III Quantum Field Theory - Quantizing the Dirac field

5Quantizing the Dirac field

III Quantum Field Theory

5.1 Fermion quantization

Quantization

We now start to quantize our spinor field! As it will turn out, this is complicated.

Recall that for a complex scalar field, classically a solution of momentum

can be written as

−ip·x

+ c

∗

ip·x

for some constants

, c

. Here the first term is the positive frequency solution,

and the second term is the negative frequency solution. To quantize this field,

we then promoted the b

, c

to operators, and we had

φ(x) =

(2π)

ip·x

+ c

†

−ip·x

Similarly, classically, every positive frequency solution to Dirac’s equation of

momentum p can be written as

+ b

−ip·x

for some b

, and similarly a negative frequency solution can be written as

+ c

ip·x

for some

. So when we quantize our field, we could promote the

and

operators and obtain

ψ(x) =

s=1

(2π)



ip·x

+ c

s†

−ip·x



†

(x) =

s=1

(2π)



s†

−ip·x

+ c

s†

ip·x



In these expressions,

and

are operators, and

and

are the elements of

the spinor space we have previously found, which are concrete classical objects.

We can compute the conjugate momentum to be

π =

∂L

∂

= i

ψγ

= iψ

†

The Hamiltonian density is then given by

H = π

ψ − L

= iψ

†

ψ − i

ψγ

ψ − i

ψγ

∂

ψ + m

ψψ

ψ(−iγ

∂

+ m)ψ.

What are the appropriate commutation relations to impose on the

, or, equiva-

lently, b

and c

s†

? The naive approach would be to impose

[ψ

(x), ψ

(y)] = [ψ

†

(x), ψ

†

(y)] = 0

and

[ψ

(x), ψ

†

(y)] = δ

αβ

(x − y).

These are in turn equivalent to requiring

, b

s†

] = (2π)

(p − q),

, c

s†

] = −(2π)

(p − q),

and all others vanishing.

If we do this, we can do the computations and find that (after normal

ordering) we obtain a Hamiltonian of

H =

(2π)

s†

− c

s†

Note the minus sign before

s†

. This is a disaster! While we can define a

vacuum

|0i

such that

|0i

= 0, this isn’t really a vacuum, because we

can keep applying c

†

to get lower and lower energies. This is totally bad.

What went wrong? The answer comes from actually looking at particles in

the Real World

. In the case of scalar fields, the commutation relation

†

, a

†

] = 0.

tells us that

†

|0i = |p, qi = |q, pi.

This means the particles satisfy Bose statistics, namely swapping two particles

gives the same state.

However, we know that fermions actually satisfy Fermi statistics. Swapping

two particles gives us a negative sign. So what we really need is

= −b

In other words, instead of setting the commutator zero, we want to set the

anticommutator to zero:

, b

} = 0.

In general, we are going to replace all commutators with anti-commutators. It

turns out this is what fixes our theory. We require that

Axiom. The spinor field operators satisfy

{ψ

(x), ψ

(y)} = {ψ

†

(x), ψ

†

(y)} = 0,

and

{ψ

(x), ψ

†

(y)} = δ

αβ

(x − y).

Proposition. The anti-commutation relations above are equivalent to

, c

s†

} = {b

, b

s†

} = (2π)

(p − q),

and all other anti-commutators vanishing.

Note that from a computational point of view, these anti-commutation

relations are horrible. When we manipulate our operators, we will keep on

introducing negative signs to our expressions, and as we all know, keeping track

of signs is the hardest thing in mathematics. Indeed, these negative signs will

come and haunt us all the time, and we have to insert funny negative signs

everywhere.

If we assume these anti-commutation relations with the same Hamiltonian

as before, then we would find

Proposition.

H =

(2π)



s†

+ c

s†



We now define the vacuum as in the bosonic case, where it is annihilated by

the b and c operators:

|0i = c

|0i = 0.

Although the

’s and

’s satisfy the anti-commutation relations, the Hamiltonian

satisfies commutator relations

[H, b

r†

] = Hb

r†

− b

r†

(2π)

s†

+ c

s†

†

− b

r†

(2π)

s†

+ c

s†

)

= E

r†

Similarly, we have

[H, b

] = −E

and the corresponding relations for c

and c

r†

Heisenberg picture

As before, we would like to put our Dirac field in the Heisenberg picture. We

have a spinor operator at each point x

ψ(x) = ψ(x, t).

The time evolution of the field is given by

∂ψ

∂t

= i[H, ψ],

which is solved by

ψ(x) =

s=1

(2π)



−ip·x

+ c

s†

ip·x



†

(x) =

s=1

(2π)



s†

ip·x

+ c

s†

−ip·x



We now look at the anti-commutators of the fields in the Heisenberg picture.

We define

Definition.

αβ

(x − y) = {ψ

(x),

(y)}.

Dropping the indices, we write

iS(x − y) = {ψ(x),

ψ(y)}.

If we drop the indices, then we have to remember that

is a 4

4 matrix with

indices α, β.

Substituting ψ and

ψ in using the expansion, we obtain

iS(x − y) =

s,r

p d

(2π)



, b

r†

¯u

−i(p·x−q·y)

+ {c

s†

, c

¯v

i(p·x−q·y)



(2π)



¯u

−p·(x−y)

+ v

¯v

ip·(x−y)



(2π)

(

p + m)e

−ip·(x−y)

+ (

p − m)e

ip·(x−y)

Recall that we had a scalar propagator

D(x −y) =

(2π)

−ip·(x−y)

So we can write the above quantity in terms of the D(x − y) by

iS(x − y) = (i

∂

+ m)(D(x −y) − D(y − x)).

Some comments about this: firstly, for spacelike separations where

(x − y)

< 0,

we had D(x − y) −D(y − x) = 0.

For bosonic fields, we made a big deal of this, since it ensured that

[

(

)

, φ

(

)] = 0. We then interpreted this as saying the theory was causal.

For spinor fields, we have anti-commutation relations rather than commutation

relations. What does this say about causality? The best we can say is that all

our observables are bilinear (or rather quadratic) in fermions, e.g.

H =

x ψ

†

(−iγ

∂

+ m)ψ.

So these observables will commute for spacelike separations.

Propagators

The next thing we need to figure out is the Feynman propagator. By a similar

computation to that above, we can determine the vacuum expectation value

h0|ψ(x)

ψ(y) |0i =

(2π)

(

p + m)e

−ip·(x−y)

h0|

ψ(y)ψ(x) |0i =

(2π)

(

p − m)e

ip·(x−y)

We define the Feynman propagator by

Definition

(Feynman propagator)

The Feynman propagator of a spinor field

is the time-ordered product

(x − y) = h0|T ψ

(x)

(y) |0i =

(

h0|ψ

(x)

(y) |0i x

> y

−h0|

(y)ψ

(x) |0i y

> x

Note that we have a funny negative sign! This is necessary for Lorentz

invariance — when (

x − y

)

0, then there is no invariant way to determine

whether one time is bigger than the other. So we need the expressions for the two

cases to agree. In the case of a boson, we had

h0|φ

(

)

(

)

|0i

h0|φ

(

)

(

)

|0i

but here we have anti-commutation relations, so we have

ψ(x)

ψ(y) = −

ψ(y)ψ(x).

So we need to insert the negative sign to ensure time-ordered product as defined

is Lorentz invariant.

For normal ordered products, we have the same behaviour. Fermionic opera-

tors anti-commute, so

:ψ

: = −:ψ

: .

As before, the Feynman propagator appears in Wick’s theorem as the contraction:

Proposition.

ψ(x)

ψ(y) = T (ψ(x)

ψ(y)) −:ψ(x)

ψ(y): = S

(x − y).

When we apply Wick’s theorem, we need to remember the minus sign. For

example,

:ψ

: = −:ψ

: = −ψ

:ψ

: .

Again, S

can be expressed as a 4-momentum integral

(x − y) = i

(2π)

−ip·(x−y)

p + m

− m

+ iε

As in the case of a real scalar field, it is a Green’s function of Dirac’s equation:

∂

− m)S

(x − y) = iδ

(x − y).