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Last modified: May 6, 2020

Relativistic Quantum Field Theory
---------------------------------

Canonical Representation
------------------------

One of the key insights of quantum field theory is to do away with 
classical mathematical forms of the wave function by listing how 
many particles (quanta) are in each momentum state, k.  These
occupation numbers are the analog of the Slater determinant in QM.
This referred to as the CANONICAL REPRESENTATION of QFT.

We can specify the state of a system as:

    |n(p1) n(p2) n(p2) .... n(p∞)>

where,

n(pi) = number of quanta with momentum pi

These states are referred to as FOCK States. 

Examples:

|0> = |0,0,0,0,0>

a1†|0> = |1,0,0,0,0>

a2†a1|0> = |0,1,0,0,0> 

|0> = |0,0,0,0,0>

a1†a1†|0> = |2,0,0,0,0>

Consider the periodic loop. Momentum (and frequency) are quantized where
p = 2πn/L.  The waves moving on the loop are equivalent to a collection 
of harmonic oscillators - one harmonic oscillator for each value of n.  
For each harmonic there is a number of quanta n(pn) represented by the
occupation number.

Field Operators
---------------

In QM, the general solution to the wave equation is a superposition 
of plane waves (Fourier expansion of basis vectors):

ψ(x) = ΣpApexp(ipx) where Ap is an amplitude term.

and,

ψ*(x) = ΣpAp*exp(-ipx)

By analogy, in QFT we can write:

Ψ†(x) = Σpap†exp(-ipx) where ap† is the creation operator

and

Ψ(x) = Σpapexp(ipx) where ap is the annihilation operator

The field operator is not the same thing as a single-particle 
wavefunction.  The former is an operator acting on the Fock 
space, and the latter is a quantum-mechanical amplitude for 
finding a particle in some position.  However, they are closely 
related.  We use capitals to denotes the field operators.  
Ψ†(x)/Ψ(x) create/destroy a particle at a particular point in 
space.  ap†/ap create/annihilate particles with certain momenta.  
Field operators act by applying the Fourier transform to the 
creation and annihilation operators.  

Examples:

Pre-existing particle, n5 = 1
                   
                    |0 0 0 0 1 0 0> 

Create a new particle in momentum state p = 3 at x.  Thus, 
n3 -> 1, n5 = 1

Σpa3†exp(-i3x)|0 0 0 0 1 0 0> => e-i3x|0 0 1 0 1 0 0>

Create another new particle in momentum state p = 5 at the same x. 
Thus, n5 -> 2

Σpa5†exp(-i5x)|0 0 1 0 1 0 0> => √2e-i5x|0 0 1 0 2 0 0>  

The factor of √2 comes from a†|n> = √(n+1)|n+1>

Creation and Annihilation Operators on Bra Vectors
--------------------------------------------------

The rule for the creation and annihilation operators when they 
operate on bra vectors is the opposite of the case for ket 
vectors.  Thus, the creation operator acts on a bra as an 
annihilation operator and vice versa.

<n|a† => <n-1|√n

and

<n|a => <n+1|√n+1

What is the meaning of ∫Ψ†(x)Ψ(x)dx ?
-------------------------------------

Expand in terms of basis states.

 = ∫Σpap†e-ipxΣqaqeiqxdx

 = ∫Σpap†Σqaqei(q-p)xdx

 = ∫Σpap†Σqaqδpqdx

 = ∫Σpap†ap  - the occupation number is a†pap

 = the number of particles at x.

Therefore, Ψ†(x)Ψ(x) is the density of particles at x.

Time Dependent Schrodinger Equation
-----------------------------------

Differentiate, Ψ w.r.t. t and x:

∂Ψ/∂t = Σp(-iω)apexp(i(px - ωt))

∂Ψ/∂x = Σp(ip)apexp(i(px - ωt))

∂2Ψ/∂x2 = Σp(ip)2apexp(i(px - ωt))

So we can write:

∂Ψ/∂t/∂2Ψ/∂x2 = Σp(-iω)apexp(i(px - ωt))/Σp(ip)2apexp(i(px - ωt))

              = (-iω)/(ip)2

Which leads to:

   (ip)2∂Ψ/∂t = (-iω)∂2Ψ/∂x2

      p2∂Ψ/∂t = (iω)∂2Ψ/∂x2

Now E = hω = p2/2m 

Therefore p2 = 2mω/h

(2mω/h)∂Ψ/∂t/∂ = (iω)∂2Ψ/∂x2

∂Ψ/∂t =(ih/2m)∂2Ψ/∂x2

Note: In both cases we could have equally integrated over 
position/time to demonstrate conservation of momentum and 
conservation of energy respectively.

Heisenberg Uncertainty Principle in QFT
---------------------------------------

Analagous to the relationship in QM, the creation and annihilation 
for position and momentum are Fourier transforms of each other.  
Thus: 

Ψ(x)† = Σpap†exp(-ipx) and Ψ(x) = Σpapexp(ipx)

and

ap† = ΣxΨ(x)†exp(ipx) and ak = ΣxΨ(x)exp(-ipx)

This shows that to create a state of definite position, it is 
necessary to sum over many momentum states. Conversely, to create 
a state of definite momentum it is necessary to sum over many 
position states.  

Free Field Quantization
-----------------------

It is well known that the standard equations in field theory for 
scalar and fermionic fields are respectively the Klein-Gordon and 
Dirac equations.

Massive Spin 0 Real Scalar Fields
---------------------------------

Consider the Klein-Gordon equation (h = c = 1):

∂2φ/∂t2 - ∇2φ + m2φ = 0

This can be derived using the relativistic relationship (h = c = 1):

E2 - p2 - m2 = 0 <=> -ημνpμpν + m2 = 0 where ημν = (-, +, +, +)

and

pμ -> i∂/∂xμ  (p0 -> i∂/∂t)

In QM solutions to the K-G equation are of the form φ = exp(i(kx - ωt)) 
with the constraint that:

ω = √(p2 + m2)  {from E = hω = √(p2c2 + m2c4)}

Canonical Commutation Relations
-------------------------------

Let's look at the commutator, [x,p] = ihδij.  In QM this is referred 
to as the CANONICAL COMMUTATION RELATION and defines the relationship 
between canonical conjugate quantities (quantities which are related 
through the Fourier transform).  The quantity p is often written as π 
and is referred to as the momentum conjugate to x.  By analogy with 
this we can write a similar relationship for the commutators for 
quantum fields:

[φ(x),π(y)] = ihδ(3)(x - y)

[φ(x),φ(y)] = [π(x),π(y)] = 0

The Lagrangian for the K-G equation is (h = c = 1):

L = (1/2)∂μφ∂μφ - (1/2)m2φ2

The quantity, π, for the fields is obtained using the Euler-Lagrange 
equations generally written as:

∂μ(∂L/∂(∂μφ)) - ∂L/∂φ = 0

The momenta canonically conjugate to the coordinates of the
field (or simply the canonical momentum) is defined as:
        .
π = ∂L/∂φ

For the K-G equation we get:
        .   .               
π = ∂L/∂φ = φ

Therefore:
                    .
[φ(x),π(y)] ≡ [φ(x),φ(y)] = ihδ3(x - y)

The quantization of the harmonic oscillator can now be applied to the 
free scalar field by writing φ as a linear sum (Fourier expansion) of 
creation and annihilation operators. Using the expessions for the 
creation and annihilation operators for the harmonic oscillator, we 
can write: 

ap† = √(ω/2)(φ + iπ/√(2ω) or (1/√(2ω)(ωφ + iπ)

and

ap = √(ω/2)(φ - iπ/√(2ω) or (1/√(2ω)(ωφ - iπ)

Therefore,

φ = (1/√(2ω))(apexp(i(px - ωt) + ap†exp(-i(px - ωt))

and
    .
π = φ = -i√(ω/2)(apexp(i(px - ωt) - ap†exp(-i(px - ωt)) 

Thus, the field operator equations are:

φ(x) = ∫(d3p/(2π)3) (1/√(2ω))[apexp(ipx) + ap†exp(-ipx)]

and
       .
π(x) = φ(x) = ∫(d3p/(2π)3) (-i√(ω/2))[apexp(ipx) - ap†exp(-ipx)]

The commutator [ap,aq†] is a little tricky to derive.  It can be 
shown to be:

[ap,aq†] = (2π)3δ(3)(p - q)

These creation and annihilation operators represent a more abstract 
form of the operators that we derived for the Quantum Harmonic 
oscillator.

We can summarize the commutators for the real scalar field as 
follows:

[ap,aq†] = (2π)3δ(3)(p - q)       
[φ(x),π(y)] = ihδ(3)(x - y)  
                             
[ap,aq] = [ap†,aq†] = 0    
[φ(x),φ(y)] = [π(x),π(y)] = 0   

The operators φ(x) and π(y) depend on space and not time.  This is 
the SCHRODINGER PICTURE.  In the HEISENBERG PICTURE the operators 
change with time. However, the operators in the two pictures agree 
at a fixed time, say, t = 0. The commutation relations become EQUAL 
TIME commutation relations in the Heisenberg picture. 

[φ(x,t),π(y,t)] = ihδ(3)(x - y)  

[φ(x,t),φ(y,t)] = [π(x),π(y)] = 0  

We can show this as follows:

ap(t) = exp(iHt)ap(0)exp(-iHt)  

Now, from before we had:

H = ωap†ap

Therefore,

i[H,ap(t)] = -iωap(t) 

[H,ap(t)] = -ωap(t) 

Similarly,

[H,ap†(t)] = ωap†(t) 

Plugging these into the Heisenberg equation, i[H,ap(t)] = ∂ap(t)/∂t, 
yields:

∂ap(t)/∂t = -ωap(0) 

with solution,

ap(t) = ap(0)exp(-iωt)

and, 

∂ap†(t)/∂t = ωap†(t)

with solution,

ap†(t) = ap†(0)exp(iωt)

Thus, the field operator equations in the Heisenberg picture are:

φ(x,t) = ∫(d3p/(2π)3) (1/√(2ω))[exp(-iωt)apexp(ipx) + exp(iωt)ap†exp(-ipx)]

       = ∫(d3p/(2π)3) (1/√(2ω))[apexp(ip.x) + ap†exp(-ip.x)]

Where p is now a 4-vector instead of a 3-vector.

p.x = px - ωt

It is easy to that this satisfies the K-G equation.
                                  .
Now, we said from before that π = φ.  Therefore,
.
φ(x,t) = ∫(d3p/(2π)3) (1/√(2ω))[-iωapexp(ip.x) + iωap†exp(-ip.x)]

       = ∫(d3p/(2π)3) (1/√(2ω))(-iω)[apexp(ip.x) - ap†exp(-ip.x)]

       = ∫(d3p/(2π)3) (-i√(ω/2))[apexp(ip.x) - ap†exp(-ip.x)]

       = π

Which is what we had before, so everything checks out!

Non Relativistic Limit
----------------------

In the non-relativistic limit the K-G equation yields the Schrodinger 
equation.  We can see this by replacing E = √(p2 + m2) with the 
non-relativistic relationship E = p2/2m.  We get:

ih∂tφ - (1/2m)∇2φ = 0

The Lagrangian for the SE is:

L = φ†(ih∂t - (h2/2m)∇2)φ

Where,
        .                 
π = ∂L/∂φ = iφ†

Therefore:
           
[φ(x),π(y)] ≡ [φ(x),iψ(y)]

Now,

[φ(x),iψ(y)] = φ(x)iφ(y) - iφ(y)φ(x) 

              = i(φ(x)φ(y) - φ(y)φ(x))

              = i[φ(x),φ(y)]

Therefore, by comparison:

[φ(x),φ(y)] = hδ(x - y)

Relativistic Normalization
--------------------------

The formulation involving 4-vectors is clearly Lorentz invariant.  
But what about 3-vectors?

The Dirac Delta function in 3 dimensions is defined as:

δ(3)(p - q) = (1/2π)3∫ d3x exp(i(p - q).x)

Likewise in spacetime it is defined as:

δ(4)(p - q) = (1/2π)4∫ d4x exp(i(p - q).x)

δ(4) is Lorentz invariant but δ(3) is not.  How do we make it so?  
Consider:

δ(4)(p - q) = δ(Ep - Eq)δ(3)(p - q)

Now consider;

δ(Ep2 - Eq2)

Using the following property of the delta function:

δ(x2 - α2) = (1/|2α|)[δ(x + α) + δ(x - α)]

We can write:

δ(Ep2 - Eq2) = (1/|2Eq|)[δ(Ep + Eq) + δ(Ep - Eq)]

If we disregard the negative energies we get:

δ(Ep - Eq) = 2Eqδ(Ep2 - Eq2)

Therefore,

δ(4)(p - q) = 2Eqδ(Ep2 - Eq2)δ(3)(p - q)

Now δ(Ep2 - Eq2) is Lorentz invariant so the implication is that the 
invariant form of δ(3)(p - q) is 2Eqδ(3)(p - q) 

Now by definition ∫d3p δ(3)(p - q) = 1.  Therefore, to compensate for 
the Lorentz invariant for we need to divide by 2Eq.  Thus,

∫(d3p/2Eq) 2Eqδ(3)(p - q) = 1

Therefore, the relativistically normalized momentum states are 
given by:

<p|q> = (2π)32ωδ(3)(p - q)

|p> = √(2ω)|p> = √(2ω)ap†|0>

<q| = <q|√(2ω) = <0|√(2ω)ap

Note:  Some texts also define relativistically normalized creation 
operators as:

ap = a(p)/√(2ω). 

This results in the 1/2ω factor seen in these texts.  We will not 
assume this throughout this document.


The Scalar Propagator
---------------------

We can use the canonical approach to write the propagator as:

G(x,y) = <0|annihilate particle at x, create particle at y|0>

Thus, y -> x

The propagator can now be written as:

G(x,y) = <0|φ(x)φ(y)|0>

By plugging in the equations for φ(x), φ(y) we end up with:

<0|φ(x)φ(y)|0> = ∫(d3pd3q/(2π)6) (1/√(4ωpωq))<0|ap,aa†|0>exp(-ipx + iqy)

Which, after some math, yields:

<0|φ(x)φ(y)|0> = ∫(d3p/(2π)3) (1/2ω)exp(-ip(x - y))

Until now, we integrated only over 3-momentum, with p0 fixed by the 
'on-shell' relationship, E2 = ω2 = p02 = p2 + m2.  To accomodate 'off-shell' 
virtual particles we need to integrate over the 4-momentum.  This 
also makes the propagator manifestly Lorentz invariant.  Therefore, 
we now integrate over 4-dimensional Minkowski spacetime.  Thus,

G(x,y) = ∫(d4p/(2π)4) exp(-ip(x-y))/(p2 - m2)

However, as it stands this integral is ill-defined because the 
integrand has 2 poles at p0 = ±√(p2 + m2).  The final expression 
for the propagator will depend on the choice of contour.  A 
contour going under the left pole and over the right pole gives 
the FEYNMAN PROPAGATOR.  Thus, the contour looks like:



Instead of specifying the contour, it is standard to write the 
Feynman propagator as:

G(x,y) = ∫(d4p/(2π)4) exp(-ip(x-y))/(p2 - m2 + iε)

This way of writing the propagator is called the "iε prescription".
The iε term has the effect of shifting the poles slightly off 
the real axis, so the integral is equivalent to the contour along 
the real p0 axis.

The Feynman propagator satisfies:

(∂2φ/∂t2 - ∇2φ + m2)G(x,y) = -iδ(4)(x - y)

So G(x,y) is a Green's function for the Klein-Gordon operator.

Solutions to G(x,y) over the chosen contour yield:

GF(x,y) = limε->0∫(d4p/(2π)4) exp(-ip(x-y)/(p2 - m2 + iε)

        = -(1/4π)δ(s) + (m/8π√s)H1(2)(m√s) for s ≥ 0

        = -(im/4π2√(-s)K1(m√(-s)) for s < 0

Where s := (x0 -  y0) - (x - y), H1(2)(m√s) is a Hankel function 
and K1(m√(-s)) is a modified Bessel function.

The Fourier transform of the position space propagators can be 
thought of as propagators in momentum space.  This is simpler 
than the position space form.

GF(p) = -i/(p2 - m2 + iε)


Massive Spin 0 Complex Scalar Fields
------------------------------------

Complex fields are required to accomodate charge. φ carries a charge 
of +1.  φ† carries a charge of -1

We can write φ in terms of real and imaginary parts.

φ = (1/√2)(φR + iφI)

The K-G Lagrangian becomes (h = c = 1):

L = ∂μφ∂μφ† - m2φφ†

  = (1/2)∂μφR∂μφR + (1/2)∂μφL∂μφL - (1/2)m2φR2 - (1/2)m2φL2

Which is 2 uncoupled real scalar fields.
        .   .               .   .
π = ∂L/∂φ = φ† and π† = ∂L/∂φ† = φ

Therefore:
                    .
[φ(x),π(y)] ≡ [φ(x),φ†(y)]

The equation of motion associated to the above Lagrangian is the 
very same Klein-Gordon equation.  However, now complex solutions 
of φ(x) are allowed. 

In order to describe boson particles with electric charge we need
to introduce a new pair of creation and annihilation operators, b†
and b which create and annihilate antiparticles. Therefore,

bp† creates a particle of momentum p.
cp† creates an antiparticle of momentum p.

The field operator equations are:

φ(x) = ∫(d3p/(2π)3) (1/√(2ω))[bpexp(ip.x) + cp†exp(-ip.x)]

φ†(x) = ∫(d3p/(2π)3) (1/√(2ω))[bp†exp(-ip.x) + cpexp(ip.x)]
       .
π(x) = φ†(x) = ∫(d3p/(2π)3) (i(√(ω/2))[bp†exp(-ip.x) - cpexp(ip.x)]
        .
π†(x) = φ(x) = ∫(d3p/(2π)3) (-i(√(ω/2))[bpexp(ip.x) - cp†exp(-ip.x)]

We can summarize the same-time commutators for the complex field 
as follows:

[bp,bq†] = [cp,cq†] = (2π)3δ(3)(p - q) 
[φ(x),π(y)] = [φ†(x),π†(y)] = ihδ(x - y)  

All other combinations of bp, bp†, cp, cp† = 0 
All other combinations of φ(x), φ†(x), π(y), π(y)† = 0 

Conserved Quantities
--------------------

It is easy to see that the Lagrangian is invariant under the 
transformation (φ -> φexp(iθ) and φ* -> φ*exp(-iθ)).  This 
implies there is a symmetry and a conserved quantity.  

Now eiε = 1 + iε   ... Taylor series

Thus,

φ -> φ + iεφ -> φ + εfφ where fφ = iφ

φ* -> φ* - iεφ* -> φ* - εfφ* where fφ* = -iφ*

Look for a conserved quantity using Noether's Theorem. 
         .   .
πφ = ∂L/∂φ = φ*
         .    .
πφ* = ∂L/∂φ* = φ

Conserved quantity = ∫[πφfφ + πφ*fφ*]d3x
                        .     .
                   = i∫[φ*φ - φφ*]d3x
                             
                   = i∫[πφ - π*φ*]d3x

The quantity in [] is the charge density of the field, ρ.  

Another way to write this is:
        .
ρ =  Im(φφ*)

∴ Q = ∫ρd3x

We can also write a 4-vector for current, jμ, as:
   
jμ = Im(φ*∂μφ)

∂μjμ = Im(∂μ(φ*∂μφ))

∂μjμ = Im(∂μφ*∂μφ + φ*∂μ∂μφ)

The first term is real so the imaginary part is 0.  The second
term can be written as:

-φ*m2φ since ∂μ∂μφ + m2φ = 0

Again this is real.  We can therefore conclude that:

∂μjμ = 0

This is the CONTINUITY EQUATION.

Substituting the equations for the field operators yields:

Q = ∫d3p/(2π)3 (cp†cp - bp†bp)

The Complex Scalar Propagator
-----------------------------

The real scalar propagator describes the creation of a particle 
at y and its annihilation at x.  The question is how how to deal 
with antiparticles*.  Feynman decided that the propagator must 
be made up of 2 parts - particles are represented by the case 
x0 > y0 and antiparticles are represented by x0 < y0.  In order 
to include the latter it is necessary to introduce the WICK TIME
0RDERING symbol, T.  T is defined as:

Tφ(x)φ(y) = φ(x)φ(y) for x0 > y0
          = φ(y)φ(x) for x0 < y0

The propagator can now be written as:

G(x,y) = <0|Tφ(x)φ(y)|0>

       = <0|Θ(x0 - y0)φ(x)φ(y) + Θ(y0 - x0)φ(y)φ(x)|0>

* Although we need complex fields to represent antiparticles, this 
interpretation is also there for a real scalar field because the 
particle is its own antiparticle.

For a complex field:

G(x,y) = <0|Tφ(x)φ†(y)|0>

Where,
                   
Tφ(x)φ(y)† = φ(x)φ(y)† for x0 > y0
           = φ†(y)φ(x) for x0 < y0

Therefore,

G(x,y) = <0|Θ(x0 - y0)φ(x)φ†(y) + Θ(y0 - x0)φ†(y)φ(x)|0>

The propagator of the complex scalar field can be calculated in 
the same fashion as for the non-complex case.  This can be 
seen from:

<φ(x)φ†(y)> = (1/2)[<φR(x)φR(y) - iφR(x)φL(y) + φL(x)φR(y) + φL(x)φL(y)>]

            = (1/2)[<φR(x)φR(y) + φL(x)φL(y)>]

            = <φ(x)φ(y)>

The resulting Feynman propagator in 4-dimensional Minkowski 
spacetime is:

G(x,y) = -∫(d4p/(2π)4) exp(-ip(x-y))/(p2 - m2 + iε)

The interpretation now is that the amplitude for the particle to 
propagate from y -> x cancels the amplitude for the antiparticle
to travel from x -> y.   In fact, this interpretation is also there 
for the real scalar field because, as stated, the particle is its own 
antiparticle.

As before, the propagator is equal to the Green's function of the 
non-complex case.

As stated previously, the Fourier transform of the position space 
propagators can be thought of as propagators in momentum space.  

GF(p) = -i/(p2 - m2 + iε)


Massive Fermionic Fields
------------------------

The Dirac equation is:

(iγμ∂μ - m)ψ = 0

The simplest solutions to the DE are plane waves u(p)exp(-ikx) and 
v(p)exp(ikx) where u(p) and v(p) are Dirac spinors.

The Dirac Lagrangian is (h = c = 1):
    _
L = ψ(iγμ∂μ - m)ψ 
    _
L = ψ(iγ0∂ψ/∂t + iγ0γj/γ0∂ψ/∂x - mψ)
_
ψ = γ0ψ†.  The ψ† by itself does not result in a Lorentz invariant 
Lagrangian and Lorentz invariant Dirac equation after the Euler-
Lagrange equations have been applied to minimize the action.
Although, the Dirac equation is first order in derivatives, it is 
the magic of the γ matrices that makes the Lagrangian Lorentz 
invariant!  
_
ψ(x) creates a Fermion.  ψ(x) annihilates a Fermion.

The canonical momenta are:
        .    _
π = ∂L/∂ψ = iψγ0 = iγ0ψ†γ0 = iψ†     (γ0)2 = 1)
         .    _
π† = ∂L/∂ψ† = iψγ0 = iγ0ψ†γ0 = 0

Therefore:
           
{ψ(x),π(y)} ≡ {ψ(x),iψ†(y)}

Now,

{ψ(x),iψ†(y)} = ψ(x)iψ†(y) + iψ†(y)ψ(x) 

              = i(ψ(x)ψ†(y) + ψ†(y)ψ(x))

              = i{ψ(x),ψ†(y)}

The anticommutator obeys the same rule as the commutator in that:

{ψ(x),π(y)} = ihδ(x - y)

Therefore, by comparison:

{ψ(x),ψ†(y)} = hδ(x - y)

The field operator equations are:

ψ(x) = ∫(d3p/(2π)3) (1/√(2ω)Σs[bpsus(p)exp(ip.x) + cps†vs(p)exp(-ip.x)]

ψ†(x) = ∫(d3p/(2π)3) (1/√(2ω)Σs[bps†us†(p)exp(-ip.x) + cpsvs†(p)exp(ip.x)]

π(x) = iψ† = ∫(d3p/(2π)3) (i/√(2ω)Σs[bps†us†(p)exp(-ip.x) + cpsvs†(p)exp(ip.x)]

π†(x) = 0

bps creates a fermion of momentum p and spin s.
cps creates an antifermion of momentum p and spin s.

We can summarize the anti-commutators for the fermionic field 
as follows:

{bpr,bqs†} = {cpr,cqs†} = (2π)3δrsδ(3)(p - q) 
{ψ(x)α,ψ(y)†β} = hδαβδ(x - y)  
                                  
All other combinations of bp, bp†, cp, cp† = 0                 
All other combinations of ψ(x) and ψ†(x) = 0 

Where r and s represents the spin and α and β are the spinor 
indeces (u1, u2, v1, v2).

Conserved Quantities
--------------------

It is easy to see that the Lagrangian is invariant under the 
                                  _    _
transformation (φ -> exp(iθ)φ and φ -> φexp(-iθ)).  This implies 
there is a symmetry and a conserved quantity.  We can derived
the conserved quantity in 2 ways.  Start with the Dirac equation: 

(iγμ∂μ + m)ψ = 0   ... 1.

Which expands to be:

(iγ0∂0ψ + iγi∂iψ + mψ) = 0 

The Hermitian conjugate is:

(i∂0ψ†γ0† + i∂iψ†γi† - mψ†) = 0

Now, γ0† = γ0 and γi† = -γi

Therefore,

(i∂0ψ†γ0 + i∂iψ†(-γi) - mψ†) = 0

Multiply by γ0 from the right and make use of the defining 
relationship:

γ0γi + γiγ0 = 2η0iI = 0 ∴ γ0γi = -γiγ0

To get:
    _       _       _
(i∂0ψγ0 + ∂iψiγi - mψ) = 0   

Or,
_       
ψ(i∂μγμ - m) = 0   ... 2.
      _
Where ψ = γ0ψ†

This is the ADJOINT DIRAC EQUATION.
                           _
Multiplying equation 1. by ψ from the left and equation 2 by 
ψ from the right and adding gives:
_               _
ψ(iγμ∂μ + m)ψ + ψ(i∂μγμ - m)ψ = 0
_           _
ψ(γμ∂μψ) + (ψ∂μγμ)ψ = 0   

Now, the 2nd term on the LHS is interpreted as being right to 
left so this is equivalent to:
_           _          _
ψγμ∂μψ + (∂μψ)γμψ = ∂μ(ψγμψ) = ∂μ(jVμ) = 0   

Again, this is the CONTINUITY EQUATION where jVμ is the 4-VECTOR 
PROBABILITY CURRENT.

The PROBABILITY DENSITY is defined as:
          _
ρ = jV0 = ψγ0ψ

Therefore we can write,

jVμ = (ρ,jVi)

The other way to get at the conserved quantity is similar to the 
method already encountered with the complex scalar field.  We 
can write:
    _
L = ψ(iγμ∂μ + m)ψ 
     _         _
  = iψγμ∂μψ + mψψ

The canonical momentum is:
        .    _
π = ∂L/∂ψ = iψγ0 
_          _      
π = ∂L/∂(∂0ψ) = 0 

The transformations are:

ψ -> ψ - iεψ -> ψ - εf where f = -iψ
_    _     _    _            _    _
ψ -> ψ + iεψ -> ψ + εf where f = iψ

Look for a conserved quantity using Noether's Theorem. 
                            __
Conserved quantity = ∫[πf + πf]d3x
                        _              
                   = ∫[iψγ0(-iψ)]d3x
                       _       
                   = ∫[ψγ0ψ]d3x

Again, the quantity in [] is the charge density of the field, ρ.  

∴ Q = ∫ρd3x

Substituting the equations for the field operators yields:

ψ(x) = ∫ d3p Σs[cp†scps - dps†dp]

Axial Vector Current
--------------------

When m = 0, the Dirac Lagrangian also has an extra internal symmetry 
which rotates left and right-handed fermions in opposite directions.  
We can go through a similar procedure to the above to get:

ψ -> exp(iαγ5)ψ
_    _
ψ -> ψexp(iαγ5)
      _
jAμ = ψγ5γμψ
                                                                                                  
The Fermionic Propagator
------------------------

For a Dirac field:
                 _    
S(x,y) = <0|Tψ(x)ψ(y)|0>

Where,
                _     
Tψ(x)ψ(y) = ψ(x)ψ(y) for x0 > y0
             _
          = -ψ(y)ψ(x) for x0 < y0

The - sign is required for Lorentz invariance.

Therefore,
                          _               _
S(x,y) = <0|Θ(x0 - y0)ψ(x)ψ(y) + Θ(y0 - x0)ψ(y)ψ(x)|0>

The propagator of the Dirac field can be calculated in the same 
fashion as before.

SF(x,y) = -∫(d4p/(2π)4) γμpμexp(-ip(x-y))/(p2 - m2 + iε)

Where γμpμ = γ0p0 + γjpj

This satisfies:

(iγμ∂μ + m)SF(x,y) = iδ(4)(x - y)

So SF(x,y) is a Green's function for the Dirac operator.

As stated previously, the Fourier transform of the position space 
propagator can be thought of as propagators in momentum space.  

SF(p) = (γμpμ + m)/(p2 - m2 + iε)


Massless Spin 1 Vector Fields (The Electromagnetic Field)
---------------------------------------------------------

There is no equivalent equation to the Klein-Gordon, Dirac
or Schrodinger equations that describes the wavefunction
of the photon.  This is because photons always behave
relativistically and can be freely emitted and absorbed.
Instead, a single photon is described quantum mechanically 
by Maxwell's equations, where the solutions are taken to be 
complex.

The electromagnetic 4 potential is defined as:

Aμ = (φ/c, -A) 

Where φ is the electric scalar potential and A is the magnetic 
vector potential. The B and E fields in terms of φ and A are:

B = ∇ x A

E = -∇φ - ∂A/∂t

We can also define an asymmetric tensor, Fμν such that,

Fμν = ∂μAν - ∂νAμ

   -                         -
  | 0      Ex/c   Ey/c   Ez/c |            
  |-Ex/c   0     -Bz     By   |  = Fμν 
  |-Ey/c   Bz     0     -Bx   |            
  |-Ez/c  -By     Bx     0    |
   -                         -

   -                         -
  | 0     -Ex/c  -Ey/c  -Ez/c |            
  | Ex/c   0     -Bz     By   |  = Fμν 
  | Ey/c   Bz     0     -Bx   |           
  | Ez/c  -By     Bx     0    |
   -                         -

From these we get Maxwell's Equations in the absence of charge
sources:

∇.B = 0

∂B/∂t = -∇ x E

∇.E = 0

∂E/∂t = -∇ x B

The Lagrangian for Maxwell's equations is:

L = (-1/4)FμνFμν - JμAμ

In the absence of charge sources is this becomes:

L = (-1/4)FμνFμν ≡ (-1/2)(B2 - E2)

  = (-1/2)(∂μAν∂μAν - ∂νAμ∂μAν)

Substituting this into the Euler-Lagrange equations for a field gives:

∂μ(∂L/∂(∂μAν) = ∂L/∂Aν 

              = 0

So the E-L equation becomes:

∂μ(∂μAν - ∂νAμ) = 0

∂μ∂μAν - ∂μ∂νAμ = 0

∂μFμν = 0

Now, the electromagnetic field is a vector field that can be decomposed 
into longitudinal and transverse components.  Longitudinal components 
do not have a curl and are parallel to the direction of motion.  Transverse 
components do not have divergence and are perpendicular to the direction 
of motion.  Thus, 

∇.FT = 0

and

∇ x FL = 0

This is referred to as the HELMHOLTZ DECOMPOSITION.  Since ∇.E = ∇.B = 0 
in the absence of sources, both E and B are transverse.

Gauge Fixing
------------

Different representative configurations of a physical state are called 
different gauges.  Picking a gauge is rather like picking coordinates 
that are adapted to a particular problem. Moreover, different gauges 
often reveal slightly different aspects of a problem.  Since the
physical fields E and B are not modified by a gauge transformation, 
we may say that the potentials φ and A contain both a physical part 
(because the physical fields can be computed from them) and a nonphysical 
part (which changes when we do a gauge transformation). We will look at 
2 gauges: the Coulomb gauge and the Lorentz gauge.  Choosing the Coulomb 
gauge makes A into a transverse field that correctly restricts the photon 
to 2 degrees of freedom corresponding to the 2 polarization states.  
However the field commutation relations are a little more involved as is 
the propagator.  The Coulomb gauge is also not Lorentz invariant although
the final result is.  Choosing the Lorentz gauge results in 4 degrees of 
freedom that need to be dealt with, but the field commutators follow the 
familiar form and the propagator is simpler.  The Lorentz gauge is also 
manifestly Lorentz invariant from the outset. 

Coulomb Gauge
-------------

E = -∇φ - ∂A/∂t.  Therefore,

∇.E = -∇2φ - (∂(∇.A)/∂t) = 0

In the Coulomb gauge ∇.A = 0 so the equation of motion becomes:

∇2φ = 0 

Let us pause for a moment to look at this.  A is a 4-vector with the 
time component set to 0 (in other words a 3-vector).  This allows 
the remaining part of the potential to satisfy the condition ∇.A = 0 
(≡ ∂iAi = 0).

We can also use ∂μ∂μAν - ∂μ∂νAμ = 0 to derive the equations of motion.
The second term is ∂ν(∇.A) = 0 leaving the first term ∂μ∂μAν ≡ ∇2A = 0.

The momentum conjugate to Aμ is:
         .
π0 = ∂L/∂A0 = 0
         .
πi = ∂L/∂Ai = -F0i = Ei

The field operator equations are similar to the real scalar field:

A(x) = ∫(d3p/(2π)3) (1/√(2ω))Σrξr[aprexp(ip.x) + apr†exp(-ip.x)]

and

πi = E(x) = ∫(d3p/(2π)3) (1/√(2ω))(-iω)Σrξr[aprexp(ip.x) - apr†exp(-ip.x)]

   = ∫(d3p/(2π)3) (-i√(ω/2))Σrξr[aprexp(ip.x) - apr†exp(-ip.x)]

Where ξr are the 2 POLARIZATION VECTORS.  These polarization 
vectors obey the rules ξ1.ξ2 = ξ1.p = ξ2.p = 0 and ξ12 = ξ22 = 1 and 
thereby satisfy the transversality condition imposed by the gauge.
r = 1 to 2.

We can summarize the commutators for the electromagnetic field 
as follows:

[apr,aqs†] = (2π)3δrsδ(3)(p - q) 

[apr,aqs] = [apr†,aqs†] = 0

Which is the same as the usual commutation rules for creation 
and anihilation operators.  However, the field commutation rules
are a little more complicated.

[Ai(x),Ej(y)] = ihδ⊥ij(x - y)

              = ih(δij - pipj/p2)δ(3)(x - y)

              = ih(δij - ∂i∂j/∇2)δ(3)(x - y) since p = -i∂/∂x 

[Ai(x),Aj(y)] = [Ei(x),Ej(y)] = 0

Where δ⊥ij(x - y) is the TRANSVERSE DELTA FUNCTION.  The TDF 
acts like a δ function on fields with 0 divergence and projects 
an arbitrary vector field onto its transverse part.  The
projection is most easily constructed in Fourier space.

FT(x) = ∫(d3p/(2π)3) (δij - pipj/p2) FL(p) exp(ipx)

The term in () also satisfies the following completeness 
relationship.

Σξirξjr = (δij - pipj/p2)

Projection Operators and Completeness
-------------------------------------

Recall from the discussion of projection operators in Quantum 
Mechanics (Bell's Theorem).

I = Σn|i><i| = ΣnPi where Pi is the projection onto i.

Also recall from QM the completeness relationship.

ψ(x) = Σnanψn(x) where an = ∫dx' ψn*(x')ψ(x')

Therefore,

ψ(x) = Σn(∫dx' ψn*(x')ψ(x'))ψn(x)

     = ∫dx' (Σn(ψn*(x')ψn(x))ψ(x')

This is an equation of the form:

ψ(x) = ∫dx' G(x,x')ψ(x') where G is Green's function.

Consider the eigenvalue equation

D|ψn> = λn|ψn> where D is a linear differential operator.

We can write:

DG(x,x') =  λnG(x,x')  

Where G(x,x') is a Green's Function such that:

DG(x,x') ≡ λnG(x,x') = δ(x - x')

Now λn is just a scalar so we can write

λnψ(x) = λn∫dx' δ(x - x') ψ(x')

By comparison we see that:

δ(x - x') = Σn(ψn*(x')ψn(x))

Returning to the completeness relationship for the polarization
vectors we see: 

Σξirξjr ≡ Σ|ξir><ξjr|

       = |ξi1><ξj1| + |ξi2><ξj2|

This is a projection onto the sub space spanned by the polarisation
vectors.  This sub space is orthogonal to the momentum. 

The Lorentz Gauge
-----------------

We now take time into consideration and replace 3-vectors with 
4-vectors associated with Minkowski spacetime.The Lorentz 
gauge is manifestly Lorentz invariant and is defined as:

∇.A + ∂φ/∂t = 0

Which can be written in space-time as:

∂μAμ = 0

The equation of motion is found from:

∂μ∂μAν - ∂μ∂νAμ = 0

Which, after imposing the gauge becomes:

∂μ∂μAν = 0

These equations of motion can also be derived from the following
Lagrangian:

L = (-1/4)FμνFμν - (1/2)(∂μAμ)2

The momentum conjugate to Aμ is:
          .
π0 = ∂L/∂A0 = -∂μAμ
         .
πi = ∂L/∂Ai = -F0i = Ei
 
The field operator equations are once again:

Aμ(x) = ∫(d3p/(2π)3) (1/√(2ω))Σrξr[aprexp(ip.x) + apr†exp(-ip.x)]

and

Eμ(x) = ∫(d3p/(2π)3) (-i√(ω/2))Σrξr[aprexp(ip.x) - apr†exp(-ip.x)]

But this time since there is no restriction regarding transverse 
and longitudinal components, ξr now represents 4 polarization 
4-vectors versus 2 3-vector polarization vectors in the Coulomb 
gauge.  These can be characterized as 2 tranverse, 1 longitudinal 
and 1 time-like.  Therefore, r = 0 to 3.

We can summarize the commutators in this gauge as follows:

[apμ,aqν†] = -(2π)3gμνδ(3)(p - q) 

[apμ,aqν] = [apμ†,aqν†] = 0

gμν is necessary to achieve Lorentz invariance.

The -ve sign is problematic because it implies negative probabilities
in the Hilbert space we are working in.  The minus is entirely due to 
the time component (note π0 = -∂μAμ. The challenge is to 'fix' this 
problem and also deal with the 2 'extra' degrees of freedom that we 
have.  The solution is to impose the gauge condition ∂μAμ = 0.  However, 
if we impose the gauge condition at the Lagrangian level we end up 
where we started in the Coulomb gauge.  The answer is to impose this 
as a constraint on the physical states associated with the tranvese 
photon.  This is done by splitting Aμ(x) into 2 separate operators.

Aμ(x) = Aμ+(x) + Aμ-(x)  

Therefore,

∂μAμ(x) = ∂μAμ+(x) + ∂μAμ-(x)  

Where,

∂μAμ+(x) = ∫(d3p/(2π)3) (1/√(2ω))(ip)Σrξr[aprexp(ip.x)] represents the +ve 
frequency (annhilation).       

and

∂μAμ-(x) = ∫(d3p/(2π)3) (1/√(2ω))(-ip)Σrξr[apr†exp(-ip.x)] represents the -ve 
frequency (creation)       

We then impose the Lorentz condition on the physical states through 
the eigenvalue equations:

∂μAμ+|Ψ> = 0|Ψ and <Ψ|∂μAμ- = <Ψ|0>

This ensures the condition <Ψ|∂μAμ|Ψ> = 0 so that the expectation 
value of ∂μAμ is 0.

This is referred to as the GUPTA-BLEULER condition.

Now, we want get rid of the 0 and 3 components since we know
these are 'unphysical' states.  Let us write:  

∂μAμ+|Ψ> = ∫(d3p/(2π)3) (1/√(2ω))(ip)Σrξr[aprexp(ip.x)]|Ψ> = 0

We can write p.ξ1 = p.ξ2 = 0 for the transverse components and 
p.ξ0 = -p.ξ3 for the time and longitudinal components.

Thus, we end up with

∫(d3p/(2π)3) (1/√(2ω))(ip)Σrξr[(ap3 - ap0)exp(ip.x)]|Ψ> = 0

Where r is now 0 and 3.  We immediately conclude that:

(ap3 - ap0)|Ψ> = 0

which leads to the crucial identity:

<Ψ|ap3†ap3|Ψ> = <Ψ|ap0†ap0|Ψ>

So only the transverse modes survive and contribute to the dynamics. 
Therefore, it is consistent to implement the Lorenz gauge condition and 
indeed decouples the unwanted degrees of freedom from the physical 
Hilbert space.  

With this condition applied the commutation relations are restored to 
their familiar form:

[apμ,aqν†] = (2π)3gμνδ(3)(p - q) 

[apμ,aqν] = [apμ†,aqν†] = 0

[Aμ(x),Eν(y)] = ihgμνδ(3)(x - y)

[Aμ(x),Aν(y)] = [Eμ(x),Eν(y)] = 0

The Photon Propagator
---------------------

In the Coulomb gauge the propagator is:

G(x,y) = <0|TAμ(x)Aν(y)|0>

       = ∫(d4p/(2π)4) i/(p2 + iε)(δij - pipj/p2)exp(-ip.(x-y))

In the Lorentz gauge the propagator is:

G(x,y) = <0|TAμ(x)Aν(y)|0>

       = ∫(d4p/(2π)4) i/(p2 + iε)exp(-ip.(x-y))

As stated previously, the Fourier transform of the position space 
propagators can be thought of as propagators in momentum space.
Thus, in the Lorentz gauge we get:

G(p) = -igμν/(p2 + iε)

Massive Spin 1 Vector Fields
----------------------------

In the discussion of Quantum Field theory we have talked about real 
and complex scalar fields, fermionic fields and the electromagnetic 
field.  The scalar fields represent massive spin 0 particles, the 
fermionic field represents massive half-integer spin particles and 
the electromagnetic field represents massless spin 1 particles.  
The question is "how do we represent massive spin 1 particles?".  
The W± and Z0 gauge bosons are examples of particles (also called 
"vector bosons") that fall into this category.  

It seems logical that somehow we should try and incorporate mass
into the spin 1 electromagnetic vector field.  However, simply adding 
a mass term to Lagrangian as follows spoils the invariance.  

L = (-1/4)FμνFμν - (1/2)m2AμAμ

This is discussed here

However, there is a way to include an extra field such that the 
resulting Lagrangian invariant and includes a massive vector field.
Consider the Lagrangian

L = (-1/4)FμνFμν - (1/2)(∂μφ + mAμ)(∂μφ + mAμ)

Where φ is a real scalar field.

This is the STUECKELBERG Lagrangian.  It is an extension to the 
Standard Model that provides a mechanism for mass generation 
in an Abelian gauge theory while preserving gauge invariance.  In
essence it provides a mass to the physical photon.  It is a special 
case of the Higgs mechanism where the Higgs excitations are so
large they can be ignored. 

The procedure for quantizing the field is the same as we have
previously encountered.  The math is similar and will not be 
repeated here.

As a fotnote it is worth noting that while the Stueckelberg 'trick' 
works for the Abelian case, the Higgs mechanism involving 
spontaneous symmetry breaking remains the only presently known 
way to give masses to non-Abelian vector fields. 

Massive Vector Field Propagator
--------------------------------

Like the electromagnetic field, the propagator for a massive 
vector field is dependant on the gauge chosen.  In the Feynman
gauge this is

G(p) = gμν/(p2 - m2 + iε)

The Field Hamiltonians
----------------------

The Legendre transformation yields:
     .
H = πφ - L
        .   
π = ∂L/∂φ

H = ∫d3x H'

Real Scalar
----------- 

L = (1/2)∂μφ∂μφ - (1/2)m2φ2

  = (1/2)∂2φ/∂t2 + (1/2)∇2φ - (1/2)m2φ2
        .   .
π = ∂L/∂φ = φ
      .
H' = πφ - L
     .         .
   = φ2 - (1/2)φ2 - (1/2)(∂φ/∂x)2 - (1/2)m2φ2
          .         
   = (1/2)φ2 - (1/2)(∂φ/∂x)2 - (1/2)m2φ2

Substituting the expression for φ and π and noting that ω2 = p2 + m2

φ(x) = ∫(d3p/(2π)3√(2ω)) (1/√(2ω))[ap†exp(-ipx) + apexp(ipx)]

and

π(x) = (-i/2)∫(d3p/(2π)3 [ap†exp(-ipx) - apexp(ipx)]

Leads to:

Hrenorm' = ∫d3p/(2π)3 ωp ap†ap

The order of the creation operators is important to achieve a vacuum 
expectation value of 0.  The creation and annihilation must satisfy 
<0|ap† = 0 and|ap|0> = 0.  This requires that when taking the product 
of quantum fields, or equivalently their creation and annihilation 
operators, all creation operators are placed to the left of all 
annihilation operators in the product.  This is referred to as NORMAL 
ORDERING.  The normal ordered string of operators is written as:

:ap†apap†ap: = ap†ap†apap

Note also that the ∫ -> ∞ unless a cutoff in the momentum is imposed.  
This is discussed in the section on Feynman diagrams.

Complex Scalar Field
--------------------

Using a similar procedure to the above we get:

Hrenorm' = (1/2)∫d3p/(2π)3 ωp (bp†bp + cp†cp)

Schrodinger
-----------  

L = ψ†(i∂/∂t + ∇2/2m)ψ 
       .
  = iψ†ψ + ψ†∇2ψ/2m
        .
π = ∂L/∂ψ = iψ†
      .
H' = πψ - L
       .      . 
   = iψ†ψ - iψ†ψ + ψ†∇2ψ/2m
                      
   = ψ†(-∇2/2m)ψ

This is just the expectation value of the operator.

<H> = <φ|H|φ>

    = ∫d3x ψ†[p2/2m]ψ

Now,

H|ψm> = Em|ψm>

      = ωm|ψm>  (h = 1)

E = ∫dx ψ† ω ψ

  = ∫d3x ω Σmam†ψm† Σnanψn

  = ∫d3x ω Σmnam†anψm†ψn

  = ∫d3x ω Σmnam†anδmn 

  = ∫d3x ω Σpap†ap 

  = ∫d3x d3/(2π)3 ωp ap†ap 

  = ∫d3x H'
                
Fermionic
---------  
    _
L = Ψ(iγμ∂μ - m)Ψ 
    _
  = Ψ(iγ0∂Ψ/∂t + iγ0γi/γ0∂Ψ/∂x - mΨ)
    _   .   _        _
  = Ψiγ0Ψ + Ψiγi∇Ψ - ΨmΨ
        .      _
π = ∂L/∂Ψ = iγ0Ψ 
      .
H' = πΨ - L
        _.   _   .   _         _
   = iγ0ΨΨ - Ψiγ0Ψ - Ψiγi∂iΨ + ΨmΨ              
      _         _   
   = -Ψiγi∂iΨ + ΨmΨ       
     _       
   = Ψ(-iγi∂i + m)Ψ       

This is just the expectation value of the operator (-iγi∂i + m)
Consider:

(iγi∂i - m)Ψ = 0

Rewrite as:

(iγ0∂0 + iγi∂i - m)u(p)exp(-ipx) = 0

Or, since p0 = i∂/∂t and  pi = i∂/∂x,

(γ0p0 + γipi - m)u(p)exp(-ipx) = 0 

Rearranging,

γ0p0u(p)exp(ip.x) = (-γipi + m)u(p)exp(ip.x)

and

-γ0p0v(p)exp(-ip.x) = (γipi + m)v(p)exp(-ip.x)

Now,

Ψ = ∫(d3p/(2π)3) (1/√(2ω))Σs[bpsus(p)exp(ip.x) + cps†vs(p)exp(-ip.x)]

Which leads to:

(-iγi∂i + m)Ψ = ∫(d3p/(2π)3) (1/√(2ω)) Σs[bps(-iγi∂i + m)us(p)exp(ip.x) 
                               + cps†(iγi∂i + m)vs(p)exp(-ip.x)]

Which leads to:

(-iγi∂i + m)Ψ = ∫(d3p/(2π)3) (1/√(2ω)) Σs[bps(γ0p0)us(p)exp(ip.x) 
                               - cps†(γ0p0)vs(p)exp(-ip.x)]
 
(-iγi∂i + m)Ψ = ∫(d3p/(2π)3) (1/√(2ω)) (γ0p0) Σs[bpsus(p)exp(ip.x) 
                               - cps†vs(p)exp(-ip.x)]

Now from Special Relativity p0 is equivalent to the energy E (= ω).  
Thus, 

(-iγi∂i + m)Ψ = ∫(d3p/(2π)3) (√(ω/2)) γ0 Σs[bpsus(p)exp(ip.x) 
                               - cps†vs(p)exp(-ip.x)]

Now, 

Ψ†(x) = ∫(d3p/(2π)3) (1/√(2ω))Σs[bps†us†(p)exp(-ip.x) + cpsvs†(p)exp(ip.x)]

Which leads to:

Ψ†(-iγi∂i + m)Ψ = ∫(d3pd3q/(2π)6) (√(ω/2))(1/√(2ω)) γ0 (Σr[bprur(p)exp(ip.x) 
                               - cpr†vr(p)exp(-ip.x)] 
                               Σs[cqs†us†(q)exp(-iq.x) + dqsvs†(q)exp(iq.x)])
 

Ψ†(-iγi∂i + m)Ψ = ∫(d3pd3q/(2π)6) (1/2) γ0 (Σr[bprur(p)exp(ip.x) 
                               - cpr†vr(p)exp(-ip.x)] 
                               Σs[cqs†us†(q)exp(-iq.x) + dqsvs†(q)exp(iq.x)])

    _
Now Ψ = γ0Ψ† and (γ0)2 = 1

Therefore,
_
Ψ(-iγi∂i + m)Ψ = ∫(d3pd3q/(2π)6) (1/2) (Σr[bprur(p)exp(ip.x) 
                               - cpr†vr(p)exp(-ip.x)] 
                               Σs[cqs†us†(q)exp(-iq.x) + dqsvs†(q)exp(iq.x)])

After manipulation and making use of the inner product relationship
between u and v we end up with:

H' = ∫(d3/(2π)3) ω(bps†bps + cps†cps)

Electromagnetic
---------------

From the electromagnetic tensor we get:

L = (-1/4)FμνFμν = (-1/2)(B2 - E2)  

In the Coulomb gauge the canonical momenta are:      

π0 = ∂L/∂A0 = 0
         .    .                                .    .
πi = ∂L/∂Ai = Ai = Ei  (same as real field with φ -> A)

H' = πAi - L
      .
H' = (Ai)2 - L
      
   = E2 + (1/2)(B2 - E2) 

   = (1/2)E2 + (1/2)B2 

We can also get this another way:
          .         
H' = (1/2)A2 + (1/2)(∂A/∂x)2 + (1/2)m2A2

Now m = 0 so ω2 = p2
          . 
H' = (1/2)A2 + (1/2)(∂A/∂x)2

Now,

A = ∫(d3p/(2π)3) (1/√(2ω))Σrξr[aprexp(ip.x) + apr†exp(-ip.x)]

Therefore,

∂A/∂x =∫(d3p/(2π)3) (i/√(2ω))(ip) Σrξr[aprexp(ip.x) - apr†exp(-ip.x)]

Now p = -i∇ ∴ ip = ∇ and B = ∇ x A.  Therefore, B = ip x A and we 
can also write: 

B(x) = ∫(d3p/(2π)3) (i/√(2ω))Σr{∇ x ξr[aprexp(ip.x) - apr†exp(-ip.x)]}

Therefore,

(∂A/∂x)2 ≡ B2

So we again get:

H' = (1/2)E2 + (1/2)B2 

After substituting in the expressions for E and B we get:

H' = ∫(d3p/(2π)3) ω Σrapr†apr
 
Where r = 1 -> 2

Note:  In the Lorentz gauge we would get:

H' = ∫(d3p/(2π)3) ω Σrapr†apr

Where r = 0 -> 3

However, the 0 and 3 components cancel leaving the Coulomb result.

 

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