4Spontaneous symmetry breaking

III The Standard Model



4.2 Continuous symmetry
We can consider a slight generalization of the above scenario. Consider an
N-component real scalar field φ = (φ
1
, φ
2
, ··· , φ
N
)
T
, with Lagrangian
L =
1
2
(
µ
φ) · (
µ
φ) V (φ),
where
V (φ) =
1
2
m
2
φ
2
+
λ
4
φ
4
.
As before, we will require that λ > 0.
This is a theory that is invariant under global O(
N
) transforms of
φ
. Again,
if
m
2
>
0, then
φ
= 0 is a global minimum, and we do not have spontaneous
symmetry breaking. Thus, we consider the case m
2
< 0.
In this case, we can write
V (φ) =
λ
4
(φ
2
v
2
)
2
+ constant,
where
v =
m
2
λ
> 0.
So any
φ
with
φ
2
=
v
2
gives a global minimum of the system, and so we have a
continuum of vacua. We call this a Sombrero potential (or wine bottle potential).
φ
1
φ
2
V (φ)
Without loss of generality, we may choose the vacuum to be
φ
0
= (0, 0, ··· , 0, v)
T
.
We can then consider steady small fluctuations about this:
φ(x) = (π
1
(x), π
2
(x), ··· , π
N1
(x), v + σ(x))
T
.
We now have a π field with N 1 components, plus a 1-component σ field.
We rewrite the Lagrangian in terms of the π and σ fields as
L =
1
2
(
µ
π) · (
µ
π) +
1
2
(
µ
σ) · (
µ
σ) V (π, σ),
where
V (π, σ) =
1
2
m
2
σ
σ
s
+ λv(σ
2
+ π
2
)σ +
λ
4
(σ
2
+ π
2
)
2
.
Again, we have dropped a constant term.
In this Lagrangian, we see that the σ field has mass
m
σ
=
2λv
2
,
but the
π
fields are massless. We can understand this as follows the
σ
field
corresponds to the radial direction in the potential, and this is locally quadratic,
and hence has a mass term. However, the
π
fields correspond to the excitations
in the azimuthal directions, which are flat.