟 (⃗𝑝 ) will become an operator of annihilation
of electron, while 𝑑*𝑟 (⃗𝑝 ) will turn into the operator of creation of positron. This is how
amplitudes of classical theory will be reinterpreted in the quantum theory.
So, it’s very similar to what was done when a general solution of the Klein-Gordon
was discussed. The amplitudes 𝑎 and 𝑎* there were only 𝑎 and 𝑎* , because in that case
a real scalar field was implied. If a complex scalar field is implied than there will be
found two independent amplitudes 𝑎 and 𝑏 or 𝑎* and 𝑏* star. And here not just complex

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scalar field was introduced, but a spinor field and as it can be seen the spinor field must
have four complex components and, indeed, it has two components, which are described
by 𝑏𝑟 like independent coefficients, which are taken values 1 and 2, and the other two
independent complex components are related to 𝑑*𝑟 , because 𝜓 (𝑥) has arbitrary four
complex components.
One more moment to discuss here is the following. For a massive particle the helicity
operator does not commute with Lorentz transformations. Besides helicity and the Hamiltonian,
in principle, what is also in the game are Lorentz transformations. Under arbitrary Lorentz
transformations neither Hamiltonian nor helicity operators are preserved even with respect
to their usual rotations. The following transformation properties are exist. So, the Dirac
Hamiltonian transforms as follows. If a Lorentz transformation is taken as

𝑆 −1 (𝑅) 𝐻 (𝑝) 𝑆 (𝑅) , (13.75)

then the result of this will be

13.75 = 𝐻 (𝑅⃗𝑝 ) . (13.76)

This is related to the fact that 𝐻 is a matrix and 𝑆 (𝑅) is also a 4x4 matrix. The (13.76)
means that, when a Lorentz frame is transformed by means of Lorentz transformation,
in general, the Hamiltonian undergoes a transformation particular, it is the Hamiltonian
evaluated at the transformed value of momentum 𝑝⃗ . Similarly, if the helicity is taken, it
transforms also under this Lorentz transformations in a similar way:

𝑆 −1 (𝑅) J (𝑝) 𝑆 (𝑅) = J (𝑅⃗𝑝 ) . (13.77)

In the case of the Klein-Gordon field there was a difference. So, the Hamiltonian was
invariant with respect to such transformations, because the Hamiltonian in that case
depends on 𝑝⃗ 2 and 𝑝⃗ is a vector, which length is invariant under rotations. So, if a vector
is taken and rotated, by the definition of the rotation group the length of this vector must
be preserved. The Dirac Hamiltonian is different, because it involves momentum not like
a 𝑝⃗ 2 , but involves it as
𝛾 𝑖 𝑝𝑖 (13.78)

and it’s a big difference. Now under rotations by Lorentz transformations result is that we
have Hamiltonian at rotated momentum 𝑝⃗ and that is a different Hamiltonian. In general,
it can be seen the only thing, which preserves actually the Hamiltonian and also helicity
are rotations, which happen in the plane orthogonal to momentum. If the momentum is

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looking in one direction of particles, then if a orthogonal plane to this momentum is taken
and rotated around the direction of 𝑝⃗ in this plane, then this will be invariant momentum
𝑝⃗ . In this case under the restricted rotations neither Hamiltonian nor helicity will be
changed (fig. 13.2).

Fig. 13.2. The plane orthogonal to momentum 𝑝⃗

The situation with helicity is actually dramatic, because helicity depends on the Lorenz
frame. In other words, a state with a positive helicity in a different Lorenz frame can
become a state with a negative helicity. So, helicity is not a Lorenz invariant quantity for
a massive particle. If particle is massive, then helicity is not a good observable, because
it depends on the choice of the Lorenz frame. But for massless particle the situation is
different. This is due to the massless Dirac equation, because if the Dirac equation for the
massless particle is written, then there is no the term with 𝑚𝑐2 and the massless Dirac
equation takes the following form:
𝐸 0
𝛾 𝑖 𝑝𝑖 = 𝛾 . (13.79)
𝑐
So, in the momentum space for a plane wave this is how the massless Dirac equation looks
like. For massless equation it is known that energy is given by

𝐸 (⃗𝑝 ) = ±𝑐 |⃗𝑝 | . (13.80)

This is a formula, which can be gotten from the general formula

√︁
2
𝐸 = 𝑝 2 𝑐2 + 
𝑚 𝑐4 , (13.81)

if the mass turns to zero. This means that the operator of helicity, in general, is equal to
1 5 0 𝑖 𝑖
J= 𝛾 𝛾 𝛾𝑝 (13.82)
|⃗𝑝 |

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and because of the relation (13.59) this can be simplified to the following form:

J = ±𝛾 5 . (13.83)

So, for massless particle helicity operator is the same as chirality operator, which was
introduced at the previous lecture. Helicity and chirality becomes the same and they
become an absolute thing, which does not depend on the Lorentz frame and this is also
something obvious, because in the Lorentz frame for a massless particle a rest frame is
not exist. Basically a particle is moving with the speed of light and since we’re a massive
persons, we can never reach it.
For a massive particle if a momentum is high enough, the observer can be placed in
the particle’s frame and in this system the sign of momentum will change. So the helicity
in this case can be changed, because the sign of 𝑝⃗ can be changed in the frame of observer
if it, for instance, will go faster than a particle.

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Lecture 14. Charge Conjugation and Anti-Particles

At this lecture the operation of quantum field theory, which acts on the Dirac equation,
will be observed. This operation of charge conjugation, which actually allows to understand
the notion of anti-particles. To do this, firstly it is needed to understand, how a Dirac
particle interacts with an electromagnetic field. And this is needed in order to be able
to introduce the notion of electric charge in the context of a Dirac particle, because the
Dirac equation didn’t involve electric charge at all. On the other hand, the properties of
the electron concerning its electric charge only shown up, when an electron is put in the
an external electromagnetic field. There should be electromagnetic field around, which
then influences the motion of an electrically charged particle.
Indeed, there is a coupling of the particle to surrounding electromagnetic field and
tracing the change of the particle trajectory. Therefore, to detect the presence of electromagnetic
field around on the one hand side and on the other hand side it is needed to say that
a particle possess a property, which can be called as electric charge. In other words, the
ability of a particle to interact with electromagnetic field is called as electric charge.
The first step would be to introduce the coupling of the Dirac spinor or a Dirac particle
to an external electromagnetic field. In this case, the four electromagnetic field has been
given. And as it known from classical electrodynamics already, the electromagnetic field
is defined by means of an electromagnetic potential 𝐴𝜇 , which can be introduced in the
Dirac equation in the following way:
[︁ (︁ 𝑒 )︁ ]︁
𝛾 𝜇 𝑖~𝜕𝜇 − 𝐴𝜇 − 𝑚𝑐 𝜓 = 0 , (14.1)
𝑐
where 𝑒 is a charge of a Dirac particle.
In other words, the introduction of the potential amounts to a procedure of, as people
sometimes say, making the usual derivative to become with long derivative, which means
that the electromagnetic potential 𝐴𝜇 was extended. In fact, the quantity in the brackets of
the (14.1) is a covariant derivative or derivative where 𝐴𝜇 is considered to be a connection
on a certain bundle. 𝐴𝜇 is an electromagnetic potential given in the whole space-time. So,
it’s a function of a space-time point 𝑥.
In natural units electric charge has a physical dimension of
[︁√ ]︁
[𝑒] = ~𝑐 . (14.2)
√
This is useful to have in mind that electric charges have the same dimension as ~𝑐
in terms of fundamental constants. And this fact can be deduced from the Coulomb force

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between two equal charges, because, as it is known, the Coulomb force between two such
charges simply equal to
𝑒2
𝐹 = , (14.3)
𝑟2
where 𝑟 is a distance between charges and charges are equal to each other.
From the (14.3) it is clearly seen that
[︃ ]︃
2
[𝑒] = 𝐹 · 𝑙2 = 𝐹 · 𝑙 ⏟·𝑡 ·⏞𝑐 ,
[︀ ]︀
(14.4)
𝑙

where 𝐹 multiplied by length is a work, which is the same as energy. And it is known
that an energy multiplied by time is a Plank constant ~. Therefore,

[𝑒]2 = [~ · 𝑐] . (14.5)

So, electromagnetic field is not a matrix, it’s just the full component vector 𝐴𝜇 (𝑥)
and the coupling is coming through the product 𝛾 𝜇 𝐴𝜇 . Then there is a constant, which
regulates the strengths of this coupling and this is what is called as an electric charge.
It should be noticed, that an introduced electromagnetic field is external, because
the dynamical equations for electromagnetic field were not added, because the dynamical
equations for an electromagnetic field would be Maxwell equations. And if electromagnetic
field interacts with a spinor particle, then this particle would serve as a source for
electromagnetic field. So, there will be nontrivial back reaction of a spinor field on an
electromagnetic field, but this back reaction for the moment is completely neglected.
That is because the given field is strong enough not to be influenced by the motion of a
Dirac particle itself.
But on the other hand, the presence of this electromagnetic field, of course, influences
the dynamics of the Dirac particle itself and that’s the reason why it is said that a particle
couples to the field.
The new Dirac equation, which was written in the (14.1) turns out that it has a new
fundamental discrete symmetry, which is a symmetry over the theory with respect to the
change of the sign of the electric charge and this is the symmetry, which is also considered
as symmetry replacing the particle by its antiparticle. To understand this a name for it
should be introduced. This symmetry is called charge conjugation. The generator of this
symmetry can be denoted as C, which is a charge conjugation. It is a new fundamental
symmetry of the Dirac equation.

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Now, existence of this symmetry implies that there is a one to one correspondence
between solutions, which was denoted as 𝜓, of the Dirac equation with a negative energy
and positive energy (a particle have the same mass). Wave functions for this cases are
connected by the following expression:

𝜓 𝑐 = C𝜓 , (14.6)

where 𝜓 𝑐 is a wave function of a particle with positive charge and 𝜓 is a wave function of
a particle with negative charge.
If there is a particle, which is an electron and it is described by solution with positive
energy, as it should be, then there will be another solution, which is constructed from side
with negative energy as 𝜓 𝑐 and this solution will carry opposite charge. It has to be treated
as a positron. So, electrons with 𝐸 less than zero should be treated as positrons with
normal situation, where energy is positive. That’s a fundamental thing in this existence of
the theory actually, in a way, returns to the Dirac equation the status of being physical,
because normally dynamical systems, which have solutions with negative energy should
be neglected. Energy must always be restricted either from above or from below and in
the Dirac equation this is not the case, because there are solutions with negative energy
and with positive energy and they are not restricted or bounded, neither from above nor
from below.
From this point of view we might say that this equation should not be interpreted as
physical. On the other hand, due to the new fundamental symmetry a new interpretation
for the solution with negative energy can be given. Namely, they can be treated as solutions
with positive energy, but opposite sign of electric charge.
In fact, the real understanding of what’s happening with this fundamental symmetry
comes only upon performing the quantization of the Dirac theory.
If second quantization in this case is performed, there will be no any more solutions
with negative energy at all, but there will be electrons and positrons solutions with
opposite sign of electric charge, but only with positive energy.
Now, it is needed to be found how the symmetry acts on 𝜓. The equation for the
charge conjugated 𝜓 should be understood in the following way. The equation for 𝜓 𝑐 have
the next form:
[︁ (︁𝑒 𝑐 )︁ ]︁
𝜇
𝛾 𝑖~𝜕𝜇 − 𝐴𝜇 − 𝑚𝑐 𝜓 𝑐 = 0 , (14.7)
𝑐
where 𝐴𝑐𝜇 is a charge conjugated potential and 𝜓 𝑐 is exactly the wave function, which is
obtained from side by application of the charge conjugation operation. The 𝐴𝑐𝜇 is simply

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amounts to change the sign of the electromagnetic potential. So, charge conjugation is
an operation, which acts on fields and the index. Since now two kinds of fields are exist:
𝜓 and 𝐴𝜇 , and the charge conjugation is operation, which acts on all fields, which are
presented in a theory, it does not touches constants.
And charge conjugation acts on the potential 𝐴𝜇 in a very simple way. In a sense, it
can be seen that changing the sign of electromagnetic potential is exactly equivalent to
changing the sign of the electric charge, because they come together:

𝐴𝑐𝜇 (𝑥) = −𝐴𝜇 (𝑥) . (14.8)

If the (14.7) is compared with the original Dirac equation, then it has exactly the same
form except that it acts on charge conjugate 𝜓 𝑐 not on 𝜓.
Once fields was transformed by symmetry transformation, the Dirac equation should
be covariant and, in other words, in terms of new transformed quantities, it should look
the same as in terms of the old or the original quantities.
Now, the following can be seen that the implementation of charge conjugation on the
electromagnetic field amounts to the change of sign minus into plus. So, if 𝐴𝜇 will be
replaced by 𝐴𝑐𝜇 in the (14.1), then it will lead to appearance of plus sign:
[︁ (︁
𝜇 𝑒 𝑐 )︁ ]︁
𝛾 𝑖~𝜕𝜇 + 𝐴𝜇 − 𝑚𝑐 𝜓 = 0 , (14.9)
𝑐
but this change of sign can be easily done by the operation of conjugation, because the
electromagnetic field is real.
In other words, what we should start from equation (14.1) and perform an operation
of complex conjugation. In this way, what will be gotten can be then written as follows:
[︁ (︁ 𝑒 )︁ ]︁
𝛾 *𝜇 −𝑖~𝜕𝜇 − 𝐴𝑐𝜇 − 𝑚𝑐 𝜓 * = 0 . (14.10)
𝑐
Then minus in the brackets of the (14.10) can be taken out and the equation can be
written in the following form:
[︁
*𝜇
(︁ 𝑒 𝑐 )︁ ]︁
𝛾 𝑖~𝜕𝜇 + 𝐴𝜇 + 𝑚𝑐 𝜓 * = 0 . (14.11)
𝑐
Then, to reach a goal, it should be assumed that the operation of charge conjugation
can be explicitly realized by applying a certain matrix to 𝜓 𝑐 , where 𝜓 𝑐 is a spinor, which
is charge conjugation of the original spinor 𝜓:
)︀−1 𝑐
𝜓 * = C𝛾 0
(︀
𝜓 , (14.12)

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where C is a 4x4 matrix and C𝛾 0 is a matrix of transformation from the 𝜓 𝑐 to 𝜓 *

In other words,
𝜓 𝑐 = C𝛾 0 · 𝜓 * . (14.13)

Now, the 𝜓 * in the (14.11) can be replaced by 𝜓 𝑐 and, therefore the following expression
will be gotten:
[︁(︀ 𝑒 )︁
)︀−1 (︁ ]︁
C𝛾 0 𝛾 *𝜇 C𝛾 0 𝑖~𝜕𝜇 − 𝐴𝑐𝜇 + 𝑚𝑐 𝜓 𝑐 = 0
)︀ (︀
(14.14)
𝑐
If we want to restore equation (14.7), then the following should be required
)︀−1
C𝛾 0 𝛾 *𝜇 C𝛾 0 = −𝛾 𝜇 .
(︀ )︀ (︀
(14.15)

The (14.15) also can be written in the following way:

)︀−1
C𝛾 0 𝛾 𝜇 C𝛾 0 = −𝛾 *𝜇 .
(︀ (︀ )︀
(14.16)

The first question, which arises here is if such a matrix C, which has the property
(14.15), exists, because this transformation formula should be valid for all 𝜇, which can
be equal to 0, 1, 2 and 3.
As can be see easily seen easily an transformation

𝛾 𝜇 → −𝛾 *𝜇 (14.17)

is automorphism of the Clifford algebra, which means that under this transformation a
Clifford algebra is not changing, it turns to itself. But, on the other hand, it is known from
the power of theorems that any of the morphism of the Clifford algebra in four dimensions
is internal. In other words there should exist a unitary matrix 𝑈 , such that

− 𝑈 𝛾 *𝜇 𝑈 −1 = 𝛾 𝜇 (14.18)

So transformation from a set 𝛾 *𝜇 to 𝛾 𝜇 is done by unitary transformation, where 𝑈 is

equal to:
𝑈 = C𝛾 0 . (14.19)

It’s form depends on the representation of 𝛾 matrices that is used. It will have one form
in the Dirac representation and it will have other forms in other representations.
It is known that a spinor transforms under the group of proper orthochronous Lorentz
transformation or, in other words under the 𝑆𝑂+ (1, 3). And an interesting point is that,
in fact, the charge conjugate spinor transforms under the same transformation, under
which 𝜓 transforms.

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If 𝜓 undergoes under Lawrence transformations, it can be written in the following

way:
𝜓 → 𝑆𝜓 , (14.20)

where 𝑆 as it known is given by

(︂ )︂
1 𝜇 𝜈
𝑆 = exp 𝛾 𝛾 𝜔𝜇𝜈 . (14.21)
4
Then for 𝜓 𝑐 the same formula will be found:

𝜓 𝑐 → 𝑆𝜓 𝑐 . (14.22)

And this is a consequence of the relation (14.15) and (14.13):

C𝛾 0 𝜓 * → C𝛾 0 𝑆 * 𝜓 * . (14.23)

Then, 𝜓 * can be replaced by 𝜓 𝑐 :

)︀−1 𝑐
14.23 = C𝛾 0 𝑆 * C𝛾 0
(︀ )︀ (︀
𝜓 . (14.24)

Then, we can look in more detail at the term (C𝛾 0 ) 𝑆 * (C𝛾 0 ) :

−1

(︂ )︂
(︀ 0 )︀ * (︀ 0 )︀−1 1 *𝜇 *𝜈 )︀−1
C𝛾 𝑆 C𝛾 = C𝛾 exp
0
𝛾 𝛾 𝜔𝜇𝜈 C𝛾 0
(︀
. (14.25)
4

Using the property of exponential term C𝛾 0 can be raised into the exponent and, therefore,
1 (︀ 0 )︀ *𝜇 (︀ 0 )︀−1 (︀ 0 )︀ *𝜈 (︀ 0 )︀−1
14.25 = exp C𝛾 𝛾 C𝛾 C𝛾 𝛾 C𝛾 𝜔𝜇𝜈 . (14.26)
4
Then according to the (14.15), formula (14.26) can be written as:
(︂ )︂
(︀ 0 )︀ * (︀ 0 )︀−1 1 𝜇 𝜈
C𝛾 𝑆 C𝛾 = exp 𝛾 𝛾 𝜔𝜇𝜈 = 𝑆 . (14.27)
4
In other words, the (14.26) can be transformed into

C𝛾 0 𝜓 * → 𝑆𝜓 𝑐 . (14.28)

Moreover, if 𝑈 is introduced as C𝛾 0 , then it’s natural to require that this transformation

𝑈 is unitary, because the equivalence of 𝛾 matrices 𝛾 𝜇 and −𝛾 *𝜇 will be then unitary
equivalence of two representations of the Clifford algebra. And this requirement as can be
seen amounts to the following.

𝑈 𝑈 + = C𝛾 0 𝛾 +0 C+ = CC+ = 1 . (14.29)

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So, matrix C itself must be unitary.

Then, for transformation of the Lawrence representation 𝑆 the following can be seen:

𝑈 𝑆 *𝑈 + = 𝑆 . (14.30)

So, this is a property of the Lawrence transformation from 𝑆𝑂+ (1, 3), which can be derived
in this way. It shows in fact that complex conjugate representation Lorentz representation
𝑆 * is related to 𝑆 by means of the formula:

𝑆 * = 𝑈 + 𝑆𝑈 . (14.31)

Mathematically this fact means that the original representation 𝑆𝑂+ (1, 3) on four-dimensional
spinors and it’s complex conjugate representation are simply a unitary equivalent.
On the other hand, now a new property of this representation was found, namely, if
he complex conjugated that in reality we don’t get new representation, but the unitary
equivalent representation is gotten and representations, for which complex conjugate are
equivalent to the original representation are called pseudoreal.
Therefore, from here the representation of 𝑆𝑂+ (1, 3) on four-component Dirac spinors
is reducible and it is pseudoreal.
If the unitary matrix 𝑈 would be one, then such representations would be called in the
representation theory as real representations. Representations for each complex conjugate
are unitary equivalent to the original representation are called pseudoreal. And finally, if,
in general, 𝑆 * is not related to 𝑆 by unitary transformation, then such representations are
called complex. If two group elements 𝐴 and 𝐵 are taken, then this is a homomorphism,
which satisfy the relation:
𝜌 (𝐴𝐵) = 𝜌 (𝐴) 𝜌 (𝐵) , (14.32)

where 𝐴 and 𝐵 are any two elements of the group.

If the representation is complex conjugated than this representation still will be a
representation:
𝜌* (𝐴𝐵) = 𝜌* (𝐴) 𝜌* (𝐵) . (14.33)

And then the questions, which can be asked is how 𝜌* is related to 𝜌. Is it unitary
equivalent to the original representation? Is it’s absolutely the same as original representation?
The answer to this questions is to give a classification of representations as real,
pseudoreal and complex. And for the case of the Lorentz representation of 𝑆𝑂+ (1, 3)
and four-component Dirac spinors will learn that such a representation is pseudoreal.

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Then it is interesting to observe what will happen if electromagnetic field is absent or

if there is no electromagnetic field at all. So, in this case, if there is no electromagnetic
field or this field is switched off, then it can be seen that the transformation 𝜓 → 𝜓 𝑐
is just a novel discrete symmetry of the Dirac equations. This is because it transforms
solutions into solutions of the Dirac equation. So, there is one solution, which can be called
as 𝜓 and, then, if the operation of charge conjugation is applied, another solution of the
same Dirac equation will be gotten. Moreover, this discrete symmetry is very similar to,
for instance, parity or time reversal. So, it acts in a similar way. It turns out that this
symmetry maps positive energy solutions into negative energy solutions and vice versa.
And it can be seen from the following. If the symmetry of charge conjugation is taken and
acts on positive energy solution, which is obtained by acting on the Dirac spinor, then
this can be written in the following way:

C (Λ+ 𝑢) = C𝛾 0 Λ*+ 𝑢* .
(︀ )︀
(14.34)

The matrix Λ*+ can be replaced by definition and the following expression can be gotten:

𝛾 *0 𝐸 − 𝑐𝛾 *𝑖 𝑝𝑖 + 𝑚𝑐2 · 1 (︀ 0 )︀−1 (︀ 0 )︀ *
C (Λ+ 𝑢) = C𝛾 0 2
C𝛾 C𝛾 𝑢 . (14.35)
⏟ 2𝑚𝑐 ⏞
Λ*+

Expression (14.35) can be simplified and as a result the following expression will be
gotten:
−𝛾 0 𝐸 + 𝑐𝛾 𝑖 𝑝𝑖 + 𝑚𝑐2 · 1 0 *
C (Λ+ 𝑢) = 2
𝑐𝛾 𝑢 = Λ− (C𝑢) . (14.36)
⏟ 2𝑚𝑐 ⏞
Λ−

In other words, the following commutation relations exist:

CΛ+ = Λ− C . (14.37)

That’s why a positive energy solution and the charge conjugation go to the negative
energy solutions and vice versa. So it’s a symmetry, which maps positive solutions into
negative solutions.
Now the same trick with helicity can be done, because for helicity projectors on solution
with positive helicity and negative helicity are exist.

𝛾 *5 𝛾 *0 𝛾 *𝑖 𝑝𝑖 (︀ 0 )︀−1 (︀ 0 )︀ *
(︂ )︂
C (J± (⃗𝑝 ) 𝑢) = C𝛾 1 ±
0
C𝛾 C𝛾 𝑢 . (14.38)
|⃗𝑝 |

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Then it can be seen, analogously to the previous case, that what expression (14.38) will
be simplified into:

𝛾 5 𝛾 0 𝛾 𝑖 𝑝𝑖
(︂ )︂
1
C (J± (⃗𝑝 ) 𝑢) = 1∓ (C𝑢) = J∓ (⃗𝑝 ) C𝑢 . (14.39)
2 |⃗𝑝 |

So, intertwining of helicity happens in the following way:

CJ± = J∓ (⃗𝑝 ) C . (14.40)

In other words, solutions with positive helicity go to solutions with negative helicity and
vice versa.
In general, now it can be concluded that this operation acts like positive energy
solution and the definite helicity goes to negative energy solution and opposite helicity
and vice versa. So, this is how this symmetry acts on solutions of the Dirac equation
even in the absence of electromagnetic field, because if there is an electromagnetic fields
and charge conjugation also implies that electromagnetic field changes the sign. But if
electromagnetic field is absent, there is nothing to charge conjugate from the point of view
of electromagnetic field and, therefore, what then is gotten is just a statement about the
charge conjugation been in 𝑈 discrete symmetry of the Dirac equation, which transforms
solutions into solutions.
Further determination of the concrete matrix 𝐶 relies on the representation chosen.
And, for instance, for the Dirac representation of 𝛾 matrices, the matrix 𝐶 can be chosen
to be equal to:
C = 𝛾 2𝛾 0 . (14.41)

Then it can be seen that this matrix exactly does the job, which is needed. So, for instance,

C𝛾 0 = 𝛾 2 . (14.42)

Then, it can be seen that the following expression is right

(︀ 2 )︀−1 𝜇 2
𝛾 𝛾 𝛾 = −𝛾 *𝜇 . (14.43)

According to the property

(︀ 2 )︀2
𝛾 = −1 , (14.44)

the (14.43) can be written as follows:

𝛾 2 𝛾 𝜇 𝛾 2 = 𝛾 *𝜇 . (14.45)

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For instance, it can be seen that in the Dirac representation a matrix 𝛾 0 is a diagonal
matrix and it’s given by
1 0
(︃ )︃
𝛾0 = . (14.46)
0 −1
So, from this point of view, 𝛾 0 is real. Then,
(︀ )︀2
𝛾 2 𝛾 0 𝛾 2 = −𝛾 0 𝛾 2 = 𝛾 0 . (14.47)

Now, for other matrices the relation (14.45) can be checked. For this purpose, 𝛾 𝑖
matrices should be recalled:
0
(︃ )︃
𝜎𝑖
𝛾𝑖 = . (14.48)
−𝜎 𝑖 0
And it should be noticed that 𝜎 1 and 𝜎 3 are real. Therefore, under complex conjugation
𝛾 1 and 𝛾 3 matrices will be unchanged
(︀ 1 )︀* (︀ )︀*
𝛾 = 𝛾1 , 𝛾3 = 𝛾3 , (14.49)

while for 𝛾 2 the following expression will be gotten:

(︀ 2 )︀*
𝛾 = −𝛾 2 . (14.50)

It is exactly seen that 𝛾 1 and 𝛾 3 matrices will behave themselves exactly like 𝛾 0 :
⎧
⎨𝛾 2 𝛾 1 𝛾 2 = −𝛾 1 (𝛾 2 )2 = 𝛾 1
(14.51)
⎩𝛾 2 𝛾 3 𝛾 2 = −𝛾 3 (𝛾 2 )2 = 𝛾 3

For 𝛾 2 the situation is different. So, we will get

𝛾 2 𝛾 2 𝛾 2 = −𝛾 2 . (14.52)
⏟ ⏞
=−1

Therefore, in the (14.52) the required expression is gotten.

Indeed, for the Dirac representation C = 𝛾 2 𝛾 0 is a matrix, which satisfies the required
property. Then it was also seen that unitarity conditions are satisfied.
In addition the matrix C has the following properties:

C2 = 1 , → 𝛾 2 𝛾 0 𝛾 2 𝛾 0 = − 𝛾 2 𝛾 2 𝛾 0 𝛾 0 = 1 , (14.53)
⏟ ⏞ ⏟ ⏞
−1 1
)︀*
C* = −C , 𝛾 2 𝛾 0 = 𝛾 *2 𝛾 0 = −𝛾 2 𝛾 0 = −C ,
(︀
(14.54)

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CC+ = CC*𝑡 = −CC𝑡 = −𝛾 2 𝛾 0 𝛾 𝑡0 𝛾 𝑡2 = −𝛾 2 𝛾 𝑡2 = 1 . (14.55)

⏟ ⏞ ⏟ ⏞
1 𝛾2

Then the charge conjugate spinor in the Dirac representation explicitly can be realized
as:
𝜓 𝑐 = C𝛾 0 𝜓 * = 𝛾 2 𝜓 * . (14.56)

The 𝜓 𝑐 can also be written in the following way. The 𝜓 should be introduced, which is
explicitly
𝜓 = 𝜓 *𝑡 𝛾 0 , (14.57)

where 𝜓 *𝑡 is a row vector. So, if then 𝜓 is transposed, then the following expression will
be gotten:
𝑡
𝜓 = 𝛾 0𝜓* . (14.58)

According to the (14.58), the (14.56) can be written as:

𝑡
𝜓 𝑐 = C𝜓 . (14.59)

One remark, also can be given here is the following:

C−1 𝛾 𝜇 C = −𝛾 0 𝛾 *𝜇 𝛾 0 , (14.60)

where the following relation was used:

)︀−1
C𝛾 0 𝛾 𝜇 C𝛾 0 = −𝛾 *𝜇
(︀ (︀ )︀
(14.61)

and the brackets (C𝛾 0 )

−1
was opened to get:

𝛾 0 C−1 𝛾 𝜇 C𝛾 0 = −𝛾 *𝜇 (14.62)

where the (14.60) can be gotten by multiplying by 𝛾 0 from the left and from the right.
But, on the other hand, what is written down in the (14.60) is nothing else in the
Dirac representation as
C−1 𝛾 𝜇 C = − (𝛾 𝜇 )𝑡 . (14.63)

So, complex conjugate matrix multiplied from the left and from the right by 𝛾 0 is the
same as 𝛾 𝜇 transposed. Alternatively, in the Dirac representation, the matrix C can be
characterized by the property that conjugating of 𝛾 𝜇 with the unitary matrix C gives the
(14.63).

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Again, this shows the unitary equivalence of 𝛾 𝜇 with (𝛾 𝜇 )𝑡 e.g. with the other representation
of the Clifford algebra, which is performed with the help of the unitary conjugation by by
the matrix C.
Also here another interesting simple consequence can be deduced that if

C−1 𝛾 𝜇 C = − (𝛾 𝜇 )𝑡 , (14.64)

then it can be checked what happens with 𝛾 5 , when it conjugated. And this simply
amounts to
(︀ )︀𝑡
C−1 𝛾 5 C = 𝛾 5 , (14.65)

because 𝛾 5 is the product

𝛾 5 = 𝑖𝛾 0 𝛾 1 𝛾 2 𝛾 3 (14.66)

and when C−1 and C are applied, then this transposes each of this matrices. So, it becomes

C−1 𝛾 0 𝛾 1 𝛾 2 𝛾 3 C = 𝛾 𝑡0 𝛾 𝑡1 𝛾 𝑡2 𝛾 𝑡3 . (14.67)

On the other hand, it is known that in the Dirac representation the 𝛾 5 matrix is
symmetric and it does not change under transposition, but then under the result (14.67)
it follows that since
C−1 𝛾 5 C = 𝛾 5 , (14.68)

C actually commutes with 𝛾 5 :

C, 𝛾 5 = 0 .
[︀ ]︀
(14.69)

Sometimes this property is also interesting and should be taken into account in some
considerations.
So in any case, this is a consequence of the explicit representation for C in the case of
Dirac representation.
In summary, the Dirac theory is invariant under charge conjugation of a spinor supplied
with a simultaneous change of electromagnetic field to minus electromagnetic field. The
physical meaning of this charge conjugation operation is that any physically realizable
state of electron in the field 𝐴𝜇 corresponds to a physically realizable state of a positron in
the field −𝐴𝜇 . So, operation of charge conjugation changes electrons with negative energy
and spin up to positrons with positive energy and spin down. If we have a solution of the
Dirac equation, for instance, with a negative energy, which can be called as 𝜓 (−) , which
is wave function corresponding to a solution with negative energy, and an electric charge,

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which can be denoted by 𝑒, then we should take as a physically realizable or physically

relevant quantity not the wave function 𝜓 (−) , because it’s a solution with negative energy,
but rather it’s a charge conjugated solution 𝜓 𝑐(+) , which has then positive energy and
given explicitly by
(−) 𝑡
𝜓 𝑐(+) = C𝜓 . (14.70)

This 𝜓 𝑐(+) will then describe a particle with a positive energy, which is a physical
particle, but with different sign in comparison to the sign of original electron.
Therefore, we can identify the positive energy plane waves 𝜓 for an electron as
𝜇 /~
𝜓𝑒𝑟 (⃗𝑥 , 𝑡) = 𝑢𝑟+ (⃗𝑝 ) 𝑒−𝑖𝑝𝜇 𝑥 (14.71)

and for a positron as

]︀𝑡
𝜓𝑝𝑟 (⃗𝑥 , 𝑡) = C 𝑢𝑟− (−⃗𝑝 ) 𝑒−𝑖𝑝𝜇 𝑥 /~ .
[︀ 𝜇
(14.72)

The (14.72) can be also written in terms of 𝑣− (⃗𝑝 ), because as it was shown in the
previous lecture:
𝑣− (⃗𝑝 ) = 𝑢− (−⃗𝑝 ) (14.73)

and the (14.73) will be linked with (14.72) by the following expression:
]︀𝑡
→ C 𝑢𝑟− (−⃗𝑝 ) 𝑒−𝑖𝑝𝜇 𝑥 /~ .
𝜇 /~ 𝜇
𝑣− (⃗𝑝 ) (⃗𝑝 ) 𝑒𝑖𝑝𝜇 𝑥
[︀
(14.74)
C

So, then it can be checked that under a charge conjugation the (14.73) will turn into the
(14.72).
Let’s return back to the general solution of the Dirac equation, which was represented
in terms of plane waves and then it would be natural to treat the amplitude, which
is given in terms of the coefficient 𝑏𝑟 in the general solution of Dirac equation as an
annihilation operator of electron, while then 𝑑*𝑟 , which appears as another amplitude
standing in front of the 𝑖~𝑝𝜇 𝑥𝜇 as a creation operator of a positron. That’s consideration
relates to the positive and negative energy solutions in the general expression for the
Dirac equation. So, negative energy solutions we would like to interpret from the point
of view of the new understanding of the charge conjugation symmetry as those, which
correspond to actually positive energy solution, but for another particle, which is called
as positron. Correspondingly, this gives an interpretation to the amplitudes, which are
existed in general solution of the Dirac equation. There are amplitudes, which introduces
𝑏𝑟 and 𝑑*𝑟 as correspondingly annihilation operator of electrons and the creation operator

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of positrons. That’s something, which can be further understood, when transition to the
quantization of the Dirac equation is performed.
The final topic, which linked with the consideration of reality properties and the charge
conjugation, concerns with the issue of so-called Majorana spinors.

Majorana spinors
We already learned that there are four-component Dirac spinors. There are also Weyl
spinors, which are two-component complex spinors. And now another kind of spinors,
which are called Majorana will be introduced. It is known that the Dirac spinor in four-
dimensional Minkowski space has four complex components, which is equivalent to 8
real degrees of freedom. It is possible to reduce a number of independent components
of this Dirac spinner by imposing certain conditions, which are compatible with Lorentz
transformations from 𝑆𝑂+ (1, 3). One of these conditions is a Weyl condition, which was
discussed and that allows to define the Weyl spinor, which has two complex components.
And the spinor transforms as it known irreducibly under 𝑆𝑂+ (1, 3).
Now, another condition can be imposed, which is called Majorana condition. This is
condition, which says to take spinor 𝜓 and identify it with its charge conjugation.

𝜓 = 𝜓𝑐 . (14.75)

In other words, Majorana condition is a condition that states that particle is its own
antiparticle.
The special Dirac spinner, which satisfies the condition (14.75) is called Majorana.
This condition is allowed to impose, because it is known that 𝜓 and 𝜓 𝑐 transform in
the same way under Lorentz transformations. So, this condition does not break Lorentz
symmetry and it is Lorentz invariant.
An explicit form of the Majorana spinner depends on the chosen representation for
𝛾 matrices, because the explicit form of the charge conjugate matrix C depends on the
representation chosen. For instance, in the Dirac and Weyl representations, C is given by

C = 𝛾 2𝑔0 (14.76)

and, therefore, when the Majorana condition is imposed, that means that we have to solve
the following equation:
𝜓 = C𝛾 0 𝜓 * = 𝛾 2 𝜓 * . (14.77)

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If 𝜓 is written in terms of two components spinors 𝜑 and 𝜒:

(︃ )︃
𝜑
𝜓= , (14.78)
𝜒

then the (14.77) can be written as:

(︃ )︃ (︃ )︃ (︃ )︃
𝜑 0 𝜎2 𝜑*
= . (14.79)
𝜒 −𝜎 2 0 𝜒*

It can be seen that from the condition (14.79) 𝜒 can be expressed and the answer will
be that such a Majorana spinor in the Dirac representation of 𝛾 matrices will have the
following form: (︃ )︃
𝜑
𝜓= . (14.80)
−𝜎 2 𝜑*
Moreover, there exists another representation of 𝛾 matrices, in which the Majorana
condition becomes very simple and transparent. This representation of 𝛾 matrices is
called Majorana representation. Majorana representation is a representation in which
all 𝛾 matrices are imaginary. So the Dirac operator contains, in fact, only real coefficients
and explicitly it can be written down straight away. In this representation:
(︃ )︃ (︃ )︃ (︃ )︃ (︃ )︃
0 0 𝜎2 1 𝑖𝜎 3 0 2 0 −𝜎 2 3 −𝑖𝜎 1 0
𝛾𝑀 = , 𝛾𝑀 = , 𝛾𝑀 = , 𝛾𝑀 = .
𝜎2 0 0 𝑖𝜎 3 𝜎2 0 0 −𝑖𝜎 1
(14.81)
It can be checked that the Clifford algebra relations for these 𝛾 matrices will be satisfied
and this means by Pauli theorem that there exists a unitary matrix that transforms Dirac
representation into this new Majorana representation by means of conjugation with the
unitary matrix 𝑈 . So,
𝜇 −1 𝜇
𝑈 𝛾𝐷 𝑈 = 𝛾𝑀 , (14.82)

where it can be checked that explicitly the matrix 𝑈 is given by

1 𝜎2
(︃ )︃
1
𝑈=√ (14.83)
2 𝜎 2 −1

and this matrix is explicitly unitary

𝑈 +𝑈 = 1 . (14.84)

Now, the charge conjugation matrix should be determined from the condition
)︀−1
C𝛾 0 𝛾 𝜇 C𝛾 0 = −𝛾 *𝜇 .
(︀ (︀ )︀
(14.85)

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But in the Majorana representation it is know that the 𝛾 *𝜇 is simply

𝛾 *𝜇 = −𝛾 𝜇 , (14.86)

because all 𝛾 matrices are purely imaginary and, therefore, the (14.85) can written as
)︀−1
C𝛾 0 𝛾 𝜇 C𝛾 0 = 𝛾 𝜇 .
(︀ (︀ )︀
(14.87)

From here, it follows that

C𝛾 0 , 𝛾 𝜇 = 0 .
[︀ ]︀
(14.88)

So, this means that C𝛾 0 is the only matrix, which commutes with all 𝛾 𝜇 in four dimensions.
So, it can be gotten that
C𝛾 0 = 1 (14.89)

and from here, it can be concluded that C is simply equals to:

C = 𝛾0 . (14.90)

Then, charge conjugate spinor in this representation is given by:

𝜓 𝑐 = C𝛾 0 𝜓 * = 𝛾 0 𝛾 0 𝜓 * = 𝜓 * . (14.91)

Therefore, we got that 𝜓 * in Majorana representation of 𝛾 matrices is equal to the

charge conjugate spinor. Therefore, in this representation Majorana condition becomes
extremely simple and transparent:
𝜓 = 𝜓* , (14.92)

which means that in the Majorana representation spinor 𝜓 is simply real. It has simply
four real components and in this representation it can be also seen that 𝑆 matrix of
Lorentz transformations must preserve reality of a spinor. So, spinor 𝜓 is equal to 𝜓 * and
that means that 𝜓 is real and, therefore, 𝑆 acting on it must transform a real matrix into
real matrix. This means that in this case 𝑆 itself is real and that’s can be easily seen from
the fact that it’s given by (︂ )︂
1 𝜇 𝜈
𝑆 = exp 𝛾 𝛾 𝜔𝜇𝜈 . (14.93)
4 𝑀 𝑀
In the Majorana representation each of the 𝛾 matrices is real and then the product of
the two imaginary 𝛾 matrices then gives a real result. So, in this representation Lorentz
transformations are realized by real matrices.

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