4Spontaneous symmetry breaking
III The Standard Model
4.4 Goldstone’s theorem
We now consider the quantum version of Goldstone’s theorem. We hope to get
something similar.
Again, suppose our Lagrangian has a symmetry group
G
, which is sponta-
neously broken into
H ≤ G
after picking a preferred vacuum
|0i
. Again, we take
a Lie algebra basis
{t
a
, θ
a
}
of
g
, where
{t
a
}
is a basis for
h
. Note that we will
assume that
a
runs from 1
, ··· , dim G
, and we use the first
dim H
labels to refer
to the t
a
, and the remaining to label the θ
a
, so that the actual basis is
{t
1
, t
2
, ··· , t
dim H
, θ
dim H+1
, ··· , θ
dim G
}.
For aesthetic reasons, this time we put the indices on the generators below.
Now recall, say from AQFT, that these symmetries of the Lagrangian give
rise to conserved currents j
µ
a
(x). These in turn give us conserved charges
Q
a
=
Z
d
3
x j
0
a
(x).
Moreover, under the action of the
t
a
and
θ
a
, the corresponding change in
φ
is
given by
δφ = −[Q
a
, φ].
The fact that the θ
a
break the symmetry of |0i tells us that, for a > dim H,
h0|[Q
a
, φ(0)] |0i 6= 0.
Note that the choice of evaluating
φ
at 0 is completely arbitrary, but we have
to pick somewhere, and 0 is easy to write. Writing out the definition of
Q
a
, we
know that
Z
d
3
x h0|[j
0
a
(x), φ(0)] |0i 6= 0.
More generally, since the label
0
is arbitrary by Lorentz invariance, we are really
looking at the relation
h0|[j
µ
a
(x), φ(0)] |0i 6= 0,
and writing it in this more Lorentz covariant way makes it easier to work with.
The goal is to deduce the existence of massless states from this non-vanishing.
For convenience, we will write this expression as
X
µ
a
= h0|[j
µ
a
(x), φ(0)] |0i.
We treat the two terms in the commutator separately. The first term is
h0|j
µ
a
(x)φ(0) |0i =
X
n
h0|j
µ
a
(x) |nihn|φ(0) |0i. (∗)
We write p
n
for the momentum of |ni. We now note the operator equation
j
µ
a
(x) = e
iˆp·x
j
µ
a
(0)e
−iˆp·x
.
So we know
h0|j
µ
a
(x) |ni = h0|j
µ
a
(0) |nie
−ip
n
·x
.
We can use this to do some Fourier transform magic. By direct verification, we
can see that (∗) is equivalent to
i
Z
d
4
k
(2π)
3
ρ
µ
a
(k)e
−ik·x
,
where
iρ
µ
a
(k) = (2π)
3
X
n
δ
4
(k −p
n
) h0|j
µ
a
(0) |nihn|φ(0) |0i.
Similarly, we define
i˜ρ
µ
a
(k) = (2π)
3
X
n
δ
4
(k −p
n
) h0|φ(0) |nihn|j
µ
a
(0) |0i.
Then we can write
X
µ
a
= i
Z
d
4
k
(2π)
3
ρ
µ
a
(k)e
−ik·x
− ˜ρ
µ
a
(k)e
+ik·x
.
This is called the K¨allen-Lehmann spectral representation.
We now claim that
ρ
µ
a
(
k
) and
˜ρ
µ
a
(
k
) can be written in a particularly nice
form. This involves a few observations (when I say
ρ
µ
a
, I actually mean
ρ
µ
a
and
˜ρ
µ
a
):
– ρ
µ
a
depends Lorentz covariantly in
k
. So it “must” be a scalar multiple of
k
µ
.
– ρ
µ
a
(k) must vanish when k
0
> 0, since k
0
≤ 0 is non-physical.
–
By Lorentz invariance, the magnitude of
ρ
µ
a
(
k
) can depend only on the
length (squared) k
2
of k.
Under these assumptions, we can write can write ρ
µ
a
and ˜ρ
µ
a
as
ρ
µ
a
(k) = k
µ
Θ(k
0
)ρ
a
(k
2
),
˜ρ
µ
a
(k) = k
µ
Θ(k
0
)˜ρ
a
(k
2
),
where Θ is the Heaviside theta function
Θ(x) =
(
1 x > 0
0 x ≤ 0
.
So we can write
X
µ
a
= i
Z
d
4
k
(2π)
3
k
µ
Θ(k
0
)(ρ
a
(k
2
)e
−ik·x
− ˜ρ
a
(k
2
)e
+ik·x
).
We now use the (nasty?) trick of hiding the
k
µ
by replacing it with a derivative:
X
µ
a
= −∂
µ
Z
d
4
k
(2π)
3
Θ(k
0
)
ρ
a
(k
2
)e
−ik·x
+ ˜ρ
a
(k
2
)e
ik·x
.
Now we might find the expression inside the integral a bit familiar. Recall that
the propagator is given by
D(x, σ) = h0|φ(x)φ(0) |0i =
Z
d
4
k
(2π)
3
Θ(k
0
)δ(k
2
− σ)e
−ik·x
.
where σ is the square of the mass of the field. Using the really silly fact that
ρ
a
(k
2
) =
Z
dσ ρ
a
(σ)δ(k
2
− σ),
we find that
X
µ
a
= −∂
µ
Z
dσ (ρ
a
(σ)D(x, σ) + ˜ρ
a
(σ)D(−x, σ)).
Now we have to recall more properties of
D
. For spacelike
x
, we have
D
(
x, σ
) =
D
(
−x, σ
). Therefore, requiring
X
µ
a
to vanish for spacelike
x
by causality, we see
that we must have
ρ
a
(σ) = −˜ρ
a
(σ).
Therefore we can write
X
µ
a
= −∂
µ
Z
dσ ρ
a
(σ)i∆(x, σ), (†)
where
i∆(x, σ) = D(x, σ) − D(−x, σ) =
Z
d
4
k
(2π)
3
δ(k
2
− σ)ε(k
0
)e
−ik·x
,
and
ε(k
0
) =
(
+1 k
0
> 0
−1 k
0
< 0
.
This ∆ is again a different sort of propagator.
We now use current conservation, which tells us
∂
µ
j
µ
a
= 0.
So we must have
∂
µ
X
µ
a
= −∂
2
Z
dσ ρ
a
(σ)i∆(x, σ) = 0.
On the other hand, by inspection, we see that ∆ satisfies the Klein-Gordon
equation
(∂
2
+ σ)∆ = 0.
So in (†), we can replace −∂
2
∆ with σ∆. So we find that
∂
µ
X
µ
a
=
Z
dσ σρ
a
(σ)i∆(x, σ) = 0.
This is true for all
x
. In particular, it is true for timelike
x
where ∆ is non-zero.
So we must have
σρ
a
(σ) = 0.
But we also know that
ρ
a
(
σ
) cannot be identically zero, because
X
µ
a
is not. So
the only possible explanation is that
ρ
a
(σ) = N
a
δ(σ),
where N
a
is a dimensionful non-zero constant.
Now we retrieve our definitions of ρ
a
. Recall that they are defined by
iρ
µ
a
(k) = (2π)
3
X
n
δ
4
(k − p
n
) h0|j
µ
a
(0) |nihn|φ(0) |0i
ρ
µ
a
(k) = k
µ
Θ(k
0
)ρ
a
(k
2
).
So the fact that
ρ
a
(
σ
) =
N
a
δ
(
σ
) implies that there must be some states, which
we shall call |B(p)i, of momentum p, such that p
2
= 0, and
h0|j
µ
a
(0) |B(p)i 6= 0
hB(p)|φ(0) |0i 6= 0.
The condition that
p
2
= 0 tell us these particles are massless, which are those
massless modes we were looking for! These are the Goldstone bosons.
We can write the values of the above as
h0|j
µ
a
(0) |B(p)i = iF
B
a
p
µ
hB(p)|φ(0) |0i = Z
B
,
whose form we deduced by Lorentz in/covariance. By dimensional analysis, we
know F
aB
is a dimension −1 constant, and Z
B
is a dimensionless constant.
From these formulas, we note that since
φ
(0)
|0i
is rotationally invariant, we
deduce that |B(p)i also is. So we deduce that these are in fact spin 0 particles.
Finally, we end with some computations that relate these different numbers
we obtained. We will leave the computational details for the reader, and just
outline what we should do. Using our formula for ρ, we find that
h0|[j
µ
a
(x), φ(0)] |0i = −∂
µ
Z
N
a
δ(σ)i∆(x, σ) dσ = −iN
a
∂
µ
∆(x, 0).
Integrating over space, we find
h0|[Q
a
, φ(0)] |0i = −iN
a
Z
d
3
x ∂
0
∆(x, 0) = iN
a
.
Then we have
h0|t
a
φ(0) |0i = i h0|[Q
a
, φ(0)] |0i = i · iN
a
= −N
a
.
So this number
N
a
sort-of measures how much the symmetry is broken, and we
once again see that it is the breaking of symmetry that gives rise to a non-zero
N
a
, hence the massless bosons.
We also recall that we had
ik
µ
Θ(k
0
)N
a
δ(k
2
) =
X
B
Z
d
3
p
2|p|
δ
4
(k − p) h0|j
µ
a
(0) |B(p)ihB(p)|φ(0) |0i.
On the RHS, we can plug in the names we gave for the matrix elements, and on
the left, we can write it as an integral
Z
d
3
p
2|p|
δ
4
(k − p)ik
µ
N
a
=
Z
d
3
p
2|p|
δ
4
(k − p)ip
µ
X
B
F
B
a
Z
B
.
So we must have
N
a
=
X
B
F
B
a
Z
B
.
We can repeat this process for each independent symmetry-breaking
θ
a
∈ g \h
,
and obtain a Goldstone boson this way. So all together, at least superficially, we
find
n = dim G − dim H
many Goldstone bosons.
It is important to figure out the assumptions we used to derive the result.
We mentioned “Lorentz invariance” many times in the proof, so we certainly
need to assume our theory is Lorentz invariant. Another perhaps subtle point is
that we also needed states to have a positive definite norm. It turns out in the
case of gauge theory, these do not necessarily hold, and we need to work with
them separately.