Introduction to Quantum Electrodynamics

Peter Presnajder

These are lecture notes devoted to introductory chapters of QuantumElectrodynamics (QED). The notes consist of two chapters:

1. The Dirac field and the relativistic invariance- The Lorentz transformations and relativistic fields- The Dirac equation and its solutions, polarization sums- Dirac field quantization, field energy and momentum, charge- Fermions, the Dirac field propagator

2. Quantum electrodynamics and Feynman rules- QED equations of motion, Gauss law, Coulomb gauge- Free transversal electromagnetic field and its quantization- The interaction picture and the perturbation theory- Self-interacting scalar field, Feynman rules- QED in Coulomb gauge, gauge invariance- The relativistic formalism and Feynman rules

This two chapters should be followed by a part devoted to simple appli-cations of Feynman perturbation technique:

3. Elementary processes in QED- Scattering amplitudes and the differential cross-section- Kinematics of binary processes, decay rate of an unstable particle- The scattering e− e+ → µ− µ+, the square of the scattering amplitude- Unpolarized scattering and its differential cross-section- The scattering e− µ− → e− µ−, the square of the scattering amplitude- Crossing symmetry, Mandelstam variables, crossed channels- Compton scattering e− γ → e− γ

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- The polarization sum for photons - Ward identity- Klein-Nishina formula for the cross-section of unpolarized e− γ scattering- The annihilation e− e− → γ γ, crossing symmetry and cross-section

1 The Dirac field and its relativistic invari-

ance

1.1. Lorentz transformations

We shall label the points of the Minkowski space-time as follows:

x = (xµ) = (x0, x1, x2, x3) = (t, ~x) , (1.1)

where t = x0 denotes time and ~x = (x1, x2, x3) labels the space position.The scalar product of two 4-vectors x = (xµ) a y = (yµ) in Minkowskispace-time is given by

x.y = xµ ηµν yν = xµ yµ , (1.2)

whereyµ = ηµν yν , yµ = ηµν yν .

We lower or rise the indices with the help of the relativistic metric tensor(ηµν) = diag(1,−1,−1,−1) or its inverse (ηµν) = diag(1,−1,−1,−1)):

ηµν ηνρ = ηρµ = δρ

µ ,

where δρµ is the Kronecker symbol defined by: δρ

µ = 1 for µ = ρ and δρµ = 0

for µ 6= ρ. We adopt the Einstein summation convention: we sum over the

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same repeated upper and lower indices, e.g. xν yν = x0 y0 + x1 y1 + x2 y2 +x3 y3.

Let us consider the linear transformation which preserves the relativisticscalar product x.y = xµ ηµν yν of any two 4-vectors x and x:

xµ 7→ Λµν xν , yµ 7→ Λµ

ν yν . (1.3)

Let us rewrite the scalar product in matrix notation x.y = xT η y, where ydenotes the column with 4 components yµ, xT is a row with 4 components xµ

and η is 4× 4 matrix with elements ηµν . Similarly the transformation law inmatrix notation can be written as follows: x 7→ x′ = Λ x a y 7→ y′ = Λ ywhere Λ is the 4× 4 matrix with elements Λµ

ν . The invariance of the scalarproducts induces a constraint on admissible matrices Λ:

xT η y = x′T η y′ = xT ΛT η Λ y ⇔ ΛT η Λ = η . (1.4)

Here, ΛT is the transposed matrix of the matrix Λ. Such matrices form a Liegroup, called Lorentz group. The elements of the Lorentz group which canbe expressed in exponential form

Λ = exp (−iωJ) ⇔ Λαβ = (exp (−iωJ))α

β (1.5)

The symbol ω in the exponent is a real number and J is 4×4 matrix satisfyingcondition

JT η + η J = 0 ⇔ JT = −η J η . (1.6)

This condition is a direct consequence of (1.4) and (1.5). There are 6 inde-pendent 4×4 matrices Jµν = Jνµ satisfying (1.6). Their αβ matrix elementsare given as:

(Jµν)αβ = i(ηµ

α ηνβ − ην

α ηµβ

). (1.7)

In this form we can freely rise and lower all indices simultaneously on bothsides.

In any proper Lorentz transformation Λ = exp (−iJ) the exponent J isgiven as a linear combination

J =1

2ωµνJ

µν µ ν = 0, 1, 2, 3 , (1.8)

specified by 6 real parameters ωµν = −ωνµ. The matrix Jµν generate Lorentztransformations in (µν)-plane in Minkowski space:

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• Three generators Jij, i, j = 1, 2, 3, generate rotations in 3-space (forits specification we need 3 parameters - 3 Euler angles θ, φ, ψ);

• Three generators J0j, j = 1, 2, 3, generate boosts (the transformationto a system moving with a speed ~v with respect to the original referenceframe - this requires again 3 parameters).

For infinitesimal Lorentz transformation, specified by infinitesimal pa-rameters ωµν we obtain

xα 7→(

exp (− i

2ωµνJ

µν)

)α

β

xβ

=

(δαβ − i

2(Jµν)α

β + . . .

)xβ = xα − i

2(Jµν)α

β xβ + . . .

xα 7→ xα + ωαβ xβ , ωαβ = −ωαβ .

In the last step we have used the explicit formula (1.7).It can be easily shown that the matrices Jµν satisfy the commutation

relations for Lorentz group generators, i.e. defining relations of Lie algebraso(3, 1):

[Jµν , Jρσ] = − i (ηνρJµσ − ηµρJ

νσ + ηµσJνρ − ηνσJ

µρ) . (1.9)

Finally, we point out that it holds

Jµν xρ = i (ηµρ xν − ηνρ xµ) . (1.10)

This formula is equivalent to the relation

xµ 7→ Λµν xν , Λ = exp (− i

2ωµν Jµν) (1.11)

which tell us that 4 real numbers x = (xµ), µ = 0, 1, 2, 3,transform asrelativistic 4-vector.

1.2. Relativistic scalar fields. The relativistic scalar field φ(x) isdescribed by a (real or complex) function defined in all points of Minkowskispace-time: x 7→ φ(x). By definition, under Lorentz transformation x 7→Λ x the field φ(x) is transforming in the following way:

φ(x) 7→ T (Λ)φ(x) = φ(Λ−1x) . (1.12)

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In (1.12) Λ−1 denotes the inverse matrix of the matrix Λ. The symbolT (Λ) represents the linear operator defined by the last equation. The as-signment Λ 7→ T (Λ) defines the Lorentz group representation because it”copies” the group product:

T (Λ1) T (Λ2) = T (Λ1Λ2) , T (1) = I .

The symbol 1 denotes the 4 × 4 unit matrix (corresponding to the unity ingroup) and the symbol I is the unit operator corresponding to the identitymap: φ(x) 7→ φ(x).

Under infinitesimal Lorentz transformation xα 7→ xα + ωαβ xβ the fieldtransforms as follows

φ(x) 7→ φ(Λ−1x) = φ(xα − ωαβ xβ)

= φ(x) − i

2ωµν (J µνφ)(x) .

Comparing the last expression with the Taylor expansion of the field on afirst line, we obtain the formula for the generator of Lorentz transformationsJ µν which acts on fields as a 1-st order differential operator:

J µν = −i (xµ ∂ν − xν ∂µ) , ∂ν = ηµν ∂ν , (1.13)

where ∂ν = ∂xν . It can be easily shown that the differential operatorsJ µν = −J νµ again satisfy the commutation relations (1.9) for Lorentz groupgenerators.

Multi-component relativistic fields. Let us consider n-component field

ψ(x) =

ψ1(x). . .

ψn(x)

with components ψa(x), a = 1, . . . , n. We shall assume that under Lorentztransformation x → Λ x the field components transform as follows:

ψa(x) 7→ Sab(Λ) ψb(Λ

−1x) ≡ (T (Λ) ψ(x))a . (1.14)

The mapping ψ(x) 7→ T (Λ) ψ(x) will generate the Lorentz group represen-tation:

T (Λ1) T (Λ2) = T (Λ1 Λ2) , T (1) = I

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exactly, when Λ 7→ Sba(Λ) will be the (n × n)-matrix representation of the

Lorentz group

Sab(Λ1) Sb

c(Λ2) = Sac(Λ1 Λ2) , Sa

b(1) = δba .

Example: As an important example of multi-component field can servethe relativistic vector field V µ(x) which under Lorentz transformations mapsas follows:

V µ(x) 7→ Λµν V ν(Λ−1x) . (1.15)

We leave as an exercise to derive, in this case, the form of the differentialoperators J µν generating Lorentz transformations.

1.3. Particles and relativistic fields. In the framework of quantumtheory, the relativistic particle with mass m and 4-momentum p = (E~p, ~p),i.e., with the 3-momentum ~p and energy E~p, is described by the de Brogliewave function

1

(2π)3/2e−ip.x =

1

(2π)3/2e− iEpt + i~p.~x , Ep ≡ E~p =

√~p2 + m2 . (1.16)

Below we shall frequently use the notation Ep instead of E~p (a similar sim-plification we shall frequently use for some other quantities too).

The set of such particles with different momenta, which do not possessother internal degrees of freedom, is described by relativistic scalar field

φ(x) =1

(2π)3

∫d3~p√2Ep

(ap e−ip.x + b∗p e+ip.x

), p = (Ep, ~p) , (1.17)

i.e. the scalar field is represented by a complex linear combination of deBroglie wave functions and complex conjugated wave functions (the numeri-cal factor in front of the integral and the factor

√2Ep in the measure repre-

sent just a convenient normalization):• The first integrand describes a system of free particles with 3-momenta

~p, the complex coefficient ap ≡ a~p is proportional to the probability ampli-tude that the a-particle with 3-momentum ~p is contained in the ensemble ofparticles;

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• The second integrand is a linear combination of complex conjugatedwave functions of b-particles with 3-momenta ~p, the coefficient bp ≡ b~p is pro-portional to the probability amplitude that the b-particle with 3-momentum~p is contained in the ensemble of particles.

Note: For complex scalar fields the coefficients ap a bp are independent(there is no relation among them). Simply, we have two sorts of particles: a-particles and b-particles which are antiparticles to a-particles. Particles andantiparticles have the same mass but they possess opposite electric charge(and all other charges they possess are opposite too).

The reality condition φ(x) = φ∗(x) for real scalar field implies constraintap = b∗p. The system contains one kind of particles: the particle a is identicalto its antiparticle, the particles possess zero charges.

The free scalar field is a solution of Klein-Gordon equation

(∂µ ∂µ + m2)φ(x) = 0 . (1.18)

The relativistic invariance of Klein-Gordon equation. Under Lorentztransformation x 7→ Λ x the scalar field φ(x) φ(x) 7→ φΛ(x) = φ(Λ−1x),kde

φ(Λ−1x) =1

(2π)3

∫d3~p√2Ep

(ap e−ip.Λ−1x + b∗p e+ip.Λ−1x

).

Taking into account, that under substitution p′ = Λp

d3~p

2Ep

=d3~p′

2Ep′, (1.19)

and that p.Λ−1x = pT η Λ−1x = pT ΛT η x = (Λp)T η x, we can write

φΛ(x) =1

(2π)3

∫d3~p′√2Ep′

(a′p′ e

−ip′.x + b′∗p′ e+ip′.x

), (1.20)

where

a′p′ =

√Ep′

Ep

ap , b′p′ =

√Ep′

E ′p

bp , p′ = Λp . (1.21)

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We see that the transformed field φΛ(x) is again a solution of the Klein-Gordon equation (1.18):

(∂µ ∂µ + m2)φΛ(x) = 0 ,

however, with expansion coefficients transformed according to (1.21).We have constructed a representation of the Lorentz group realized in

space of field configurations. In fact, we have two independent unitary rep-resentations: the first one in the space of particle configurations and theother one in the space of antiparticle configurations (the coefficients ap a bp

are independent, and the unitarity is due to the positivity of the integralmeasure).

1.4. The Dirac equationThe free particle relativistic equation, the Klein-Gordon equation, had

been first suggested by E. Schrodinger. He used the quantization rule: theenergy E and 3-momentum ~p should be replaced in the Hamiltonian (formulafor the energy) by differential operators

E 7→ i∂t , pj 7→ i∂j , (1.22)

where ∂j = ∂xj , j = 1, 2, 3. However, he found problems with relativis-tic formulation of the hydrogen atom problem. Therefore, he applied therule (1.22) within the non-relativistic formula for the electron energy mov-ing in the Coulomb field of proton. He solved and published his well-knownSchrodinger equation.

The relativistic version of the quantum equation of motion was publishedafter by Klein and Gordon (the equation was known to V. A. Fock too). Thefact, that due to relativistic invariance the Klein-Gordon equation containedsecond order time derivative (beside second order position derivatives) leadsto various complications with quantum-mechanical interpretation of the for-malism.

This fundamental problems motivated P. A. M. Dirac to search for arelativistic equation describing the free particle with mass m containing justthe first order space-time derivatives. Dirac postulated the equation in theform

(iγµ ∂µ − m) ψ(x) = 0 , ∂µ = ∂xµ , (1.23)

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with the coefficients γµ being unknown constant objects.Dirac determined γ’s from the requirement that (1.23) should contain

particle-like solutions, i.e., the solutions of the first order Dirac equation(1.23) should simultaneously satisfy the second order Klein-Gordon equation.

Multiplying (1.23) with the operator (iγν ∂ν + m) we obtain second orderequation

(iγν ∂ν + m)(iγµ ∂µ − m) ψ(x) = 0 ,

(γν γµ ∂ν ∂µ + m2) ψ(x) = 0 . (1.24)

Since the differential operator ∂µ commutes with any other ∂ν we can write

γν γµ ∂ν ∂µ =1

2{γµ, γν} ∂ν ∂µ .

Equation (1.24) will be consistent with Klein-Gordon equation (1.18) pro-vided the coefficients γµ satisfy anticommutation relations:

{γµ, γν} ≡ γµ γν + γν γµ = 2 ηµν . (1.25)

This follows directly from

γν γµ ∂ν ∂µ =1

2{γµ, γν} ∂ν ∂µ = ηνµ ∂ν ∂µ = ∂µ ∂µ .

Equations (1.25) represent the well-known defining relations for the (real)Clifford algebra generators:

• pre µ 6= ν the generators anti-commute: γµ γν = −γν γµ, and• they are normalized by: (γ0)2 = +1 and (γi)2 = −1, i = 1, 2, 3.The algebra of γ-matrices (1.25) may be realized in terms of 4×4 complex

matrices, so called Dirac matrices. In the Weyl (chiral) basis they are givenin 2× 2 block form by the formulas:

γ0 =

(0 11 0

), γj =

(0 σj

−σj 0

), j = 1, 2, 3 . (1.26)

All entering elements are 2 × 2 matrices: 0 is the zero matrix, 1 is the unitmatrix and σj, j = 1, 2, 3, are the Pauli matrices:

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

). (1.27)

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Note: Dirac used a different realization of γ-matrices:

γ0 =

(1 00 −1

), γj =

(0 σj

−σj 0

), j = 1, 2, 3 . (1.28)

This Dirac (or standard) realization of γ-matrices is equivalent to the Weylrealization. In what follows we shall use the Weylovu representation.

Products and linear combinations of matrices γµ generate the algebra ofall 4 × 4 matrices. The conventional basis of this matrix algebra is formedby the following 16 matrices:

1 , γµ , Sµν =i

4[γµ, γν ]

γµ = γ5 γµ , γ5 = iγ0γ1γ2γ3 . (1.29)

The indices µ , ν take the values 0,1,2,3:• here the symbol 1 represents the 4×4 unit matrix (we do not introduce

a new notation - from the context it will be clear the size of the unit matrixin question),

• 4 matrices γµ have been introduced earlier in (1.26) in Weyl represen-tations (or (1.28) in Dirac representation);

• since Sµν = −Sνµ we have 6 independent Sµν matrices;• we add 4 matrices γµ and the matrix γ4 ≡ −iγ5.In Weyl basis those matrices have the form

S0j = − i

2

(σj 00 −σj

), Sij =

1

2εijk

(σk 00 σk

)≡ 1

2εijk Σk ,

γ0 = −(

0 1−1 0

), γj = −

(0 σj

σj 0

), γ5 =

(1 0−0 1

). (1.30)

All indices i, j, k, . . ., take values 1, 2, 3 (the summation convention is under-stood). These matrices are chosen so that all expressions ψ(x) A ψ(x) arereal for A being some of the 16 matrices 1, γµ, Sµν , γµ and iγ5 (in fact, theyform the Lie algebra basis of the conformal group SO(4, 2)).

Exercise: Find the form of matrices Sµν , γµ and iγ5 in Dirac realization.

The properties of γ-matrices.

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• Hermitian conjugation

㵆 = γ0 γµ γ0 ⇔ γ0† = γ0 , γj† = γj , j = 1, 2, 3 .

• The properties of γ5

(γ5)2 = 1 , γ5† = γ5 , {γ5, γµ} = 0 , [γ5, Sµν ] = 0 .

Spinor representations of Lorentz group.

It can be easily shown, using the anti-commutation relations for Dirac γµ

matrices, that the matrices Sµν , µ, ν = 0, 1, 2, 3, satisfy the commutationrelations for Lorentz group Lie algebra:

[Sµν , Sρσ] = − i (ηνρSµσ − ηµρS

νσ + ηµσSνρ − ηνσS

µρ) . (1.31)

We see that we have obtained a 4-dimensional representation of the Lorentzgroup in the space C4. However, it is a reducible representation, since thegenerators Sµν are 2× 2-block diagonal:

• the 2× 2 matrices in the left upper corner form a 2-dimensional spinorrepresentation of the Lorentz group (the fundamental representation of thegroup SL(2,C) of 2× 2 matrices with the unit determinant),

• similarly, the 2×2 matrices in the right lower corner form a 2-dimensionalconjugated spinor representation of the Lorentz group (the anti-fundamentalrepresentation of the group SL(2,C)).

The linear combinations of matrices Sµν (and their exponents) can act,besides the space C4, in the space of matrices. If A is a 4 × 4 matrix, thenSµν act as commutators (so called, adjoint action or adjoint representation):

A 7→ [Sµν , A] .

In particular, we have

[Sµν ,1] = 0 , [Sµν , γ5] = 0 . (1.32)

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That means that the unit matrix 1 and the matrix iγ5 transform as scalarsunder adjoint action by Sµν .

Similarly, we have

[Sµν , γρ] = (ηµρ γν − ηνρ γµ) ,

[Sµν , γρ] = (ηµρ γν − ηνρ γµ) . (1.33)

These relations tell us that quartets γµ and γµ) transform as 4-vectors underadjoint action by Sµν .

Finally, the commutation relations (1.31) for Sµν express the fact that theset of matrices Sµν , µ ν = 1, 2, 3, transforms as an (anti-symmetric) tensor.

Transformation properties of the Dirac (spinor) field.

We postulate that under Lorentz transformation Λ the Dirac field ψ(x)transforms as follows:

ψ(x) 7→ S(Λ) ψ(Λ−1x) ,

S(Λ) = S(exp (− i

2ωµν Sµν)) , pre Λ = exp (− i

2ωµν Jµν) . (1.34)

As follows from (1.14), this is a representation of Lorentz group in the spaceof spinor fields ψ(x).

Besides ψ(x), it is convenient to introduce the Dirac conjugated spinorfield

ψ(x) = ψ(x)† γ0 = (ψ∗3(x), ψ(x)∗4, ψ∗1(x), ψ(x)∗2) (1.35)

(the last formula is valid in the Weyl representations of γ-matrices). If ψ(x)is a solution of Dirac equation the Dirac conjugated spinor field satisfies theconjugated Dirac equation:

i(∂µ ψ)(x) γµ = − mψ(x) . (1.36)

The follows directly from the z Dirac equation for ψ(x) rewritten in the form

iγµ (∂µ ψ)(x) = mψ(x) . (1.37)

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By hermitian conjugation we obtain

−i∂µ ψ†(x) 㵆 = mψ†(x) .

Multiplying this equation by γ0 from the right and using the relations

(γ0)2 = 1 , γ0 γµ γ0 = 㵆 , ψ†(x) γ0 = ψ(x) ,

we arrive directly at the conjugated Dirac equation (1.36).The conjugated Dirac field transforms under Lorentz transformations as

follows:

ψ(x) = ψ(x)† γ0 7→ ψ(Λ−1x)† S†(Λ) γ0 = ψ(Λ−1x)γ0 S†(Λ) γ0 .

Expressing S(Λ) = S(exp (− i2ωµν Sµν)) and using

γ0 Sµν † γ0 = −Sµν

we obtain γ0 S†(Λ) γ0 = S−1(Λ). This formula gives the desired transforma-tion for the conjugated Dirac spinor field:

ψ(x) 7→ ψ(Λ−1x) S−1(Λ) . (1.38)

Transformations of bilinear expressions. The transformation rules forspinor and conjugated spinor fields

ψ(x) 7→ S(Λ) ψ(Λ−1x) , ψ(x) 7→ ψ(Λ−1x) S−1(Λ) , (1.39)

allow to derive the transformation rules for bilinear expressions:

S(x) = ¯ψ(x) ψ(x) 7→ ψ(Λ−1x) ψ(Λ−1x) = S(Λ−1x)P (x) = ψ(x) iγ5 ψ(x) 7→ ψ(Λ−1x) iγ5 ψ(Λ−1x) = P (Λ−1x)V µ(x) = ψ(x) γµ ψ(x) 7→ Λµ

ν ψ(Λ−1x), γν ψ(Λ−1x) = Λµν V ν(Λ−1x)

Aµ(x) = ψ(x) γ5γµ ψ(x) 7→ Λµν ψ(Λ−1x) γ5γν ψ(Λ−1x) = Λµ

ν Aν(Λ−1x)T µν(x) = ψ(x)Sµνψ(x) 7→ Λµ

ρΛνσψ(Λ−1x)Sρσψ(Λ−1x) = Λµ

ρΛνσT

ρσ(Λ−1x)

1. The transformation rules for S(x) a P (x) tell us that the correspondingbilinear expressions transform as scalar fields. The first expression is a direct

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consequence of transformation rules (1.39), in derivation of the second onerelation γ5 S(Λ) = S(Λ) γ5 is needed.

2. The transformation rules for V µ(x) a Aµ(x) mean that the correspond-ing bilinear expressions transform as vector fields. Besides the transformationrules 1.39), we need the relation

S−1(Λ) γµ S(Λ) = Λµν γν , (1.40)

for the derivation of the rule for V µ(x), in the derivation of the rule for Aµ(x)we have to take into account the fact that S(Λ) commutes with γ5.

3. The transformation rules for T µν(x) tell us that T µν(x) transforms asa (antisymmetric) 2-nd order tensor. In the derivation we need the explicitformula (1.29) for Sµν and the relation (1.40).

Appendix A: The Fierz identities

Relativistic invariance of the Dirac equations.We shall show, that the field

ψΛ(x) = S(Λ) ψ(Λ−1x)

is a solution of the Dirac equation

(iγµ ∂µ − m) ψΛ(x) = 0 .

Using (1.40) we can put the left-hand side into the form

(iγµ ∂µ − m) S(Λ) ψ(Λ−1x) = S(Λ) (iΛµν γν ∂ν − m) ψ(Λ−1x) .

Introducing the new variable x′ = Λ−1x we can express the partial deriva-tives as follows

∂′ν ≡ ∂x′ν = (Λ−1)νµ∂µ = Λµ

ν ∂µ .

This equation is a direct consequence of the relation η Λ = Λ−1T η. We seethat the Dirac equation for ψΛ(x) is equivalent to

S(Λ) (iγν ∂′ν − m) ψ(x′) = 0 .

The last equality follows from the fact that ψ(x) is a solution of Dirac equa-tion.

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Particle solutions of Dirac equation.We search the solution of Dirac equation which is proportional to the de

Broglie wave function describing free particle with mass m, momentum ~p andenergy Ep =

√~p2 + m2 > 0. Such solution is proportional to the plane

waveψ(x) ∼ u(p) e−ip.x , p = (Ep, ~p) .

Inserting such ψ(x) into Dirac equation and the formula i∂µ e−ip.x = pµ e−ip.x

we obtain for the spinor u(p) the algebraic equation:

(γµ pµ − m) u(p) = 0 . (1.41)

Since our γ-matrices possess 2×2 block form we rewrite the 4-componentspinor u(p) in terms of two 2-component spinors uL(p) and uR(p)

u(p) =

(uL(p)uR(p)

).

Taking now γµ in Weyl representation we obtain the system of two entangledalgebraic equations (we omit the argument p in uL,R(p) and simply E insteadof Ep):

( −m E − ~p.~σE + ~p.~σ −m

) (uL

uR

)= 0 ⇔ (E − ~p.~σ) uR = muL

(E + ~p.~σ) uL = muR.

(1.42)With the help of formulas

(E − ~p.~σ) (E + ~p.~σ) = (E + ~p.~σ) (E − ~p.~σ) = E2 − ~p2 = m2

it can be easily seen that the solution of (??) is given as

uL(p) =√

E − ~p.~σ ξ , uR(p) =√

E + ~p.~σ ξ , (1.43)

where ξ is an arbitrary constant 2-component spinor. The square roots√E ∓ ~p.~σ should be understood as Taylor expansions of

√E ∓ ~p.~σ =

√E

√1∓ E−1~p.~σ =

√E

(1 ∓ 1

2

~p.~σ

E. . .

).

Since eigenvalues of the matrix E−1~p.~σ are in absolute value less than 1, theexpansion in powers of (E−1~p.~σ)n is well defined.

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There are two linear independent constant spinors ξ, we shall denote themby ξs, s = ±1/2. We choose them orthonormal, so that it holds

ξr†ξs = δrs , r, s = ±1/2 . (1.44)

In this way we obtain 2 linear independent particle solutions of Dirac equation

us(p) =

( √E − ~p.~σ ξs√E + ~p.~σ ξs

)s = ±1/2 , (1.45)

which satisfy normalization conditions

us(p) us(p) = 2mδrs . (1.46)

We search the solution proportional to conjugated de Broglie wave func-tions in the form:

ψ(x) = v(p) e+ip.x , p = (E~p, ~p) .

As we shall see, they describe antiparticles with energy E~p and momentum~p. Performing the similar steps as before, we obtain two linear independent(antiparticle) solutions

vs(p) =

( √E − ~p.~σ ηs

−√E + ~p.~σ ηs

), s = ±1/2 , (1.47)

depending on two orthonormal constant spinors ηs, s = ±1/2 satisfying

ηr†ηs = δrs , r, s = ±1/2 . (1.48)

We point out that the bases {ξs} and {ηs} are not related to each other andwe can choose them independently. The solutions vs(p) are normalized, up tosign, as us(p) and antiparticle solutions are orthogonal to particle solutions:

vs(p) vs(p) = −2mδrs ,

us(p) vs(p) = vs(p) us(p) = 0 . (1.49)

The general solution ψ(x) of Dirac equation is given as a linear combina-tion of particle and antiparticle plane wave solutions

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(as

p us(p) e−ipx + bs†p vs(p) eipx

). (1.50)

16

Similarly, can be expressed the Dirac conjugated solution

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(bsp vs(p) e−ipx + as†

p us(p) eipx)

. (1.51)

Example: Prove the following formulas

ur†(~p) us(~p) = vr†(~p) vs(~p) = 2 Ep δrs,

ur†(~p) vs(−~p) = vr†(−~p) us(~p) = 0 . (1.52)

We have used the notation us(~p) = us(p) for p = (E~p, ~p), and similarlyvs(−~p) = vs(p′) pre p′ = (E~p,−~p). In what follows we shall use suchnotation whenever it is convenient.

Spin sums.The spin sums express the completeness of the found solutions of Dirac

equation. Our goal is the calculation of the following sums∑

s=±1/2

usa(p) us

b(p) ,∑

s=±1/2

vsa(p) vs

b(p) .

The indices a, b = 1, 2, 3, 4, label the components of spinors us(p), vs(p)and Dirac conjugated spinors us(p), vs(p). Thus, the searched sums can beinterpreted as 4× 4 matrices (acting in spinor spaces).

Let us consider the first sum

∑s

usa(p) us

b(p) =∑

s

( √E − ~p.~σ ξs√E + ~p.~σ ξs

)

a

(ξs†√E + ~p.~σ, ξs†√E − ~p.~σ

)b

.

Taking into account the completeness relation∑

s ξsα ξs†

β = δαβ, α, β = 1, 2,valid for the orthonormal basis of 2-component spinor {ξs}, we obtain

∑s

usa(p) us

b(p) =

( √E − ~p.~σ

√E + ~p.~σ

√E − ~p.~σ

√E − ~p.~σ√

E + ~p.~σ√

E + ~p.~σ√

E + ~p.~σ√

E − ~p.~σ

)

ab

=

(m E − ~p.~σ

E + ~p.~σ m

)

ab

.

17

Now, taking into account the explicit form of γ-matrices we can rewrite thespin sum as follows:

∑s

usa(p) us

b(p) = (γµ pµ + m1)ab ⇔∑

s

us(p) us(p) = γµ pµ + m1 ,

(1.53)where 1 denotes the 4× 4 unit matrix. In the second formula we suppressedthe matrix (spin) indices - this for of spin sums is frequently used.

The formula for the second polarization sum can be derived analogously:

∑s

vsa(p) vs

b(p) = (γµ pµ − m1)ab ⇔∑

s

us(p) us(p) = γµ pµ − m1 .

(1.54)

Appendix B: The transformation properties of particle solu-tions.

Under Lorentz transformation Λ = exp(− i2ωµν Jµν) the Dirac field trans-

forms as follows:

ψ(x) 7→ S(Λ) ψ(Λ−1x) = e−i2

ωµν Sµν

e−i2

ωµν J µν

ψ(x)

= e−i2

ωµν (J µν + Sµν) ψ(x) .

Here J µν are generators of Lorentz transformations in Minkowski space (see(1.13)) and Sµν denote the generators in spinor representation (see (1.29)).

Let us act on a-particle part of the Dirac equation

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(as

p us(p) e−ipx + . . .)

. (1.55)

with inverse Lorentz transformation: ψ(x) 7→ S(Λ−1) ψ(Λx):• The matrix S(Λ−1) just mixes various solutions us(p), s = 1, 2,

S(Λ−1) us(p) =∑

s′Ds′s(W (Λ, p))us′(p) , (1.56)

where the summation is over s′ = 1, 2, i.e. (Dss′(W (Λ, p))) is 2× 2 matrix,its dependence on Λ and p is determined below;

18

• In integral (1.55) we perform substitution x 7→ x′ = Λx, similar stepsas by scalar field leads to the transformation rule for expansion coefficientsas

p:

asp 7→

√EΛp

Ep

∑

s′as′

Λp Ds′s(W (Λ, p)) . (1.57)

For expansion coefficients bsp a similar rule can be derived.

Wigner little group. Wigner proposed following method how to determinethe matrix (Dss′(W (Λ, p))):

• First we introduce standard rest 4-momentum k0 = (m,~0) for a particlewith m. With boost, i.e. Lorentz transformation determined by matrixLµ

ν(p) with components:

L00(p) = C(p) , Li

0(p) = L0i(p) = S(p) pi ,

Li0(j) = δi

j − pi pj + pi pj C(p)

we transform the particle to the rest-frame in which the particle possesses the4-momentum p = (Ep, ~p). The matrix elements of Lµ

ν(p) contain quantitiesdefined as follows:

S(p) =|~p|m

, C(p) =√

1 + S2(p) ~p =~p

|~p| .

• For any Lorentz transformation Λ we consider a sequence of transfor-mations

k0 7→ W (Λ, p) k0 ≡ L−1(Λp) Λ L(p) k0 = L−1(Λp) Λ p = k0 .

In the first step we used the relation L(p) k0 = p, and in the last one therelation L−1(p′) p′ = k0 valid for any p′.

• We see that it holds W (Λ, p) k0 = k0, i.e. the transformation W (Λ, p)belongs to the stability group of the standard 4-vector k0 = (m,~0). However,such transformations are just 3-rotations in space: W (Λ, p) ∈ SO(3), whereSO(3) denotes the group of spatial rotations.

• Consequently, the D(W ) is 2 × 2 unitary matrix D†(W ) = D−1(W )(because, 3-rotations in spinor space are generated by hermitian matrices

19

Sk = i2εijk Sij). It is well-known that unitary irreducible SO(3) represen-

tation by N × N matrices corresponds to spin s = (N − 1)/2. In our case,N = 2, i.e. Dirac particles possess spin s = 1/2.

1.5. The Dirac field quantization.

Dirac Hamiltonian. The Dirac Lagrangian density, which Euler-Lagrangeequations corresponds just to the Dirac equation, can be chosen as

LD(ψ, ψ) = ψ(x) (iγµ ∂µ − m) ψ(x) .

The canonical field-momentum π(x) conjugated to the field ψ(x) is obtainedby taking the derivative ψ = ∂0ψ of the Lagrangian density in question:

π =∂LD

∂ψ= i ψ γ0 = i ψ† .

We see that the canonical field-momentum π(x) is not an independent fieldvariable, but is simply proportional to the Dirac conjugated field ψ(x). Thatmeans that the system contains, rather trivial, field constraints. They donot influence the description of the system, and we skip their discussion (fordetails see, e.g. [Weinberg]).

The Hamiltonian of the system in question is given by the formula

H =

∫d3~p (π ψ − LD)

Simple calculations give the result

HD =

∫d3~x ψ(x)(−i~γ.~∂ + m)ψ(x) .

This expression is called Dirac Hamiltonian and we label it with subscriptD.

The energy operator. Let us insert into Dirac Hamiltonian the generalsolutions of the Dirac equation for ψ(x) and ψ(x):

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(as

p us(p) e−ipx + bs†p vs(p) eipx

)

20

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(bsp vs(p) e−ipx + as†

p us(p) eipx)

. (1.58)

After insertion the formula for HD reads

HD =1

(2π)6

∫d3~pd3~p′√2Ep 2Ep′

∫d3~x

∑

s,s′

(bs′p′ v

s′(p′) e−ip′x + as′†p′ us′(p′) eip′x

)

× (as

p us(p) e−ipx + bs†p vs(p) eipx

). (1.59)

Using the well-known formula∫

d3~x e±i(~p±~p′).~x = (2π)3 δ(~p± ~p′)

we can perform trivially the integration over d3~x, and then we can integratedirectly over d3~p′.

Multiplying the two integrand factors (. . .) × (. . .) in 1.60 we obtain fourterms. Taking into account the formulas 1.52 (see Example)

us′†(~p) us(~p) = vs′†(~p) vs(~p) = 2 E~p δs′s , us′†(~p) vs(−~p) = vs′†(−~p) us(~p) = 0 ,

we obtain two zero contributions and in the two remaining terms we can per-form the summation over s′. In this way we obtain a preliminary expressionfor the Dirac field energy

HD =1

(2π)3

∫d3~p√2Ep

∑s

(as†

p asp − bs

p bs†p

). (1.60)

Note: Let us remind that for scalar field an analogous calculation givesthe following result for the energy of a real scalar field

H =1

(2π)3

∫d3~p√2Ep

1

2

(a†p ap + ap a†p

). (1.61)

After quantization the coefficients ap = a~p and a†p = a†~q are replaced byoperators satisfying the canonical commutation relations for the bosonic an-nihilation and creation operators:

[ap, a†q] = (2π)3 δ(~p − ~q) ,

21

[ap, aq] = [a†p, a†q] = 0 . (1.62)

The annihilation and creation operators act in Fock space F , which is aHilbert space spanned by normalized n-particle states

|p1, p2, . . . , pn〉 ∼ a†p1a†p2

. . . a†pn|0〉 , (1.63)

where the symbol ∼ means that we do not indicate explicitly the normaliza-tion factor on right hand side. The state |0〉 is a normalized vacuum statein Fock space, 〈0|0〉 = 1, which does not contain any particle and is definedby relations

ap |0〉 = 0 , for all p = (Ep, ~p) . (1.64)

The scalar field energy given by (1.61) is ill-defined (divergent) in the Fockspace.

The consistency of QFT formalism requires that all physical field quanti-ties, such as energy, should be well-defined in Fock space. This is guaranteedwhen they are given in terms of a normal products of annihilation and cre-ation operators: in any term containing products of those operators we putall creation operators a†q to the left and all annihilation operators ap to theright:

: ap . . . a†q . . . a†q′ . . . ap′ : = a†q . . . a†q′ ap . . . ap′ , . (1.65)

The normal product is labeled by : . . . :. The right hand side contains firstthe product of all creation operators followed the product of all annihilationoperators entering the original expression. In particular,

: ap a†q : = : a†q ap : = a†q ap .

The normal ordered expression energy is well-defined in the Fock space

H =1

(2π)3

∫d3~p√2Ep

:(a†p ap + ap a†p

): =

1

(2π)3

∫d3~p√2Ep

: a†p ap : .

(1.66)The multi-particle mean values of energy are positive and in vacuum statethe mean energy vanish: 〈0|H |0〉 = 0. Without normal ordering the meanvalues of energy are all divergent.

Such approach can not be applied in the case of Dirac Hamiltonian. Con-sidering in (1.60) bosonic annihilation and creation operators we would ob-

22

tain:

HD =1

2π3

∫d3~p√2Ep

∑s

(as†

p asp − bs†

p bsp

).

Although it is a finite expression in Fock space, it is unbounded from below.However, the energy should be bounded below, otherwise the system will benot stable.

Dirac found an unexpected solution, he postulated that the spinor fieldsplane wave expansion enter fermionic annihilation and creation operatorssatisfying anti-commutation relations:

{arp, a

s†q } = {br

p, bs†q } = (2π)3 δ(~p − ~q) δrs ,

{arp, a

sq} = {ar†

p , as†q } = {br

p, bsq} = {br†

p , bs†q } = 0 .

{arp, b

s†q } = {br

p, as†q } = {ar

p, bsq} = {br†

p , as†q } = 0 . (1.67)

In particular,

(arp)

2 = (as†p )2 = (br

p)2 = (bs†

p )2 = 0 . (1.68)

Similarly as in bosonic case, the fermionic annihilation and creation op-erators act in Fock space F spanned by multi-particle orthonormal states

|(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉 ∼ ar1†p1

. . . arn†pn

bs1†q1

. . . bsm†qm|0〉 .

(1.69)State |0〉 is the normalized vacuum state in Fock space defined as

asp |0〉 = bs

p |0〉 = 0 , pre s = ±1/2 a vsetky p = (Ep, ~p) . (1.70)

Because, the squares of fermionic operators vanish the multi-particle statescannot contain 2 identical particles with same spin and momentum - fermionicparticles satisfy Pauli principle: the particle occupation numbers in (1.69)are 0 or 1.

As a next step, Dirac had modified adequately the normal product:

: arp . . . bs†

q . . . ar′†q′ . . . bs′

p′ : = ± ar′†q′ . . . bs†

q arp . . . bs′p′ . (1.71)

This definition is similar to bosonic case but there are essential differences:

23

• The right hand side contains first the product of all creation operatorsfollowed by all annihilation operators entering the expression on left handside,

• in addition the definition contains a sign factor (−1)n, where n is thenumber of neighbor transpositions needed to reshuffle the rights side to itsnormal order on left hand side.

In particular,: br†

q bsp : = − : bs

p br†q : = br†

q bsp . (1.72)

Using this normal ordering, the energy of a free Dirac field is representedby a sum of two positive terms:

HD =1

(2π)3

∫d3~p

∑s

:(Ep as†

p asp − Ep bs

p bs†p

):

=1

(2π)3

∫d3~p

∑s

(Ep as†

p asp + Ep bs†

p bsp

). (1.73)

The momentum operator. The quantum momentum operator is definedby the following expression

~P =

∫d3~x : ψ(x)(−~∂)ψ(x) : , (1.74)

where ψ(x) a ψ(x) are given as linear combinations (1.60) of particle solutionsof Dirac equation. As we are consider quantum field we introduced in (1.74)the normal product from the very beginning. Performing analogical steps asused for energy operator we obtain the following result:

~P =1

(2π)3

∫d3~p

∑s

(~p as†

p asp + ~p bs†

p bsp

). (1.75)

The conserved electric charge. The Dirac Lagrangian LD is invariantunder global gauge transformations represented by a constant change of thephase of spinor field:

ψ(x) 7→ eiα ψ(x) , ψ(x) 7→ e−iα ψ(x) , α − realna konstanta .

24

On a classical (non-quantized) level this invariance leads to the continuityequation for the current density:

• Let ψ(x) and ψ(x) are solutions of the Dirac equations

iγµ (∂µ ψ)(x) = mψ(x) ,

i(∂µ ψ)(x) γµ = − mψ(x) .

Then,

∂µ (ψ(x) γµ ψ(x)) = (∂µ ψ)(x) γµ ψ(x) + ψ(x) γµ (∂µ ψ)(x)

= im ψ(x) ψ(x) − im ψ(x) ψ(x) = 0 . (1.76)

• Using (1.76) it follows directly that the current density

jµ(x) = e ψ(x) γµ ψ(x) , (1.77)

satisfies continuity equation

∂µ jµ(x) = 0 . (1.78)

• The total sa charge of particles Q(t = x0) corresponding to the currentjµ(x))

Q(t) = e

∫d3~x j0(x) = e

∫d3~x ψ(x) γ0 ψ(x) , (1.79)

is conserved provided the positions of particles are restricted to a finite do-main in the space, i.e. ψ(x) = 0 and ψ(x) = 0 for |~x| ≥ R. The parametere in(1.79) is the charge of Dirac particle, as we shall see the charge of anti-particle is −e. Using continuity equation for jµ(x) it can be proved easilythat the total charge Q(t) is conserved in time:

Q(t) = e

∫

|~x|<R

d3~x ∂0 j0(x)

= −e

∫

|~x|<R

d3~x ∂i ji(x) = e

∫

|~x|=R

dSiji(x) = 0 .

In the last step we used the Gauss theorem: we integrated the functiondiv~j(t, ~x) over 3-ball |~x| ≤ R and used the fact that the 3-current vanish on3-sphere with radius ~x| = R.

25

In the quantum case the field charge is defined by same expression butwith normal ordered charge density (we skip variable t):

Q = e

∫d3~x : ψ(x) γ0 ψ(x) : , (1.80)

We insert here the ψ(x) a ψ(x) particle solutions of Dirac equation with ex-pansion coefficients being corresponding annihilation and creation operators.Performing similar steps as used for energy operator we obtain the result:

Q =e

(2π)3

∫d3~p

∑s

(as†

p asp − bs†

p bsp

). (1.81)

In what follows we shall identify the conserved current jµ(x) = e :ψ(x) γµ ψ(x) : with the electromagnetic current: the a-particles we identifywith electrons with charge e and b-particles with positrons with charge −e.The quantity Q describes the total charge of the field.

The interpretation of the free Dirac field. The state of the field with nparticles (electrons) and m anti-particles (positrons) with given momentaand spins is given by the following vector normalized vector in Fock space:

|(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉 ∼ ar1†p1

. . . arn†pn

bs1†q1

. . . bsm†qm|0〉 .

(1.82)We shall show that (1.82) are eigenstates of the field energy operator HD,

the field momentum operator ~P and the field charge operator Q:

HD |(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉

=

(n∑

i=1

Epi+

m∑j=1

Eqj

)|(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉 ,

(1.83)~P |(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉

=

(n∑

i=1

~pi +m∑

j=1

~qj

)|(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉 , (1.84)

Q |(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉

26

= e (n− ) |(p1, r1), . . . , (pn, rn); (q1, s1), . . . , (qm, sm)〉 . (1.85)

These eigenvalue equations are a direct consequence of formulas

[as†p as

p, as′†p′ ] = δ(~p− ~p′) δss′ as†

p , [as†p as

p, as′p′ ] = −δ(~p− ~p′) δss′ as′†

p′ ,

[bs†p bs

p, bs′†p′ ] = δ(~p− ~p′) δss′ bs†

p , [bs†p bs

p, as′p′ ] = −δ(~p− ~p′) δss′ bs†

p . (1.86)

The first relation can be obtained as follows:

[as†p as

p, as′†p′ ] = as†

p asp, as′†

p′ − as′†p′ as†

p asp, =

as†p as

p, as′†p′ + as†

p as′†p′ as

p, = as†p {as

p, as′†p′ } = as†

p δ(~p− ~p′) .

Last expression is equivalent to the desired formula (we used the definitionof the commutator, then we anti-commutate two creation operators, andfinally we used the canonical anti-commutation relation among annihilationand creation operators). The remaining relations can be proved similarly.

The operator identity [A, BC] = [A,B]C + B[A,C] gives more generalformulas

[ar†p ar

p, ar1†p1

. . . arn†pn

] =n∑

i=1

δ(~p− ~pi) δrri ar1†p1

. . . arn†pn

,

[bs†q bs

q, bs1†q1

. . . bsm†qn

] =m∑

j=1

δ(~q − ~qj) δssj bs1†q1

. . . bsn†qm

. (1.87)

Equations (1.87) lead directly lead to the eigenvalue equations (1.83)-(1.85) for the energy, momentum and charge operators:

• It follows from (1.83) and (1.85) that the total energy and total momen-tum of the system are conserved (time independent). Further it follows thatthe total energy and the total momentum are sums of individual energies andmomenta of all individual particles in the state in question. Both propertiesare typical for ensembles of non-interacting particles: there are no bindingenergies and the individual energies and momenta of particles do not changein time (there no mutual interaction among particles).

• Similarly, the total charge of the system is conserved. Moreover, thetotal charge is equal to difference of charges shared by particles (electrons)and that shared by anti-particles (positrons): electrons and positrons havethe same mass (the relation between Ep a ~p is the same for both) but the

27

possess opposite electric charge (the electron charge e and the positron chargeis −e).

Equal-time canonical anti-commutation relations for Dirac field. Our goalis to prove the anti-commutation relations among spinor fields ψ(X) andψ†(y) at equal time x0 = y0 = t:

{ψa(t, ~x), ψb(t, ~y)} = {ψ†a(t, ~x), ψ†b(t, ~y)} = 0 .

{ψa(t, ~x), ψ†b(t, ~y)} = δ(~x− ~y) δab . (1.88)

We shall prove the last relations, the first two can be proved along similarlines. We insert the field expansions

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(as

p us(p) e−ipx + bs†p vs(p) eipx

), x = (t, ~x)

ψ†(y) =1

(2π)3

∫d3~p′√2Ep′

∑

s′

(bs′p′ v

s′†(p′) e−ip′x′ + as′†p′ us′†(′p) eip′y

), y = (t, ~y) ,

into the last anti-commutator in (1.88):

{ψa(t, ~x), ψ†b(t, ~y)} =1

(2π)6

∫d3~pd3~p′√2Ep 2Ep′

∑

s,s′

{asp us

a(p) e−ipx + bs†p vs

a(p) eipx, bs′p′ v

s′†b (p′) e−ip′x′ + as†

p us′†b (p′) eip′x′}

=1

(2π)6

∫d3~pd3~p′√2Ep 2Ep′

∑

s,s′({as

p, as′†p′ }us

a(p) us′†b (p′) e−ipx+ip′x′ + {bs

p, bs′†p′ } vs

a(p) vs′†b (p′) e+ipx−ip′x′

)

=1

(2π)3

∫d3~p

2Ep(∑s

usa(p) us′†

b (p′) ei~p.(~x−~y) +∑

s

vsa(p) vs′†

b (p′) e−i~p.(~x−~y)

)

=1

(2π)3

∫d3~p

2Ep

28

((Ep γ0 − ~γ.~p + m)γ0 ei~p.(~x−~y) + (Ep γ0 − ~γ.~p − m)γ0 e−i~p.(~x−~y)

)ab

=1

(2π)3

∫d3~p ei~p.(~x−~y) δab = δ(~x− ~y) δab .

During calculations we have used the following steps:• The anti-commutation canonical relations for annihilation and creation

operators{as

p, as′†p′ } = {bs

p, bs′†p′ } = (2π)3 δ(~p− ~q′) δss′ ,

• the sum rules

∑s

usa(p) us′†

b (p′) = (Ep γ0 − ~γ.~p + m)γ0 ,

∑s

vsa(p) vs′†

b (p′) = (Ep γ0 − ~γ.~p − m)γ0 ,

• we replaced ~p 7→ −~p in the integral containing exp(−i~p.(~x − ~y)), andfinally we have used the formula

∫d3~p exp(i~p.(~x− ~y)) = (2π)3 δ(~x− ~y) .

The Dirac field propagator. We shall see later that the free fieldpropagators have a key role when the mutual interaction among particlesis described within the framework of perturbation theory. First we brieflyremind the form of the scalar field propagator, and then we derive its formfor the free Dirac field.

The scalar field propagator. Let us consider the free real scalar field

φ(x) =1

(2π)3

∫d3~p√2Ep

(ap e−ip.x + a†p e+ip.x

), p = (Ep, ~p) . (1.89)

Its propagator is defined as the vacuum mean value of the T-product of fields:

DF (x− y) = 〈0|T [φ(x)φ(y)] |0〉 . (1.90)

29

The symbol T [φ(x)φ(y)] denotes the time-ordered product of fields definedby:

T [φ(x)φ(y)] = φ(x)φ(y) pre x0 > y0 ,

T [φ(x)φ(y)] = φ(y)φ(x) pre y0 > x0 . (1.91)

Let us consider first the case x0 > y0, then

DF (x− y) = 〈0|φ(x)φ(y) |0〉 .Inserting here the plane wave expansions (1.89) of φ(x) and φ(y) we obtain

DF (x− y) =1

(2π)6

∫d3~pd3~p′√2Ep 2Ep′

〈0| ap a†p′ |0〉 e−ipx + ip′y . (1.92)

Since,〈0| ap ap′ |0〉 = 〈0| a†p a†p′ |0〉 = 〈0| a†p ap′ |0〉 = 0 .

the potential other three terms do not contribute. Now we use in the formula(1.92) for the propagator the canonical anti-commutation relation written inthe form

ap a†p′ = (2π)3 δ(~p − ~p′) + a†p′ ap .

The first term on r.h.s. does not contribute (again, 〈0| ap ap′ |0〉 = 0), thecontribution from the second term is proportional to δ-function. This allowsto perform the integration over d3~p′, and after that we obtain

DF (x− y) =1

(2π)3

∫d3~p

2Ep

e−ip(x−y) , p = (Ep, ~p) . (1.93)

Using the same steps for y0 > x0, we obtain

DF (x− y) =1

(2π)3

∫d3~p

2Ep

e+ip(x−x′) , p = (Ep, ~p) . (1.94)

It can be shown that equations (1.93) and (1.94) can be expressed interms of a one explicitly relativistic invariant formula

DF (x− y) =i

(2π)4

∫d4p

e−ip(x−y)

p2 − m2 + iε. (1.95)

Here ε > 0 is a ”small” positive number, and at the end of calculations thelimit ε → 0+ is understood.

30

The explicit relativistic invariance of the final propagator formula has itsprice. The 3 dimensional integration (1.93) and (1.94) is replaced by the 4dimensional one: the zeroth component p0 is no more fixed to Ep and theintegration over dp0 is added. Therefore, in the relativistic formula (1.95)the 4-momentum p is off the mass shell (or simply, off-shell): p2 6= m2. Thisis in contrast with the free particle 4-momentum p = (Ep, ~p) which is onthe mass shell (or simply, on-shell): p2 = m2.

The propagator for the Dirac field. The Dirac field propagator is againdefined as the vacuum mean value of the T-product of fields:

SF ab(x− y) = 〈0|T [ψa(x)ψb(y)] |0〉 . (1.96)

Here the symbol T [ψa(x)ψb(y)] denotes the time-ordered product of fermionicDirac fields which is defined similarly as the T-product of bosonic, but withthe distinction that transposition of two fermionic fields changes the sign ofthe product:

T [ψa(x)ψb(y)] = ψa(x)ψb(y) pre x0 > y0 ,

T [ψa(x)ψb(y)] = −ψa(x)ψb(y) pre y0 > x0 . (1.97)

Let us consider the case x0 > y0. We insert into formula (1.96) for thepropagator the Dirac field plane wave expansions:

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(as

p us(p) e−ipx + bs†p vs(p) eipx

),

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(bsp vs(p) e−ipx + as†

p us(p) eipx)

. (1.98)

There will be again only one contribution to the vacuum mean value whichcontains annihilation operator on the left and the creation on the right:

〈0|T [ψa(x)ψb(y)] |0〉

=1

(2π)6

∫d3~pd3~p′√2Ep 2Ep′

∑

s,s′〈0| as

p as′†p′ |0〉us

a(p) us′b (p′) e−ipx + ip′x′ .

31

In the next step we use the canonical anti-commutation relation

asp as′†

p′ = (2π)3 δ(~p − ~p′)δss′ − as′†p′ as

p .

The non-vanishing contribution proportional comes from the first term: thepresence of δ(~p − ~p′) allows a direct integration over d3~p′ and δss′ allows asummation over s′. Performing these steps we obtain

〈0|T [ψa(x)ψb(y)] |0〉 =1

(2π)3

∫d3~p

2Ep

∑s

usa(p) us

b(p) e−ip(x−x′) .

Here we recognize the polarization sum, so that we can write

〈0|T [ψa(x)ψb(y)] |0〉 =1

(2π)3

∫d3~p

2Ep

(γµpµ + m)ab e−ip(x−y)

=(iγµ∂x

µ + m)

ab

1

(2π)3

∫d3~p

2Ep

e−ip(x−y) =(iγµ∂x

µ + m)

abDF (x− y) ,

(1.99)where DF (x − y) denotes the expression valid for scalar field propagatorfor x0 > y0 (we have used the simple formula i∂x

µ exp(−ip(x − y)) =pµ exp(−ip(x− y))).

Similarly, for y0 > x0 we obtain

〈0|T [ψa(x)ψb(y)] |0〉

= − 1

(2π)6

∫d3~pd3~p′√2Ep 2Ep′

∑

s,s′〈0|bs′†

p′ bsp |0〉 vs

a(p) vs′b (p′) e+ipx− ip′y

= − 1

(2π)3

∫d3~p

2Ep

∑s

usa(p) us

b(p) e−ip(x−y)

− 1

(2π)3

∫d3~p

2Ep

(γµpµ − m)ab e+ip(x−y) . (1.100)

Using again the rule i∂xµ exp(+ip(x− y)) = −pµ exp(−ip(x− y)) we obtain

〈0|T [ψa(x)ψb(y)] |0〉

=(iγµ∂x

µ + m)

ab

1

(2π)3

∫d3~p

2Ep

e+ip(x−y) =(iγµ∂x

µ + m)

abDF (x− y) ,

(1.101)

32

where DF (x− y) denotes the expression valid for scalar field propagator fory0 > x0.

We see that in both cases we obtain the identical expression for the Diracpropagator in terms of DF (x− x′). Finally, we can write

SF (x− x′) =(iγµ∂x

µ + m1)

DF (x− x′)

=i

(2π)4

∫d4p

γµpµ + m

p2 − m2 + iεe−ip(x−x′) . (1.102)

In this final formula we suppressed the matrix indices of the Dirac propagator.

33

2 The electromagnetic field

With start with Lagrangian describing electromagnetic field acting with anexternal given source. This is important because the system contains con-straints besides equations of motion. The external source is essential for thebetter understanding of the origin of constraints and their relation to equa-tions of motion. Without external source various important features wouldbe simply lost. Moreover, the obtained results allow a natural generaliza-tion to the situation when electromagnetic fields interacts with the systemof charged particles. In this case, the electromagnetic field and the ensembleof charged particles will mutually interact - the current of charged particleswill represents dynamical source (and not a given external source).

2.1. The electromagnetic field Lagrangian.

In quantum theory the electromagnetic field is specified by 4-potentialAµ(x), µ = 0, 1, 2, 3. The Lagrangian for electromagnetic field Aµ(x) inter-acting with an external electromagnetic current density Jµ(x) is given as:

L(A, J) = − 1

4Fµν F µν − e Jµ Aν . (2.103)

Here Fµν = ∂µAν − ∂µAν is the electromagnetic field strength which is

related, in standard way, to the electric field strength ~E and magnetic fieldstrength ~B:

Ei = F i0 = ∂0A

i − ∂iA0 ⇔ ~E = ∇A0 + ~A ,

Bk =1

2εijkFij ⇔ ~B = ∇ × ~A . (2.104)

The indices i, j, k, take values 1, 2, 3 (the summation convention is under-

stood), the dot over the symbol means the time derivative, i.e., ~A = ∂t~A,

the symbol ∇ denotes the gradient which is the vector differential operator∇ = (∂x1 , ∂x2 , ∂x3) = −(∂1, ∂2, ∂3).

34

The charge density ρ and the space current density ~J form the 4-vectorof the current density: Jµ = (ρ, ~J).

The Maxwell equations

∂µ F µν = − e Jν , (2.105)

can be obtained as Euler-Lagrange equations following from the LagrangianL(A, J). Since the partial derivatives commute ∂µ ∂ν = ∂ν ∂µ and F µν =−F νµ, from Maxwell equation (2.105) we immediately obtain the continuityequation for the electromagnetic current

∂µ ∂ν F µν = 0 , ⇒ ∂ν Jν = 0 . (2.106)

Thus, the continuity equation for the electromagnetic current is necessaryconsistency condition with Maxwell equations (2.105). From the continuityequation directly follows the electric charge conservation - the term ”cur-rent conservation” is frequently used instead of ”continuity equation”. Inwhat follows, we always assume, that the external electromagnetic current isconserved.

We shall now investigate the Maxwell equations in more detail:• The Gauss law. Let us consider first the time component ν = 0 in

(2.105). Since the Lagrangian L(A, J) does not contain ∂0 A0 = A0, thecorresponding equation does not correspond to equation of motion but aconstraint - the Gauss law in differential form:

∇. ~E = 4A0 + ∇. ~A = −e ρ , (2.107)

where 4 = ∇.∇. Using the well-known formula

4 1

|~x − ~y| = − 4π δ(~x) − ~y)

we can express from the Gauss law (2.107) the 0th component of electromag-netic potential (under assumption that the charge density vanish for large|~x|):

A0(x) ≡ A0(t, ~x) =1

4π

∫d3~y

eρ(t, ~x) + ∇. ~A(t, ~y)

|~x − ~y| . (2.108)

35

We see that the Gauss law determines the time-component A0 of the elec-tromagnetic field potential.

The Maxwell equations represent true equations of motion for the spacecomponents Ai, i = 1, 2, 3, of the electromagnetic field potential. TheMaxwell equations for vector potential ~A reads:

~E −∇ × ~B = ~J , (2.109)

where the electric field strength ~E = ~A + ∇A0 is the conjugated fieldmomentum to ~A and ~B = ∇ × ~A is the magnetic field strength.

Field energy - Hamiltonian. The electromagnetic field Hamiltonian isrelated to the corresponding Lagrangian in a usual way

H =

∫d3~x [ ~E. ~A − L(A, J)] ~A→ ~E

=

∫d3~x [ ~E. ~A − 1

2~E2 +

1

2(∇× ~A)2 + e ρ A0 − e ~J. ~A] . (2.110)

Here the field component A0 should be replaced by the Gauss law solution(2.108).

Let us decompose the vector potential ~A to the longitudinal and transver-sal parts:

~A = ∇λ + ~A⊥ , (2.111)

where λ = λ(t, ~x) is chosen so that ~A⊥ is just the transversal potential:

∇. ~A⊥ − 0, whereas ∇λ represents longitudinal part. Inserting this decom-position of ~A the formula (2.108) for A0 takes the form:

A0(t, ~x) = −λ(t, ~x) +e

4π

∫d3~y

ρ(t, ~x)

|~x) − ~y| . (2.112)

For the electric field strength we obtain a similar decomposition to thelongitudinal part (first two terms) and transversal part (third term):

~E = ∇A0 + ∇λ + ~A⊥ , (2.113)

Here, A0 should be replaced by (2.112).

Now, we insert into Hamiltonian H the decompositions of ~A and ~E intothe transversal and longitudinal parts (see equations (2.111) and (2.113)):

H =

∫d3~x [

1

2~A

2

⊥ +1

2(∇× ~A⊥)2 − e ~J. ~A⊥

36

+1

2(∇A0 + ∇λ)2 − (∇A0 + ∇λ).∇A0 + e ρ A0 − e ~J.∇λ, , (2.114)

• The first line contains just the transversal part of ~A⊥ and its interactionwith the space part ~J of the electromagnetic current density.

• Using the formula (2.112) for A0 we can rewrite the first term on thesecond line in the following way

1

2

∫d3~x (∇A0 + ∇λ)2 =

e2

8π

∫d3~xd3~y

ρ(t, ~x) ρ(t, ~y)

|~x − ~y| . (2.115)

This contribution just represents the Coulomb energy of external charges.• Due the Gauss law the second term on second line vanish. Still there

is the third term proportional to λ. As we shall show below, the gaugeinvariance allow us choose λ = 0 i.e., we can choose the Coulomb gauge inwhich the vector potential is transversal:

~A = ~A⊥ ⇔ ∇. ~A = 0 . (2.116)

In Coulomb gauge the Hamiltonian has the form

H =

∫d3~x [

1

2~A

2

⊥ +1

2(∇× ~A⊥)2] (2.117)

− e

∫d3~x ~J. ~A⊥ +

e2

8π

∫d3~xd3~y

ρ(t, ~x) ρ(t, ~y)

|~x − ~y| , (2.118)

• The first line contains just the free transversal electromagnetic field ~A⊥(since, the Hamiltonian is quadratic);

• the second line contain the interaction of transversal field ~A⊥ withexternal electromagnetic current ~J and the Coulomb energy external charges.

Quantization of the free electromagnetic field - photons

We shall show that the quantized free transversal electromagnetic field~A, ∇. ~A = 0, describes a system of photons - noninteracting particles withvanishing mass.

Let us consider free field case, i.e., with vanishing external current Jµ =(ρ, ~J) = 0. In Coulomb gauge the photon field is given as

A = (A0, ~A) , ∇. ~A = 0 . (2.119)

37

Then the Maxwell equations of motion for Ai reduce to the Klein-Gordonequations

∂µ F µi = ¤ Ai − ∂i∂µ Ai = ¤ Ai ¤ = ∂µ∂µ ,

for a particle with vanishing mass m. We should supplement it by thetransversality of the field ~A. Thus, the equations of motion for the pho-ton field reads

¤ ~A = 0 , ∇. ~A = 0 . (2.120)

We expand the solution of (2.119) into plane waves

~A(x) =1

(2π)3

∫d3~k√2ωk

∑σ=1,2

[aσ(k)~eσ(k) e−ikx + a†σ(k)~e∗σ(k) e+ikx] ,

(2.121)

with the photon 4-momentum given as k = (ωk, ~k), ωk = |~k| - this ex-presses the fact that photons possess vanishing mass m = 0. Further, thesymbol ~eσ(k), σ′, = 1, 2, represents two complex polarization vectors, ~e∗σ(k),σ′, = 1, 2, are conjugated polarization vectors - choose them normalized andperpendicular to the photon 3-momentum ~k:

~eσ(k).~e∗σ(k′) = δσσ′ , ~k.~eσ(k) = ~k.~e∗σ(k′) = 0 . (2.122)

The last condition guarantees the transversality of the photon field ~A.The Quantization. We replace the expansion coefficients aσ(k) and a†σ(k)

in (2.120) by annihilation operators aσ(k) and creation operator a†σ(k), re-spectively. We postulate for them the boson commutation relations:

[aσ(k), a†σ′(k′)] = (2π)3 δ(~k − ~k,) δσ,σ′

[aσ(k), aσ′(k′)] = [a†σ(k), a†σ′(k

′)] = 0 . (2.123)

The annihilation and creation operators act in the Fock space generated bythe action of photon creation operators on vacuum:

|(k1, σ1), . . . (kn, σn)〉 ∼ a†σ1(k1) . . . †σn

(kn) |0〉 . (2.124)

The vacuum state |0〉 is defined in as usual: aσ(k) 0〉 = 0, 〈0|0〉 = 1.

38

Polarization sum. The polarization vectors ~eσ(k) and ~e∗σ(k), σ = 1, 2,supplemented by vector ~e0(k) = ~e

|~e| form an orthonormal base in the 3-

dimensional space of 3-momenta ~k. Therefore,

ei0(k) ej

0(k) +∑σ=1,2

eiσ(k) ej∗

σ (k) = δij .

From this relation we obtain directly the polarization sum formula

∑σ=1,2

eiσ(k) = δij − kikj

~k2. (2.125)

The photon propagator. The photon propagator is defined as the vacuummean value of the T -product of photon fields (transversal electromagneticpotential):

〈0|T [Ai⊥(x) Aj

⊥(x′)]0〉 = DijC (x− x′) . (2.126)

• Inserting for x0 > x′0 into (2.124) the plane wave expansion (2.120) weobtain:

DijC (x− x′) =

1

(2π)6

∫d3~k d3~k′√2ωk 2ωk′

×∑

σ σ′ei

σ(k) e∗jσ′(k′) e−ikx−+ik′x′ 〈0| aσ(k) a†σ′(k

′) |0〉

=1

(2π)3

∫d3~k√2ωk

∑σ

eiσ(k) ej∗

σ (k) e−ikx(−x′) .

Here we have used the relation

〈0| aσ(k) a†σ′(k′) |0〉 = δσσ′ δ(~k − ~k′) ,

then we integrated over d3~k′ and sum over σ′. Taking into account thepolarization sum (2.124) we obtain, in the case x0 > x′0, the formula forpropagator

DijC (x− x′) =

1

(2π)3

∫d3~k

2ωk

(δij − kikj

~k2

)e−ikx(−x′)

=(δij − 4−1 ∂i∂j

) 1

(2π)3

∫d3~k

2ωk

e−ikx(−x′) . (2.127)

39

• Similarly, for x′0 > x0, we obtain:

DijC (x− x′) =

(δij − 4−1 ∂i∂j

) 1

(2π)3

∫d3~k

2ωk

e−ikx(−x′) . (2.128)

In both cases the last integral exactly corresponds the expression forthe scalar field propagator. Therefore, introducing off-shell 4-momentumk = (k0, ~k) with arbitrary k0 (instead of on-shell k = (ωk, ~k)) we obtain,irrespective time-ordering, one common formula for the photon propagator:

DijC (x− x′) =

(δij − 4−1 ∂i∂j

) i

(2π)4

∫d4k

e−ik(x−x′)

k2 + iε

=1

(2π)4

∫d4k

(δij − kikj

~k2

)i

k2 + iεe−ikx(−x′) . (2.129)

2.2. Lagrangian in quantum electrodynamics (QED)

The quantum electrodynamics (QED) usually means a system chargedparticles interacting with a bunch of photons. For us, the charged particleswill be electrons and positrons described by the Dirac fields ψ(x) and ψ(x).We shall take the QED Lagrangian in the form

L(A,ψ, ψ) = − 1

4Fµν F µν + iψ γµ ∂µψ − m ψψ

− e ψ γν ψ Aν , (2.130)

where Fµν = ∂µAν − ∂µAν is the electromagnetic field strength.• The first line is a free Lagrangian describing noninteracting electromag-

netic and Dirac fields, whereas• the second line describes interaction of the electromagnetic field with

the current of charged particles

Jν = e ψ γν ψ . (2.131)

The Euler-Lagrange equations are:

40

• The Maxwell equations for electromagnetic field

∂µ F µν = − e Jν , (2.132)

with current Jν given in (2.129), and• the Dirac equations for the fields ψ and ψ interacting with electromag-

netic fieldiγµ (∂µ ψ)(x) = mψ(x) + e γµ ψ(x) Amu(x) , (2.133)

i(∂µ ψ)(x) γµ = − m ψ(x) − e ψ(x)γµ Amu(x) . (2.134)

The continuity ∂µ Jµ(x) = 0, which is necessary for the consistencywith Maxwell equations (2.130), follows directly from equations (2.131) and(2.132) for fields ψ(x) and ψ(x). The continuity equation is equivalent to thefundamental conservation law of the total electric charge

Q(t) = e

∫d3~x ψ(x) γ0 ψ(x) . (2.135)

Thus, Q(t) is constant in time t = x0, i.e. Q(t) = 0.

The QED Hamiltonian. The form of the Hamiltonian can be derived alongsame lines as we did above in the case of electromagnetic field interactingwith an external current (generated by charged particles):

H =

∫d3~x [

1

2~A

2

⊥ +1

2(∇× ~A⊥)2] + HD

− e

∫d3~x ~J. ~A⊥ +

e2

8π

∫d3~xd3~y

ρ(t, ~x) ρ(t, ~y)

|~x − ~y| − e

∫d3~x ~J.∇λ . (2.136)

Here HD is the free Dirac field Hamiltonian

HD =

∫d3~x ψ(−i~γ.∇ + m)ψ (2.137)

and ~J denotes the 3-density of the electromagnetic current of Dirac particles

~J(x) = e ψ(x)~γ ψ(x) . (2.138)

41

We can eliminate the last term in Hamiltonian depending on a longitudi-nal part ∇λ(x) of the magnetic field by changing properly the phase of Diracfield. Putting

ψ(x) = e−ieλ(x)ψ′(x) , ψ(x) = eieλ(x)ψ′(x) . (2.139)

we modify the free Dirac Hamiltonian

HD =

∫d3~x ψ′(−i~γ.∇ + m)ψ′ + e

∫d3~x ψ′~γψ′.∇λ .

However, the additional last terms just compensates the last ∇(x) dependingterm in (2.133).

We see that the QED Hamiltonian can be always chosen in the form (wewrite ψ and ψ instead of ψ′ and ψ′):

H =

∫d3~x [

1

2~A

2

⊥(x) +1

2(∇× ~A⊥)2(x)] +

∫d3~x ψ(x)(−i~γ.∇ + m)ψ(x)

− e

∫d3~x ~J(x). ~A⊥(x) +

e2

8π

∫d3~xd3~y

ρ(t, ~x) ρ(t, ~y)

|~x − ~y| , (2.140)

The terms in H have the following interpretation:• The first line in (2.140) corresponds to the Hamiltonian describing free

transversal photons and free charged particles (electrons and positrons),• The second line in (2.140) describes interactions: the first term corre-

sponds to the interaction of photons with charged particles and second onerepresents the mutual Coulomb interaction of charged particles.

The gauge invariance. The remarkable success with the eliminationof the longitudinal magnetic field in QED Hamiltonian, so that were leftcharged particles and photons, is closely related to the local gauge invarianceof QED Lagrangian.

Inn fact, it can easily shown that the QED Lagrangian L(A,ψ, ψ) isinvariant with respect to the local gauge transformations

Aµ(x) 7→ A′µ(x) = Aµ(x) + ∂µα(x) .

ψ(x) 7→ ψ′(x) = eieα(x) ψ(x) , ψ(x) 7→ ψ′(x) = ψ(x) e−ieα(x) , (2.141)

42

where α(x) generates a real local (x-depending) change of the phase of Diracfield:

• The terms 14Fµν F µν and m ψψ are evidently gauge invariant, and

• the changes of the terms iψ γµ ∂µψ and −e ψ γν ψ Aν mutually cancelunder gauge transformations.

Note. A special choice of the phase function α(x) may guarantee that elec-tromagnetic field potential satisfies some additional gauge condition. Mostoften used gauges are the following:

• The Coulomb gauge ∇. ~A = 0 in which the electromagnetic potentialis transversal. This we were considering above, when we eliminated thelongitudinal part of the potential;

• The Lorentz gauge ∂µ Aµ = 0. This is a relativistic invariant gauge con-dition (which remains same in all inertial systems). Later we shall considerthe electromagnetic field propagator in relativistic gauges.

2.3. Perturbation approach to QED

QED in the Coulomb gauge

Below we shall apply perturbation method to the field-theoretic systemdescribed by the QED Hamiltonian in Coulomb gague:

H =

∫d3~x ψ(x)(−i~γ.∇ + m)ψ(x) +

∫d3~x [

1

2~A

2

⊥(x) +1

2(∇× ~A⊥)2(x)]

− e

∫d3~x ~J(x). ~A⊥(x) +

e2

8π

∫d3~xd3~y

ρ(t, ~x) ρ(t, ~y)

|~x − ~y| . (2.142)

In the first line we have the Hamiltonian for free Dirac electrons and positronswith mass m and charge ±e and the Hamiltonian for free transversal pho-tons with vanishing mass. The first term on the second line describes theinteraction of charged particles with photons and the second one representsthe mutual Coulomb interaction of charged particles.

43

The fermions in interaction picture. In the interaction picture the elec-tron and positrons are described by free Dirac fields ψ(x) and ψ(x) whichcan be expanded into plane waves

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(bsp us(p) e−ipx + cs†

p vs(p) eipx)

, (2.143)

ψ(x) =1

(2π)3

∫d3~p√2Ep

∑s

(csp vs(p) e−ipx + bs†

p us(p) eipx)

, (2.144)

where bsp and bs†

p are fermionic annihilation and creation operators in Fock

describing electrons with 4-momentum p = (Ep, ~p), Ep =√

~p2 + m2, andspin s = ±1/2. Similarly, cs

p and cs†p are fermionic annihilation and creation

operators describing positrons.

External fermion links

Electron in initial state:

ψ(x)√

2Epbs†p =

√2Ep〈0|ψ(x) bs†p |0〉 = us(p) eipx . (2.145)

Electron in final state:√

2Epbsp ψ(x) =

√2Ep〈0| bs

p ψ(x) |0〉 = us(p) e−ipx . (2.146)

Positron in initial state:

ψ(x)√

2Epcs†p =

√2Ep〈0| ψ(x) cs†

p |0〉 = vs(p) e−ipx . (2.147)

Positron in final state:√

2Epcsp ψ(x) =

√2Ep〈0| cs

p ψ(x) |0〉 = vs(p) eipx . (2.148)

Internal fermion links are given by Feynman propagator:

SF (x− y) = 〈0|T [ψ(x)ψ(y)] |0〉 =i

(2π)4

∫d4p

p.γ + m

p2 −m2 + iεe−ip(x−y) .

(2.149)

44

The fermion contraction 〈0|T [ψ(x)ψ(y)]|0〉 and 〈0|T [ψ(x)ψ(y)] |0〉 vanish.

Photons in interaction picture. Photons are described as the transversalvector field of zero mass bosons:

~A⊥(x) =1

(2π)3

∫d3~k√2ωk

∑σ

(aσ

k ~eσ(k) e−ikx + aσ†k ~e∗σ(k) eikx

), (2.150)

where the photon 4-momentum is given as k = (ωk, ~k), ωk = |~k|. The

complex polarization vectors are orthonomal and transversal ~eσ(k) . ~k =

~e∗σ(k) . ~k = 0. The photon annihilation and creation operators satisfyingbosonic commutation relations:

[aσ(k), a†σ′(k′)] = (2π)3 δ(~k − ~k,) δσσ′ . (2.151)

The annihilation operators mutually commute, and similarly the creationoperators.

External photon links

Photon in initial state:

~A⊥(x)√

2ωkaσ†k =

√ωk〈0| ~A⊥(x) aσ†

k |0〉 = ~eσ(k) e−ikx . (2.152)

Photon in final state:√

2ωkaσk

~A⊥(x) =√

ωk〈0| aσk

~A⊥(x) |0〉 = ~e∗σ(k) eikx . (2.153)

Internal photon links are given by the propagator of transversal photons:

DijC (x− y) = 〈0|T [Ai

⊥(x) Aj⊥(y)] |0〉

=1

(2π)4

∫d4k

i

k2 + iε

(δij − kikj

~k2

)e−ip(x−y) . (2.154)

Vertices follow from the interaction Hamiltonian:

Hint = − e

∫d3~x ~J(x). ~A⊥(x) +

e2

8π

∫d3~xd3~y

ρ(t, ~x) ρ(t, ~y)

|~x − ~y| , (2.155)

45

where ~J(x) = ψ(x)~γψ(x) and ρ(x) = ψ(x)γ0ψ(x). The first term describesthe interaction of electrons and positrons with photons and the second termthe Coulomb interaction among charged particles.

Vertex 1. To the first term in interaction Hamiltonian we assign theexpression

− e γj

∫d4x . . . . (2.156)

The corresponding diagram is determined by a 1 point vertex at x: to thevertex are attached 2 arrowed fermion links (one arrow oriented into vertexand the other out of vertex) and a photon link with index j. At the sametime we indicated the needed integration over positions of the d4x.

Vertex 2. To the second term in interaction Hamiltonian we assign theexpression

e2

8πe

∫dt d3~x d3~y

1

|~x− ~y| . . . =e2

8π

∫d4x d4y

δ(x0 − y0)

|~x− ~y| . . . . (2.157)

The corresponding diagram is specified by 2 vertices x and y, to each one aretwo attached 2 arrowed fermion links (one arrow oriented into vertex and theother out of vertex). The vertices x and y are connected by correspondingto the integral kernel δ(x0 − y0)/|~x − ~y| responsible for the instantaneousCoulomb interaction. Again we indicated the integrations over d4x a d4y.

The full electromagnetic propagator

A way how to simplify considerably the QED diagrammatic rules inCoulomb gauge can be seen by a closer look to the electron elastic scat-tering amplitude up to the power e2 in the electric charge - the lowest orderof perturbation theory.

We are interesting in the process with 2 electrons both in initial andstates:

|i〉 = |p1, s1; p2, s2〉 −→ |f〉 = |p′1, s′1; p′2, s′2〉 . (2.158)

We point out that there is no contribution proportional to e. There are totypes of contribution proportional to e2:

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• The second order contribution from the first term in interaction Hamil-tonian proportional to e e which describes the interaction of charged particleswith photons, and

• the first order contribution from the second term in interaction Hamil-tonian proportional to e2 which describes the mutual Coulomb interaction ofcharged particles.

The Wick theorem gives the following contribution of the order e2 to thescattering amplitude in question:

S(2)fi =

′∑ (ie)2

2〈f |

∫d4xd4y : ψ(x)γiψ(x) : Dij

C (x− y) : ψ(y)γjψ(y) : |i〉

+′∑

(−i)e2

8π〈f |

∫d4xd4y : ψ(x)γ0ψ(x) :

x0 − y0

|~x− ~y| : ψ(y)γ0ψ(y) : |i〉 .(2.159)

In the first term we have explicitly indicated the 2 photon contraction (prop-agator) Ai

⊥(x)Aj⊥(y) = Dij

C (x − y). The symbol∑′ denotes the sum over

Dirac field contractions between electrons in initial/final states and those incurrents (represented by external lines). We stress that those summationsare identical in both interaction terms:

• The contributions from the first term have the form,

e2 : ψ(x)γiψ(x) : DijC (x− y) : ψ(y)γjψ(y) : , (2.160)

• whereas, in the second term we a very similar expression

e2 : ψ(x)γ0ψ(x) : D00C (x− y) : ψ(y)γ0ψ(y) : , (2.161)

in which γi and γj are replaced by two γ0’s and DijC (x− y) is replaced by

D00C (x− y) ≡ −i

4π

δ(x0 − y0)

|~x− ~y| . , (2.162)

From the structure of QED interaction Hamiltonian (2.155) is evident,that regardless the process in question and the order of perturbation expan-sion, it holds: the contribution with s Dij

F (x − y) is supplemented by thesame expression with D00

F (x− y).

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The photon propagator in Coulomb gauge. The analysis presented abovesuggest the following modification of Feynman rules:

Internal photon links. We join the transversal photon propagator DijC (x−

y) with D00C (x − y) into one common Feynman photon propagator which in

x-representation in Coulomb gauge possesses the DµνC (x−y), µ, ν = 0, 1, 2, 3

given as follows:

D00C (x− y) =

−i

4π

δ(x0 − y0)

|~x− ~y| =1

(2π)4

∫d4k

−i

~k2e−ik(x−y) ,

D0jC (x− y) = Dj0

C (x− y) = 0 ,

DijC (x− y) =

1

(2π)4

∫d4k

(δij − kikj

~k2

)i

k2 + iεe−ik(x−y) . (2.163)

In p-representation the photon propagator in Coulomb gauge is given as:

D00C (k) =

−i

~k2, D0j

C (k) = Dj0C (k) = 0 ,

DijC (k) =

i

k2 + iε

(δij − kikj

~k2

). (2.164)

Note: In the process of the derivation of integral representation of D00C (x−

y) we have used the well-known formula

δ(x0 − y0) =1

2′pi

∫dk0 e−ik0(x0−y)

,1

|~x− ~y| =1

2π2

∫d3~k

1

~k2ei~k.(~x−~y)

Vertex v QED. The propagator (2.163) (or (2.164) in prepresentation)corresponds to one vertex

− e γµ

∫d4x . . . . (2.165)

We assign to this vertex the diagram specified by one position x: to vertexare attached 2 arrowed fermion links (one arrow oriented into vertex and theother out of vertex) and a photon link with index µ = 0, 1, 2, 3. At the sametime we indicated the needed integration over positions of the d4x.

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The vertex (2.165) would follow from the interaction Lagrangian density

Lint = −e : ψ(x)γµψ(x) Aµ(x) : . (2.166)

Effectively, the Coulomb interaction among charged particles was replacedby a ”new” interaction mediated by longitudinal electromagnetic field. Thisfield just generates the Coulomb interaction among charged particles andthere no ”new” particles - longitudinal photons in initial and final states.The remaining Feynman rules stay unchanged:

• The rule for external and internal fermion links, and• the rule for external transversal photon links.

Gauge transformation of photon propagator. We prove that the perturba-tion contributions to scattering amplitude are unchanged under gauge trans-formation of photon propagator:

DµνC (x− y) 7→ Dµν(x− y) ≡ Dµν

C (x− y) + ∂µχν(x− y) + ∂µχν(x− y) ,(2.167)

where χµ(z) is an arbitrary function of the variable z = x− y.This is a consequence of the fact that the propagator Dµν

C (x− y) alwaysenters the perturbation contributions in the following way:

∫d4x . . . d4y : ψ(x)γµψ(x) : . . . Dµν

C (x− y) . . . : ψ(y)γjψ(y) : . (2.168)

Performing the gauge transformation (2.167) the integral (2.168) gains twosimilar terms:

• The first one vanishes∫d4x . . . d4y : ψ(x)γµψ(x) : ∂µχν(x− y) . . . : ψ(y)γνψ(y) :

= −∫

d4x . . . d4y : ∂µ(ψ(x)γµψ(x)) : χν(x− y) . . . : ψ(y)γνψ(y) : = 0 .

Here we have performed per-partes integration over d4x (under assumptionthat the boundary x → ∞ does not contribute) and then we used thecontinuity equation for the electromagnetic current ∂µ(ψ(x)γµψ(x)) = 0.

• The first one vanishes due to similar reasons∫d4x . . . d4y : ψ(x)γµψ(x) : ∂νχµ(x− y) . . . : ψ(y)γνψ(y) :

49

= −∫

d4x . . . d4y : ψ(x)γµψ(x) : χµ(x− y) . . . : ∂ν(ψ(y)γνψ(y)) : = 0 .

In p-representation the gauge transformation (2.167) of the photon prop-agator is given as

DµνC (k) → Dµν(k) = Dµν

C (k) + kµ χν(k) + kν χµ(k) , (2.169)

where χµ(k) is the Fourier transform of the function χµ(z).

The photon propagator in Feynman gauge. Let us choose the functionχµ(k) as follows:

χ0(k) =ik0

k2 ~k2, χj(k) =

−kj

k2 ~k2. (2.170)

Performing the corresponding p-representation gauge transformation (2.169),the propagator Dµν

C (k) is mapped to a simple fully relativistic form - thephoton propagator in Feynman gauge:

DµνF (k) =

−i ηµν

k2 + iε. (2.171)

In x-representation the Feynman propagator represented by the function

DµνF (x− y) =

1

(2π)4

∫, d4k

−i ηµν

k2 + iεe−ik(x−y) . (2.172)

In Feynman gauge the vertex is represented by the diagram with 1 vertexto which are attached 2 arrowed fermion links (one arrow oriented into vertexand the other out of vertex) and a photon link with index µ. To vertex weassign the expression In p-representation

−i e γµ . (2.173)

• In p-representation we assign particle 4-momentum to any link attachedto the vertex - these momenta satisfy the 4-momentum conservation law inthe vertex,

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• In x-representation we assign to vertex its position x - the integrationover d4x is assumed.

The Feynman gauge represented a great progress: the formalism is rel-ativistic invariant and the rules for diagram construction are much simplerthan those in Coulomb gauge.

Note: One can use other relativistic gauges for photon propagator. Quitepopular is the relativistic gauge depending on one parameter ξ:

Dµνξ (k) =

−i

k2 + iε

(ηµν − ξ

kµkν

k2

). (2.174)

The case ξ = 0 corresponds to Feyman gauge, whereas ξ = 1 corresponds toLandau gauge.

The rules for the calculation of Feynman diagrams for self-interactingscalar field and QED in Feynman gauge are summarized in attached Table.

51