Quantum Electrodynamics

PH 333 Supervised Learning Project

Gopal Parwani
Department of Physics
Under the Supervision of
Prof. Asmita Mukherjee
Department of Physics
Indian Institute of Technology Bombay
7 May, 2024
Presentation Outline 2

Non-relativistic Quantum Mechanics

Relativistic Quantum Mechanics

QED of Spinless Particles

The Dirac Equation

QED of Spin− 21 Particles

Thank You
Non-relativistic Quantum Mechanics
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
▶ Schrödinger equation: i ∂ψ
∂t +
1 2
2m ∇ ψ = 0; ℏ ≡ 1
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
▶ Schrödinger equation: i ∂ψ 1 2
∂t + 2m ∇ ψ = 0; ℏ ≡ 1
▶ ψ(x, t), the wavefunction: the solution
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
▶ Schrödinger equation: i ∂ψ 1 2
∂t + 2m ∇ ψ = 0; ℏ ≡ 1
▶ ψ(x, t), the wavefunction: the solution
▶ Born’s Interpretation: ρ = |ψ|2
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
▶ Schrödinger equation: i ∂ψ 1 2
∂t + 2m ∇ ψ = 0; ℏ ≡ 1
▶ ψ(x, t), the wavefunction: the solution
▶ Born’s Interpretation: ρ = |ψ|2
▶ Continuity equation: ∂ρ
∂t + ∇ · j = 0
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
▶ Schrödinger equation: i ∂ψ 1 2
∂t + 2m ∇ ψ = 0; ℏ ≡ 1
▶ ψ(x, t), the wavefunction: the solution
▶ Born’s Interpretation: ρ = |ψ|2
▶ Continuity equation: ∂ρ
∂t + ∇ · j = 0
i
▶ The current density, j = − 2m (ψ ∗ ∇ψ − ψ∇ψ ∗ )
Schrödinger Equation 4

▶ Non-relativistic energy-momentum relation: E = p2

2m
∂
▶ E −→ iℏ ∂t , and p −→ −iℏ∇
▶ Schrödinger equation: i ∂ψ 1 2
∂t + 2m ∇ ψ = 0; ℏ ≡ 1
▶ ψ(x, t), the wavefunction: the solution
▶ Born’s Interpretation: ρ = |ψ|2
▶ Continuity equation: ∂ρ
∂t + ∇ · j = 0
i
▶ The current density, j = − 2m (ψ ∗ ∇ψ − ψ∇ψ ∗ )
▶ Free-particle solution, ψ = N ei(p·x−Et)
p
ρ = |N |2 , and current density, j = m |N |2
Relativistic Quantum Mechanics
Klein-Gordon Equation 6

▶ Relativistic energy-momentum relation: E 2 = p2 + m2

Klein-Gordon Equation 6

▶ Relativistic energy-momentum relation: E 2 = p2 + m2

2
▶ The Klein-Gordon Equation: − ∂∂t2ϕ + ∇2 ϕ = m2 ϕ
Klein-Gordon Equation 6

▶ Relativistic energy-momentum relation: E 2 = p2 + m2

2
▶ The Klein-Gordon Equation: − ∂∂t2ϕ + ∇2 ϕ = m2 ϕ
▶ Continuity equation 
by comparing:
∂ϕ∗
 
∗ ∂ϕ
∂
∂t i ϕ ∂t − ϕ ∂t +∇ · [−i (ϕ∗ ∇ϕ − ϕ∇ϕ∗ )] = 0
| {z }
| {z } j
ρ
Klein-Gordon Equation 6

▶ Relativistic energy-momentum relation: E 2 = p2 + m2

2
▶ The Klein-Gordon Equation: − ∂∂t2ϕ + ∇2 ϕ = m2 ϕ
▶ Continuity equation 
by comparing:
∂ϕ∗
 
∗ ∂ϕ
∂
∂t i ϕ ∂t − ϕ ∂t +∇ · [−i (ϕ∗ ∇ϕ − ϕ∇ϕ∗ )] = 0
| {z }
| {z } j
ρ
▶ Free particle solution: ϕ = N eip·x−iEt
Klein-Gordon Equation 6

▶ Relativistic energy-momentum relation: E 2 = p2 + m2

2
▶ The Klein-Gordon Equation: − ∂∂t2ϕ + ∇2 ϕ = m2 ϕ
▶ Continuity equation 
by comparing:
∂ϕ∗
 
∗ ∂ϕ
∂
∂t i ϕ ∂t − ϕ ∂t +∇ · [−i (ϕ∗ ∇ϕ − ϕ∇ϕ∗ )] = 0
| {z }
| {z } j
ρ
▶ Free particle solution: ϕ = N eip·x−iEt
▶ ρ = i(−2iE)|N |2 = 2E|N |2
j = −i(2ip)|N |2 = 2p|N |2
Klein-Gordon Equation 6

▶ Relativistic energy-momentum relation: E 2 = p2 + m2

2
▶ The Klein-Gordon Equation: − ∂∂t2ϕ + ∇2 ϕ = m2 ϕ
▶ Continuity equation 
by comparing:
∂ϕ∗
 
∗ ∂ϕ
∂
∂t i ϕ ∂t − ϕ ∂t +∇ · [−i (ϕ∗ ∇ϕ − ϕ∇ϕ∗ )] = 0
| {z }
| {z } j
ρ
▶ Free particle solution: ϕ = N eip·x−iEt
▶ ρ = i(−2iE)|N |2 = 2E|N |2
j = −i(2ip)|N |2 = 2p|N |2
▶ In four-vector form:
KG equation: □2 + m2 ϕ = 0

Current and flux density: j µ = (ρ, j) = i (ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ )

Continuity relation: ∂µ j µ = 0
Klein-Gordon Equation 7

▶ For free particle

ϕ = N e−ip·x
j µ = 2pµ |N |2
Klein-Gordon Equation 7

▶ For free particle

ϕ = N e−ip·x
j µ = 2pµ |N |2

1/2
▶ Energy eigenvalues: E = ± p2 + m2
Klein-Gordon Equation 7

▶ For free particle

ϕ = N e−ip·x
j µ = 2pµ |N |2
▶ Energy eigenvalues: E = ± p2 + m2 1/2

▶ The issue: Negative Energy solution implies

Klein-Gordon Equation 7

▶ For free particle

ϕ = N e−ip·x
j µ = 2pµ |N |2
▶ Energy eigenvalues: E = ± p2 + m2 1/2

▶ The issue: Negative Energy solution implies

1. Lower and lower energy levels always available
Klein-Gordon Equation 7

▶ For free particle

ϕ = N e−ip·x
j µ = 2pµ |N |2
▶ Energy eigenvalues: E = ± p2 + m2 1/2

▶ The issue: Negative Energy solution implies

1. Lower and lower energy levels always available
2. Negative probability density
Interpretation of Negative Energy solutions 8
▶ Dirac(1928-29):
All -ve states are filled
+ve energy spin- 21 electrons can’t fill them
Can excite a -ve energy electron to +ve energy
Absence of (-e) ≡ Presence of (+e)
Absence of (-E) ≡ Presence of (+E)
Excitation results in pair production: e− (E ′ ) + e+ (E)
Interpretation of Negative Energy solutions 8
▶ Dirac(1928-29):
All -ve states are filled
+ve energy spin- 21 electrons can’t fill them
Can excite a -ve energy electron to +ve energy
Absence of (-e) ≡ Presence of (+e)
Absence of (-E) ≡ Presence of (+E)
Excitation results in pair production: e− (E ′ ) + e+ (E)
▶ Pauli and Weisskopf (1934)
Wrote j µ = −ie (ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ )
Makes ρ = j 0 , charge density and not a probability density
E > 0, ρ < 0: electron, and E < 0, ρ > 0: positron
Can be applied to bosons
Interpretation of Negative Energy solutions 8
▶ Dirac(1928-29):
All -ve states are filled
+ve energy spin- 21 electrons can’t fill them
Can excite a -ve energy electron to +ve energy
Absence of (-e) ≡ Presence of (+e)
Absence of (-E) ≡ Presence of (+E)
Excitation results in pair production: e− (E ′ ) + e+ (E)
▶ Pauli and Weisskopf (1934)
Wrote j µ = −ie (ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ )
Makes ρ = j 0 , charge density and not a probability density
E > 0, ρ < 0: electron, and E < 0, ρ > 0: positron
Can be applied to bosons
▶ Feynman(1948)-Stückelberg(1941)
Negative energy solutions: particles that propagate backward in time or
positive energy antiparticles that propagate forward in time
Because e−i(−E)(−t) = e−iEt
Scattering 9

▶ Transition amplitude up to first order

Z T /2 Z
∗
dt d3 x ϕf (x)e−iEf t V (x, t) ϕi (x)e−iEi t
  
Tf i = −i
−T /2
= Probability of transition i → f
Scattering 9

▶ Transition amplitude up to first order

Z T /2 Z
∗
dt d3 x ϕf (x)e−iEf t V (x, t) ϕi (x)e−iEi t
  
Tf i = −i
−T /2
= Probability of transition i → f
▶ For time-independent perturbation: Tf i = −2πiVf i δ (Ef − Ei )
where Vfi ≡ d3 xϕ∗f (x)V (x)ϕi (x)
R
Scattering 9

▶ Transition amplitude up to first order

Z T /2 Z
∗
dt d3 x ϕf (x)e−iEf t V (x, t) ϕi (x)e−iEi t
  
Tf i = −i
−T /2
= Probability of transition i → f
▶ For time-independent perturbation: Tf i = −2πiVf i δ (Ef − Ei )
where Vfi ≡ d3 xϕ∗f (x)V (x)ϕi (x)
R
▶ Transition rate:
|Tf i |2
W = lim
T →∞ T
= 2π |Vf i |2 δ (Ef − Ei )
Fermi’s Golden Rule
Scattering 9

▶ Transition amplitude up to first order

Z T /2 Z
∗
dt d3 x ϕf (x)e−iEf t V (x, t) ϕi (x)e−iEi t
  
Tf i = −i
−T /2
= Probability of transition i → f
▶ For time-independent perturbation: Tf i = −2πiVf i δ (Ef − Ei )
where Vfi ≡ d3 xϕ∗f (x)V (x)ϕi (x)
R
▶ Transition rate:
|Tf i |2
W = lim
T →∞ T
= 2π |Vf i |2 δ (Ef − Ei )
Fermi’s Golden Rule
Vf n Vni
▶ Second order corrections: Tf i = · · · − 2πi n̸=i
P
Ei −En +iε δ (Ef − Ei )
Vf i → Vf i + n̸=i Vf n Ei −E1n +iε Vni + · · ·
P
Feynman Diagrams 10

ϕ∗outgoing V ϕingoing d4 x
R
▶ Tf i ∝
Feynman Diagrams 10

ϕ∗outgoing V ϕingoing d4 x
R
▶ Tf i ∝

∗
e−iEf t e−iωt e−iEi t dt = 2πδ (Ef − ω − Ei )
R
▶ Tf i ∝
Feynman Diagrams 10

ϕ∗outgoing V ϕingoing d4 x
R
▶ Tf i ∝

R −iE t
∗ −iωt −iE t

▶ Tf i ∝ e f e e i dt = 2πδ (Ef − ω − Ei )
R −i(−E )t
∗ −iωt −i(−E )t
∝ e i e e f
dt = 2πδ (−Ei − ω + Ef )
QED of Spinless Particles
Spinless Electron in EM field 12

▶ Electromagnetic potential Aµ
Spinless Electron in EM field 12

▶ Electromagnetic potential Aµ
▶ pµ → pµ + eAµ , or i∂ µ → i∂ µ + eAµ
Spinless Electron in EM field 12

▶ Electromagnetic potential Aµ
▶ pµ → pµ + eAµ , or i∂ µ → i∂ µ + eAµ
▶ ∂µ ∂ µ + m2 ϕ = −V ϕ

where V = −ie (∂µ Aµ + Aµ ∂µ ) − e2

A2

Spinless Electron in EM field 12

▶ Electromagnetic potential Aµ
▶ pµ → pµ + eAµ , or i∂ µ → i∂ µ + eAµ
▶ ∂µ ∂ µ + m2 ϕ = −V ϕ

where V = −ie (∂µ Aµ + Aµ ∂µ ) −  e2

A2

R ∗
▶ Tf i = −i ϕf (x)V (x)ϕi (x)d4 x
= −i jµf i Aµd4 x
R

where jµf i (x) ≡ −ie ϕ∗f ∂µ ϕi − ϕi ∂µ ϕ∗f , charge current density
Spinless Electron in EM field 12

▶ Electromagnetic potential Aµ
▶ pµ → pµ + eAµ , or i∂ µ → i∂ µ + eAµ
▶ ∂µ ∂ µ + m2 ϕ = −V ϕ

where V = −ie (∂µ Aµ + Aµ ∂µ ) −  e2

A2

R ∗
▶ Tf i = −i ϕf (x)V (x)ϕi (x)d4 x
= −i jµf i Aµd4 x
R

where jµf i (x) ≡ −ie ϕ∗f ∂µ ϕi − ϕi ∂µ ϕ∗f , charge current density
▶ Substitute ϕi (x) = Ni e−ipi ·x , and ϕf (x) = Nf e−ipf ·x to get
jµf i = −eN N (p + p ) ei(pf −pi )·x
i f i f µ
Spinless Electron-Muon Scattering 13

▶ Take muon as the source of EM field

Spinless Electron-Muon Scattering 13

▶ Take muon as the source of EM field

µ
▶ Muon current: j(2) = −eNB ND (pD + pB )µ ei(pD −pB )·x
Spinless Electron-Muon Scattering 13

▶ Take muon as the source of EM field

µ
▶ Muon current: j(2) = −eNB ND (pD + pB )µ ei(pD −pB )·x
µ
▶ According to Maxwell’s Equations: □2 Aµ = j(2)
Spinless Electron-Muon Scattering 13

▶ Take muon as the source of EM field

µ
▶ Muon current: j(2) = −eNB ND (pD + pB )µ ei(pD −pB )·x
µ
▶ According to Maxwell’s Equations: □2 Aµ = j(2)
▶ We also have □2 eiq·x = −q 2 eiq·x with q = pD − pB
Spinless Electron-Muon Scattering 13

▶ Take muon as the source of EM field

µ
▶ Muon current: j(2) = −eNB ND (pD + pB )µ ei(pD −pB )·x
µ
▶ According to Maxwell’s Equations: □2 Aµ = j(2)
▶ We also have □2 eiq·x = −q 2 eiq·x with q = pD − pB
µ
▶ ∴ Aµ = − q12 j(2)
Spinless Electron-Muon Scattering 13

▶ Take muon as the source of EM field

µ
▶ Muon current: j(2) = −eNB ND (pD + pB )µ ei(pD −pB )·x
µ
▶ According to Maxwell’s Equations: □2 Aµ = j(2)
▶ We also have □2 eiq·x = −q 2 eiq·x with q = pD − pB
µ
▶ ∴ Aµ = − q12 j(2)
 
▶ Transition amplitude: Tf i = −i jµ(1) (x) − q12 j(2)
µ
(x)d4 x
R
Spinless Electron-Muon Scattering 14

▶ After substituting jµ(1) (x):

Tf i = −iNA NB NC ND (2π)4 δ (4)
 (pD +pC − pB − pA ) ℜ
gµν
where −iℜ = (ie (pA + pC )µ ) −i q2
(ie (pB + pD )ν )
Spinless Electron-Muon Scattering 14

▶ After substituting jµ(1) (x):

Tf i = −iNA NB NC ND (2π)4 δ (4)
 (pD +pC − pB − pA ) ℜ
gµν
where −iℜ = (ie (pA + pC )µ ) −i q2
(ie (pB + pD )ν )
▶ ℜ is known as the invariant amplitude
Spinless Electron-Muon Scattering 14

▶ After substituting jµ(1) (x):

Tf i = −iNA NB NC ND (2π)4 δ (4)
 (pD +pC − pB − pA ) ℜ
gµν
where −iℜ = (ie (pA + pC )µ ) −i q2
(ie (pB + pD )ν )
▶ ℜ is known as the invariant amplitude
▶ Kronecker-delta =⇒ the energy-momentum conservation in the interaction
Feynman Diagrams 15
Feynman Diagrams 15

▶ For more vertices or propagators: multiply corresponding factors to get ℜ

Scattering Cross-section 16

▶ Transition rate per unit volume:

2
|T |
Wf i = TfVi
Scattering Cross-section 16

▶ Transition rate per unit volume:

2
|T |
Wf i = TfVi
Wf i
▶ Cross-section = (initial flux) (number of final states)
Scattering Cross-section 16

▶ Transition rate per unit volume:

2
|T |
Wf i = TfVi
Wf i
▶ Cross-section = (initial flux) (number of final states)
▶ For 2E particles in volume V, No. of final states = V d 3 pC V d 3 pD
(2π)3 2EC (2π)3 2ED
Scattering Cross-section 16

▶ Transition rate per unit volume:

2
|T |
Wf i = TfVi
Wf i
▶ Cross-section = (initial flux) (number of final states)
V d pC 3 3
V d pD
▶ For 2E particles in volume V, No. of final states = (2π) 3 2E (2π)3 2E
C D
▶ Initial flux = No. of beam particles passing through a unit area per unit time
× No. of target particles per
V d 3 pC V d 3 pD
No. of final states = (2π) 3 2E (2π)3 2E
C D
Scattering Cross-section 16

▶ Transition rate per unit volume:

2
|T |
Wf i = TfVi
Wf i
▶ Cross-section = (initial flux) (number of final states)
V d pC 3
V d pD 3
▶ For 2E particles in volume V, No. of final states = (2π) 3 2E (2π)3 2E
C D
▶ Initial flux = No. of beam particles passing through a unit area per unit time
× No. of target particles per
V d 3 pC V d 3 pD
No. of final states = (2π) 3 2E (2π)3 2E
C D
4 3
V2 2 (2π) (4) (p d3 pD 2
▶ dσ = 1
|vA |2EA 2EB V 4 |ℜ| (2π)6 δ C + pD − pA − pB ) d2EpCC 2ED V
Scattering Cross-section 16

▶ Transition rate per unit volume:

2
|T |
Wf i = TfVi
Wf i
▶ Cross-section = (initial flux) (number of final states)
V d pC 3
V d pD 3
▶ For 2E particles in volume V, No. of final states = (2π) 3 2E (2π)3 2E
C D
▶ Initial flux = No. of beam particles passing through a unit area per unit time
× No. of target particles per
V d 3 pC V d 3 pD
No. of final states = (2π) 3 2E (2π)3 2E
C D
2 4 3 3
▶ dσ = |v |2EV 1 2 (2π) (4) (p + p − p − p ) d pC d pD V 2
4 |ℜ| (2π)6 δ C D A B 2EC 2ED
A A 2EB V
▶ It’s an intrinsic quantity and depends only on the process we are considering
Electron-Electron Scattering 17

 
e2 (pA +pC )µ (pB +pD )µ e2 (pA +pD )µ (pB +pC )µ
▶ −iℜe− e− = −i − −
(p −p )2
D B (pC −pB )2
Electron-Electron Scattering 17

 
e2 (pA +pC )µ (pB +pD )µ e2 (pA +pD )µ (pB +pC )µ
▶ −iℜe− e− = −i − −
(p −p )2
D B (pC −pB )2
▶ symmetric under pA ↔ pB
Electron-Positron Scattering 18

(p +p ) (−p −p )µ (p −p ) (−p +p )µ
 
▶ −iRe− e+ = −i −e2 A C µ D2 B − e2 A B µ D2 C
(p −p )
D B (p +p )
C D
Electron-Positron Scattering 18

(p +p ) (−p −p )µ (p −p ) (−p +p )µ
 
▶ −iRe− e+ = −i −e2 A C µ D2 B − e2 A B µ D2 C
(p −p )
D B (p +p )
C D
▶ Symmetric under pC ↔ −pB
Electron-Positron Scattering 18

(p +p ) (−p −p )µ (p −p ) (−p +p )µ
 
▶ −iRe− e+ = −i −e2 A C µ D2 B − e2 A B µ D2 C
(p −p ) D B (p +p ) C D
▶ Symmetric under pC ↔ −pB
▶ Re− e+ →e− e+ (pA , pB , pC , pD ) = Re− e− →e− e− (pA , −pD , pC , −pB )
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
▶ Can write ℜ in terms of Mandelstam variables given as
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
▶ Can write ℜ in terms of Mandelstam variables given as
s = (pA + pB )2
t=(pA − pC )2
u=(pA − pD )2
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
▶ Can write ℜ in terms of Mandelstam variables given as
s = (pA + pB )2
t=(pA − pC )2
u=(pA − pD )2
▶ Relation: s + t + u = m2A + m2B + m2C + m2D
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
▶ Can write ℜ in terms of Mandelstam variables given as
s = (pA + pB )2
t=(pA − pC )2
u=(pA − pD )2
▶ Relation: s + t + u = m2A + m2B + m2C + m2D
▶ f-channel process if f = COM energy or q 2 where f = s,t, or u
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
▶ Can write ℜ in terms of Mandelstam variables given as
s = (pA + pB )2
t=(pA − pC )2
u=(pA − pD )2
▶ Relation: s + t + u = m2A + m2B + m2C + m2D
▶ f-channel process if f = COM energy or q 2 where f = s,t, or u

▶ Re− e+ (s, t, u) = e2 s−u t−u

t + s
Mandelstam Variables 19
▶ Two independent variables for a scattering: incident energy and the scattering
angle
▶ Can write ℜ in terms of Mandelstam variables given as
s = (pA + pB )2
t=(pA − pC )2
u=(pA − pD )2
▶ Relation: s + t + u = m2A + m2B + m2C + m2D
▶ f-channel process if f = COM energy or q 2 where f = s,t, or u

▶ Re− e+ (s, t, u) = e2 s−u t−u

t + s
▶ Re− e+ (s, t, u) = Re− e− (u, t, s)
The Dirac Equation
The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
▶ Dirac set out for an equation linear in ∂/∂t
The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
▶ Dirac set out for an equation linear in ∂/∂t
▶ Hψ = (α · P + βm)ψ
The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
▶ Dirac set out for an equation linear in ∂/∂t
▶ Hψ = (α · P + βm)ψ
▶ Must satisfy the relativistic energy-momentum relation:
H 2 ψ = P2 + m2 ψ


The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
▶ Dirac set out for an equation linear in ∂/∂t
▶ Hψ = (α · P + βm)ψ
▶ Must satisfy the relativistic energy-momentum relation:
H 2 ψ = P2 + m2 ψ

▶ H 2 ψ = (αi Pi + βm) (αj Pj + βm) ψ

=(αi2 Pi2 + (αi αj + αj αi ) Pi Pj + (αi β + βαi ) Pi m + β 2 m2 )ψ; where i > j
| {z } | {z }
0 0
The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
▶ Dirac set out for an equation linear in ∂/∂t
▶ Hψ = (α · P + βm)ψ
▶ Must satisfy the relativistic energy-momentum relation:
H 2 ψ = P2 + m2 ψ

▶ H 2 ψ = (αi Pi + βm) (αj Pj + βm) ψ

=(αi2 Pi2 + (αi αj + αj αi ) Pi Pj + (αi β + βαi ) Pi m + β 2 m2 )ψ; where i > j
| {z } | {z }
0 0
▶ α1 , α2 , α3 , β all anti-commute with each other =⇒ matrices and ψ is a 4
component Dirac Spinor
The Dirac Equation 21

▶ Getting -ve energy solutions because of the square of the energy operator
▶ Dirac set out for an equation linear in ∂/∂t
▶ Hψ = (α · P + βm)ψ
▶ Must satisfy the relativistic energy-momentum relation:
H 2 ψ = P2 + m2 ψ

▶ H 2 ψ = (αi Pi + βm) (αj Pj + βm) ψ

=(αi2 Pi2 + (αi αj + αj αi ) Pi Pj + (αi β + βαi ) Pi m + β 2 m2 )ψ; where i > j
| {z } | {z }
0 0
▶ α1 , α2 , α3 , β all anti-commute with each other =⇒ matrices and ψ is a 4
component Dirac Spinor
▶ In Dirac-Pauli representation they are given as:
   
0 σ I 0
α= , β=
σ 0 0 −I
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂µ ψ̄γ µ + mψ̄ = 0
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
▶ ρ ≡ j 0 = ψ̄γ 0 ψ = ψ † ψ = 4i=1 |ψi |2
P
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
▶ ρ ≡ j 0 = ψ̄γ 0 ψ = ψ † ψ = 4i=1 |ψi |2
P
▶ Ansatz for free-particle ψ = u(p)e−ip·x
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
▶ ρ ≡ j 0 = ψ̄γ 0 ψ = ψ † ψ = 4i=1 |ψi |2
P
▶ Ansatz for free-particle ψ = u(p)e−ip·x
▶ Hu = (α · p + βm)u = Eu
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
▶ ρ ≡ j 0 = ψ̄γ 0 ψ = ψ † ψ = 4i=1 |ψi |2
P
▶ Ansatz for free-particle ψ = u(p)e−ip·x
▶ Hu = (α · p + βm)u = Eu
▶ If p = 0,  
mI 0
Hu = βmu = u
0 −mI
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
▶ ρ ≡ j 0 = ψ̄γ 0 ψ = ψ † ψ = 4i=1 |ψi |2
P
▶ Ansatz for free-particle ψ = u(p)e−ip·x
▶ Hu = (α · p + βm)u = Eu
▶ If p = 0,  
mI 0
Hu = βmu = u
0 −mI
Eigenvalues of E are m, m, −m, −m, and eigenvectors are
       
1 0 0 0
 0   1   0   0 
 ,  ,  ,  .
 0   0   1   0 
0 0 0 1
The Dirac Equation 22
▶ Adjoint spinor: ψ̄ ≡ ψ † γ 0
which satisfies i∂ µ + mψ̄ = 0

µ ψ̄γ
▶ ψ̄γ ∂µ ψ + ∂µ ψ̄ γ ψ = ∂µ ψ̄γ µ ψ = 0
µ µ


▶ j µ = ψ̄γ µ ψ satisfies the continuity equation
▶ ρ ≡ j 0 = ψ̄γ 0 ψ = ψ † ψ = 4i=1 |ψi |2
P
▶ Ansatz for free-particle ψ = u(p)e−ip·x
▶ Hu = (α · p + βm)u = Eu
▶ If p = 0,  
mI 0
Hu = βmu = u
0 −mI
Eigenvalues of E are m, m, −m, −m, and eigenvectors are
       
1 0 0 0
 0   1   0   0 
 ,  ,  ,  .
 0   0   1   0 
0 0 0 1
E > 0 solutions describes an electron, E < 0 solutions describe an antiparticle
The Dirac Equation 23
▶ For p ̸= 0, we have
    
m σ·p uA uA
Hu = =E
σ · p −m uB uB
The Dirac Equation 23
▶ For p ̸= 0, we have
    
m σ·p uA uA
Hu = =E
σ · p −m uB uB
▶ σ · puB = (E − m)uA ,
σ ·puA = (E + m)uB
(σ ·p)2 uA = (E 2 − m2 )uA
(σ · p)2 = p2 = E 2 − m2
The Dirac Equation 23
▶ For p ̸= 0, we have
    
m σ·p uA uA
Hu = =E
σ · p −m uB uB
▶ σ · puB = (E − m)uA ,
σ ·puA = (E + m)uB
(σ ·p)2 uA = (E 2 − m2 )uA
(σ · p)2 = p2 = E 2 − m2
▶ For E > 0 : u(s)
A =χ
(s) where
   
(1) 1 (2) 0
χ = , χ =
0 1
(s) σ·p (s)
and uB = E+m χ
The Dirac Equation 23
▶ For p ̸= 0, we have
    
m σ·p uA uA
Hu = =E
σ · p −m uB uB
▶ σ · puB = (E − m)uA ,
σ ·puA = (E + m)uB
(σ ·p)2 uA = (E 2 − m2 )uA
(σ · p)2 = p2 = E 2 − m2
▶ For E > 0 : u(s)
A =χ
(s) where
   
(1) 1 (2) 0
χ = , χ =
0 1
(s) σ·p (s)
and uB = E+m χ

χ(s) √
 
u(s) = N σ·p (s) , where s = 1, 2 and N = E+m
E+m χ
The Dirac Equation 24

▶ For E < 0 : u(s) (s)

B = χ , and

(s) σ · p (s) σ · p (s)

uA = uB = − χ
E−m |E| + m

Then, we get !
−σ·p (s)
u(s+2) = N |E|+m χ
χ(s)
The Dirac Equation 24

▶ For E < 0 : u(s) (s)

B = χ , and

(s) σ · p (s) σ · p (s)

uA = uB = − χ
E−m |E| + m

Then, we get !
−σ·p (s)
u(s+2) = N |E|+m χ
χ(s)
▶ For a given momentum there are 4 orthogonal solutions: u(1,2) , with positive
energy, and u(3,4) , with negative energy.
Helicity 25

▶ There is a two-fold degeneracy in positive and negative energy solutions

Helicity 25

▶ There is a two-fold degeneracy in positive and negative energy solutions

▶ So there must be an operator which commutes with H and p, and whose
eigenvalues can be used to label the states
Helicity 25

▶ There is a two-fold degeneracy in positive and negative energy solutions

▶ So there must be an operator which commutes with H and p, and whose
eigenvalues can be used to label the states
▶ One such operator is Helicity Operator
Helicity 25

▶ There is a two-fold degeneracy in positive and negative energy solutions

▶ So there must be an operator which commutes with H and p, and whose
eigenvalues can be used to label the states
▶ One such operator is Helicity Operator
 1 
1 σ · p̂ 0
h = 2 Σ · p̂ ≡ 2
1 = 12 (σ · p̂) ⊗ I
0 2 σ · p̂
Helicity 25

▶ There is a two-fold degeneracy in positive and negative energy solutions

▶ So there must be an operator which commutes with H and p, and whose
eigenvalues can be used to label the states
▶ One such operator is Helicity Operator
 1 
1 σ · p̂ 0
h = 2 Σ · p̂ ≡ 2
1 = 12 (σ · p̂) ⊗ I
0 2 σ · p̂
▶ Helicity is spin in the direction of motion or eigenvalues of 21 σ · p̂ , λ
Intrinsic Angular Momentum 26

▶ OAM, L = r × P doesn’t commute with Hamiltonian Hψ = (α · P + βm)ψ

[H, L] = −i(α × P)
i.e. L is not conserved.
Intrinsic Angular Momentum 26

▶ OAM, L = r × P doesn’t commute with Hamiltonian Hψ = (α · P + βm)ψ

[H, L] = −i(α × P)
i.e. L is not conserved.
▶ Some other angular momentum must exist in order to make total angular
momentum a conserved quantity.
Intrinsic Angular Momentum 26

▶ OAM, L = r × P doesn’t commute with Hamiltonian Hψ = (α · P + βm)ψ

[H, L] = −i(α × P)
i.e. L is not conserved.
▶ Some other angular momentum must exist in order to make total angular
momentum a conserved quantity.
▶ We notice that  
σ 0
[H, Σ] = +2i(α × P), where Σ ≡
0 σ
Intrinsic Angular Momentum 26

▶ OAM, L = r × P doesn’t commute with Hamiltonian Hψ = (α · P + βm)ψ

[H, L] = −i(α × P)
i.e. L is not conserved.
▶ Some other angular momentum must exist in order to make total angular
momentum a conserved quantity.
▶ We notice that  
σ 0
[H, Σ] = +2i(α × P), where Σ ≡
0 σ
▶ J = L + 21 Σ
[H, J] = 0
Intrinsic Angular Momentum 26

▶ OAM, L = r × P doesn’t commute with Hamiltonian Hψ = (α · P + βm)ψ

[H, L] = −i(α × P)
i.e. L is not conserved.
▶ Some other angular momentum must exist in order to make total angular
momentum a conserved quantity.
▶ We notice that  
σ 0
[H, Σ] = +2i(α × P), where Σ ≡
0 σ
▶ J = L + 21 Σ
[H, J] = 0
∴ J is the total angular momentum
Intrinsic Angular Momentum 26

▶ OAM, L = r × P doesn’t commute with Hamiltonian Hψ = (α · P + βm)ψ

[H, L] = −i(α × P)
i.e. L is not conserved.
▶ Some other angular momentum must exist in order to make total angular
momentum a conserved quantity.
▶ We notice that  
σ 0
[H, Σ] = +2i(α × P), where Σ ≡
0 σ
▶ J = L + 21 Σ
[H, J] = 0
∴ J is the total angular momentum
▶ 1
2Σ denotes the intrinsic angular momentum
QED of Spin− 21 Particles
Electron in an Electromagnetic Field 28

▶ Dirac Equation:(γµ pµ − m) ψ = 0
Electron in an Electromagnetic Field 28

▶ Dirac Equation:(γµ pµ − m) ψ = 0
▶ Solution: ψ = u(p)e−ip·x
Electron in an Electromagnetic Field 28

▶ Dirac Equation:(γµ pµ − m) ψ = 0
▶ Solution: ψ = u(p)e−ip·x
▶ In an electromagnetic field, Aµ ,

pµ → pµ + eAµ

(γµ pµ − m) ψ = γ 0 V ψ
where γ 0 V = −eγµ Aµ is electromagnetic perturbation
Electron in an Electromagnetic Field 28

▶ Dirac Equation:(γµ pµ − m) ψ = 0
▶ Solution: ψ = u(p)e−ip·x
▶ In an electromagnetic field, Aµ ,

pµ → pµ + eAµ

(γµ pµ − m) ψ = γ 0 V ψ
where γ 0 V = −eγµ Aµ is electromagnetic perturbation
▶ Tf i = −i ψf† (x)V (x)ψi (x)d4 x
R

=i e ψ̄f γµ Aµ ψi d4 x
R

=-i jµf i Aµ d4 x
R
Electron in an Electromagnetic Field 28

▶ Dirac Equation:(γµ pµ − m) ψ = 0
▶ Solution: ψ = u(p)e−ip·x
▶ In an electromagnetic field, Aµ ,

pµ → pµ + eAµ

(γµ pµ − m) ψ = γ 0 V ψ
where γ 0 V = −eγµ Aµ is electromagnetic perturbation
▶ Tf i = −i ψf† (x)V (x)ψi (x)d4 x
R

=i e ψ̄f γµ Aµ ψi d4 x
R

=-i jµf i Aµ d4 x
R

where jµf i ≡ −eψ̄f γµ ψi

= −eūf γµ ui ei(pf −pi )·x
Electron in an Electromagnetic Field 29

▶ Comparing with the spinless electron case where jµf i = −e (pf + pi )µ ei(pf −pt )·x
Electron in an Electromagnetic Field 29

▶ Comparing with the spinless electron case where jµf i = −e (pf + pi )µ ei(pf −pt )·x
Vertex factor is a 4 × 4 matrix sandwiched between u(s) (pi ) and row ū(r) (pf )
spinors
Electron in an Electromagnetic Field 29

▶ Comparing with the spinless electron case where jµf i = −e (pf + pi )µ ei(pf −pt )·x
Vertex factor is a 4 × 4 matrix sandwiched between u(s) (pi ) and row ū(r) (pf )
spinors

1
▶ ūf γ µ ui = 2m ūf (pf + pi )µ + iσ µν (pf − pi )v ui

i.e. the physical spin- 1/2 electron interacts via both its charge and its
magnetic moment, unlike ‘spinless’ electron, which gives coupling (pf + pi) as
it interacts with charge only
Møller Scattering 30
▶ The transition amplitude for Møller scattering, e− e− → e− e− is
Z  
(1) 1 µ
Tf i = −i jµ (x) − 2 j(2) (x)d4 x
q
 
1
= −i (−eūC γµ uA ) − 2 (−eūD γ µ uB ) (2π)4 δ (4) (pA + pB − pC − pD )
q

where q = pA − pC
Møller Scattering 30
▶ The transition amplitude for Møller scattering, e− e− → e− e− is
Z  
(1) 1 µ
Tf i = −i jµ (x) − 2 j(2) (x)d4 x
q
 
1
= −i (−eūC γµ uA ) − 2 (−eūD γ µ uB ) (2π)4 δ (4) (pA + pB − pC − pD )
q

where q = pA − pC
▶ The total lowest-order amplitude for Møller scattering is
(ū γ µ u )(ū γ u ) (ū γ µ u )(ū γ u )
R = −e2 C (p A−p D)2 µ B + e2 D (p A−p C)2µ B
A C A D
Propagators 31

▶ Perturbative expansionof the transition amplitude 

P 1
Tf i = −i2πδ (Ef − Ei ) ⟨f |V |i⟩ + n̸=i ⟨f |V |n⟩ Ei −En
⟨n|V |i⟩ + · · ·
Propagators 31

▶ Perturbative expansionof the transition amplitude 

P 1
Tf i = −i2πδ (Ef − Ei ) ⟨f |V |i⟩ + n̸=i ⟨f |V |n⟩ Ei −En
⟨n|V |i⟩ + · · ·
▶ ⟨f |v|n⟩ are regarded as vertices and the associated propagator for interaction
is 1/ (Ei − En )
Propagators 31

▶ Perturbative expansionof the transition amplitude 

P 1
Tf i = −i2πδ (Ef − Ei ) ⟨f |V |i⟩ + n̸=i ⟨f |V |n⟩ Ei −En
⟨n|V |i⟩ + · · ·
▶ ⟨f |v|n⟩ are regarded as vertices and the associated propagator for interaction
is 1/ (Ei − En )
▶ Unperturbed Hamiltonian
D H0 |n⟩ = En |n⟩ gives E
i
Tf i = 2πδ (Ef − Ei ) f (−iV ) + (−iV ) Ei −H 0
(−iV ) + · · · i
Propagators 31

▶ Perturbative expansionof the transition amplitude 

P 1
Tf i = −i2πδ (Ef − Ei ) ⟨f |V |i⟩ + n̸=i ⟨f |V |n⟩ Ei −En
⟨n|V |i⟩ + · · ·
▶ ⟨f |v|n⟩ are regarded as vertices and the associated propagator for interaction
is 1/ (Ei − En )
▶ Unperturbed Hamiltonian
D H0 |n⟩ = En |n⟩ gives E
i
Tf i = 2πδ (Ef − Ei ) f (−iV ) + (−iV ) Ei −H 0
(−iV ) + · · · i
▶ Now, −iV is the vertex factor
Propagators 31

▶ Perturbative expansionof the transition amplitude 

P 1
Tf i = −i2πδ (Ef − Ei ) ⟨f |V |i⟩ + n̸=i ⟨f |V |n⟩ Ei −En
⟨n|V |i⟩ + · · ·
▶ ⟨f |v|n⟩ are regarded as vertices and the associated propagator for interaction
is 1/ (Ei − En )
▶ Unperturbed Hamiltonian
D H0 |n⟩ = En |n⟩ gives E
i
Tf i = 2πδ (Ef − Ei ) f (−iV ) + (−iV ) Ei −H 0
(−iV ) + · · · i
▶ Now, −iV is the vertex factor
▶ Associated propagator: inverse of the −i(Ei − H0 )
Propagators 31

▶ Perturbative expansionof the transition amplitude 

P 1
Tf i = −i2πδ (Ef − Ei ) ⟨f |V |i⟩ + n̸=i ⟨f |V |n⟩ Ei −En
⟨n|V |i⟩ + · · ·
▶ ⟨f |v|n⟩ are regarded as vertices and the associated propagator for interaction
is 1/ (Ei − En )
▶ Unperturbed Hamiltonian
D H0 |n⟩ = En |n⟩ gives E
i
Tf i = 2πδ (Ef − Ei ) f (−iV ) + (−iV ) Ei −H 0
(−iV ) + · · · i
▶ Now, −iV is the vertex factor
▶ Associated propagator: inverse of the −i(Ei − H0 )
▶ Propagators can be identified by comparing with
−i (Ei − H0 ) ψ = −iV ψ
Propagator of Spinless Particles 32

▶ Compare Klein-Gordon equation

i □2 + m2 ϕ = −iV ϕ

with −i (Ei − H0 ) ψ = −iV ψ

Propagator of Spinless Particles 32

▶ Compare Klein-Gordon equation

i □2 + m2 ϕ = −iV ϕ

with −i (Ei − H0 ) ψ = −iV ψ

▶ Propagator for an intermediate state of momentum p is
1 i
i(−p2 +m2 )
= p2 −m 2
Electron Propagator 33

▶ In an electromagnetic field, an electron obeys the equation

(p − m)ψ = −eγ µ Aµ ψ
where p = γ µ pµ i □2 + m2 ϕ = −iV ϕ

with −i (Ei − H0 ) ψ = −iV ψ

Electron Propagator 33

▶ In an electromagnetic field, an electron obeys the equation

(p − m)ψ = −eγ µ Aµ ψ
where p = γ µ pµ i □2 + m2 ϕ = −iV ϕ

with −i (Ei − H0 ) ψ = −iV ψ

▶ Propagator for an intermediate state of momentum p is
1 i
i(−p2 +m2 )
= p2 −m 2
Thank You