Ph 406: Elementary Particle Physics

Problem Set 4
K.T. McDonald
kirkmcd@princeton.edu
Princeton University
Due Monday, October 13, 2014 (updated August 11, 2017)

1. The form,  
iδ θ θ θ
U=e cos I + i sin û · σ = eiδ ei 2 û·σ, (1)
2 2
of a general 2 × 2 unitary matrix [(Set 2, eq. (12)] suggests that these matrices have
something to do with rotations. Certainly, a matrix that describes the rotation of a
vector is a unitary transformation.
 2  2
   
A general 2-component (spinor) state |ψ = ψ+ |+ + ψ − |−, where ψ +  + ψ−  = 1,
can also be written as,
 
|ψ = eiδ cos θ|+ + eiφ sin θ|− . (2)

The overall phase δ has no meaning to a measurement of |ψ. So, it is tempting

to interpret parameters θ and φ as angles describing the orientation in a spherical
coordinate system (r, θ, φ) of a unit 3-vector that is associated with the state |ψ. The
state |+ might then correspond to the unit 3-vector ẑ that points up along the z-axis,
while |− ↔ −ẑ.
However, this doesn’t work! The suggestion is that the state |+ corresponds to angles
θ = 0, φ = 0 and state |− to angles θ = π, φ = 0. With this hypothesis, eq. (2) gives a
satisfactory representation of a spin-up state as |+, but it implies that the spin-down
state would be −|+ = eiπ times the spin-up state, which is not really distinct from
the spin-up state.
We fix up things be writing,
 
iδ θ θ
|ψ = e cos |+ + eiφ sin |− , (3)
2 2

and identifying angles θ and φ with the polar and azimuthal angles of a unit 3-vector
in an abstract 3-space (sometimes called the Bloch sphere). That is, we associate the
state |ψ with the unit 3-vector whose components are ψx = sin θ cos φ, ψ y = sin θ sin φ
and ψz = cos θ. Now, the associations,

spin up ↔ (θ = 0, φ = 0) ↔ |+, spin down ↔ (θ = π, φ = 0) ↔ |−, (4)

given by eq. (3) are satisfactory.

1
 We then infer from eq. (3) that the spin-up and spin-down states in the direction (θ, φ)
are, to within an overall phase factor,
⎛ ⎞ ⎛ ⎞
cos θ2 sin θ2
| + (θ, φ) ∝ ⎜
⎝
⎟
⎠, | − (θ, φ) ∝ | + (π − θ, φ + π) = ⎜
⎝
⎟
⎠. (5)
sin θ2 eiφ − cos θ2 eiφ

The standard form of the spin-up/down states is,

⎛ ⎞ ⎛ ⎞
θ −iφ/2 θ −iφ/2
⎜ cos e2 ⎟ ⎜ sin e
2 ⎟
| + (θ, φ) = ⎝ ⎠, | − (θ, φ) = ⎝ ⎠, (6)
θ iφ/2 θ iφ/2
sin e2
− cos e2

which is consistent with eq. (5), but perhaps does not obviously follow from it.

The Problem: Deduce the up and down 2-component spinor states along direction
(θ, φ) in a spherical coordinate system via rotation matrices (where first a rotation is
made by angle θ and then by angle φ).

Rotation Matrices

A general rotation in 3-space is characterized by 3 angles. We follow Euler in naming

these angles as in the figure above.1 The rotation takes the axis (x, y, z) into the axes
(x, y , z  ) in 3 steps:

(a) A rotation by angle α about the z-axis, which brings the y-axis to the y1 axis.
(b) A rotation by angle β about the y1 -axis, which brings the z-axis to the z -axis.
(c) A rotation by angle γ about the z  -axis, which brings the y1-axis to the y -axis
(and the x-axis to the x-axis).
1
From sec. 58 of Landau and Lifshitz, Quantum Mechanics.

2
 The 2 × 2 unitary matrix that corresponds to this rotation (of coordinate axes) is,
⎛ ⎞
β i(α+γ)/2 β i(−α+γ)/2
cos e sin e
R(α, β, γ) = ⎜
⎝
2 ⎟
⎠
2
β i(α−γ)/2 β −i(α+γ)/2
− sin 2 e cos 2 e
⎛ ⎞⎛ ⎞⎛ ⎞
β β iα/2
iγ/2
0 ⎠⎜ cos sin ⎟⎜ e 0 ⎟
= ⎝ e ⎝
2 2
⎠⎝ ⎠
0 e−iγ/2 − sin β
cos β
0 e −iα/2
2 2
= Rz (γ)Ry1 (β)Rz (α), (7)

where the decomposition into the product of 3 rotation matrices2 follows from the
particular rules,
⎛ ⎞
φ
⎜ cos 2 i sin φ2 ⎟
Rx (φ) = ⎝ ⎠, (8)
i sin φ2 cos φ2
⎛ ⎞
φ
⎜ cos 2 sin φ2 ⎟
Ry (φ) = ⎝ ⎠, (9)
φ φ
− sin 2
cos 2
⎛ ⎞
iφ/2
⎜ e 0 ⎟
Rz (φ) = ⎝ ⎠. (10)
−iφ/2
0 e

Convince yourself that the combined rotation (7) could also be achieved if first a
rotation is made by angle γ about the z axis, then a rotation is made by angle β about
the original y axis, and finally a rotation is made by angle α about the original z axis.
There is unfortunately little consistency among various authors as to the conventions
used to describe rotations. I follow the notation of Barenco et al.,3 who appear to write
eq. (7) simply as,
R(α, β, γ) = Rz (γ)Ry (β)Rz (α). (11)
Occasionally one needs to remember that in eq. (11) the axes of the second and third
rotations are the results of the previous rotation(s).
Note that according to eqs. (8)-(10),

σ x = σ 1 = −iRx (180◦ ), σ y = σ 2 = −iRy (180◦ ), σ z = σ 3 = −iRz (180◦ ),

(12)
and also,

σ x = iRx (−180◦ ), σ y = iRy (−180◦ ), σ z = iRz (−180◦ ), (13)

so that the Pauli spin matrices are equivalent to the formal matrices for 180◦ rotations
only up to a phase factor i.
2
The order of operations is that the rightmost rotation in eq. (7) is to be performed first.
3
http://physics.princeton.edu/~mcdonald/examples/QM/barenco_pra_52_3457_95.pdf

3
 Show that a more systematic relation between the Pauli spin matrices and the rotation
matrices is that eqs. (8)-(10) can be written as,
φ
Ru (φ) = ei 2 û·σ , (14)

which describes a rotation of the coordinate axes in Bloch space by angle φ about the
û axis (in a right-handed convention).
Rather than rotating the coordinate axes, we may wish to rotate vectors in
Bloch space by an angle φ about a given axis û, while leaving the coordinate
axes fixed. The operator,
φ
Ru (−φ) = e−i 2 û·σ (15)

performs this type of rotation. With this in mind, you can finally solve the main
problem posed on p. 2.

2. Helicity Conservation in High-Energy Electromagnetic Interactions of point-

like spin-1/2 particles.
Recalling pp. 86 and 88 of Lecture 6 of the Notes, general (spin-1/2) particle 4-spinors
u for plane-wave states,
μ
ψ = u e−ipx = u e−ipμx , (16)
√ 2
with rest mass m, 3-momentum p and energy E = p + m2 , can be written as,
⎛ ⎞ ⎛ √ ⎞ ⎛ √ ⎞
√ ⎜ χ ⎟ ⎜ E + mχ ⎟ ⎜ E + mχ ⎟
u= E + m⎝ ⎠ =⎝ ⎠ =⎝ √ ⎠, (17)
p·σ √ p
E+m
χ E+m
p̂ · σ χ E + m p̂ · σ χ

where the 2-spinor χ obeys χ†χ = 1. Similarly, antiparticle 4-spinors v are associated
with plane-wave states,4,5
ψ̃ = v eipx , (18)
(note the sign change with respect to the form (16)), that can be written as,
⎛ ⎞ ⎛ √ ⎞ ⎛⎞
p·σ √ p
√ ⎜ E+m
χ̃ ⎟ ⎜ p̂ · σ χ̃ ⎟ ⎜ E − m p̂ · σ χ̃ ⎟
E+m
v= E + m⎝ ⎠=⎝ √ ⎠=⎝ √ ⎠, (19)
χ̃ E + m χ̃ E + m χ̃

where χ̃ is a 2-spinor with χ̃† χ̃ = 1.

These states obey the Dirac equations i∂ μ γ μ ψ = p/ψ = mψ and i∂ μγ μ ψ̃ = −/pψ̃ = mψ̃,
which imply the 4-spinor equations p/u = mu and −/pv = mv.
4
The antiparticle of particle a is often denoted as ā, but as ū is the adjoint of a Dirac 4-spinor u, we
write ã for the antiparticle of state a.
5
Dirac interpreted his negative-energy solutions as related to “anti-electrons” on p. 52 of Quantised
Singularities in the Electromagnetic Field, Proc. Roy. Soc. London A 133, 60 (1931),
http://physics.princeton.edu/~mcdonald/examples/QED/dirac_prsla_133_60_31.pdf.
That paper is also noteworthy for relating the possible existence of a magnetic monopole of pole strength p
to the electric charge e by ep = h̄/2.

4
 The positive and negative helicity spinor states for a particle with 3-momentum p
in direction (θ, φ) are χ+ = | + (θ, φ) and χ− = | − (θ, φ), respectively, recalling
eq. (6), while the helicity states of an antiparticle are χ̃+ = | − (θ, φ) = χ− and
χ̃− = −| + (θ, φ) = −χ+ . In all cases, positive helicity means spin in the direction of
momentum p.
In the high-energy limit, these 4-spinors simplify to,
⎛ ⎞ ⎛ ⎞
√ ⎜ χ ⎟ √ ⎜ p̂ · σ χ̃ ⎟
u→ E⎝ ⎠, v → E⎝ ⎠, (20)
p̂ · σ χ χ̃

Give explicit forms of the helicity spinors u+ (θ, φ), u− (θ, φ), v+ (θ, φ) and v− (θ, φ)for
(anti)particles moving and at angles (θ, φ) to the +z-axis, and also their simplification
to u+ (0), u− (0), v+ (0) and v− (0) for motion along the z-axis in the high-energy limit.
If these are pointlike particles of charge e, their electromagnetic interaction is described
by the 4-current jμ = e γ μ . Verify that the matrix elements ū− (θ)|γ μ |u+ (0) vanish
for μ = 0, 1, 2, 3, and similarly that v̄+ (θ)|γ μ |u+ (0) = 0. Remember that v̄ = v † γ 0,
etc.

Digression: Electric Charge Conjugation. The above claim that the antiparticle
helicity 2-spinors χ̃± are related to the particle helicity 2-spinors χ± by χ̃± = ±χ∓
can be justified by considerations of a transformation, called electric charge conjugation
with symbol C, between particles and their antiparticles (with respect to their elec-
tromagnetic interactions), such that ψ̃ = Cψ is the antiparticle state of a spin-1/2
particle ψ.6
6
That ψ̃ = Cψ and not ψ̃ = Cψ follows from the sign change in the spacetime waveform between
eqs. (16) and (18).
Charge conjugation leaves mass unchanged, such that a particle and its antiparticle have the same rest
mass m. This was not initially understood by Dirac, who first speculated that the antiparticle of an electron
is a proton, A Theory of Electrons and Protons, Proc. Roy. Soc. London A 126, 360 (1930),
http://physics.princeton.edu/~mcdonald/examples/QED/dirac_prsla_126_360_30.pdf.
The charge-conjugation operator C was discussed (in a different representation, and not given a name) on
p. 130 of W. Pauli, Contributions mathématique à la théorie des matrices de Dirac, Ann. Inst. H. Poincaré
6, 109 (1936), http://physics.princeton.edu/~mcdonald/examples/QED/pauli_aihp_6_109_36.pdf.
The term “charge conjugation” (but with the symbol L) may have been first used in H.A. Kramers, The use
of charge conjugated wavefunctions in the hole theory of the electron, Proc. Roy. Neder. Acad. Sci. 40, 814
(1937), http://physics.princeton.edu/~mcdonald/examples/neutrinos/kramers_pknaw_40_814_37.pdf.
The term antimatter was introduced by Schuster in 1898, but in his vision antimatter had negative mass;
Potential Matter—A Holiday Dream, Nature 58, 367, 618 (1898),
http://physics.princeton.edu/~mcdonald/examples/GR/schuster_nature_58_367_98.pdf
http://physics.princeton.edu/~mcdonald/examples/GR/schuster_nature_58_618_98.pdf.
The present vision of antiparticles via electric charge conjugation of particles is perhaps closer to Kelvin’s
image method for a planar conductor, p. 288 of W. Thomson, Effects of Electrical Influence on Internal
Spherical and on Plane Conducting Surfaces, Camb. Dublin Math. J. 4, 276 (1849),
http://physics.princeton.edu/~mcdonald/examples/EM/thomson_cdmj_4_276_49.pdf.

5
 One way to do this starts with the Dirac equation for a spin-1/2 particle state ψ,7

i∂ μγ μ ψ = mψ. (21)

We expect that the antiparticle state ψ̃ also satisfies the Dirac equation,

i∂ μγ μ ψ̃ = mψ̃. (22)

A clever step is to take the complex conjugate of eq. (21),

− i∂ μγ ∗μ ψ∗ = mψ ∗. (23)

Applying the desired charge-conjugation operator C to this, we have,

− i∂ μ Cγ ∗μ ψ∗ = mCψ∗ = mψ̃. (24)

For this to be the Dirac equation (22),8 we require that,

− Cγ ∗μ = γ μ C. (25)

You can verify that this implies the electric-charge-conjugation matrix operator to be,9
⎛ ⎞
⎜ 0 0 0 1 ⎟
⎛ ⎞
⎜ ⎟
⎜
⎜ 0 0 −1 0 ⎟
⎟ ⎜ 0 iσ2 ⎟
C = iγ 2 = ⎜
⎜
⎟
⎟ =⎝ ⎠. (26)
⎜ 0 −1 0 0 ⎟ −iσ2 0
⎜ ⎟
⎝ ⎠
1 0 0 0

Then, applying the electric-charge-conjugation transformation √ to the particle 4-spinor

u of eq. (17), we obtain (on suppression of the overall factor E + m) the antiparticle
spinor,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
p·σ p·σ p·σ
⎜ χ ⎟ ⎜ iσ2 E+m χ ⎟ ⎜ E+m
(−iσ2χ ) ⎟ ⎜ E+m
χ̃ ⎟
ũ = iγ 2 ⎝ ⎠ =⎝ ⎠ =⎝ ⎠ =⎝ ⎠ = v, (27)
p·σ
E+m
χ −iσ2 χ −iσ2χ χ̃

using that fact (verify it!) that σ 2σ ∗ = −σσ2 . Hence, the antiparticle 2-spinor χ̃ is
related to its corresponding particle 2-spinor χ by,

χ̃ = −iσ2 χ , χ = iσ2 χ̃∗ . (28)

7
This argument follows sec. 5.4, p. 107 of F. Halzen and A.D. Martin, Quarks and Leptons (Wiley, 1984),
http://physics.princeton.edu/~mcdonald/examples/EP/halzen_martin_84.pdf.
8
For ψ̃ = v eipx , eqs. (24)-(25) lead to the spinor form of the Dirac equation for antiparticles, −/pv = mv.
9
Warning: Many people write Cγ 0 for the matrix C of eq. (26).

6
 In particular, the helicity 2-spinors of eq. (6) transform under electric-charge conjuga-
tion as,
⎛ ⎞ ⎛ ⎞
θ −iφ/2 θ −iφ/2
⎜ cos e
2 ⎟ ⎜ − sin e 2 ⎟
χ+ = ⎝ ⎠ → χ̃+ = −iσ2 χ∗+ = ⎝ ⎠ = χ− , (29)
θ iφ/2 θ iφ/2
sin e 2
cos e2
⎛ ⎞ ⎛ ⎞
θ −iφ/2 θ −iφ/2
⎜ − sin e2 ⎟ ⎜ − cos e 2 ⎟
χ− = ⎝ ⎠ → χ̃− = −iσ2 χ∗− = ⎝ ⎠ = −χ+ , (30)
θ iφ/2 θ iφ/2
cos e
2
− sin e 2

as claimed above.
3. The cross section for inelastic scattering of electrons off some target can be expressed
in terms of two generalized structure functions W1,2(q 2, ν) where q = pei − pef and
ν = q0 = Ei − Ef , as on p. 131, Lecture 8 of the Notes. If the inelastic scattering is
due to the interaction of the virtual photon emitted by the incident electron with a
spin-1/2, charge Q, mass m constituent of the target, such that the rest of the target is
a “spectator” to this interaction, then the cross section is that given on p. 99, Lecture
6 of the Notes, and we infer that,10
   
−q 2 2 q2 2 2q2
W1 (q ∗ 2, ν) = Q δ ν + , W2 (q , ν) = Q δ ν + . (31)
4m2 2m 2m
An argument of Bjorken11 is that the lab-frame energy difference between the initial
and final electron can be written as,
qP
Ei − Ef = ν = q0 = , (32)
M
where P is the energy-momentum 4-vector of the target (of rest mass M), which is
just P = (M, 0, 0, 0) in the lab frame. Then, in a frame in which the target has very
high momentum, the 4-vector p of a constituent which carries (scalar) fraction x of the
target’s 3-momentum can be written approximately as p ≈ xP . A consequence of this
approximation is that the constituent mass m is related by m2 = p2 ≈ x2 P 2 = x2M 2 ,
i.e., that m ≈ xM (as appropriate for consideration of very high-energy scattering).
This permits us to rewrite eq. (31) as12
   
−q 2 2 q2 2 q2
W1 = Q δ ν + , W2 = Q δ ν + . (33)
4M 2 x2 2Mx 2Mx
Supposing the constituents are distributed with the target (as viewed from a frame
in which the target has high speed) with probability f(x) dx, give expressions for the
generalized structure functions W1 and W2 in terms of a single variable x.
10
C.G. Callan, Jr and D.J. Gross, High-Energy Electroproduction and the Constitution of the Electric
Current, Phys. Rev. Lett. 22, 156 (1969),
http://physics.princeton.edu/~mcdonald/examples/EP/callan_prl_22_156_69.pdf.
11
J.D. Bjorken and E.A. Paschos, Inelastic Electron-Proton and γ-Proton Scattering and the Structure
of the Nucleon, Phys. Rev. 185, 1975 (1969),
http://physics.princeton.edu/~mcdonald/examples/EP/bjorken_pr_185_1975_69.pdf.
12
A different version of this argument is given on p. 139, Lecture 8 of the Notes, where a Breit frame is
used.

7
Solutions
1. Rotation and Pauli Spin Matrices.
The rotations (8)-(10) are readily seen to be exponentials of the Pauli matrices,
⎛ ⎞

Rx (φ) = ⎝ cos φ2 i sin φ2 ⎠

φ φ φ
= cos I + i sin σ x = ei 2 σx , (34)
i sin φ2 cos φ2 2 2
⎛ ⎞
φ φ φ φ φ
Ry (φ) = ⎝ cos φ sin 2 2
φ
⎠ = cos I + i sin σ y = ei 2 σy , (35)
− sin 2 cos 2
2 2
⎛ ⎞
iφ/2 φ φ φ
Rz (φ) = ⎝ e 0
−iφ/2
⎠ = cos I + i sin σ z = ei 2 σz , (36)
0 e 2 2

recalling eq. (1).

Digression: Pauli Spin Matrices and Rotations.
The NOT operation, X = σ x , that “flips” a bit can be interpreted as a rotation by 180◦
of the Bloch-sphere state vector about the x-axis. Thus,
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
α α iβ α
σx ⎝ cos 2 ⎠ = ⎝ 0 1 ⎠ ⎝ cos ⎠ = ⎝ e sin 2 2 ⎠, (37)
eiβ sin α2 1 0 eiβ sin α2 cos α2

while a rotation Rx (180◦ ) by 180◦ about the x-axis in our abstract spherical coordinate
system takes α to π − α and β to −β,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
α
Rx(180◦ ) ⎝ cos ⎠=⎝2
cos π−α
2 ⎠
iβ
= e−iβ ⎝ e sinα
α
2 ⎠. (38)
eiβ sin α2 e−iβ sin π−α
2
cos 2

Since the overall phase of a state does not affect its meaning, our prescription can be
considered satisfactory thus far.
Can we interpret the operation σ y as a rotation by 180◦ about the y-axis? On one
hand, ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
α α iβ α
cos
σ y ⎝ iβ 2 α ⎠ = ⎝ 0 −i ⎠⎝ cos 2 ⎠=⎝ −ie sin 2 ⎠, (39)
e sin 2 i 0 eiβ sin α2 i cos α2
while a rotation Ry (180◦ ) by 180◦ about the y-axis in our abstract spherical coordinate
system takes α to π − α and β to π − β,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
α
Ry (180◦ ) ⎝ cos ⎠=⎝
2
cos π−α
2 ⎠ = ie−iβ ⎝ −ie sin
iβ α
2 ⎠. (40)
e sin α2
iβ
ei(π−β) sin π−α
2
i cos α2

Similarly, we interpret the operation σ z as a rotation by 180◦ about the z-axis:

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
α α α
σz ⎝ cos 2 ⎠ = ⎝ 1 0 ⎠ ⎝ cos ⎠=⎝ cos ⎠, 2 2 (41)
eiβ sin α2 0 −1 eiβ sin α2 −eiβ sin α2

8
while a rotation Rz (180◦ ) by 180◦ about the z-axis in our abstract spherical coordinate
system takes α to α and β to π + β,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
α α α
Rz (180◦ ) ⎝ cos
2 ⎠=⎝ cos 2 ⎠=⎝ cos ⎠. 2 (42)
eiβ sin α2 ei(π+β) sin α2 −eiβ sin α2

Solution to the Main Problem.

We desire to rotate states in Bloch space, rather than the coordinate axes thereof, so
we must heed eq. (15).
Referring to the figure on p. 3 (and reproduced below), rotation of the z-axis to the
direction (θ, φ), keeping the x-axis in the original x-y plane, could be accomplished
with rotation angles α = 0, β = θ, γ = φ in the general rotation (7). Hence, rotation
of a Bloch vector along the z-axis to one along the direction (θ, φ) can be accomplished
by the rotation operator,

⎛ ⎞
θ −iφ/2
⎜ cos 2 e − sin 2θ e−iφ/2 ⎟
R(0, −θ, −φ) = ⎝ ⎠. (43)
sin 2θ eiφ/2 cos 2θ eiφ/2

Then,
⎛ ⎞ ⎛ ⎞
θ −iφ/2
⎜ 1 ⎟ ⎜ cos 2 e ⎟
| + (θ, φ) = R(0, −θ, −φ) ⎝ ⎠ =⎝ ⎠, (44)
0 sin θ2 eiφ/2
⎛ ⎞ ⎛ ⎞
θ −iφ/2
⎜ 0 ⎟ ⎜ − sin 2 e ⎟
| − (θ, φ) = R(0, −θ, −φ) ⎝ ⎠ =⎝ ⎠, (45)
1 cos θ2 eiφ/2

as on p. 113, Lecture 7 of the Notes, and in agreement with eq. (6).

9
2. From eq. (6), Prob. 1, the helicity 2-spinors for a particle are,
⎛ ⎞ ⎛ ⎞
θ −iφ/2 θ −iφ/2
⎜ cos e2 ⎟ ⎜ − sin e
2 ⎟
χ+ = | + (θ, φ) = ⎝ ⎠, χ− = | − (θ, φ) = ⎝ ⎠. (46)
θ iφ/2 θ iφ/2
sin e
2
cos e
2

The operator p̂(θ, φ) · σ has the form,

⎛ ⎞
⎜ cos θ sin θ e−iφ ⎟
p̂(θ, φ) · σ = ⎝ ⎠, such that p̂(θ, φ) · σ χ± = ±χ± . (47)
sin θ eiφ − cos θ

Digression: Helicity Projection Operator for 2-Spinors. If a general 2-spinor is

written as χ = a+ χ+ + a− χ− , in terms of helicity spinors for the (θ, φ) direction, then
[I ± p̂(θ, φ) · σ]χ = 2a± χ± . Hence,
I ± p̂(θ, φ) · σ
(48)
2
are 2-spinor helicity projection operators for the direction (θ, φ).
The particle and antiparticle helicity
√ 4-spinors
√ u± and
√ v± are,
√ recalling eqs.
√(17)-(19),
with χ̃± = ±χ∓ , and defining E+ = E + m and E− = E − m = p/ E + m,
⎛ √ ⎞
⎜ E+ cos θ2 e−iφ/2 ⎟
⎛ √ ⎞ ⎛ ⎞ ⎜ ⎟
√ ⎜ √
⎜ E + m χ+ ⎟ ⎜ E χ
+ + ⎟ ⎜ E+ sin θ2 eiφ/2 ⎟⎟
u+ (θ, φ) = ⎝ ⎠=⎝ √ ⎠ = ⎜
⎜ √
⎟ , (49)
⎟
√ p p̂ · σ χ+ E− χ+ ⎜ θ −iφ/2 ⎟
E− cos 2 e
E+m ⎜ ⎟
⎝ √ ⎠
E− sin θ2 eiφ/2
⎛ √ ⎞
θ −iφ/2
⎜ − E+ sin 2 e ⎟
⎛ √ ⎞ ⎛ ⎞ ⎜ ⎟
√ ⎜ √ θ iφ/2 ⎟
⎜ E + m χ− ⎟ ⎜ E+ χ− ⎟ ⎜ E+ cos 2 e ⎟
u− (θ, φ) = ⎝ ⎠ =⎝ ⎠ = ⎜ ⎟ (, 50)
√ ⎜ √ ⎟
√ p p̂ · σ χ− − E− χ− ⎜ E− sin θ2 e−iφ/2 ⎟
E+m ⎜ ⎟
⎝ √ ⎠
− E− cos θ2 eiφ/2
⎛ √ ⎞
⎜ E− sin θ2 e−iφ/2 ⎟
⎛ ⎞ ⎛ ⎞ ⎜ ⎟
√ ⎜ √
⎜
√ p
p̂ · σ χ̃+
E+m ⎟ ⎜ − E− χ− ⎟ ⎜ − E− cos θ2 eiφ/2 ⎟ ⎟
v+ (θ, φ) = ⎝ √ ⎠=⎝ √ ⎠ = ⎜
⎜ √
⎟ (, 51)
⎟
E + m χ̃+ E+ χ− ⎜ − E+ sin 2 eθ −iφ/2 ⎟
⎜ ⎟
⎝ √ ⎠
θ iφ/2
E+ cos 2 e
⎛ √ ⎞
⎜ − E− cos θ2 e−iφ/2 ⎟
⎛ ⎞ ⎛ ⎞ ⎜ ⎟
√ ⎜ √
⎜
√ p
p̂ · σ χ̃− ⎟ ⎜ − E χ
− + ⎟ ⎜ − E− sin θ2 eiφ/2 ⎟ ⎟
⎠ = ⎜ ⎟(52)
E+m
v− (θ, φ) = ⎝ √ ⎠=⎝ √ ⎜ √ ⎟.
E + m χ̃− − E+ χ+ ⎜ − E+ cos 2 e θ −iφ/2 ⎟
⎜ ⎟
⎝ √ ⎠
− E+ sin θ2 eiφ/2

10
Note that v± = Cu∗±, using the electric-charge-conjugation operator C = iγ 2 found in
eq. (26).
In case of high-speed motion (E+ ≈ E− ≈ E) along the +z-axis the 4-spinors are,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ 1 ⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ −1 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ 0 ⎟ ⎜ 1 ⎟ ⎜ −1 ⎟ ⎜ 0 ⎟
u+ (0) → ⎜
⎜
⎟,
⎟ u− (0) → ⎜
⎜
⎟,
⎟ v+ (0) → ⎜
⎜
⎟,
⎟ v− (0) → ⎜
⎜
⎟.
⎟ (53)
⎜ 1 ⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ −1 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 −1 1 0

Recalling that,
⎛ ⎞ ⎛ ⎞
⎜ 1 0 0 0 ⎟ ⎜ 0 0 0 1 ⎟
⎜ ⎟ ⎜ ⎟
⎜
⎜ 0 1 0 0 ⎟
⎟
⎜
⎜ 0 0 1 0 ⎟
⎟
γ0 = ⎜
⎜
⎟,
⎟ γ1 = ⎜
⎜
⎟,
⎟
⎜ 0 0 −1 0 ⎟ ⎜ 0 −1 0 0 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0 0 0 −1 −1 0 0 0
⎛ ⎞ ⎛ ⎞
⎜ 0 0 0 −i ⎟ ⎜ 0 0 1 0 ⎟
⎜ ⎟ ⎜ ⎟
⎜
⎜ 0 0 i 0 ⎟⎟
⎜
⎜ 0 0 0 −1 ⎟
⎟
γ2 = ⎜
⎜
⎟,
⎟ γ3 = ⎜
⎜
⎟,
⎟ (54)
⎜ 0 i 0 0 ⎟ ⎜ −1 0 0 0 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
−i 0 0 0 0 1 0 0

we have that,
⎛ ⎞ ⎛ ⎞
⎜ 1 ⎟ ⎜ 0 ⎟
⎜ ⎟ ⎜ ⎟
⎜
⎜ 0 ⎟
⎟
⎜
⎜ 1 ⎟
⎟
γ 0u+ (0) = ⎜ ⎟, γ 1u+ (0) = ⎜ ⎟,
⎜ ⎟ ⎜ ⎟
⎜ −1 ⎟ ⎜ 0 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0 −1
⎛ ⎞ ⎛ ⎞
⎜ 0 ⎟ ⎜ 1 ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ i ⎟ ⎜ 0 ⎟
γ 2u+ (0) = ⎜
⎜
⎟,
⎟ γ 3 u+ (0) = ⎜
⎜
⎟.
⎟ (55)
⎜ 0 ⎟ ⎜ −1 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
−i 0

To evaluate matrix elements such as ūf |γ μ |ui  we recall that this equals u†f γ 0 γ μ ui , so

11
we multiply eq. (55) by γ 0 to obtain,
⎛ ⎞ ⎛ ⎞
⎜ 1 ⎟ ⎜ 0 ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ 0 ⎟ ⎜ 1 ⎟
γ 0γ 0 u+ (0) = ⎜
⎜
⎟,
⎟ γ 0γ 1 u+ (0) = ⎜
⎜
⎟,
⎟
⎜ 1 ⎟ ⎜ 0 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0 1
⎛ ⎞ ⎛ ⎞
⎜ 0 ⎟ ⎜ 1 ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ i ⎟ ⎜ 0 ⎟
γ 0 γ 2u+ (0) = ⎜
⎜
⎟,
⎟ γ 0 γ 3u+ (0) = ⎜
⎜
⎟.
⎟ (56)
⎜ 0 ⎟ ⎜ 1 ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
i 0

Now, we can use eqs. (50)-(52) to see that,

ū− (θ, φ)|γ μ |u+ (0) = u†− (θ, φ)γ 0 γ μ u+ (0) = 0 = v̄+ (θ, φ)|γ μ |u+ (0). (57)
Hence, a high-energy pointlike spin-1/2 particle of a given helicity cannot couple a
particle of the opposite helicity via the electromagnetic interaction, nor can it annhilate
with an antiparticle of the same helicity. It is possible for a high-energy spin-1/2
particle of a given helicity to scatter into a particle of the same helicity, or annihilate
with an antiparticle of opposite helicity, via single-photon emission.
Examples where helicity conservation in the high-energy limit is useful in providing
a simplified understanding include e+ e− annihilation into a pair of spin-0 or spin-1/2
particles, as well as elastic scattering of electrons off spin-0 and spin-1/2 particles, as
discussed on p. 118 ff, Lecture 7 of the Notes.

Digression: Orthogonality. If we label the four components of a general spinor ψ

as ψi , i = 1, 4, then,
φ† ψ = φ∗1 φ1 + φ∗2 φ2 + φ∗3 φ3 + φ∗4φ4 , (58)
while,
φ̄ψ = φ† γ 0 ψ = φ∗1 φ1 + φ∗2 φ2 − φ∗3 φ3 − φ∗4 φ4 . (59)
Then, from eqs. (49)-(52),
u†+ u+ = u†− u− = v+
† †
v+ = v− v− = 2E, (60)
† † † †
u+ u− = u+ v + = u− v − = v + v− = 0, (61)
u+ v− = −2p = −u†− v+ ,
†
(62)
while,
ū+ u+ = ū− u− = v̄+ v+ = v̄− v− = 2m, (63)
ū+ u− = ū+ v+ = ū− v− = v̄+ v− = ū+ v− = −ū− v+ = 0. (64)

12
 That is, the u and v spinors are orthogonal with respect to the scalar product φ̄ψ, but
not with respect to φ† ψ.

Digression: Helicity Projection Operator for 4-Spinors. The generalization to

4-spinors of the 2-spinor helicity projection operators (48) is,13
⎛ ⎞
1 ⎜ I ± p̂(θ, φ) · σ 0 ⎟ 1 ± γ 0 p̂(θ, φ) · γ γ 5
⎝ ⎠= , (65)
2 0 I ± p̂(θ, φ) · σ 2

recalling from eq. (68) of Set 3 that,

⎛ ⎞
⎜ σi 0 ⎟
γ 0γ iγ 5 = ⎝ ⎠. (66)
0 σi
Then, recalling eqs. (49)-(52), the operator 1 + γ 0 p̂(θ, φ) · γ γ 5 /2 projects positive
helicity for particles, but negative helicity for antiparticles, while 1 − γ 0 p̂(θ, φ) · γ γ 5 /2
projects negative helicity for particles, but positive helicity for antiparticles,
Note that γ μ γ 0p̂ · γ γ 5 = γ 0 p̂ · γ γ 5 γ μ , which has the consequence that the helicity
4-spinor states ψ± satisfy the Dirac equation, i∂ μ γ μ ψ± = m ψ± ,
1 ± γ 0 p̂ · γ γ 5 μ 1 ± γ 0 p̂ · γ γ 5 1 ± γ 0 p̂ · γ γ 5
i∂ γ μ ψ = i∂ μγ μ ψ = i∂ μ γ μ ψ ± = m ψ = m ψ ± .(67)
2 2 2

Digression: Chirality vs. Helicity States.

On p. 116, Lecture 7 of the Notes, it was argued that in the high-energy limit we can
consider a different form of the helicity projection operator, more properly called the
chirality projection operator,
⎛ ⎞
1 ± γ5 1 ⎜ I ±I ⎟
= ⎝ ⎠, (68)
2 2 ±I I

which leads, for momentum p = p ẑ along the z-axis, to,

⎛ ⎞ ⎛ ⎞
 
1 + γ5 χ± p ⎜ χ± ⎟
√ u± = (1 + γ 5 ) ⎜
⎝
⎟
⎠= 1± ⎝ ⎠ (69)
E+m p E+m
± E+m χ± χ±
⎛ ⎞ ⎛ ⎞
 
1 − γ5 ⎜ ⎟ χ± p ⎜ χ± ⎟
√ u± = (1 − γ 5) ⎝ ⎠= 1∓ ⎝ ⎠ (70)
E+m p E+m
± E+m χ± −χ±
⎛ ⎞ ⎛ ⎞
p  
1 + γ5 ⎜ − E+m χ∓ ⎟ p ⎜ ±χ∓ ⎟
√ v± = (1 + γ 5) ⎝ ⎠ = 1∓ ⎝ ⎠ (71)
E+m ±χ∓ E+m ±χ∓
⎛ ⎞ ⎛ ⎞
p  
1 − γ5 ⎜ − E+m χ∓ ⎟ p ⎜ ∓χ∓ ⎟
√ v± = (1 − γ 5 ) ⎝ ⎠= 1± ⎝ ⎠ (72)
E+m ±χ∓ E+m ±χ∓
13
We now follow the usual convention in writing the unit 4 × 4 matrix as 1.

13
Only in the high-energy limit do the chirality and the helicity projection operators
produce that same results.
The result of the positive-(negative-)chirality operator on a particle 4-spinor is called
a righthanded-(lefthanded-)chirality spinor, and conversely for antiparticles,
1 + γ5 1 − γ5 1 + γ5 1 − γ5
u ≡ uR , u ≡ uL , v ≡ vL , v ≡ vR , (73)
2 2 2 2
Then, eq. (70) reminds us the a lefthanded-chirality particle is not precisely a negative-
helicity state, etc. That is, uL = (1 − γ 5 )u/2 contains a positive-helicity component
of amplitude (E + m − p)/(E + m + p) ≈ m/2E relative to the nominal negative-
helicity component. A famous application of this in the decays π → μν μ vs. π → eν e
is discussed on p. 293, Lecture 16 of the Notes. See also Set 9, Prob. 1b.
Similarly, a righthanded-chirality antiparticle state, (1−γ 5 )v/2, is nominally a positive-
helicity state (with 2-spinor χ̃+ = χ− ), but has a negative-helicity component of am-
plitude ≈ m/E.
Suppose that the antiparticle of state u is v = ũ. We can decompose u and v into chi-
rality states, u = uR + uL , v = vR + vL . Then, recalling the electric-charge-conjugation
operator (26) and that γ 2 γ 5 = −γ 5γ 2 ,
1 + γ5 ∗ 1 − γ5 1 − γ5 1 − γ5
ũR = iγ 2u∗R = iγ 2 u = (iγ 2 u∗) = ũ = v = vR . (74)
2 2 2 2
Similarly, ũL = vL. Note that the antiparticle of uR (in the sense of electric-charge
conjugation) is vR and not vL .
A peculiarity is that right- and lefthanded-chirality states uR,L (and vR,L) do not strictly
satisfy the Dirac equation i∂ μ γ μ u = m u, but rather,
i∂ μγ μ uR,L = m uL,R, (75)
(and similarly i∂ μγ μ vR,L = m vR,L), since γ 5 γ μ = −γ μ γ 5 :
1 ± γ5 μ 1 ∓ γ5 1 ± γ5
i∂ γ μ u = i∂ μγ μ u = i∂ μγ μ uL,R = mu = m uR,L . (76)
2 2 2

Digression. Helicity Conservation when Chirality Approximates Helicity.

For relativistic spin-1/2 particles, with E m, their chilarity and helicity states are
essentially identical, as noted in eqs. (69)-(72). Then, for example, a matrix element
between helicity states such as eq. (57), ū− |γ μ |u+  is well approximated by the matrix
element of chirality states,
ū− |γ μ |u+  ≈ ūL |γ μ |uR
1 1
= [(1 − γ 5 )u]†γ 0|γ μ |(1 + γ 5 )u = u†(1 − γ 5)γ 0 |γ μ |(1 + γ 5)u
4 4
1 † 1
= u γ 0|(1 + γ 5)γ μ (1 + γ 5 )|u = ū|γ μ (1 − γ 5 )(1 + γ 5 )|u
4 4
= 0, (77)

14
recalling that γ 5γ μ = −γ μ γ 5 and γ 25 = 1.
That is, only helicity-conserving matrix elements of the operator γ μ are nonzero for
relativistic spin-1/2 states.

Digression: Orthogonality of Chirality States.

For general particle and antiparticle 4-spinors,
⎛ ⎞ ⎛ ⎞
σ·p
√ ⎜ χ ⎟ √ ⎜ E+m
χ̃ ⎟
u= E + m⎝ ⎠, v= E + m⎝ ⎠, (78)
σ·p
E+m
χ χ̃

with 2-spinors χ and χ̃ = −iσ2χ∗ (from eq. (28)) that obey χ† χ = 1 = χ̃† χ̃, we have
that u† u = 2E = v †v and ūu = 2m = −v̄v. The the right- and lefthanded 4-spinors of
eqs. (73) and (78) are then,
⎛   ⎞ ⎛   ⎞
√ σ·p √ σ·p
E +m⎜ 1 ± E+m
χ ⎟ E+m⎜ E+m
∓ 1 χ̃ ⎟
uR,L = ⎝   ⎠, vR,L = ⎝   ⎠, (79)
2 σ·p 2 σ·p
E+m
±1 χ 1∓ E+m
χ̃

These 4-spinors have normalizations,

u†RuR = E + χ† σ · p χ, u†LuL = E − χ† σ · p χ, (80)

vR† vR = E − χ̃† σ · p χ̃, vL† vL = E + χ̃†σ · p χ̃, (81)

while,

ūRuR = ūL uL = v̄RvR = v̄L vL = 0. (82)

They also satisfy the relations,

u†RuL = u†RvR = u†L vL = vR† vL = 0, (83)

u†R vL = χ†(E + σ · p) χ̃, u†L vR = χ† (E − σ · p) χ̃, (84)

while,

ūR vL = ūL vR = 0, ūR uL = −v̄RvL = m, −ūR vR = ūL vL = m χ† χ̃. (85)

Digression: Antiparticles of Chirality States. Also note that the antiparticles

(in the sense of electric-charge conjugation) of the chirality states (79) are, recalling
that σ2 σ ∗ = −σσ2 ,
⎛ ⎞⎛   ⎞
√ σ∗ ·p ∗
E+ ⎜ 0 iσ2 ⎟ ⎜ 1 ± E+m χ ⎟
ũR,L = iγ 2 u∗R,L = ⎝ ⎠⎝  ∗  ⎠
2 σ ·p ∗
−iσ2 0 E+m
± 1 χ
⎛   ⎞
√ σ·p ∗
E+ ⎜ E+m
∓ 1 (−iσ2 χ ) ⎟
= ⎝   ⎠ = vR,L, ṽR,L = uR,L. (86)
2 σ·p
1 ∓ E+m (−iσ2 χ∗ )

15
 Digression: Two-Component Theory of Massless Fermions.
Until relatively recently, experimental evidence was consistent with neutrinos being
massless. The character of Dirac 4-spinors for massless spin-1/2 states was considered
by Weyl,14 who formulated a “two component” theory.
For example, the four helicity states (49)-(52) reduce to two independent state when
p
m = 0, since then v± = −u∓ . Also, when m = 0 then E+m = 1, and eqs. (69)-(72)
indicate that the helicity spinors and the chirality spinors are identical, uR = u+ =
−v− = −vL and uL = u− = −v+ = −vR.
As discussed in Lecture 16 of the Notes, in the so-called V -A theory, only lefthanded
particle (righthanded antiparticle) states participate in the weak interaction. Since the
neutrino has no strong or electromagnetic interaction (presuming that the neutrino has
no magnetic moment as well as no electric charge), then a righthanded neutrino (left-
handed antineutrino) would have no interactions (except gravity) and could be called
sterile.15 While a massless, sterile neutrino is a somewhat trivial concept, the possi-
bility of a sterile neutrino with mass has led to considerable discussion/controversy,
despite lack of clear experimental evidence for such a particle.16

3. Convoluting the generalized structure functions (31) with a longitudinal-momentum

distribution f(x) of constituents of a target particle of mass M, we have,
 1  
  2 q2 Q2f(x) Q2xf(x)
W2 (x) = f(x ) dx Q δ ν + = = , (87)
0 2Mx −q 2/2Mx2 ν

where we recall that,

 
dy f(x)
f(x) dx δ[y(x)] = f(x) δ(y) = , for x = y −1 (0). (88)
y  (x) g  (x)

so that in the present case,

q2
ν=− . (89)
2Mx
The result (87) is usually recast as,

νW2 (x) = Q2 xf(x) ≡ F2(x). (90)

14
H. Weyl, Elektron und Gravitation. I, Z. Phys. 56, 330 (1929),
http://physics.princeton.edu/~mcdonald/examples/EP/weyl_zp_56_330_29.pdf.
Gravitation and the Electron, Proc. Nat. Acad. Sci. 15, 323 (1929),
http://physics.princeton.edu/~mcdonald/examples/GR/weyl_pnas_15_323_29.pdf.
15
The notion of a sterile neutrino seems to have been introduced on p. 986 of B. Pontecorvo, Neutrino
Experiments and the Problem of Conservation of Leptonic Charge, Sov. Phys. JETP 26, 984 (1968),
http://physics.princeton.edu/~mcdonald/examples/neutrinos/pontecorvo_spjetp_26_984_68.pdf .
16
A recent experimental limit on the existence of sterile neutrinos is P. Adamson et al., Limits on Ac-
tive to Sterile Neutrino Oscillations from Disappearance Searches in the MINOS, Daya Bay, and Bugey-3
Experiments, https://arxiv.org/abs/1607.01177.

16
Similarly,
 1  
  −q 2 2 q2 −q 2 Q2 f(x) Q2f(x)
W1(x) = f(x ) dx Q δ ν+ = = , (91)
0 4M 2 x2 2Mx 4M 2 x2 −q 2/2Mx2 2M

which is usually recast as,

2MW1 (x) = Q2f(x) ≡ 2F1(x). (92)

17