Download Introduction to Quantum Electrodynamics Peter Prešnajder

Introduction to Quantum Electrodynamics
Peter Prešnajder
These are lecture notes devoted to introductory chapters of Quantum
Electrodynamics (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 applications 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
1
- 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 invariance
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 = (y µ ) in Minkowski
space-time is given by
x.y = xµ ηµν y ν = xµ yµ ,
(1.2)
where
yµ = ηµν y ν , yµ = ηµν y ν .
We lower or rise the indices with the help of the relativistic metric tensor
(ηµν ) = diag(1, −1, −1, −1) or its inverse (η µν ) = diag(1, −1, −1, −1)):
ηµν η νρ = ηµρ = δµρ ,
where δµρ is the Kronecker symbol defined by: δµρ = 1 for µ = ρ and δµρ = 0
for µ 6= ρ. We adopt the Einstein summation convention: we sum over the
2
same repeated upper and lower indices, e.g. xν y ν = x0 y 0 + x1 y 1 + x2 y 2 +
x3 y 3 .
Let us consider the linear transformation which preserves the relativistic
scalar product x.y = xµ ηµν y ν of any two 4-vectors x and x:
xµ 7→ Λµ ν xν ,
y µ 7→ Λµ ν y ν .
(1.3)
Let us rewrite the scalar product in matrix notation x.y = xT η y, where y
denotes the column with 4 components y µ , xT is a row with 4 components xµ
and η is 4 × 4 matrix with elements ηµν . Similarly the transformation law in
matrix notation can be written as follows: x 7→ x0 = Λ x a y 7→ y 0 = Λ y
where Λ is the 4 × 4 matrix with elements Λµ ν . The invariance of the scalar
products induces a constraint on admissible matrices Λ:
xT η y = x0T η y 0 = 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 Lie
group, called Lorentz group. The elements of the Lorentz group which can
be expressed in exponential form
Λ = exp (−iωJ)
Λα β = (exp (−iωJ))α β
⇔
(1.5)
The symbol ω in the exponent is a real number and J is 4×4 matrix satisfying
condition
J T η + η J = 0 ⇔ J T = −η J η .
(1.6)
This condition is a direct consequence of (1.4) and (1.5). There are 6 independent 4 × 4 matrices Jµν = Jνµ satisfying (1.6). Their αβ matrix elements
are given as:
¡
¢
(J µν )αβ = i ηαµ ηβν − ηαν ηβµ .
(1.7)
In this form we can freely rise and lower all indices simultaneously on both
sides.
In any proper Lorentz transformation Λ = exp (−iJ) the exponent J is
given as a linear combination
J =
1
ωµν J µν µ ν = 0, 1, 2, 3 ,
2
(1.8)
specified by 6 real parameters ωµν = −ωνµ . The matrix J µν generate Lorentz
transformations in (µν)-plane in Minkowski space:
3
• Three generators Ji j, i, j = 1, 2, 3, generate rotations in 3-space (for
its specification we need 3 parameters - 3 Euler angles θ, φ, ψ);
• Three generators J0 j, j = 1, 2, 3, generate boosts (the transformation
to a system moving with a speed ~v with respect to the original reference
frame - this requires again 3 parameters).
For infinitesimal Lorentz transformation, specified by infinitesimal parameters ωµν we obtain
µ
¶α
i
µν
xα 7→ exp (− ωµν J )
xβ
2
β
µ
¶
i
i
= δβα − (J µν )α β + . . . xβ = xα − (J µν )α β xβ + . . .
2
2
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 algebra
so(3, 1):
[J µν , J ρσ ] = − i (ηνρ J µσ − ηµρ J νσ + ηµσ J νρ − ηνσ J µρ ) .
(1.9)
Finally, we point out that it holds
J µν xρ = i (η µρ xν − η νρ xµ ) .
(1.10)
This formula is equivalent to the relation
i
xµ 7→ Λµ ν xν , Λ = exp (− ωµν J µν )
2
(1.11)
which tell us that 4 real numbers x = (xµ ), µ = 0, 1, 2, 3,transform as
relativistic 4-vector.
1.2. Relativistic scalar fields. The relativistic scalar field φ(x) is
described by a (real or complex) function defined in all points of Minkowski
space-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−1 x) .
4
(1.12)
In (1.12) Λ−1 denotes the inverse matrix of the matrix Λ. The symbol
T (Λ) represents the linear operator defined by the last equation. The assignment Λ 7→ T (Λ) defines the Lorentz group representation because it
”copies” the group product:
T (Λ1 ) T (Λ2 ) = T (Λ1 Λ2 ) ,
T (1) = I .
The symbol 1 denotes the 4 × 4 unit matrix (corresponding to the unity in
group) and the symbol I is the unit operator corresponding to the identity
map: φ(x) 7→ φ(x).
Under infinitesimal Lorentz transformation xα 7→ xα + ωαβ xβ the field
transforms as follows
φ(x) 7→ φ(Λ−1 x) = φ(xα − ωαβ xβ )
i
ωµν (J µν φ)(x) .
2
Comparing the last expression with the Taylor expansion of the field on a
first line, we obtain the formula for the generator of Lorentz transformations
J µν which acts on fields as a 1-st order differential operator:
= φ(x) −
J µν = −i (xµ ∂ ν − xν ∂ µ ) ,
∂ ν = η µν ∂ν ,
(1.13)
where ∂ν = ∂xν . It can be easily shown that the differential operators
J µν = −J νµ again satisfy the commutation relations (1.9) for Lorentz group
generators.
Multi-component relativistic fields. Let us consider n-component field


ψ1 (x)
ψ(x) =  . . . 
ψn (x)
with components ψa (x), a = 1, . . . , n. We shall assume that under Lorentz
transformation x → Λ x the field components transform as follows:
ψa (x) 7→ Sa b (Λ) ψb (Λ−1 x) ≡ (T (Λ) ψ(x))a .
(1.14)
The mapping ψ(x) 7→ T (Λ) ψ(x) will generate the Lorentz group representation:
T (Λ1 ) T (Λ2 ) = T (Λ1 Λ2 ) , T (1) = I
5
exactly, when Λ 7→ Sab (Λ) will be the (n × n)-matrix representation of the
Lorentz group
Sa b (Λ1 ) Sb c (Λ2 ) = Sa c (Λ1 Λ2 ) ,
Sa b (1) = δab .
Example: As an important example of multi-component field can serve
the relativistic vector field V µ (x) which under Lorentz transformations maps
as follows:
V µ (x) 7→ Λµ ν V ν (Λ−1 x) .
(1.15)
We leave as an exercise to derive, in this case, the form of the differential
operators J µν generating Lorentz transformations.
1.3. Particles and relativistic fields. In the framework of quantum
theory, the relativistic particle with mass m and 4-momentum p = (Ep~ , p~),
i.e., with the 3-momentum p~ and energy Ep~ , is described by the de Broglie
wave function
p
1
1
e−ip.x =
e− iEp t + i~p.~x , Ep ≡ Ep~ = p~2 + m2 . (1.16)
3/2
3/2
(2π)
(2π)
Below we shall frequently use the notation Ep instead of Ep~ (a similar simplification we shall frequently use for some other quantities too).
The set of such particles with different momenta, which do not possess
other internal degrees of freedom, is described by relativistic scalar field
Z
¢
1
d3 p~ ¡
−ip.x
∗ +ip.x
p
φ(x) =
a
e
+
b
e
, p = (Ep , p~) , (1.17)
p
p
(2π)3
2Ep
i.e. the scalar field is represented by a complex linear combination of de
Broglie wave functions and complex conjugated p
wave functions (the numerical factor in front of the integral and the factor 2Ep in the measure represent just a convenient normalization):
• The first integrand describes a system of free particles with 3-momenta
p~, the complex coefficient ap ≡ ap~ is proportional to the probability amplitude that the a-particle with 3-momentum p~ is contained in the ensemble of
particles;
6
• The second integrand is a linear combination of complex conjugated
wave functions of b-particles with 3-momenta p~, the coefficient bp ≡ bp~ is proportional 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: aparticles and b-particles which are antiparticles to a-particles. Particles and
antiparticles 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 constraint
ap = b∗p . The system contains one kind of particles: the particle a is identical
to 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 Lorentz
transformation x 7→ Λ x the scalar field φ(x) φ(x) 7→ φΛ (x) = φ(Λ−1 x),
kde
Z
´
1
d3 p~ ³
−ip.Λ−1 x
∗ +ip.Λ−1 x
−1
p
ap e
+ bp e
.
φ(Λ x) =
(2π)3
2Ep
Taking into account, that under substitution p0 = Λp
d3 p~
d3 p~0
=
,
2Ep
2Ep0
(1.19)
and that p.Λ−1 x = pT η Λ−1 x = pT ΛT η x = (Λp)T η x, we can write
Z
´
1
d3 p~0 ³ 0 −ip0 .x
0∗ +ip0 .x
p
a
e
+
b
e
,
(1.20)
φΛ (x) =
0
0
p
p
(2π)3
2Ep0
where
s
a0p0 =
E
ap ,
Ep
p0
s
b0p0 =
7
Ep0
bp ,
Ep0
p0 = Λp .
(1.21)
We see that the transformed field φΛ (x) is again a solution of the KleinGordon 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 representations: the first one in the space of particle configurations and the
other one in the space of antiparticle configurations (the coefficients ap a bp
are independent, and the unitarity is due to the positivity of the integral
measure).
1.4. The Dirac equation
The free particle relativistic equation, the Klein-Gordon equation, had
been first suggested by E. Schrödinger. He used the quantization rule: the
energy E and 3-momentum p~ should be replaced in the Hamiltonian (formula
for 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 relativistic formulation of the hydrogen atom problem. Therefore, he applied the
rule (1.22) within the non-relativistic formula for the electron energy moving in the Coulomb field of proton. He solved and published his well-known
Schrödinger equation.
The relativistic version of the quantum equation of motion was published
after by Klein and Gordon (the equation was known to V. A. Fock too). The
fact, that due to relativistic invariance the Klein-Gordon equation contained
second order time derivative (beside second order position derivatives) leads
to various complications with quantum-mechanical interpretation of the formalism.
This fundamental problems motivated P. A. M. Dirac to search for a
relativistic equation describing the free particle with mass m containing just
the first order space-time derivatives. Dirac postulated the equation in the
form
(iγ µ ∂µ − m) ψ(x) = 0 , ∂µ = ∂xµ ,
(1.23)
8
with the coefficients γ µ 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 (iγ ν ∂ν + m) we obtain second order
equation
(iγ ν ∂ν + m)(iγ µ ∂µ − m) ψ(x) = 0 ,
(γ ν γ µ ∂ν ∂µ + m2 ) ψ(x) = 0 .
(1.24)
Since the differential operator ∂µ commutes with any other ∂ν we can write
γ ν γ µ ∂ν ∂µ =
1 µ ν
{γ , γ } ∂ν ∂µ .
2
Equation (1.24) will be consistent with Klein-Gordon equation (1.18) provided the coefficients γ µ satisfy anticommutation relations:
{γ µ , γ ν } ≡ γ µ γ ν + γ ν γ µ = 2 η µν .
(1.25)
This follows directly from
γ ν γ µ ∂ν ∂µ =
1 µ ν
{γ , γ } ∂ν ∂µ = η νµ ∂ν ∂µ = ∂ µ ∂µ .
2
Equations (1.25) represent the well-known defining relations for the (real)
Clifford algebra generators:
• pre µ 6= ν the generators anti-commute: γ µ γ ν = −γ ν γ µ , and
• they are normalized by: (γ 0 )2 = +1 and (γ i )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 given
in 2 × 2 block form by the formulas:
µ
¶
µ
¶
0 1
0 σj
0
j
γ =
, γ =
, j = 1, 2, 3 .
(1.26)
1 0
−σ j 0
All entering elements are 2 × 2 matrices: 0 is the zero matrix, 1 is the unit
matrix and σ j , j = 1, 2, 3, are the Pauli matrices:
µ
¶
µ
¶
µ
¶
0 1
0 −i
1 0
1
2
3
σ =
, σ =
, σ =
.
(1.27)
1 0
i 0
0 −1
9
Note: Dirac used a different realization of γ-matrices:
µ
¶
µ
¶
1 0
0 σj
j
0
, γ =
, j = 1, 2, 3 .
γ =
0 −1
−σ j 0
(1.28)
This Dirac (or standard) realization of γ-matrices is equivalent to the Weyl
realization. In what follows we shall use the Weylovu representation.
Products and linear combinations of matrices γ µ generate the algebra of
all 4 × 4 matrices. The conventional basis of this matrix algebra is formed
by the following 16 matrices:
1 , γ µ , S µν =
i µ ν
[γ , γ ]
4
γ̃ µ = γ 5 γ µ , γ 5 = iγ 0 γ 1 γ 2 γ 3 .
(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 matrix
in question),
• 4 matrices γ µ have been introduced earlier in (1.26) in Weyl representations (or (1.28) in Dirac representation);
• since S µν = −S νµ we have 6 independent S µν matrices;
• we add 4 matrices γ̃ µ and the matrix γ̃ 4 ≡ −iγ 5 .
In Weyl basis those matrices have the form
µ j
¶
µ k
¶
i
1 ijk
1
σ
0
σ
0
0j
ij
S = −
, S = ε
≡ εijk Σk ,
j
k
0 −σ
0 σ
2
2
2
µ
¶
µ
¶
µ
¶
j
0 1
0
σ
1
0
0
5
γ̃ = −
, γ˜j = −
, γ =
. (1.30)
−1 0
σj 0
−0 1
All indices i, j, k, . . ., take values 1, 2, 3 (the summation convention is understood). These matrices are chosen so that all expressions ψ̄(x) A ψ(x) are
real for A being some of the 16 matrices 1, γ µ , S µν , γ̃ µ and iγ 5 (in fact, they
form the Lie algebra basis of the conformal group SO(4, 2)).
Exercise: Find the form of matrices S µν , γ̃ µ and iγ 5 in Dirac realization.
The properties of γ-matrices.
10
• Hermitian conjugation
γ µ† = γ 0 γ µ γ 0
⇔
γ 0† = γ 0 , γ j† = γ j , j = 1, 2, 3 .
• The properties of γ 5
(γ 5 )2 = 1 ,
γ 5† = γ 5 ,
{γ 5 , γ µ } = 0 , [γ 5 , S µν ] = 0 .
Spinor representations of Lorentz group.
It can be easily shown, using the anti-commutation relations for Dirac γ µ
matrices, that the matrices S µν , µ, ν = 0, 1, 2, 3, satisfy the commutation
relations for Lorentz group Lie algebra:
[S µν , S ρσ ] = − i (ηνρ S µσ − ηµρ S νσ + ηµσ S νρ − ηνσ S µρ ) .
(1.31)
We see that we have obtained a 4-dimensional representation of the Lorentz
group in the space C4 . However, it is a reducible representation, since the
generators S µν are 2 × 2-block diagonal:
• the 2 × 2 matrices in the left upper corner form a 2-dimensional spinor
representation of the Lorentz group (the fundamental representation of the
group SL(2, C) of 2 × 2 matrices with the unit determinant),
• similarly, the 2×2 matrices in the right lower corner form a 2-dimensional
conjugated spinor representation of the Lorentz group (the anti-fundamental
representation of the group SL(2, C)).
The linear combinations of matrices S µν (and their exponents) can act,
besides the space C4 , in the space of matrices. If A is a 4 × 4 matrix, then
S µν act as commutators (so called, adjoint action or adjoint representation):
A 7→ [S µν , A] .
In particular, we have
[S µν , 1] = 0 ,
[S µν , γ 5 ] = 0 .
11
(1.32)
That means that the unit matrix 1 and the matrix iγ 5 transform as scalars
under adjoint action by S µν .
Similarly, we have
[S µν , γ ρ ] = (ηµρ γ ν − ηνρ γ µ ) ,
[S µν , γ̃ ρ ] = (ηµρ γ̃ ν − ηνρ γ̃ µ ) .
(1.33)
These relations tell us that quartets γ µ and γ̃ µ ) transform as 4-vectors under
adjoint action by S µν .
Finally, the commutation relations (1.31) for S µν express the fact that the
set of matrices S µν , 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−1 x) ,
i
i
S(Λ) = S(exp (− ωµν S µν )) , pre Λ = exp (− ωµν J µν ) .
(1.34)
2
2
As follows from (1.14), this is a representation of Lorentz group in the space
of spinor fields ψ(x).
Besides ψ(x), it is convenient to introduce the Dirac conjugated spinor
field
ψ̄(x) = ψ(x)† γ 0 = (ψ3∗ (x), ψ(x)∗4 , ψ1∗ (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 the
conjugated Dirac equation:
i(∂µ ψ̄)(x) γ µ = − m ψ̄(x) .
(1.36)
The follows directly from the z Dirac equation for ψ(x) rewritten in the form
iγ µ (∂µ ψ)(x) = m ψ(x) .
12
(1.37)
By hermitian conjugation we obtain
−i∂µ ψ † (x) γ µ† = m ψ † (x) .
Multiplying this equation by γ 0 from the right and using the relations
(γ 0 )2 = 1 ,
γ 0 γ µ γ 0 = γ µ† ,
ψ † (x) γ 0 = ψ̄(x) ,
we arrive directly at the conjugated Dirac equation (1.36).
The conjugated Dirac field transforms under Lorentz transformations as
follows:
ψ̄(x) = ψ(x)† γ 0 7→ ψ(Λ−1 x)† S † (Λ) γ 0 = ψ̄(Λ−1 x)γ 0 S † (Λ) γ 0 .
Expressing S(Λ) = S(exp (− 2i ωµν S µν )) and using
γ 0 S µν † γ 0 = − S µν
we obtain γ 0 S † (Λ) γ 0 = S −1 (Λ). This formula gives the desired transformation for the conjugated Dirac spinor field:
ψ̄(x) 7→ ψ̄(Λ−1 x) S −1 (Λ) .
(1.38)
Transformations of bilinear expressions. The transformation rules for
spinor and conjugated spinor fields
ψ(x) 7→ S(Λ) ψ(Λ−1 x) ,
ψ̄(x) 7→ ψ̄(Λ−1 x) S −1 (Λ) ,
(1.39)
allow to derive the transformation rules for bilinear expressions:
¯ ψ(x)
S(x) = ψ(x)
P (x) = ψ̄(x) iγ 5 ψ(x)
V µ (x) = ψ̄(x) γ µ ψ(x)
Aµ (x) = ψ̄(x) γ 5 γ µ ψ(x)
T µν (x) = ψ̄(x)S µν ψ(x)
7→
ψ̄(Λ−1 x) ψ(Λ−1 x) = S(Λ−1 x)
7→
ψ̄(Λ−1 x) iγ 5 ψ(Λ−1 x) = P (Λ−1 x)
7→
Λµ ν ψ̄(Λ−1 x), γ ν ψ(Λ−1 x) = Λµ ν V ν (Λ−1 x)
7→
Λµ ν ψ̄(Λ−1 x) γ 5 γ ν ψ(Λ−1 x) = Λµ ν Aν (Λ−1 x)
7→ Λµ ρ Λν σ ψ̄(Λ−1 x)S ρσ ψ(Λ−1 x) = Λµ ρ Λν σ T ρσ (Λ−1 x)
1. The transformation rules for S(x) a P (x) tell us that the corresponding
bilinear expressions transform as scalar fields. The first expression is a direct
13
consequence of transformation rules (1.39), in derivation of the second one
relation γ 5 S(Λ) = S(Λ) γ 5 is needed.
2. The transformation rules for V µ (x) a Aµ (x) mean that the corresponding bilinear expressions transform as vector fields. Besides the transformation
rules 1.39), we need the relation
S −1 (Λ) γ µ S(Λ) = Λµ ν γ ν ,
(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 γ 5 .
3. The transformation rules for T µν (x) tell us that T µν (x) transforms as
a (antisymmetric) 2-nd order tensor. In the derivation we need the explicit
formula (1.29) for S µ ν 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−1 x)
is a solution of the Dirac equation
(iγ µ ∂µ − m) ψΛ (x) = 0 .
Using (1.40) we can put the left-hand side into the form
(iγ µ ∂µ − m) S(Λ) ψ(Λ−1 x) = S(Λ) (iΛµ ν γ ν ∂ν − m) ψ(Λ−1 x) .
Introducing the new variable x0 = Λ−1 x we can express the partial derivatives as follows
µ
∂ν0 ≡ ∂x0ν = (Λ−1 )ν ∂µ = Λµ ν ∂µ .
This equation is a direct consequence of the relation η Λ = Λ−1T η. We see
that the Dirac equation for ψΛ (x) is equivalent to
S(Λ) (iγ ν ∂ν0 − m) ψ(x0 ) = 0 .
The last equality follows from the fact that ψ(x) is a solution of Dirac equation.
14
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~ and
p
2
energy Ep =
p~ + m2 > 0. Such solution is proportional to the plane
wave
ψ(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:
(γ µ pµ − m) u(p) = 0 .
(1.41)
Since our γ-matrices possess 2 × 2 block form we rewrite the 4-component
spinor u(p) in terms of two 2-component spinors uL (p) and uR (p)
µ
¶
uL (p)
u(p) =
.
uR (p)
Taking now γ µ in Weyl representation we obtain the system of two entangled
algebraic equations (we omit the argument p in uL,R (p) and simply E instead
of Ep ):
¶µ
¶
µ
uL
−m
E − p~.~σ
(E − p~.~σ ) uR = m uL
= 0
⇔
.
uR
E + p~.~σ
−m
(E + p~.~σ ) uL = m uR
(1.42)
With the help of formulas
(E − p~.~σ ) (E + p~.~σ ) = (E + p~.~σ ) (E − p~.~σ ) = E 2 − p~2 = m2
it can be easily seen that the solution of (??) is given as
p
p
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
µ
¶
p
√ p
√
1
p
~
.~
σ
E ∓ p~.~σ = E 1 ∓ E −1 p~.~σ = E 1 ∓
... .
2 E
Since eigenvalues of the matrix E −1 p~.~σ are in absolute value less than 1, the
expansion in powers of (E −1 p~.~σ )n is well defined.
15
There are two linear independent constant spinors ξ, we shall denote them
by ξ s , s = ±1/2. We choose them orthonormal, so that it holds
ξ r† ξ s = δ rs ,
r, s = ±1/2 .
(1.44)
In this way we obtain 2 linear independent particle solutions of Dirac equation
µ √
¶
E − p~.~σ ξ s
s
√
u (p) =
s = ±1/2 ,
(1.45)
E + p~.~σ ξ s
which satisfy normalization conditions
ūs (p) us (p) = 2m δ rs .
(1.46)
We search the solution proportional to conjugated de Broglie wave functions in the form:
ψ(x) = v(p) e+ip.x , p = (Ep~ , p~) .
As we shall see, they describe antiparticles with energy Ep~ and momentum
p~. Performing the similar steps as before, we obtain two linear independent
(antiparticle) solutions
µ √
¶
s
E
−
p
~
.~
σ
η
s
√
v (p) =
, s = ±1/2 ,
(1.47)
− E + p~.~σ η s
depending on two orthonormal constant spinors η s , s = ±1/2 satisfying
η r† η s = δ rs ,
r, s = ±1/2 .
(1.48)
We point out that the bases {ξ s } and {η s } are not related to each other and
we can choose them independently. The solutions v s (p) are normalized, up to
sign, as us (p) and antiparticle solutions are orthogonal to particle solutions:
v̄ s (p) v s (p) = −2m δ rs ,
ūs (p) v s (p) = v̄ s (p) us (p) = 0 .
(1.49)
The general solution ψ(x) of Dirac equation is given as a linear combination of particle and antiparticle plane wave solutions
Z
¢
d3 p~ X ¡ s s
1
ipx
−ipx
s† s
p
.
(1.50)
v
(p)
e
u
(p)
e
+
b
ψ(x) =
a
p
p
(2π)3
2Ep s
16
Similarly, can be expressed the Dirac conjugated solution
Z
¢
1
d3 p~ X ¡ s s
−ipx
s† s
ipx
p
b
v̄
(p)
e
+
a
ū
(p)
e
.
ψ̄(x) =
p
p
(2π)3
2Ep s
(1.51)
Example: Prove the following formulas
ur† (~p) us (~p) = v r† (~p) v s (~p) = 2 Ep δ rs ,
ur† (~p) v s (−~p) = v r† (−~p) us (~p) = 0 .
(1.52)
We have used the notation us (~p) = us (p) for p = (Ep~ , p~), and similarly
v s (−~p) = v s (p0 ) pre p0 = (Ep~ , −~p). In what follows we shall use such
notation 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
X
X
vas (p) v̄bs (p) .
usa (p) ūsb (p) ,
s=±1/2
s=±1/2
The indices a, b = 1, 2, 3, 4, label the components of spinors us (p), v s (p)
and Dirac conjugated spinors ūs (p), v̄ s (p). Thus, the searched sums can be
interpreted as 4 × 4 matrices (acting in spinor spaces).
Let us consider the first sum
´
X
X µ √E − p~.~σ ξ s ¶ ³ p
p
s†
s
s
s†
√
ua (p) ūb (p) =
ξ
E
+
p
~
.~
σ
,
ξ
E
−
p
~
.~
σ
.
E + p~.~σ ξ s a
b
s
s
P
Taking into account the completeness relation s ξαs ξβs† = δαβ , α, β = 1, 2,
valid for the orthonormal basis of 2-component spinor {ξ s }, we obtain
µ √
¶
√
√
√
X
E
−
p
~
.~
σ
E
+
p
~
.~
σ
E
−
p
~
.~
σ
E
−
p
~
.~
σ
s
s
√
√
√
√
ua (p) ūb (p) =
E + p~.~σ E + p~.~σ
E + p~.~σ E − p~.~σ ab
s
µ
=
m
E − p~.~σ
E + p~.~σ
m
17
¶
.
ab
Now, taking into account the explicit form of γ-matrices we can rewrite the
spin sum as follows:
X
X
usa (p) ūsb (p) = (γ µ pµ + m 1)ab ⇔
us (p) ūs (p) = γ µ pµ + m 1 ,
s
s
s
s
(1.53)
where 1 denotes the 4 × 4 unit matrix. In the second formula we suppressed
the matrix (spin) indices - this for of spin sums is frequently used.
The formula for the second polarization sum can be derived analogously:
X
X
us (p) ūs (p) = γ µ pµ − m 1 .
vas (p) v̄bs (p) = (γ µ pµ − m 1)ab ⇔
(1.54)
Appendix B: The transformation properties of particle solutions.
Under Lorentz transformation Λ = exp(− 2i ωµν J µν ) the Dirac field transforms as follows:
i
µν
+ S µν )
ψ(x) .
ψ(x) 7→ S(Λ) ψ(Λ−1 x) = e− 2 ωµν S
i
= e− 2 ωµν (J
µν
i
e− 2 ωµν J
µν
ψ(x)
Here J µν are generators of Lorentz transformations in Minkowski space (see
(1.13)) and S µν denote the generators in spinor representation (see (1.29)).
Let us act on a-particle part of the Dirac equation
Z
¢
1
d3 p~ X ¡ s s
−ipx
p
ψ(x) =
a
u
(p)
e
+
.
.
.
.
(1.55)
p
(2π)3
2Ep s
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,
X
0
S(Λ−1 ) us (p) =
Ds0 s (W (Λ, p))us (p) ,
(1.56)
s0
where the summation is over s0 = 1, 2, i.e. (Dss0 (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→ x0 = Λx, similar steps
as by scalar field leads to the transformation rule for expansion coefficients
asp :
s
EΛp X s0
a Ds0 s (W (Λ, p)) .
(1.57)
asp 7→
Ep s0 Λp
For expansion coefficients bsp a similar rule can be derived.
Wigner little group. Wigner proposed following method how to determine
the matrix (Dss0 (W (Λ, p))):
• First we introduce standard rest 4-momentum k0 = (m, ~0) for a particle
with m. With boost, i.e. Lorentz transformation determined by matrix
Lµ ν (p) with components:
L0 0 (p) = C(p) ,
Li 0 (p) = L0 i (p) = S(p) p̂i ,
Li 0 (j) = δji − p̂i p̂j + p̂i p̂j C(p)
we transform the particle to the rest-frame in which the particle possesses the
4-momentum p = (Ep , p~). The matrix elements of Lµ ν (p) contain quantities
defined as follows:
S(p) =
|~p|
,
m
C(p) =
p
1 + S 2 (p)
p~
pˆ~ =
.
|~p|
• For any Lorentz transformation Λ we consider a sequence of transformations
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 the
relation L−1 (p0 ) p0 = k0 valid for any p0 .
• 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), where
SO(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 = 2i εijk S ij ). It is well-known that unitary irreducible SO(3) representation 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-Lagrange
equations corresponds just to the Dirac equation, can be chosen as
LD (ψ̄, ψ) = ψ̄(x) (iγ µ ∂µ − m) ψ(x) .
The canonical field-momentum π(x) conjugated to the field ψ(x) is obtained
by taking the derivative ψ̇ = ∂0 ψ of the Lagrangian density in question:
π =
∂LD
= i ψ̄ γ 0 = i ψ † .
∂ ψ̇
We see that the canonical field-momentum π(x) is not an independent field
variable, but is simply proportional to the Dirac conjugated field ψ̄(x). That
means that the system contains, rather trivial, field constraints. They do
not influence the description of the system, and we skip their discussion (for
details see, e.g. [Weinberg]).
The Hamiltonian of the system in question is given by the formula
Z
H =
d3 p~ (π ψ̇ − LD )
Simple calculations give the result
Z
HD =
d3~x ψ̄(x)(−i~γ .∂~ + m)ψ(x) .
This expression is called Dirac Hamiltonian and we label it with subscript
D.
The energy operator. Let us insert into Dirac Hamiltonian the general
solutions of the Dirac equation for ψ(x) and ψ̄(x):
Z
¢
d3 p~ X ¡ s s
1
ipx
−ipx
s† s
p
v
(p)
e
u
(p)
e
+
b
ψ(x) =
a
p
p
(2π)3
2Ep s
20
1
ψ̄(x) =
(2π)3
Z
¢
d3 p~ X ¡ s s
s
ipx
p
bp v̄ (p) e−ipx + as†
.
p ū (p) e
2Ep s
(1.58)
After insertion the formula for HD reads
Z
Z
´
X³ 0 0
1
d3 p~d3 p~0
s0 † s0 0
0
−ip0 x
ip0 x
s s
3
p
v̄
(p
)
e
+
a
ū
(p
)
e
b
HD =
d
~
x
p0
p0
(2π)6
2Ep 2Ep0
s,s0
¡
¢
s
ipx
× asp us (p) e−ipx + bs†
.
p v (p) e
(1.59)
Using the well-known formula
Z
0
d3~x e±i(~p±~p ).~x = (2π)3 δ(~p ± p~0 )
we can perform trivially the integration over d3~x, and then we can integrate
directly over d3 p~0 .
Multiplying the two integrand factors (. . .) × (. . .) in 1.60 we obtain four
terms. Taking into account the formulas 1.52 (see Example)
0
0
0
0
0
us † (~p) us (~p) = v s † (~p) v s (~p) = 2 Ep~ δ s s , us † (~p) v s (−~p) = v s † (−~p) us (~p) = 0 ,
we obtain two zero contributions and in the two remaining terms we can perform the summation over s0 . In this way we obtain a preliminary expression
for the Dirac field energy
Z
¢
1
d3 p~ X ¡ s† s
p
HD =
ap ap − bsp bs†
.
(1.60)
p
3
(2π)
2Ep s
Note: Let us remind that for scalar field an analogous calculation gives
the following result for the energy of a real scalar field
Z
¢
1
d3 p~ 1 ¡ †
p
H =
ap ap + ap a†p .
(1.61)
3
(2π)
2Ep 2
After quantization the coefficients ap = ap~ and a†p = a†q~ are replaced by
operators satisfying the canonical commutation relations for the bosonic annihilation 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 a
Hilbert space spanned by normalized n-particle states
|p1 , p2 , . . . , pn i ∼ a†p1 a†p2 . . . a†pn |0i ,
(1.63)
where the symbol ∼ means that we do not indicate explicitly the normalization factor on right hand side. The state |0i is a normalized vacuum state
in Fock space, h0|0i = 1, which does not contain any particle and is defined
by relations
ap |0i = 0 , for all p = (Ep , p~) .
(1.64)
The scalar field energy given by (1.61) is ill-defined (divergent) in the Fock
space.
The consistency of QFT formalism requires that all physical field quantities, such as energy, should be well-defined in Fock space. This is guaranteed
when they are given in terms of a normal products of annihilation and creation operators: in any term containing products of those operators we put
all creation operators a†q to the left and all annihilation operators ap to the
right:
: ap . . . a†q . . . a†q0 . . . ap0 : = a†q . . . a†q0 ap . . . ap0 , .
(1.65)
The normal product is labeled by : . . . :. The right hand side contains first
the product of all creation operators followed the product of all annihilation
operators 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
Z
Z
¡ †
¢
1
d3 p~
1
d3 p~
†
p
p
H =
: ap ap + ap ap : =
: a†p ap : .
3
3
(2π)
(2π)
2Ep
2Ep
(1.66)
The multi-particle mean values of energy are positive and in vacuum state
the mean energy vanish: h0| H |0i = 0. Without normal ordering the mean
values of energy are all divergent.
Such approach can not be applied in the case of Dirac Hamiltonian. Considering in (1.60) bosonic annihilation and creation operators we would ob22
tain:
HD
1
=
2π 3
Z
¢
d3 p~ X ¡ s† s
s
p
ap ap − bs†
p bp .
2Ep s
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 be
not stable.
Dirac found an unexpected solution, he postulated that the spinor fields
plane wave expansion enter fermionic annihilation and creation operators
satisfying anti-commutation relations:
r s†
3
{arp , as†
p − ~q) δ rs ,
q } = {bp , bq } = (2π) δ(~
s†
r s
r† s†
{arp , asq } = {ar†
p , aq } = {bp , bq } = {bp , bq } = 0 .
r
s†
r s
r† s†
{arp , bs†
q } = {bp , aq } = {ap , bq } = {bp , aq } = 0 .
(1.67)
In particular,
2
r 2
s† 2
(arp )2 = (as†
p ) = (bp ) = (bp ) = 0 .
(1.68)
Similarly as in bosonic case, the fermionic annihilation and creation operators act in Fock space F spanned by multi-particle orthonormal states
m†
|(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i ∼ apr11† . . . arpnn† bsq11 † . . . bsqm
|0i .
(1.69)
State |0i is the normalized vacuum state in Fock space defined as
asp |0i = bsp |0i = 0 ,
pre s = ±1/2 a vsetky p = (Ep , p~) .
(1.70)
Because, the squares of fermionic operators vanish the multi-particle states
cannot contain 2 identical particles with same spin and momentum - fermionic
particles 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:
0
0
0
r†
r†
s
s† r
0
: arp . . . bs†
q . . . aq 0 . . . bp0 : = ± aq 0 . . . bq ap . . . bs p0 .
(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 operators
followed by all annihilation operators entering the expression on left hand
side,
• in addition the definition contains a sign factor (−1)n , where n is the
number of neighbor transpositions needed to reshuffle the rights side to its
normal order on left hand side.
In particular,
s
s r†
r† s
: br†
(1.72)
q bp : = − : bp bq : = bq bp .
Using this normal ordering, the energy of a free Dirac field is represented
by a sum of two positive terms:
Z
X ¡
¢
1
3
s† s
s s†
HD =
a
−
E
b
b
:
d
p
~
:
E
a
p
p
p
p
p
p
(2π)3
s
1
=
(2π)3
Z
d3 p~
X¡
¢
s
s† s
Ep as†
p ap + Ep bp bp .
(1.73)
s
The momentum operator. The quantum momentum operator is defined
by the following expression
Z
~
~
P =
d3~x : ψ̄(x)(−∂)ψ(x)
:,
(1.74)
where ψ̄(x) a ψ(x) are given as linear combinations (1.60) of particle solutions
of Dirac equation. As we are consider quantum field we introduced in (1.74)
the normal product from the very beginning. Performing analogical steps as
used for energy operator we obtain the following result:
Z
X¡
¢
1
3
s† s
s† s
P~ =
d
p
~
p
~
a
a
+
p
~
b
b
.
(1.75)
p
p
p
p
(2π)3
s
The conserved electric charge. The Dirac Lagrangian LD is invariant
under global gauge transformations represented by a constant change of the
phase of spinor field:
ψ(x) 7→ eiα ψ(x) ,
ψ̄(x) 7→ e−iα ψ̄(x) ,
24
α − realna konstanta .
On a classical (non-quantized) level this invariance leads to the continuity
equation for the current density:
• Let ψ(x) and ψ̄(x) are solutions of the Dirac equations
iγ µ (∂µ ψ)(x) = m ψ(x) ,
i(∂µ ψ̄)(x) γ µ = − m ψ̄(x) .
Then,
∂µ (ψ̄(x) γ µ ψ(x)) = (∂µ ψ̄)(x) γ µ ψ(x) + ψ̄(x) γ µ (∂µ ψ)(x)
= im ψ̄(x) ψ(x) − im ψ̄(x) ψ(x) = 0 .
(1.76)
• Using (1.76) it follows directly that the current density
j µ (x) = e ψ̄(x) γ µ ψ(x) ,
(1.77)
satisfies continuity equation
∂µ j µ (x) = 0 .
(1.78)
• The total sa charge of particles Q(t = x0 ) corresponding to the current
j (x))
Z
Z
3
0
Q(t) = e
d ~x j (x) = e
d3~x ψ̄(x) γ 0 ψ(x) ,
(1.79)
µ
is conserved provided the positions of particles are restricted to a finite domain in the space, i.e. ψ(x) = 0 and ψ̄(x) = 0 for |~x| ≥ R. The parameter
e in(1.79) is the charge of Dirac particle, as we shall see the charge of antiparticle is −e. Using continuity equation for j µ (x) it can be proved easily
that the total charge Q(t) is conserved in time:
Z
Q̇(t) = e
d3~x ∂0 j 0 (x)
|~
x|<R
Z
Z
3
= −e
i
dS i j i (x) = 0 .
d ~x ∂i j (x) = e
|~
x|<R
|~
x|=R
In the last step we used the Gauss theorem: we integrated the function
div~j(t, ~x) over 3-ball |~x| ≤ R and used the fact that the 3-current vanish on
3-sphere with radius ~x| = R.
25
In the quantum case the field charge is defined by same expression but
with normal ordered charge density (we skip variable t):
Z
Q = e
d3~x : ψ̄(x) γ 0 ψ(x) : ,
(1.80)
We insert here the ψ̄(x) a ψ(x) particle solutions of Dirac equation with expansion coefficients being corresponding annihilation and creation operators.
Performing similar steps as used for energy operator we obtain the result:
Z
X¡
¢
e
3
s
s† s
Q =
d
p
~
as†
(1.81)
p ap − bp bp .
3
(2π)
s
In what follows we shall identify the conserved current j µ (x) = e :
ψ̄(x) γ µ ψ(x) : with the electromagnetic current: the a-particles we identify
with 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 n
particles (electrons) and m anti-particles (positrons) with given momenta
and spins is given by the following vector normalized vector in Fock space:
m†
|(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i ∼ apr11† . . . arpnn† bsq11 † . . . bsqm
|0i .
(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 )i
à n
!
m
X
X
=
Epi +
Eqj |(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i ,
i=1
j=1
(1.83)
~
P |(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i
!
à n
m
X
X
~qj |(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i , (1.84)
p~i +
=
i=1
j=1
Q |(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i
26
= e (n − ) |(p1 , r1 ), . . . , (pn , rn ); (q1 , s1 ), . . . , (qm , sm )i .
(1.85)
These eigenvalue equations are a direct consequence of formulas
0
0
0
0
0
s†
s
p − p~0 ) δ ss bs†
[bs†
p ,
p bp , bp0 ] = δ(~
0
0
s
s
[as†
p − p~0 ) δ ss asp0† ,
p ap , ap0 ] = −δ(~
s†
s
[as†
p − p~0 ) δ ss as†
p ,
p ap , ap0 ] = δ(~
0
0
s
s
p − p~0 ) δ ss bs†
[bs†
p . (1.86)
p bp , ap0 ] = −δ(~
The first relation can be obtained as follows:
0
0
0
s†
s † s† s
s†
s
s† s
[as†
p ap , ap0 ] = ap ap , ap0 − ap0 ap ap , =
0
0
0
s†
s†
s†
s
s†
s
s† s † s
p − p~0 ) .
as†
p ap , ap0 + ap ap0 ap , = ap {ap , ap0 } = ap δ(~
Last expression is equivalent to the desired formula (we used the definition
of the commutator, then we anti-commutate two creation operators, and
finally we used the canonical anti-commutation relation among annihilation
and creation operators). The remaining relations can be proved similarly.
The operator identity [A, BC] = [A, B]C + B[A, C] gives more general
formulas
r
[ar†
p ap ,
arp11†
...
arpnn† ]
=
n
X
δ(~p − p~i ) δ rri apr11† . . . arpnn† ,
i=1
s
[bs†
q bq ,
bsq11 †
...
bsqnm † ]
=
m
X
n†
δ(~q − ~qj ) δ ssj bsq11 † . . . bsqm
.
(1.87)
j=1
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 momentum of the system are conserved (time independent). Further it follows that
the total energy and the total momentum are sums of individual energies and
momenta of all individual particles in the state in question. Both properties
are typical for ensembles of non-interacting particles: there are no binding
energies and the individual energies and momenta of particles do not change
in time (there no mutual interaction among particles).
• Similarly, the total charge of the system is conserved. Moreover, the
total charge is equal to difference of charges shared by particles (electrons)
and that shared by anti-particles (positrons): electrons and positrons have
the 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 charge
is −e).
Equal-time canonical anti-commutation relations for Dirac field. Our goal
is to prove the anti-commutation relations among spinor fields ψ(X) and
ψ † (y) at equal time x0 = y 0 = t:
{ψa (t, ~x), ψb (t, ~y )} = {ψa† (t, ~x), ψb† (t, ~y )} = 0 .
{ψa (t, ~x), ψb† (t, ~y )} = δ(~x − ~y ) δab .
(1.88)
We shall prove the last relations, the first two can be proved along similar
lines. We insert the field expansions
Z
¢
d3 p~ X ¡ s s
1
−ipx
s† s
ipx
p
ψ(x) =
a
u
(p)
e
+
b
v
(p)
e
, x = (t, ~x)
p
p
(2π)3
2Ep s
1
ψ (y) =
(2π)3
†
Z
´
0
d3 p~0 X ³ s0 s0 † 0 −ip0 x0
0
0
p
bp0 v (p ) e
+ asp0† us † (0 p) eip y , y = (t, ~y ) ,
2Ep0 s0
into the last anti-commutator in (1.88):
{ψa (t, ~x), ψb† (t, ~y )}
1
=
(2π)6
0
Z
0
d3 p~d3 p~0 X
p
2Ep 2Ep0 s,s0
0 0
0
0 0
s† 0
s
ipx
ip x
{asp usa (p) e−ipx + bs†
, bsp0 vbs † (p0 ) e−ip x + as†
}
p va (p) e
p ub (p ) e
Z
d3 p~d3 p~0 X
1
p
=
(2π)6
2Ep 2Ep0 s,s0
³
´
0
0
0
0
0 0
0 0
{asp , asp0† } usa (p) usb † (p0 ) e−ipx+ip x + {bsp , bsp0† } vas (p) vbs † (p0 ) e+ipx−ip x
Z 3
1
d p~
=
3
(2π)
2Ep
!
Ã
X
X
0
0
vas (p) vbs † (p0 ) e−i~p.(~x−~y)
usa (p) usb † (p0 ) ei~p.(~x−~y) +
s
=
1
(2π)3
28
Z
s
d3 p~
2Ep
¡
(Ep γ 0 − ~γ .~p + m)γ 0 ei~p.(~x−~y) + (Ep γ 0 − ~γ .~p − m)γ 0 e−i~p.(~x−~y)
Z
1
d3 p~ ei~p.(~x−~y) δab = δ(~x − ~y ) δab .
=
(2π)3
¢
ab
During calculations we have used the following steps:
• The anti-commutation canonical relations for annihilation and creation
operators
0
0
0
{asp , asp0† } = {bsp , bsp0† } = (2π)3 δ(~p − ~q0 ) δ ss ,
• the sum rules
X
0
usa (p) usb † (p0 ) = (Ep γ 0 − ~γ .~p + m)γ 0 ,
s
X
0
vas (p) vbs † (p0 ) = (Ep γ 0 − ~γ .~p − m)γ 0 ,
s
• we replaced p~ 7→ −~p in the integral containing exp(−i~p.(~x − ~y )), and
finally we have used the formula
Z
d3 p~ exp(i~p.(~x − ~y )) = (2π)3 δ(~x − ~y ) .
The Dirac field propagator. We shall see later that the free field
propagators have a key rôle when the mutual interaction among particles
is described within the framework of perturbation theory. First we briefly
remind the form of the scalar field propagator, and then we derive its form
for the free Dirac field.
The scalar field propagator. Let us consider the free real scalar field
Z
¢
1
d3 p~ ¡
p
φ(x) =
ap e−ip.x + a†p e+ip.x , p = (Ep , p~) . (1.89)
3
(2π)
2Ep
Its propagator is defined as the vacuum mean value of the T-product of fields:
DF (x − y) = h0| T [φ(x)φ(y)] |0i .
29
(1.90)
The symbol T [φ(x)φ(y)] denotes the time-ordered product of fields defined
by:
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) = h0| φ(x)φ(y) |0i .
Inserting here the plane wave expansions (1.89) of φ(x) and φ(y) we obtain
Z
1
d3 p~d3 p~0
0
p
DF (x − y) =
h0| ap a†p0 |0i e−ipx + ip y .
(1.92)
6
(2π)
2Ep 2Ep0
Since,
h0| ap ap0 |0i = h0| a†p a†p0 |0i = h0| a†p ap0 |0i = 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 in
the form
ap a†p0 = (2π)3 δ(~p − p~0 ) + a†p0 ap .
The first term on r.h.s. does not contribute (again, h0| ap ap0 |0i = 0), the
contribution from the second term is proportional to δ-function. This allows
to perform the integration over d3 p~0 , and after that we obtain
Z 3
1
d p~ −ip(x−y)
DF (x − y) =
e
, p = (Ep , p~) .
(1.93)
(2π)3
2Ep
Using the same steps for y0 > x0 , we obtain
Z 3
1
d p~ +ip(x−x0 )
DF (x − y) =
e
, p = (Ep , p~) .
3
(2π)
2Ep
(1.94)
It can be shown that equations (1.93) and (1.94) can be expressed in
terms of a one explicitly relativistic invariant formula
Z
e−ip(x−y)
i
4
d
p
.
(1.95)
DF (x − y) =
(2π)4
p2 − m2 + iε
Here ε > 0 is a ”small” positive number, and at the end of calculations the
limit ε → 0+ is understood.
30
The explicit relativistic invariance of the final propagator formula has its
price. The 3 dimensional integration (1.93) and (1.94) is replaced by the 4
dimensional one: the zeroth component p0 is no more fixed to Ep and the
integration 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 . This
is in contrast with the free particle 4-momentum p = (Ep , p~) which is on
the mass shell (or simply, on-shell): p2 = m2 .
The propagator for the Dirac field. The Dirac field propagator is again
defined as the vacuum mean value of the T-product of fields:
S F ab(x − y) = h0| T [ψa (x)ψ̄b (y)] |0i .
(1.96)
Here the symbol T [ψa (x)ψ̄b (y)] denotes the time-ordered product of fermionic
Dirac fields which is defined similarly as the T-product of bosonic, but with
the distinction that transposition of two fermionic fields changes the sign of
the product:
T [ψa (x)ψ̄b (y)] = ψa (x)ψ̄b (y) pre x0 > y0 ,
T [ψa (x)ψ̄b (y)] = − ψa (x)ψ̄b (y) pre y0 > x0 .
(1.97)
Let us consider the case x0 > y0 . We insert into formula (1.96) for the
propagator the Dirac field plane wave expansions:
Z
¢
1
d3 p~ X ¡ s s
s
ipx
p
ψ(x) =
ap u (p) e−ipx + bs†
,
p v (p) e
3
(2π)
2Ep s
1
ψ̄(x) =
(2π)3
Z
¢
d3 p~ X ¡ s s
s
ipx
p
bp v̄ (p) e−ipx + as†
.
p ū (p) e
2Ep s
(1.98)
There will be again only one contribution to the vacuum mean value which
contains annihilation operator on the left and the creation on the right:
h0| T [ψa (x)ψ̄b (y)] |0i
1
=
(2π)6
Z
X
0
d3 p~d3 p~0
0
0 0
p
h0| asp asp0† |0i usa (p) ūsb (p0 ) e−ipx + ip x .
2Ep 2Ep0 s,s0
31
In the next step we use the canonical anti-commutation relation
0
0
0
asp asp0† = (2π)3 δ(~p − p~0 )δ ss − asp0† asp .
The non-vanishing contribution proportional comes from the first term: the
0
presence of δ(~p − p~0 ) allows a direct integration over d3 p~0 and δ ss allows a
summation over s0 . Performing these steps we obtain
Z 3 X
1
d p~
0
h0| T [ψa (x)ψ̄b (y)] |0i =
usa (p) ūsb (p) e−ip(x−x ) .
3
(2π)
2Ep s
Here we recognize the polarization sum, so that we can write
Z 3
1
d p~ µ
h0| T [ψa (x)ψ̄b (y)] |0i =
(γ pµ + m)ab e−ip(x−y)
3
(2π)
2Ep
Z 3
¡
¢
¡ µ x
¢
1
d p~ −ip(x−y)
= iγ µ ∂µx + m ab
e
=
iγ
∂
+
m
DF (x − y) ,
µ
ab
(2π)3
2Ep
(1.99)
where DF (x − y) denotes the expression valid for scalar field propagator
for x0 > y0 (we have used the simple formula i∂µx exp(−ip(x − y)) =
pµ exp(−ip(x − y))).
Similarly, for y0 > x0 we obtain
h0| T [ψa (x)ψ̄b (y)] |0i
Z
0
d3 p~d3 p~0 X
0
0
p
h0|bsp0† bsp |0i vas (p) v̄bs (p0 ) e+ipx − ip y
2Ep 2Ep0 s,s0
Z 3 X
1
d p~
= −
usa (p) ūsb (p) e−ip(x−y)
3
(2π)
2Ep s
Z 3
1
d p~ µ
−
(γ pµ − m)ab e+ip(x−y) .
(1.100)
(2π)3
2Ep
1
= −
(2π)6
Using again the rule i∂µx exp(+ip(x − y)) = −pµ exp(−ip(x − y)) we obtain
=
¡
¢
iγ µ ∂µx + m ab
h0| T [ψa (x)ψ̄b (y)] |0i
Z 3
¡
¢
1
d p~ +ip(x−y)
e
= iγ µ ∂µx + m ab DF (x − y) ,
3
(2π)
2Ep
(1.101)
32
where DF (x − y) denotes the expression valid for scalar field propagator for
y0 > x 0 .
We see that in both cases we obtain the identical expression for the Dirac
propagator in terms of DF (x − x0 ). Finally, we can write
¢
¡
SF (x − x0 ) = iγ µ ∂µx + m1 DF (x − x0 )
Z
i
γ µ pµ + m
0
4
=
d
p
e−ip(x−x ) .
(1.102)
4
2
2
(2π)
p − m + iε
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 an
external given source. This is important because the system contains constraints besides equations of motion. The external source is essential for the
better understanding of the origin of constraints and their relation to equations of motion. Without external source various important features would
be simply lost. Moreover, the obtained results allow a natural generalization to the situation when electromagnetic fields interacts with the system
of charged particles. In this case, the electromagnetic field and the ensemble
of charged particles will mutually interact - the current of charged particles
will 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-potential
Aµ (x), µ = 0, 1, 2, 3. The Lagrangian for electromagnetic field Aµ (x) interacting with an external electromagnetic current density J µ (x) is given as:
L(A, J) = −
1
Fµν F µν − e J µ Aν .
4
(2.103)
Here Fµν = ∂µ Aν − ∂µ Aν is the electromagnetic field strength which is
~ and magnetic field
related, in standard way, to the electric field strength E
~
strength B:
~ = ∇A0 + A
~˙ ,
E i = F0i = ∂0 Ai − ∂ i A0 ⇔ E
1 ijk
~ = ∇×A
~.
ε Fij ⇔ B
(2.104)
2
The indices i, j, k, take values 1, 2, 3 (the summation convention is under~
~˙ = ∂t A,
stood), the dot over the symbol means the time derivative, i.e., A
the symbol ∇ denotes the gradient which is the vector differential operator
∇ = (∂x1 , ∂x2 , ∂x3 ) = −(∂ 1 , ∂ 2 , ∂ 3 ).
Bk =
34
The charge density ρ and the space current density J~ form the 4-vector
~
of the current density: J µ = (ρ, J).
The Maxwell equations
∂µ F µν = − e J ν ,
(2.105)
can be obtained as Euler-Lagrange equations following from the Lagrangian
L(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 continuity
equation for the electromagnetic current
∂µ ∂ν F µν = 0 ,
⇒ ∂ν J ν = 0 .
(2.106)
Thus, the continuity equation for the electromagnetic current is necessary
consistency condition with Maxwell equations (2.105). From the continuity
equation directly follows the electric charge conservation - the term ”current conservation” is frequently used instead of ”continuity equation”. In
what follows, we always assume, that the external electromagnetic current is
conserved.
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 = Ȧ0 , the
corresponding equation does not correspond to equation of motion but a
constraint - the Gauss law in differential form:
~ = 4A0 + ∇.A
~˙ = −e ρ ,
∇.E
(2.107)
where 4 = ∇.∇. Using the well-known formula
4
1
= − 4π δ(~x) − ~y )
|~x − ~y |
we can express from the Gauss law (2.107) the 0th component of electromagnetic potential (under assumption that the charge density vanish for large
|~x|):
1
A0 (x) ≡ A0 (t, ~x) =
4π
Z
d3 ~y
35
~˙ ~y )
eρ(t, ~x) + ∇.A(t,
.
|~x − ~y |
(2.108)
We see that the Gauss law determines the time-component A0 of the electromagnetic field potential.
The Maxwell equations represent true equations of motion for the space
components Ai , i = 1, 2, 3, of the electromagnetic field potential. The
~ reads:
Maxwell equations for vector potential A
~˙ − ∇ × B
~ = J~ ,
E
(2.109)
~ = A
~˙ + ∇A0 is the conjugated field
where the electric field strength E
~ and B
~ = ∇×A
~ is the magnetic field strength.
momentum to A
Field energy - Hamiltonian. The electromagnetic field Hamiltonian is
related to the corresponding Lagrangian in a usual way
Z
~ A
~˙ − L(A, J)] ˙
H =
d3~x [E.
~ →E
~
A
Z
~ A
~˙ − 1 E
~ 2 + 1 (∇ × A)
~ 2 + e ρ A0 − e J.
~ A]
~ .
d3~x [E.
(2.110)
2
2
Here the field component A0 should be replaced by the Gauss law solution
(2.108).
~ to the longitudinal and transverLet us decompose the vector potential A
sal parts:
~ = ∇λ + A
~⊥ ,
A
(2.111)
~ ⊥ is just the transversal potential:
where λ = λ(t, ~x) is chosen so that A
~ ⊥ − 0, whereas ∇λ represents longitudinal part. Inserting this decom∇.A
~ the formula (2.108) for A0 takes the form:
position of A
Z
e
ρ(t, ~x)
A0 (t, ~x) = −λ̇(t, ~x) +
d3 ~y
.
(2.112)
4π
|~x) − ~y |
=
For the electric field strength we obtain a similar decomposition to the
longitudinal part (first two terms) and transversal part (third term):
~ = ∇A0 + ∇λ̇ + A
~˙ ⊥ ,
E
(2.113)
Here, A0 should be replaced by (2.112).
~ and E
~ into
Now, we insert into Hamiltonian H the decompositions of A
the transversal and longitudinal parts (see equations (2.111) and (2.113)):
Z
1 ~˙ 2
1
~ ⊥ )2 − e J.
~A
~⊥
H =
d3~x [ A
(∇ × A
⊥ +
2
2
36
1
~
(∇A0 + ∇λ̇)2 − (∇A0 + ∇λ̇).∇A0 + e ρ A0 − e J.∇λ,
, (2.114)
2
~ ⊥ and its interaction
• The first line contains just the transversal part of A
~
with the space part J of the electromagnetic current density.
• Using the formula (2.112) for A0 we can rewrite the first term on the
second line in the following way
Z
Z
1
e2
ρ(t, ~x) ρ(t, ~y )
3
2
d ~x (∇A0 + ∇λ̇) =
d3~xd3 ~y
.
(2.115)
2
8π
|~x − ~y |
+
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 gauge
invariance allow us choose λ = 0 i.e., we can choose the Coulomb gauge in
which the vector potential is transversal:
~ = A
~⊥
A
⇔
~ = 0.
∇.A
In Coulomb gauge the Hamiltonian has the form
Z
1 ~˙ 2
1
~ ⊥ )2 ]
H =
d3~x [ A
(∇ × A
⊥ +
2
2
Z
Z
2
ρ(t, ~x) ρ(t, ~y )
~A
~⊥ + e
−e
d3~x J.
d3~xd3 ~y
,
8π
|~x − ~y |
(2.116)
(2.117)
(2.118)
~⊥
• The first line contains just the free transversal electromagnetic field A
(since, the Hamiltonian is quadratic);
~ ⊥ with
• the second line contain the interaction of transversal field A
~
external 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
~
~ = 0, describes a system of photons - noninteracting particles with
A, ∇.A
vanishing mass.
Let us consider free field case, i.e., with vanishing external current J µ =
~ = 0. In Coulomb gauge the photon field is given as
(ρ, J)
~ , ∇.A
~ = 0.
A = (A0 , A)
37
(2.119)
Then the Maxwell equations of motion for Ai reduce to the Klein-Gordon
equations
∂µ F µi = ¤ Ai − ∂ i ∂µ Ai = ¤ Ai
¤ = ∂µ ∂ µ ,
for a particle with vanishing mass m. We should supplement it by the
~ Thus, the equations of motion for the photransversality of the field A.
ton field reads
~ = 0 , ∇.A
~ = 0.
¤A
(2.120)
We expand the solution of (2.119) into plane waves
Z
d3~k X
√
[aσ (k) ~eσ (k) e−ikx + a†σ (k) ~e∗σ (k) e+ikx ] ,
2ωk σ=1,2
(2.121)
with the photon 4-momentum given as k = (ωk , ~k), ωk = |~k| - this expresses the fact that photons possess vanishing mass m = 0. Further, the
symbol ~eσ (k), σ 0 , = 1, 2, represents two complex polarization vectors, ~e∗σ (k),
σ 0 , = 1, 2, are conjugated polarization vectors - choose them normalized and
perpendicular to the photon 3-momentum ~k:
~
A(x)
=
1
(2π)3
~eσ (k).~e∗σ (k 0 ) = δσσ0 ,
~k.~eσ (k) = ~k.~e∗ (k 0 ) = 0 .
σ
(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), respectively. We postulate for them the boson commutation relations:
[aσ (k), a†σ0 (k 0 )] = (2π)3 δ(~k − ~k,) δσ,σ0
[aσ (k), aσ0 (k 0 )] = [a†σ (k), a†σ0 (k 0 )] = 0 .
(2.123)
The annihilation and creation operators act in the Fock space generated by
the action of photon creation operators on vacuum:
|(k1 , σ1 ), . . . (kn , σn )i ∼ a†σ1 (k1 ) . . .
†
σn (kn ) |0i .
(2.124)
The vacuum state |0i is defined in as usual: aσ (k) 0i = 0, h0|0i = 1.
38
Polarization sum. The polarization vectors ~eσ (k) and ~e∗σ (k), σ = 1, 2,
supplemented by vector ~e0 (k) = |~~ee| form an orthonormal base in the 3dimensional space of 3-momenta ~k. Therefore,
ei0 (k) ej0 (k) +
X
ij
eiσ (k) ej∗
σ (k) = δ .
σ=1,2
From this relation we obtain directly the polarization sum formula
X
eiσ (k) = δ ij −
σ=1,2
kikj
.
~k 2
(2.125)
The photon propagator. The photon propagator is defined as the vacuum
mean value of the T -product of photon fields (transversal electromagnetic
potential):
(2.126)
h0|T [Ai⊥ (x) Aj⊥ (x0 )]0i = DCij (x − x0 ) .
• Inserting for x0 > x00 into (2.124) the plane wave expansion (2.120) we
obtain:
Z
1
d3~k d3~k 0
ij
0
√
DC (x − x ) =
(2π)6
2ωk 2ωk0
X
0
0
0
−ikx − +ik x
×
eiσ (k) e∗j
h0| aσ (k) a†σ0 (k 0 ) |0i
σ 0 (k ) e
σ σ0
1
=
(2π)3
Z
d3~k X i
−ikx(−x0 )
√
eσ (k) ej∗
.
σ (k) e
2ωk σ
Here we have used the relation
h0| aσ (k) a†σ0 (k 0 ) |0i = δσσ0 δ(~k − ~k 0 ) ,
then we integrated over d3~k 0 and sum over σ 0 . Taking into account the
polarization sum (2.124) we obtain, in the case x0 > x00 , the formula for
propagator
¶
Z 3~ µ
dk
kikj
1
0
ij
ij
0
δ −
e−ikx(−x )
DC (x − x ) =
3
2
~
(2π)
2ωk
k
Z 3~
¡ ij
¢ 1
d k −ikx(−x0 )
−1 i j
= δ −4 ∂∂
e
.
(2.127)
3
(2π)
2ωk
39
• Similarly, for x00 > x0 , we obtain:
DCij (x
¡
¢
− x ) = δ ij − 4−1 ∂ i ∂ j
0
1
(2π)3
Z
d3~k −ikx(−x0 )
e
.
2ωk
(2.128)
In both cases the last integral exactly corresponds the expression for
the scalar field propagator. Therefore, introducing off-shell 4-momentum
k = (k0 , ~k) with arbitrary k0 (instead of on-shell k = (ωk , ~k)) we obtain,
irrespective time-ordering, one common formula for the photon propagator:
Z
−ik(x−x0 )
¡ ij
¢ i
ij
0
−1 i j
4 e
DC (x − x ) = δ − 4 ∂ ∂
dk 2
(2π)4
k + iε
µ
¶
Z
1
i
kikj
0
4
ij
=
dk δ −
e−ikx(−x ) .
(2.129)
4
2
2
~k
(2π)
k + iε
2.2. Lagrangian in quantum electrodynamics (QED)
The quantum electrodynamics (QED) usually means a system charged
particles interacting with a bunch of photons. For us, the charged particles
will be electrons and positrons described by the Dirac fields ψ(x) and ψ̄(x).
We shall take the QED Lagrangian in the form
L(A, ψ, ψ̄) = −
1
Fµν F µν + iψ̄ γ µ ∂µ ψ − m ψ̄ψ
4
− e ψ̄ γ ν ψ Aν ,
(2.130)
where Fµν = ∂µ Aν − ∂µ Aν is the electromagnetic field strength.
• The first line is a free Lagrangian describing noninteracting electromagnetic and Dirac fields, whereas
• the second line describes interaction of the electromagnetic field with
the current of charged particles
J ν = e ψ̄ γ ν ψ .
The Euler-Lagrange equations are:
40
(2.131)
• The Maxwell equations for electromagnetic field
∂µ F µν = − e J ν ,
(2.132)
with current J ν given in (2.129), and
• the Dirac equations for the fields ψ and ψ̄ interacting with electromagnetic field
iγ µ (∂µ ψ)(x) = m ψ(x) + e γ µ ψ(x) Am u(x) ,
(2.133)
i(∂µ ψ̄)(x) γ µ = − m ψ̄(x) − e ψ̄(x)γ µ Am u(x) .
(2.134)
The continuity ∂µ J µ (x) = 0, which is necessary for the consistency
with Maxwell equations (2.130), follows directly from equations (2.131) and
(2.132) for fields ψ(x) and ψ̄(x). The continuity equation is equivalent to the
fundamental conservation law of the total electric charge
Z
Q(t) = e
d3~x ψ̄(x) γ 0 ψ(x) .
(2.135)
Thus, Q(t) is constant in time t = x0 , i.e. Q̇(t) = 0.
The QED Hamiltonian. The form of the Hamiltonian can be derived along
same lines as we did above in the case of electromagnetic field interacting
with an external current (generated by charged particles):
Z
1 ~˙ 2
1
~ ⊥ )2 ] + HD
H =
d3~x [ A
(∇ × A
⊥ +
2
2
Z
Z
Z
e2
x) ρ(t, ~y )
3 ~ ~
3
3 ρ(t, ~
~
−e
d ~x J.A⊥ +
d ~xd ~y
− e
d3~x J.∇λ
. (2.136)
8π
|~x − ~y |
Here HD is the free Dirac field Hamiltonian
Z
HD =
d3~x ψ̄(−i~γ .∇ + m)ψ
(2.137)
and J~ denotes the 3-density of the electromagnetic current of Dirac particles
~
J(x)
= e ψ̄(x) ~γ ψ(x) .
41
(2.138)
We can eliminate the last term in Hamiltonian depending on a longitudinal part ∇λ(x) of the magnetic field by changing properly the phase of Dirac
field. Putting
ψ(x) = e−ieλ(x) ψ 0 (x) , ψ̄(x) = eieλ(x) ψ̄ 0 (x) .
(2.139)
we modify the free Dirac Hamiltonian
Z
Z
3
0
0
HD =
d ~x ψ̄ (−i~γ .∇ + m)ψ + e
d3~x ψ̄ 0~γ ψ 0 .∇λ .
However, the additional last terms just compensates the last ∇(x) depending
term in (2.133).
We see that the QED Hamiltonian can be always chosen in the form (we
write ψ and ψ̄ instead of ψ 0 and ψ̄ 0 ):
Z
Z
1
1 ~˙ 2
3
2
~
H =
d ~x [ A⊥ (x) + (∇ × A⊥ ) (x)] +
d3~x ψ̄(x)(−i~γ .∇ + m)ψ(x)
2
2
Z
2 Z
e
ρ(t, ~x) ρ(t, ~y )
3 ~
~ ⊥ (x) +
−e
d ~x J(x).A
d3~xd3 ~y
,
(2.140)
8π
|~x − ~y |
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 corresponds to the interaction of photons with charged particles and second one
represents the mutual Coulomb interaction of charged particles.
The gauge invariance. The remarkable success with the elimination
of the longitudinal magnetic field in QED Hamiltonian, so that were left
charged particles and photons, is closely related to the local gauge invariance
of QED Lagrangian.
Inn fact, it can easily shown that the QED Lagrangian L(A, ψ, ψ̄) is
invariant with respect to the local gauge transformations
Aµ (x) 7→ A0µ (x) = Aµ (x) + ∂µ α(x) .
ψ(x) 7→ ψ 0 (x) = eieα(x) ψ(x) , ψ̄(x) 7→ ψ̄ 0 (x) = ψ̄(x) e−ieα(x) , (2.141)
42
where α(x) generates a real local (x-depending) change of the phase of Dirac
field:
• The terms 14 Fµν F µν and m ψ̄ψ are evidently gauge invariant, and
• the changes of the terms iψ̄ γ µ ∂µ ψ and −e ψ̄ γ ν ψ Aν mutually cancel
under gauge transformations.
Note. A special choice of the phase function α(x) may guarantee that electromagnetic field potential satisfies some additional gauge condition. Most
often used gauges are the following:
~ = 0 in which the electromagnetic potential
• The Coulomb gauge ∇.A
is transversal. This we were considering above, when we eliminated the
longitudinal part of the potential;
• The Lorentz gauge ∂µ Aµ = 0. This is a relativistic invariant gauge condition (which remains same in all inertial systems). Later we shall consider
the 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 system
described by the QED Hamiltonian in Coulomb gague:
Z
Z
1 ~˙ 2
1
3
~ ⊥ )2 (x)]
H =
d ~x ψ̄(x)(−i~γ .∇ + m)ψ(x) +
d3~x [ A
(∇ × A
⊥ (x) +
2
2
Z
Z
e2
ρ(t, ~x) ρ(t, ~y )
3 ~
~
−e
d ~x J(x).A⊥ (x) +
d3~xd3 ~y
.
(2.142)
8π
|~x − ~y |
In the first line we have the Hamiltonian for free Dirac electrons and positrons
with mass m and charge ±e and the Hamiltonian for free transversal photons with vanishing mass. The first term on the second line describes the
interaction of charged particles with photons and the second one represents
the mutual Coulomb interaction of charged particles.
43
The fermions in interaction picture. In the interaction picture the electron and positrons are described by free Dirac fields ψ(x) and ψ̄(x) which
can be expanded into plane waves
Z
¢
1
d3 p~ X ¡ s s
−ipx
s† s
ipx
p
b
u
(p)
e
+
c
v
(p)
e
,
(2.143)
ψ(x) =
p
p
(2π)3
2Ep s
Z
¢
1
d3 p~ X ¡ s s
ipx
−ipx
s† s
p
ū
(p)
e
, (2.144)
ψ̄(x) =
v̄
(p)
e
+
b
c
p
p
(2π)3
2Ep s
where bsp and bs†
in Fock
p are fermionic annihilation and creation operators
p
describing electrons with 4-momentum p = (Ep , p~), Ep =
p~2 + m2 , and
s†
s
spin s = ±1/2. Similarly, cp and cp are fermionic annihilation and creation
operators describing positrons.
External fermion links
Electron in initial state:
p
p
ψ(x) 2Ep bs†
2Ep h0| ψ(x) bs†p |0i = us (p) eipx .
p =
(2.145)
Electron in final state:
p
p
2Ep bsp ψ̄(x) =
2Ep h0| bsp ψ̄(x) |0i = ūs (p) e−ipx .
(2.146)
Positron in initial state:
p
p
−ipx
ψ̄(x) 2Ep cs†
=
2Ep h0| ψ̄(x) cs†
.
p
p |0i = v̄s (p) e
(2.147)
Positron in final state:
p
p
2Ep csp ψ(x) =
2Ep h0| csp ψ(x) |0i = vs (p) eipx .
(2.148)
Internal fermion links are given by Feynman propagator:
Z
i
p.γ + m
4
SF (x − y) = h0| T [ψ(x)ψ̄(y)] |0i =
d
p
e−ip(x−y) .
(2π)4
p2 − m2 + iε
(2.149)
44
The fermion contraction h0|T [ψ(x)ψ(y)]|0i and h0|T [ψ̄(x)ψ̄(y)] |0i vanish.
Photons in interaction picture. Photons are described as the transversal
vector field of zero mass bosons:
Z
´
1
d3~k X ³ σ
σ† ∗
−ikx
ikx
~ ⊥ (x) =
√
A
a
~
e
(k)
e
+
a
~
e
(k)
e
, (2.150)
k σ
σ
k
(2π)3
2ωk σ
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 satisfying
bosonic commutation relations:
[aσ (k), a†σ0 (k 0 )] = (2π)3 δ(~k − ~k,) δσσ0 .
(2.151)
The annihilation operators mutually commute, and similarly the creation
operators.
External photon links
Photon in initial state:
√
~ ⊥ (x) 2ωk aσ† = √ωk h0| A
~ ⊥ (x) aσ† |0i = ~eσ (k) e−ikx .
A
k
k
(2.152)
Photon in final state:
√
~ ⊥ (x) = √ωk h0| aσ A
~ ⊥ (x) |0i = ~e∗ (k) eikx .
2ωk aσk A
k
σ
(2.153)
Internal photon links are given by the propagator of transversal photons:
DCij (x − y) = h0| T [Ai⊥ (x) Aj⊥ (y)] |0i
µ
¶
Z
1
i
kikj
4
ij
=
dk 2
δ −
e−ip(x−y) .
~k 2
(2π)4
k + iε
Vertices follow from the interaction Hamiltonian:
Z
2 Z
e
ρ(t, ~x) ρ(t, ~y )
3 ~
~ ⊥ (x) +
Hint = − e
d ~x J(x).A
d3~xd3 ~y
,
8π
|~x − ~y |
45
(2.154)
(2.155)
~
where J(x)
= ψ̄(x)~γ ψ(x) and ρ(x) = ψ̄(x)γ 0 ψ(x). The first term describes
the interaction of electrons and positrons with photons and the second term
the Coulomb interaction among charged particles.
Vertex 1. To the first term in interaction Hamiltonian we assign the
expression
Z
j
−eγ
d4 x . . . .
(2.156)
The corresponding diagram is determined by a 1 point vertex at x: to the
vertex are attached 2 arrowed fermion links (one arrow oriented into vertex
and the other out of vertex) and a photon link with index j. At the same
time we indicated the needed integration over positions of the d4 x.
Vertex 2. To the second term in interaction Hamiltonian we assign the
expression
Z
Z
δ(x0 − y 0 )
e2
1
e2
3
3
d4 x d4 y
e
dt d ~x d ~y
... =
... .
(2.157)
8π
|~x − ~y |
8π
|~x − ~y |
The corresponding diagram is specified by 2 vertices x and y, to each one are
two attached 2 arrowed fermion links (one arrow oriented into vertex and the
other out of vertex). The vertices x and y are connected by corresponding
to the integral kernel δ(x0 − y 0 )/|~x − ~y | responsible for the instantaneous
Coulomb interaction. Again we indicated the integrations over d4 x a d4 y.
The full electromagnetic propagator
A way how to simplify considerably the QED diagrammatic rules in
Coulomb gauge can be seen by a closer look to the electron elastic scattering amplitude up to the power e2 in the electric charge - the lowest order
of perturbation theory.
We are interesting in the process with 2 electrons both in initial and
states:
(2.158)
|ii = |p1 , s1 ; p2 , s2 i −→ |f i = |p01 , s01 ; p02 , s02 i .
We point out that there is no contribution proportional to e. There are to
types of contribution proportional to e2 :
46
• The second order contribution from the first term in interaction Hamiltonian proportional to e e which describes the interaction of charged particles
with photons, and
• the first order contribution from the second term in interaction Hamiltonian proportional to e2 which describes the mutual Coulomb interaction of
charged particles.
The Wick theorem gives the following contribution of the order e2 to the
scattering amplitude in question:
(2)
Sf i
Z
0
X
(ie)2
=
hf |
d4 xd4 y : ψ̄(x)γ i ψ(x) : DCij (x − y) : ψ̄(y)γ j ψ(y) : |ii
2
0
X
Z
x0 − y 0
: ψ̄(y)γ 0 ψ(y) : |ii .
|~x − ~y |
(2.159)
In the first term we have explicitly indicated the 2 photon
contraction (propP0
j
ij
i
denotes the sum over
agator) A⊥ (x)A⊥ (y) = DC (x − y). The symbol
Dirac field contractions between electrons in initial/final states and those in
currents (represented by external lines). We stress that those summations
are identical in both interaction terms:
• The contributions from the first term have the form,
+
e2
(−i) hf |
8π
d4 xd4 y : ψ̄(x)γ 0 ψ(x) :
e2 : ψ̄(x)γi ψ(x) : DCij (x − y) : ψ̄(y)γj ψ(y) : ,
(2.160)
• whereas, in the second term we a very similar expression
e2 : ψ̄(x)γ0 ψ(x) : DC00 (x − y) : ψ̄(y)γ0 ψ(y) : ,
(2.161)
in which γi and γj are replaced by two γ0 ’s and DCij (x − y) is replaced by
DC00 (x − y) ≡
−i δ(x0 − y 0 )
.,
4π |~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 expansion, it holds: the contribution with s DFij (x − y) is supplemented by the
same expression with DF00 (x − y).
47
The photon propagator in Coulomb gauge. The analysis presented above
suggest the following modification of Feynman rules:
Internal photon links. We join the transversal photon propagator DCij (x −
y) with DC00 (x − y) into one common Feynman photon propagator which in
x-representation in Coulomb gauge possesses the DCµν (x − y), µ, ν = 0, 1, 2, 3
given as follows:
Z
−i δ(x0 − y 0 )
1
−i −ik(x−y)
00
DC (x − y) =
=
d4 k
e
,
4
~k 2
4π |~x − ~y |
(2π)
DC0j (x − y) = DCj0 (x − y) = 0 ,
µ
¶
Z
1
kikj
i
ij
4
ij
dk δ −
e−ik(x−y) .
DC (x − y) =
4
2
~k 2
(2π)
k + iε
(2.163)
In p-representation the photon propagator in Coulomb gauge is given as:
D̃C00 (k) =
D̃Cij (k)
−i
,
~k 2
D̃C0j (k) = D̃Cj0 (k) = 0 ,
i
= 2
k + iε
µ
δ
ij
kikj
−
~k 2
¶
.
(2.164)
Note: In the process of the derivation of integral representation of DC00 (x−
y) we have used the well-known formula
Z
Z
1
1
1 i~k.(~x−~y)
1
0 −ik0 (x0 −y )
0
0
dk e
,
=
d3~k
δ(x − y ) = 0
e
2
~
2 pi
|~x − ~y |
2π
k2
Vertex v QED. The propagator (2.163) (or (2.164) in prepresentation)
corresponds to one vertex
Z
µ
−eγ
d4 x . . . .
(2.165)
We assign to this vertex the diagram specified by one position x: to vertex
are attached 2 arrowed fermion links (one arrow oriented into vertex and the
other out of vertex) and a photon link with index µ = 0, 1, 2, 3. At the same
time we indicated the needed integration over positions of the d4 x.
48
The vertex (2.165) would follow from the interaction Lagrangian density
Lint = −e : ψ̄(x)γ µ ψ(x) Aµ (x) : .
(2.166)
Effectively, the Coulomb interaction among charged particles was replaced
by a ”new” interaction mediated by longitudinal electromagnetic field. This
field just generates the Coulomb interaction among charged particles and
there 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 perturbation contributions to scattering amplitude are unchanged under gauge transformation of photon propagator:
DCµν (x − y) 7→ Dµν (x − y) ≡ DCµν (x − y) + ∂ µ χν (x − y) + ∂ µ χν (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 DCµν (x − y) always
enters the perturbation contributions in the following way:
Z
d4 x . . . d4 y : ψ̄(x)γµ ψ(x) : . . . DCµν (x − y) . . . : ψ̄(y)γ j ψ(y) : . (2.168)
Performing the gauge transformation (2.167) the integral (2.168) gains two
similar terms:
• The first one vanishes
Z
d4 x . . . d4 y : ψ̄(x)γµ ψ(x) : ∂ µ χν (x − y) . . . : ψ̄(y)γν ψ(y) :
Z
= −
d4 x . . . d4 y : ∂ µ (ψ̄(x)γµ ψ(x)) : χν (x − y) . . . : ψ̄(y)γν ψ(y) : = 0 .
Here we have performed per-partes integration over d4 x (under assumption
that the boundary x → ∞ does not contribute) and then we used the
continuity equation for the electromagnetic current ∂ µ (ψ̄(x)γµ ψ(x)) = 0.
• The first one vanishes due to similar reasons
Z
d4 x . . . d4 y : ψ̄(x)γµ ψ(x) : ∂ ν χµ (x − y) . . . : ψ̄(y)γν ψ(y) :
49
Z
= −
d4 x . . . d4 y : ψ̄(x)γµ ψ(x) : χµ (x − y) . . . : ∂ ν (ψ̄(y)γν ψ(y)) : = 0 .
In p-representation the gauge transformation (2.167) of the photon propagator is given as
D̃Cµν (k) → D̃µν (k) = D̃Cµν (k) + k µ χ̃ν (k) + k ν χ̃µ (k) ,
(2.169)
where χ̃µ (k) is the Fourier transform of the function χµ (z).
The photon propagator in Feynman gauge. Let us choose the function
χ̃ (k) as follows:
µ
χ̃0 (k) =
ik 0
,
k 2 ~k 2
χ̃j (k) =
−k j
.
k 2 ~k 2
(2.170)
Performing the corresponding p-representation gauge transformation (2.169),
the propagator D̃Cµν (k) is mapped to a simple fully relativistic form - the
photon propagator in Feynman gauge:
D̃Fµν (k) =
−i ηµν
.
k 2 + iε
(2.171)
In x-representation the Feynman propagator represented by the function
Z
1
−i ηµν −ik(x−y)
µν
DF (x − y) =
, d4 k 2
e
.
(2.172)
4
(2π)
k + iε
In Feynman gauge the vertex is represented by the diagram with 1 vertex
to which are attached 2 arrowed fermion links (one arrow oriented into vertex
and the other out of vertex) and a photon link with index µ. To vertex we
assign the expression In p-representation
−i e γ µ .
(2.173)
• In p-representation we assign particle 4-momentum to any link attached
to the vertex - these momenta satisfy the 4-momentum conservation law in
the vertex,
50
• In x-representation we assign to vertex its position x - the integration
over d4 x is assumed.
The Feynman gauge represented a great progress: the formalism is relativistic invariant and the rules for diagram construction are much simpler
than those in Coulomb gauge.
Note: One can use other relativistic gauges for photon propagator. Quite
popular is the relativistic gauge depending on one parameter ξ:
µ
¶
−i
kµkν
µν
ηµν − ξ 2
.
(2.174)
D̃ξ (k) = 2
k + iε
k
The case ξ = 0 corresponds to Feyman gauge, whereas ξ = 1 corresponds to
Landau gauge.
The rules for the calculation of Feynman diagrams for self-interacting
scalar field and QED in Feynman gauge are summarized in attached Table.
51