# higgs - page 12

right vanishes, and hence the current generated this way is
conserved. This argument is a variant of what is known as a
Noether theorem.
From the Noether theorem, we have
J
= ·
J x
,
t
+
/
tJ
0
x
,
t
= 0,
9a
0 =
d
3
x J
.
9b
If nothing strange happens at the boundary in space, Eq.
reduces to
0 =
/
t d
3
xJ x
,
t
0
,
10
where
Q
is the spatial integral of the charge density
J
o
x
,
t
.
We now want to quantize this theory. Thus we write
x
=
d
3
k b
k k
x
+
a
k k
+
x
.
11
Here,
in the integral can be taken as a shorthand for the
Fourier transform coefficients, say, exp
ikx
, with some suit-
able normalization. The
a
and
b
coefficients are annihilation
operators, and their conjugates are creation operators. The
only nonvanishing commutators of the
a
and
b
operators are
a
k
,
a
k
=
3
k
k
=
b
k
,
b
k
.
12
The momentum
x
conjugate to is given by
x
=
t
x
= −
i
d
3
k
k
b
k k
x
a
k
k
+
x
.
13
Equation
yields the commutation relation
x
,
y
=
i
x
y
.
14
The charge can be written in terms of the
a
and
b
operators
as
Q
=
d
3
p a
p
a
p
b
p
b
p
.
15
From Eq.
it follows that
Q
,
=
16
so that
exp
i Q
exp −
i Q
= exp
i
.
17
Thus the charge generates the global gauge transformation,
that is,
transforms under unitary transformations generated
by the operator exp
i Q
.
The Hamiltonian in normal ordered form with
a
and
b
to
the left of
a
and
b
is given by
H
=
d
3
k
k
a
k
a
k
+
b
k
b
k
.
18
the Hamiltonian do not change the physics because we are
concerned with energy differences and not with absolute en-
ergies. In contrast, an additive constant to the Lagrangian
changes the action, and thus the transition amplitudes that
are calculated with path integrals. We must make sure that
additive constants do not change the physics by eliminating
any constants.
The vacuum state 0 is that state for which the energy is
minimum. If it has the property that 0
H
0 =0, it follows
that from equations such as 0
a
k
a
k
0 =0 that
a
k
0 =
b
k
0 = 0,
19
and thus both the Hamiltonian and the charge operators act-
ing on this state are zero. We also have the obvious property
0 0 = 0.
20
So far the particle mass has been specified without any
explanation for its origin. We now want to introduce mass
generation through spontaneous symmetry breaking. We in-
troduce a new Lagrangian
L
= 1
/
2
x
x
+
m
2
/
2
x
x
/
4
x x
2
.
21
Several properties of this Lagrangian are evident. First, the
term proportional to
m
2
is not a mass term. Compare the sign
self-interaction term. Second, the Lagrangian is invariant un-
der global gauge transformations but not local ones. There is
a conserved current as before and a conserved charge. But
does this charge annihilate the vacuum, that is,
Q
0 =0 as
before, and if not what does this mean? Here we run into the
question of what is the vacuum.
We recall the equation
Q
,
= ,
22
which also holds here. Equation
implies that if
Q
does
not annihilate the vacuum, then it must be that
0 0 0.
23
Equation
means that
cannot have a particle interpre-
tation because we cannot build up the single particle states
by the creation operators acting on the zero particle vacuum
state.
We recall from the discussion following Eq.
that the
energy is minimized for constant fields, and it is therefore
determined by minimizing the potential. Let us consider the
classical potential
V
= −
m
2
/
2
+
/
4
† 2
.
24
Clearly one extremal is when =0. Quantum mechanically
we want to replace this condition by the condition that the
vacuum expectation value of the potential is a minimum. We
shall see that in this case, there is no unique answer.
Let us warm up with a simpler case, which will illustrate
the issues. We consider a real field and the potential
V
= −
m
2
/
2
2
+
/
4
4
.
25
The potential and are all functions of the spacetime point
x
. At the minima of the vacuum expectation value of the
energy, they are constants, and hence we can find a condition
27
27
Am. J. Phys., Vol. 79, No. 1, January 2011
Jeremy Bernstein
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