4Spinors
III Quantum Field Theory
4.8 Symmetries and currents
We now look at what Noether’s theorem tells us about spinor fields. We will
only briefly state the corresponding results because the calculations are fairly
routine. The only interesting one to note is that the Lagrangian is given by
L =
¯
ψ(i
/
∂ − m)ψ,
while the equations of motion say
(i
/
∂ − m)ψ = 0.
So whenever the equations of motion are satisfied, the Lagrangian vanishes. This
is going to simplify quite a few of our results.
Translation symmetry
As usual, spacetime translation is a symmetry, and the infinitesimal transforma-
tion is given by
x
µ
7→ x
µ
− ε
µ
ψ 7→ ψ + ε
µ
∂
µ
ψ.
So we have
δψ = ε
µ
∂
µ
ψ, δ
¯
ψ = ε
µ
∂
µ
¯
ψ.
Then we find a conserved current
T
µν
= i
¯
ψγ
µ
∂
ν
ψ −η
µν
L = i
¯
ψγ
µ
∂
ν
ψ.
In particular, we have a conserved energy of
E =
Z
T
00
d
3
x
=
Z
d
3
x i
¯
ψγ
0
˙
ψ
=
Z
d
3
x
¯
ψ(−iγ
i
∂
i
+ m)ψ,
where we used the equation of motion in the last line.
Similarly, we have a conserved total momentum
P
i
=
Z
d
3
x T
0i
=
Z
d
3
x iψ
†
∂
i
ψ.
Lorentz transformations
We can also consider Lorentz transformations. This gives a transformation
ψ 7→ S[Λ]ψ(x
µ
− ω
µ
ν
x
ν
).
Taylor expanding, we get
δψ = −ω
µ
ν
x
ν
∂
µ
ψ +
1
2
ω
ρσ
(S
ρσ
)ψ
= −ω
µν
x
ν
∂
µ
ψ −
1
2
(S
µν
)ψ
.
Similarly, we have
δ
¯
ψ = −ω
µν
x
ν
∂
µ
¯
ψ +
1
2
¯
ψ(S
µν
)
.
The change is sign is because
¯
ψ
transforms as
S
[Λ]
−1
, and taking the inverse of
an exponential gives us a negative sign.
So we can write this as
j
µ
= −ω
ρσ
i
¯
ψγ
µ
(x
σ
∂
ρ
ψ −S
ρσ
ψ)
+ ω
µ
ν
x
ν
L
= −ω
ρσ
i
¯
ψγ
µ
(x
σ
∂
ρ
ψ −S
ρσ
ψ)
.
So if we allow ourselves to pick different
ω
ρσ
, we would have one conserved
current for each choice:
(J
µ
)
ρσ
= x
σ
T
µρ
− x
ρ
T
µσ
− i
¯
ψγ
µ
S
ρσ
ψ.
In the case of a scalar field, we only have the first two terms. We will later see
that the extra term will give us spin
1
2
after quantization. For example, we have
(J
0
)
ij
= −i
¯
ψS
ij
ψ =
1
2
ε
ijk
ψ
†
σ
k
0
0 σ
k
ψ.
This is our angular momentum operator.
Internal vector symmetry
We also have an internal vector symmetry
ψ 7→ e
iα
ψ.
So the infinitesimal transformation is
δψ = iαψ.
We then obtain
j
µ
V
=
¯
ψγ
µ
ψ.
We can indeed check that
∂
µ
j
µ
V
= (∂
µ
¯
ψ)γ
µ
ψ +
¯
ψγ
µ
(∂
µ
ψ) = im
¯
ψψ −im
¯
ψψ = 0.
The conserved quantity is then
Q =
Z
d
3
x j
0
V
=
Z
d
3
x
¯
ψγ
0
ψ =
Z
d
3
x ψ
†
ψ.
We will see that this is electric charge/particle number.
Axial symmetry
When
m
= 0, we have an extra internal symmetry obtained by rotating left-
handed and right-handed fermions in opposite signs. These symmetries are called
chiral. We consider the transformations
ψ 7→ e
iαγ
5
ψ,
¯
ψ 7→
¯
ψe
iαγ
5
.
This gives us an axial current
j
µ
A
=
¯
ψγ
µ
γ
5
ψ.
This is an axial vector. We claim that this is conserved only when
m
= 0. We
have
∂
µ
j
µ
A
= ∂
µ
¯
ψγ
µ
γ
5
ψ +
¯
ψγ
µ
γ
5
∂
µ
ψ
= 2im
¯
ψγ
5
ψ.
This time the two terms add rather than subtract. So this vanishes iff m = 0.
It turns out that this is an anomalous symmetry, in the sense that the
classical axial symmetry does not survive quantization.