3Discrete symmetries
III The Standard Model
3.4 Time reversal
Finally, we get to time reversal. This is more messy. Under T, we have
x
µ
= (x
0
, x
i
) 7→ x
µ
T
= (−x
0
, x
i
).
Momentum transforms in the opposite way, with
p
µ
= (p
0
, p
i
) 7→ p
µ
T
= (p
0
, −p
i
).
Theories that are invariant under
ˆ
T
look the same when we run them backwards.
For example, Newton’s laws are time reversal invariant. We will also see that
the electromagnetic and strong interactions are time reversal invariant, but the
weak interaction is not.
Here our theory begins to get more messy.
Assumption. For any field φ, we have
ˆ
T φ(x)
ˆ
T
−1
∝ φ(x
T
).
Assumption. For bosonic fields with creator a
†
, we have
ˆ
T a(p)
ˆ
T
−1
= ηa(p
T
)
for some η. For Dirac fields with creator b
s†
, we have
ˆ
T b
s
(p)
ˆ
T
−1
= η(−1)
1/2−s
b
−s
(p
T
)
for some η.
Why that complicated rule for Dirac fields? We first justify why we want
to swap spins when we perform time reversal. This is just the observation that
when we reverse time, we spin in the opposite direction. The factor of (
−
1)
1
2
−s
tells us spinors of different spins transform differently.
Boson field
Again, bosonic fields are easy. The creation and annihilation operators transform
as
ˆ
T a(p)
ˆ
T
−1
= η
T
a(p
T
)
ˆ
T c
†
(p)
ˆ
T
−1
= η
T
c
†
(p
T
).
Note that the relative phases are fixed using the same argument as for
ˆ
P
and
ˆ
C
. When we do the derivations, it is very important to keep in mind that
ˆ
T
is
anti-linear.
Then we have
ˆ
T φ(x)
ˆ
T
−1
= η
T
φ(x
T
).
Dirac fields
As mentioned, for Dirac fields, the annihilation and creation operators transform
as
ˆ
T b
s
(p)
ˆ
T
−1
= η
T
(−1)
1/2−s
b
−s
(p
T
)
ˆ
T d
s†
(p)
ˆ
T
−1
= η
T
(−1)
1/2−s
d
−s†
(p
T
)
It can be shown that
(−1)
1
2
−s
u
−s∗
(p
T
) = −Bu
s
(p)
(−1)
1
2
−s
v
−s∗
(p
T
) = −Bv
s
(p),
where
B = γ
5
C =
iσ
2
0
0 iσ
2
in the chiral representation.
Then, we have
ˆ
T ψ(x)
ˆ
T
−1
= η
T
X
p,s
(−1)
1
2
−s
b
−s
(p
T
)u
s∗
(p)e
+ip·x
+ d
−s†
v
s∗
(p)e
−ip·x
.
Doing the standard manipulations, we find that
ˆ
T ψ(x)
ˆ
T
−1
= η
T
X
p,s
(−1)
1
2
−s+1
b
s
(p)u
−s∗
(p
T
)e
−ip·x
T
+ d
s†
(p)v
−s∗
(p
T
)e
+ip·x
T
= +η
T
X
p,s
b
s
(p)BU
s
(p)e
−ip·x
T
+ d
s†
(p)BV
s
(p)e
+ip·x
T
= η
T
Bψ(x
T
).
Similarly,
ˆ
T
¯
ψ(x)
ˆ
T
−1
= η
∗
T
¯
ψ(x
T
)B
−1
.
Fermion bilinears
Similarly, we have
¯
ψ(x)ψ(x) 7→
¯
ψ(x
T
)ψ(x
T
)
¯
ψ(x)γ
µ
ψ(x) 7→ −T
µ
ν
¯
ψ(x
T
)γ
µ
ψ(x
T
)
This uses the fact that
B
−1
γ
µ∗
B = −T
µ
ν
γ
ν
.
So we see that in the second case, the 0 component, i.e. charge density, is
unchanged, while the spacial component, i.e. current density, gets a negative
sign. This makes physical sense.