III The Standard Model - Discrete symmetries

3Discrete symmetries

III The Standard Model

3.3 Charge conjugation

We will do similar manipulations, and figure out how different fields transform

under charge conjugation. Unlike parity, this is not a spacetime symmetry. It

transforms particles to anti-particles, and vice versa. This is a unitary operator

C, and we make the following assumption:

Assumption.

is an annihilation operator of a particle, and

is the annihi-

lation operator of the anti-particle, then we have

Ca(p)

−1

= ηc(p)

for some η.

As before, this is motivated by the requirement that

C |pi

∗

|¯pi

. We will

also assume the following:

Assumption.

Let

be any field. Then

Cφ

(

)

−1

is a multiple of the conjugate

, where “a multiple” can be multiplication by a linear map in the case where

has more than 1 component. Here the interpretation of “the conjugate” depends

on the kind of field:

–

is a bosonic field, then the conjugate of

∗

= (

†

)

. Of course, if

this is a scalar field, then the conjugate is just φ

†

– If φ is a spinor field, then the conjugate of φ is

Scalar and vector fields

Scalar and vector fields behave in a manner very similar to that of parity

operators. We simply have

Cφ

−1

= η

†

Cφ

†

−1

= η

∗

for some

. In the case of a real field, we have

†

. So we must have

which is known as the intrinsic

-parity of the field. For complex fields, we can

introduce a global phase change of the field so as to set η

= 1.

This has some physical significance. For example, the photon field transforms

(x)

−1

= −A

(x).

Experimentally, we see that

only decays to 2 photons, but not 1 or 3. Therefore,

assuming that c-parity is conserved, we infer that

= (−1)

= +1.

Dirac fields

Dirac fields are more complicated. As before, we can compute

Cψ(x)

−1

= η

p,s



(p)u

(p)e

−ip·x

+ b

s†

(p)v

(p)e

+ip·x



We compare this with

(x) =

p,s



s†

(p)¯u

(p)e

+ip·x

+ d

(p)¯v

(p)e

−ip·x



We thus see that we have to relate ¯u

and v

somehow; and ¯v

with u

Recall that we constructed

and

in terms of elements

, η

∈ R

. At

that time, we did not specify what

and

are, or how they are related, so we

cannot expect

and

to have any relation at all. So we need to make a specific

choice of ξ and η. We will choose them so that

= iσ

s∗

Now consider the matrix

C = −iγ



iσ

0 −iσ



The reason we care about this is the following:

Proposition.

(p) = C ¯u

, u

(p) = C¯v

(p).

From this, we infer that

Proposition.

(x) ≡

Cψ(x)

−1

= η

(x)

(x) ≡

ψ(x)

−1

= η

∗

(x)C = −η

∗

(x)C

−1

It is convenient to note the following additional properties of the matrix C:

Proposition.

(Cγ

)

= Cγ

C = −C

= −C

†

= −C

−1

(γ

)

= −Cγ

−1

, (γ

)

= +Cγ

−1

Note that if ψ(x) satisfies the Dirac equation, then so does ψ

(x).

Apart from Dirac fermions, there are also things called Majorana fermions.

They have

(

) =

(

). This means that the particle is its own antiparticle,

and for these, we have

(x) = ψ(x).

These fermions have to be neutral. A natural question to ask is — do these

exist? The only type of neutral fermions in the standard model is neutrinos, and

we don’t really know what neutrinos are. In particular, it is possible that they

are Majorana fermions. Experimentally, if they are indeed Majorana fermions,

then we would be able to observe neutrino-less double

decay, but current

experiments can neither observe nor rule out this possibility.

Fermion bilinears

We can look at how fermion bilinears change. For example,

ψ(x)γ

ψ(x).

Then we have

(x)

−1

Cψ

−1

= −η

∗

−1

We now notice that the

∗

and

cancel each other. Also, since this is a scalar,

we can take the transpose of the whole thing, but we will pick up a minus sign

because fermions anti-commute

ψ(C

−1

ψC

µT

−1

)

= −

ψγ

= −j

(x).

Therefore A

(x)j

(x) is invariant under

C. Similarly,

ψγ

ψ 7→ +

ψγ

ψ.