4Spontaneous symmetry breaking
III The Standard Model
4.3 General case
What we just saw is a completely general phenomenon. Suppose we have an
N
-dimensional field
φ
= (
φ
1
, ··· , φ
N
), and suppose our theory has a Lie group
of symmetries
G
. We will assume that the action of
G
preserves the potential
and kinetic terms individually, and not just the Lagrangian as a whole, so that
G will send a vacuum to another vacuum.
Suppose there is more than one choice of vacuum. We write
Φ
0
= {φ
0
: V (φ
0
) = V
min
}
for the set of all vacua. Now we pick a favorite vacuum
φ
0
∈
Φ
0
, and we want
to look at the elements of G that fix this vacuum. We write
H = stab(φ
0
) = {h ∈ G : hφ
0
= φ
0
}.
This is the invariant subgroup, or stabilizer subgroup of
φ
0
, and is the symmetry
we are left with after the spontaneous symmetry breaking.
We will make some further simplifying assumptions. We will suppose
G
acts
transitively on Φ
0
, i.e. given any two vacua
φ
0
, φ
0
0
, we can find some
g ∈ G
such
that
φ
0
0
= gφ
0
.
This ensures that any two vacua are “the same”, so which
φ
0
we pick doesn’t really
matter. Indeed, given two such vacua and
g
relating them, it is a straightforward
computation to check that
stab(φ
0
0
) = g stab(φ
0
)g
−1
.
So any two choices of vacuum will result in conjugate, and in particular isomor-
phic, subgroups H.
It is a curious, and also unimportant, observation that given a choice of
vacuum φ
0
, we have a canonical bijection
G/H Φ
0
g gφ
0
∼
.
So we can identify G/H and Φ
0
. This is known as the orbit-stabilizer theorem.
We now try to understand what happens when we try to do perturbation
theory around our choice of vacuum. At φ = φ
0
+ δφ, we can as usual write
V (φ
0
+ δφ) − V (φ
0
) =
1
2
δφ
r
∂
2
V
∂φ
r
∂φ
s
δφ
s
+ O(δφ
3
).
This quadratic ∂
2
V term is now acting as a mass. We call it the mass matrix
M
2
rs
=
∂
2
V
∂φ
r
∂φ
s
.
Note that we are being sloppy with where our indices go, because (
M
2
)
rs
is ugly.
It doesn’t really matter much, since
φ
takes values in
R
N
(or
C
n
), which is a
Euclidean space.
In our previous example, we saw that there were certain modes that went
massless. We now want to see if the same happens. To do so, we pick a basis
{
˜
t
a
} of h (the Lie algebra of H), and then extend it to a basis {t
a
, θ
a
} of g.
We consider a variation
δφ = εθ
a
φ
0
around φ
0
.
Note that when we write
θ
a
φ
0
, we mean the resulting infinitesimal change of
φ
0
when
θ
a
acts on field. For a general Lie algebra element, this may be zero.
However, we know that
θ
a
does not belong to
h
, so it does not fix
φ
0
. Thus
(assuming sensible non-degeneracy conditions) this implies that
θ
a
φ
0
is non-zero.
At this point, the uncareful reader might be tempted to say that since
G
is a
symmetry of V , we must have V (φ
0
+ δφ) = V (φ
0
), and thus
(θ
a
φ)
r
M
2
rs
(θ
a
φ)
s
= 0,
and so we have found zero eigenvectors of
M
rs
. But this is wrong. For example,
if we take
M =
1 0
0 −1
, θ
a
φ =
1
1
,
then our previous equation holds, but
θ
a
φ
is not a zero eigenvector. We need to
be a bit more careful.
Instead, we note that by definition of
G
-invariance, for any field value
φ
and
a corresponding variation
φ 7→ φ + εθ
a
φ,
we must have
V (φ + δφ) = V (φ),
since this is what it means for G to be a symmetry.
Expanding to first order, this says
V (φ + δφ) − V (φ) = ε(θ
a
φ)
r
∂V
∂φ
r
= 0.
Now the trick is to take a further derivative of this. We obtain
0 =
∂
∂φ
s
(θ
a
φ)
r
∂V
∂φ
r
=
∂
∂φ
s
(θ
a
φ)
r
∂V
∂φ
r
+ (θ
a
φ)
r
M
2
rs
.
Now at a vacuum φ = φ
0
, the first term vanishes. So we must have
(θ
a
φ
0
)
r
M
2
rs
= 0.
By assumption,
θ
a
φ
0
6
= 0. So we have found a zero vector of
M
rs
. Recall that
we had
dim G − dim H
of these
θ
a
in total. So we have found
dim G − dim H
zero eigenvectors in total.
We can do a small sanity check. What if we replaced
θ
a
with
t
a
? The same
derivations would still follow through, and we deduce that
(t
a
φ
0
)
r
M
2
rs
= 0.
But since
t
a
is an unbroken symmetry, we actually just have
t
a
φ
0
= 0, and this
really doesn’t say anything interesting.
What about the other eigenvalues? They could be zero for some other reason,
and there is no reason to believe one way or the other. However, generally, in
the scenarios we meet, they are usually massive. So after spontaneous symmetry
breaking, we end up with
dim G−dim H
massless modes, and
N−
(
dim G−dim H
)
massive ones.
This is Goldstone’s theorem in the classical case.
Example.
In our previous
O
(
N
) model, we had
G
=
O
(
N
), and
H
=
O
(
N −
1).
These have dimensions
dim G =
N(N − 1)
2
, dim H =
(N − 1)(N − 2)
2
.
Then we have
Φ
0
∼
=
S
N−1
,
the (N − 1)-dimensional sphere. So we expect to have
N − 1
2
(N − (N − 2)) = N − 1
massless modes, which are the
π
fields we found. We also found
N −
(
N −
1) = 1
massive mode, namely the σ field.
Example.
In the first discrete case, we had
G
=
Z/
2
Z
and
H
=
{e}
. These are
(rather degenerate) 0-dimensional Lie groups. So we expect 0
−
0 = 0 massless
modes, and 1 − 0 = 1 massive modes, which is what we found.