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

= ·

,

+

/

0

,

= 0,

9a

0 =

3

.

9b

If nothing strange happens at the boundary in space, Eq.

reduces to

0 =

/

3

,

0

,

10

where

is the spatial integral of the charge density

,

.

We now want to quantize this theory. Thus we write

=

3

†

−

+

+

.

11

Here,

in the integral can be taken as a shorthand for the

Fourier transform coefficients, say, exp

, with some suit-

able normalization. The

and

coefficients are annihilation

operators, and their conjugates are creation operators. The

only nonvanishing commutators of the

and

operators are

,

†

=

3

−

=

,

†

.

12

The momentum

conjugate to is given by

=

†

= −

3

−

−

†

+

.

13

Equation

yields the commutation relation

,

=

−

.

14

The charge can be written in terms of the

and

operators

as

=

3

†

−

†

.

15

From Eq.

it follows that

,

=

16

so that

exp

exp −

= exp

.

17

Thus the charge generates the global gauge transformation,

that is,

transforms under unitary transformations generated

by the operator exp

.

The Hamiltonian in normal ordered form with

†

and

†

to

the left of

and

is given by

=

3

†

+

†

.

18

We have dropped an additive constant. Additive constants to

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

0 =0, it follows

that from equations such as 0

†

0 =0 that

0 =

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

= 1

/

2

†

+

2

/

2

†

−

/

4

†

2

.

21

Several properties of this Lagrangian are evident. First, the

term proportional to

2

is not a mass term. Compare the sign

to the sign in the mass term in Eq.

Instead, it is a

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,

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

,

= ,

22

which also holds here. Equation

implies that if

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

= −

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

= −

2

/

2

2

+

/

4

4

.

25

The potential and are all functions of the spacetime point

. 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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