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
ˆ
C, and we make the following assumption:
Assumption.
If
a
is an annihilation operator of a particle, and
c
is the annihi-
lation operator of the anti-particle, then we have
ˆ
Ca(p)
ˆ
C
1
= ηc(p)
for some η.
As before, this is motivated by the requirement that
ˆ
C |pi
=
η
|¯pi
. We will
also assume the following:
Assumption.
Let
φ
be any field. Then
ˆ
Cφ
(
x
)
ˆ
C
1
is a multiple of the conjugate
of
φ
, where “a multiple” can be multiplication by a linear map in the case where
φ
has more than 1 component. Here the interpretation of “the conjugate” depends
on the kind of field:
If
φ
is a bosonic field, then the conjugate of
φ
is
φ
= (
φ
)
T
. 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
¯
φ
T
.
Scalar and vector fields
Scalar and vector fields behave in a manner very similar to that of parity
operators. We simply have
ˆ
Cφ
ˆ
C
1
= η
c
φ
ˆ
Cφ
ˆ
C
1
= η
c
φ
for some
η
c
. In the case of a real field, we have
φ
=
φ
. So we must have
η
c
=
±
1,
which is known as the intrinsic
c
-parity of the field. For complex fields, we can
introduce a global phase change of the field so as to set η
c
= 1.
This has some physical significance. For example, the photon field transforms
like
ˆ
CA
µ
(x)
ˆ
C
1
= A
µ
(x).
Experimentally, we see that
π
0
only decays to 2 photons, but not 1 or 3. Therefore,
assuming that c-parity is conserved, we infer that
η
π
0
c
= (1)
2
= +1.
Dirac fields
Dirac fields are more complicated. As before, we can compute
ˆ
Cψ(x)
ˆ
C
1
= η
c
X
p,s
d
s
(p)u
s
(p)e
ip·x
+ b
s
(p)v
s
(p)e
+ip·x
.
We compare this with
¯
ψ
T
(x) =
X
p,s
b
s
(p)¯u
sT
(p)e
+ip·x
+ d
s
(p)¯v
sT
(p)e
ip·x
.
We thus see that we have to relate ¯u
sT
and v
s
somehow; and ¯v
sT
with u
s
.
Recall that we constructed
u
s
and
v
s
in terms of elements
ξ
s
, η
s
R
2
. At
that time, we did not specify what
ξ
and
η
are, or how they are related, so we
cannot expect
u
and
v
to have any relation at all. So we need to make a specific
choice of ξ and η. We will choose them so that
η
s
=
2
ξ
s
.
Now consider the matrix
C =
0
γ
2
=
2
0
0
2
.
The reason we care about this is the following:
Proposition.
v
s
(p) = C ¯u
sT
, u
s
(p) = C¯v
sT
(p).
From this, we infer that
Proposition.
ψ
c
(x)
ˆ
Cψ(x)
ˆ
C
1
= η
c
C
¯
ψ
T
(x)
¯
ψ
c
(x)
ˆ
C
¯
ψ(x)
ˆ
C
1
= η
c
ψ
T
(x)C = η
c
ψ
T
(x)C
1
.
It is convenient to note the following additional properties of the matrix C:
Proposition.
(Cγ
µ
)
T
= Cγ
µ
C = C
T
= C
= C
1
(γ
µ
)
T
= Cγ
µ
C
1
, (γ
5
)
T
= +Cγ
5
C
1
.
Note that if ψ(x) satisfies the Dirac equation, then so does ψ
c
(x).
Apart from Dirac fermions, there are also things called Majorana fermions.
They have
b
s
(
p
) =
d
s
(
p
). This means that the particle is its own antiparticle,
and for these, we have
ψ
c
(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
β
decay, but current
experiments can neither observe nor rule out this possibility.
Fermion bilinears
We can look at how fermion bilinears change. For example,
j
µ
=
¯
ψ(x)γ
µ
ψ(x).
Then we have
ˆ
Cj
µ
(x)
ˆ
C
1
=
ˆ
C
¯
ψ
ˆ
C
1
γ
µ
ˆ
Cψ
ˆ
C
1
= η
c
ψ
T
C
1
γ
µ
C
¯
ψ
T
η
c
We now notice that the
η
c
and
η
c
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
γ
µ
C)
T
ψ
=
¯
ψC
T
γ
µT
(C
1
)
T
ψ
=
¯
ψγ
µ
ψ
= j
µ
(x).
Therefore A
µ
(x)j
µ
(x) is invariant under
ˆ
C. Similarly,
¯
ψγ
µ
γ
5
ψ 7→ +
¯
ψγ
µ
γ
5
ψ.