2Chiral and gauge symmetries
III The Standard Model
2.1 Chiral symmetry
Chiral symmetry is something that manifests itself when we have spinors. Since
all matter fields are spinors, this is clearly important.
The notion of chirality is something that exists classically, and we shall begin
by working classically. A Dirac spinor is, in particular, a (4-component) spinor
field. As usual, the Dirac matrices
γ
µ
are 4
×
4 matrices satisfying the Clifford
algebra relations
{γ
µ
, γ
ν
} = 2g
µν
I,
where g
µν
is the Minkowski metric. For any operator A
µ
, we write
/
A = γ
µ
A
µ
.
Then a Dirac fermion is defined to be a spinor field satisfying the Dirac equation
(i
/
∂ − m)ψ = 0.
We define
γ
5
= +iγ
0
γ
1
γ
2
γ
3
,
which satisfies
(γ
5
)
2
= I, {γ
5
, γ
µ
} = 0.
One can do a lot of stuff without choosing a particular basis/representation for
the
γ
-matrices, and the physics we get out must be the same regardless of which
representation we choose, but sometimes it is convenient to pick some particular
representation to work with. We’ll generally use the chiral representation (or
Weyl representation), with
γ
0
=
0 1
1 0
, γ
i
=
0 σ
i
−σ
i
0
, γ
5
=
−1 0
0 1
,
where the σ
i
∈ Mat
2
(C) are the Pauli matrices.
Chirality
γ
5
is a particularly interesting matrix to consider. We saw that it satisfies
(
γ
5
)
2
= 1. So we know that
γ
5
is diagonalizable, and the eigenvalues of
γ
5
are
either +1 or −1.
Definition
(Chirality)
.
A Dirac fermion
ψ
is right-handed if
γ
5
ψ
=
ψ
, and
left-handed if γ
5
ψ = −ψ.
A left- or right-handed fermion is said to have definite chirality.
In general, a fermion need not have definite chirality. However, as we know
from linear algebra, the spinor space is a direct sum of the eigenspaces of
γ
5
. So
given any spinor ψ, we can write it as
ψ = ψ
L
+ ψ
R
,
where ψ
L
is left-handed, and ψ
R
is right-handed.
It is not hard to find these ψ
L
and ψ
R
. We define the projection operators
P
R
=
1
2
(1 + γ
5
), P
L
=
1
2
(1 − γ
5
).
It is a direct computation to show that
γ
5
P
L
= −P
L
, γ
5
P
R
= P
R
, (P
R,L
)
2
= P
R,L
, P
L
+ P
R
= I.
Thus, we can write
ψ = (P
L
+ P
R
)ψ = (P
L
ψ) + (P
R
ψ),
and thus we have
Notation.
ψ
L
= P
L
ψ, ψ
R
= P
R
ψ.
Moreover, we notice that
P
L
P
R
= P
R
P
L
= 0.
This implies
Lemma. If ψ
L
is left-handed and φ
R
is right-handed, then
¯
ψ
L
φ
L
=
¯
ψ
R
φ
R
= 0.
Proof. We only do the left-handed case.
¯
ψ
L
φ
L
= ψ
†
L
γ
0
φ
L
= (P
L
ψ
L
)
†
γ
0
(P
L
φ
L
)
= ψ
†
L
P
L
γ
0
P
L
φ
L
= ψ
†
L
P
L
P
R
γ
0
φ
L
= 0,
using the fact that {γ
5
, γ
0
} = 0.
The projection operators look very simple in the chiral representation. They
simply look like
P
L
=
I 0
0 0
, P
R
=
0 0
0 I
.
Now we have produced these
ψ
L
and
ψ
R
, but what are they? In particular, are
they themselves Dirac spinors? We notice that the anti-commutation relations
give
/
∂γ
5
ψ = −γ
5
/
∂ψ.
Thus,
γ
5
does not commute with the Dirac operator (
i
/
∂ −m
), and one can check
that in general,
ψ
L
and
ψ
R
do not satisfy the Dirac equation. But there is one
exception. If the spinor is massless, so
m
= 0, then the Dirac equation becomes
/
∂ψ = 0.
If this holds, then we also have
/
∂γ
5
ψ
=
−γ
5
/
∂ψ
= 0. In other words,
γ
5
ψ
is still
a Dirac spinor, and hence
Proposition. If ψ is a massless Dirac spinor, then so are ψ
L
and ψ
R
.
More generally, if the mass is non-zero, then the Lagrangian is given by
L =
¯
ψ(i
/
∂ − m)ψ =
¯
ψ
L
i
/
∂ψ
L
+
¯
ψ
L
i
/
∂ψ
R
− m(
¯
ψ
L
ψ
R
+
¯
ψ
R
ψ
L
).
In general, it “makes sense” to treat ψ
L
and ψ
R
as separate objects only if the
spinor field is massless. This is crucial. As mentioned in the overview, the weak
interaction only couples to left-handed fermions. For this to “make sense”, the
fermions must be massless! But the electrons and fermions we know and love
are not massless. Thus, the masses of the electron and fermion terms cannot
have come from a direct mass term in the Lagrangian. They must obtain it via
some other mechanism, namely the Higgs mechanism.
To see an example of how the mass makes such a huge difference, by staring
at the Lagrangian, we notice that if the fermion is massless, then we have a
U(1)
L
×
U(1)
R
global symmetry — under an element (
α
L
, α
R
)
∈
U(1)
L
×
U(1)
R
,
the fermion transforms as
ψ
L
ψ
R
7→
e
iα
L
ψ
L
e
iα
R
ψ
R
.
The adjoint field transforms as
¯
ψ
L
¯
ψ
R
7→
e
−iα
L
¯
ψ
L
e
−iα
R
¯
ψ
R
,
and we see that the Lagrangian is invariant. However, if we had a massive
particle, then we would have the cross terms in the Lagrangian. The only way
for the Lagrangian to remain invariant is if
α
L
=
α
R
, and the symmetry has
reduced to a single U(1) symmetry.
Quantization of Dirac field
Another important difference the mass makes is the notion of helicity. This is
a quantum phenomenon, so we need to describe the quantum theory of Dirac
fields. When we quantize the Dirac field, we can decompose it as
ψ =
X
s,p
b
s
(p)u
s
(p)e
−ip·x
+ d
s†
(p)v
s
(p)e
−ip·x
.
We explain these things term by term:
– s is the spin, and takes values s = ±
1
2
.
– The summation over all p is actually an integral
X
p
=
Z
d
3
p
(2π)
3
(2E
p
)
.
– b
†
and
d
†
operators that create positive and negative frequency particles
respectively. We use relativistic normalization, where the states
|pi = b
†
(p) |0i
satisfy
hp|qi = (2π)
3
2E
p
δ
(3)
(p − q).
–
The
u
s
(
p
) and
v
s
(
p
) form a basis of the solution space to the (classical)
Dirac equation, so that
u
s
(p)e
−ip·x
, v
s
(p)e
ip·x
are solutions for any
p
and
s
. In the chiral representation, we can write
them as
u
s
(p) =
√
p · σξ
s
√
p · ¯σξ
s
, v
s
(p) =
√
p · ση
s
−
√
p · ¯ση
s
,
where as usual
σ
µ
= (I, σ
i
), ¯σ
µ
= (I, −σ
i
),
and {ξ
±
1
2
} and {η
±
1
2
} are bases for R
2
.
We can define a quantum operator corresponding to the chirality, known as
helicity.
Definition
(Helicity)
.
We define the helicity to be the projection of the angular
momentum onto the direction of the linear momentum:
h = J ·
ˆ
p = S ·
ˆ
p,
where
J = −ir × ∇ + S
is the total angular momentum, and S is the spin operator given by
S
i
=
i
4
ε
ijk
γ
j
γ
k
=
1
2
σ
i
0
0 σ
i
.
The main claim about helicity is that for a massless spinor, it reduces to the
chirality, in the following sense:
Proposition. If we have a massless spinor u, then
hu(p) =
γ
5
2
u(p).
Proof. Note that if we have a massless particle, then we have
/
pu = 0,
since quantumly,
p
is just given by differentiation. We write this out explicitly
to see
γ
µ
p
µ
u
s
= (γ
0
p
0
− γ · p)u = 0.
Multiplying it by γ
5
γ
0
/p
0
gives
γ
5
u(p) = γ
5
γ
0
γ
i
p
i
p
0
u(p).
Again since the particle is massless, we know
(p
0
)
2
− p · p = 0.
So
ˆ
p = p/p
0
. Also, by direct computation, we find that
γ
5
γ
0
γ
i
= 2S
i
.
So it follows that
γ
5
u(p) = 2hu(p).
In particular, we have
hu
L,R
=
γ
5
2
u
L,R
= ∓
1
2
u
L,R
.
So u
L,R
has helicity ∓
1
2
.
This is another crucial observation. Helicity is the spin in the direction of the
momentum, and spin is what has to be conserved. On the other hand, chirality is
what determines whether it interacts with the weak force. For massless particles,
these to notions coincide. Thus, the spin of the particles participating in weak
interactions is constrained by the fact that the weak force only couples to left-
handed particles. Consequently, spin conservation will forbid certain interactions
from happening.
However, once our particles have mass, then helicity is different from spin.
In general, helicity is still quite closely related to spin, at least when the mass is
small. So the interactions that were previously forbidden are now rather unlikely
to happen, but are actually possible.