5Electroweak theory
III The Standard Model
5.2 Coupling to leptons
We now look at how gauge bosons couple to matter. In general, for a particle
with hypercharge Y , the covariant derivative is
D
µ
= ∂
µ
+ igW
a
µ
τ
a
+ ig
0
Y B
µ
In terms of the W
±
and Z bosons, we can write this as
D
µ
= ∂
µ
+
ig
√
2
(W
+
µ
τ
+
+ W
−
µ
τ
−
) +
igZ
µ
cos θ
W
(cos
2
θ
W
τ
3
− sin
2
θ
W
Y )
+ ig sin θ
W
A
µ
(τ
3
+ Y ),
where
τ
±
= τ
1
± iτ
2
.
By analogy, we interpret the final term
ig sin θ
W
A
µ
(
τ
3
+
Y
) is the usual coupling
with the photon. So we can identify the (magnitude of the) electron charge as
e = g sin θ
W
,
while the U(1)
EM
charge matrix is
Q = U(1)
EM
= τ
3
+ Y.
If we want to replace all occurrences of the hypercharge
Y
with
Q
, then we need
to note that
cos
2
θ
W
τ
3
− sin
2
θ
W
Y = τ
3
− sin
2
θ
W
Q.
We now introduce the electron field. The electron field is given by a spinor field
e(x). We will decompose it as left and right components
e(x) = e
L
(x) + e
R
(x).
There is also a neutrino field
ν
e
L
(
x
). We will assume that neutrinos are massless,
and there are only left-handed neutrinos. We know this is not entirely true,
because of neutrinos oscillation, but the mass is very tiny, and this is a very
good approximation.
To come up with an actual electroweak theory, we need to specify a represen-
tation of SU(2) × U(1). It is convenient to group our particles by handedness:
R(x) = e
R
(x), L(x) =
ν
e
L
(x)
e
L
(x)
.
Here
R
(
x
) is a single spinor, while
L
(
x
) consists of a pair, or (
SU
(2)) doublet of
spinors.
We first look at
R
(
x
). Experimentally, we find that
W
±
only couples to
left-handed leptons. So
R
(
x
) will have the trivial representation of
SU
(2). We
also know that electrons have charge
Q
=
−
1. Since
τ
3
acts trivially on
R
, we
must have
Q = Y = −1
for the right-handed leptons.
The left-handed particles are more complicated. We will assert that
L
has
the “fundamental” representation of SU(2), by which we mean given a matrix
g =
a b
−
¯
b ¯a
∈ SU(2),
it acts on L as
gL =
a b
−
¯
b ¯a
ν
e
L
e
L
=
aν
e
L
+ be
L
−
¯
bν
e
L
+ ¯ae
L
.
We know the electron has charge
−
1 and the neutrino has charge 0. So we need
Q =
0 0
0 −1
.
Since Q = τ
3
+ Y , we must have Y = −
1
2
.
Using this notation, the gauge part of the lepton Lagrangian can be written
concisely as
L
EW
lepton
=
¯
Li
/
DL +
¯
Ri
/
DR.
Note that
¯
L
means we take the transpose of the matrix
L
, viewed as a matrix
with two components, and then take the conjugate of each spinor individually,
so that
¯
L =
¯ν
e
L
(x) ¯e
L
(x)
.
How about the mass? Recall that it makes sense to deal with left and right-
handed fermions separately only if the fermion is massless. A mass term
m
e
(¯e
L
e
R
+ ¯e
R
e
L
)
would be very bad, because our SU(2) action mixes e
L
with ν
e
L
.
But we do know that the electron has mass. The answer is that the mass
is again granted by the spontaneous symmetry breaking of the Higgs boson.
Working in unitary gauge, we can write the Higgs boson as
φ(x) =
1
√
2
0
v + h(x)
,
with v, h(x) ∈ R.
The lepton-Higgs interactions is given by
L
lept,φ
= −
√
2λ
e
(
¯
LφR +
¯
Rφ
†
L),
where
λ
e
is the Yukawa coupling. It is helpful to make it more explicit what we
mean by this expression. We have
¯
LφR =
¯ν
e
L
(x) ¯e
L
(x)
1
√
2
0
v + h(x)
e
R
(x) =
1
√
2
(v + h(x))¯e
L
(x)e
R
(x).
Similarly, we can write out the second term and obtain
L
lept,φ
= −λ
e
(v + h)(¯e
L
e
R
+ ¯e
R
e
L
)
= −m
e
¯ee − λ
e
h¯ee,
where
m
e
= λ
e
v.
We see that while the interaction term is a priori gauge invariant, once we have
spontaneously broken the symmetry, we have magically obtained a mass term
m
e
. The second term in the Lagrangian is the Higgs-fermion coupling. We see
that this is proportional to m
e
. So massive things couple more strongly.
We now return to the fermion-gauge boson interactions. We can write it as
L
EM,int
lept
=
g
2
√
2
(J
µ
W
+
µ
+ J
µ†
W
−
µ
) + eJ
µ
EM
A
µ
+
g
2 cos θ
W
J
µ
n
Z
µ
.
where
J
µ
EM
= −¯eγ
µ
e
J
µ
= ¯ν
e
L
γ
µ
(1 − γ
5
)e
J
µ
n
=
1
2
¯ν
e
L
γ
µ
(1 − γ
5
)ν
e
L
− ¯eγ
µ
(1 − γ
5
− 4 sin
2
θ
w
)e
Note that
J
µ
EM
has a negative sign because the electron has negative charge.
These currents are known as the EM current, charge weak current and neutral
weak current respectively.
There is one thing we haven’t mentioned so far. We only talked about
electrons and neutrinos, but the standard model has three generations of leptons.
There are the muon (
µ
) and tau (
τ
), and corresponding left-handed neutrinos.
With these in mind, we introduce
L
1
=
ν
e
L
e
L
, L
2
=
ν
µ
L
µ
L
, L
3
=
ν
τ
L
τ
L
.
R
1
= e
R
, R
2
= µ
R
, R
3
= τ
R
.
These couple and interact in exactly the same way as the electrons, but have
heavier mass. It is straightforward to add these into the
L
EW,int
lept
term. What is
more interesting is the Higgs interaction, which is given by
L
lept,φ
= −
√
2
λ
ij
¯
L
i
φR
j
+ (λ
†
)
ij
¯
R
i
φ
†
L
j
,
where
i
and
j
run over the different generations. What’s new now is that we
have the matrices λ ∈ M
3
(C). These are not predicted by the standard model.
This
λ
is just a general matrix, and there is no reason to expect it to be
diagonal. However, in some sense, it is always diagonalizable. The key insight
is that we contract the two indices
λ
with two different kinds of things. It is a
general linear algebra fact that for any matrix
λ
at all, we can find some unitary
matrices U and S such that
λ = UΛS
†
,
where Λ is a diagonal matrix with real entries. We can then transform our fields
by
L
i
7→ U
ij
L
j
R
i
7→ S
ij
R
j
,
and this diagonalizes L
lept,φ
.
It is also clear that this leaves
L
EW
lept
invariant, especially if we look at the
expression before symmetry breaking. So the mass eigenstates are the same as
the “weak eigenstates”. This tells us that after diagonalizing, there is no mixing
within the different generations.
This is important. It is not possible for quarks (or if we have neutrino mass),
and mixing does occur in these cases.