17.5. Spontaneously broken global U(1) symmetry: the Goldstone model 211
But this expression is independent of x 0 . Massive states |n› will produce
±iMnx0
explicit x 0 -dependent factors e
(p n0 → M n as the δ-function constrains
p = 0), hence the matrix elements of ˆ j 0 between |0› and such a massive state
n
must vanish, and such states contribute zero to (17.67). Equally, if we take
|n› = |0›, (17.67) vanishes identically. But it has been assumed to be not zero.
Hence some state or states must exist among |n› such that ‹0|j 0 |n› / = 0 and
yet (17.67) is independent of x 0 . The only possibility is states whose energy
p n0 goes to zero as their 3-momentum does (from δ
3 (p )). Such states are,
n
of course, massless; they are called generically Goldstone modes. Thus the
existence of a non-vanishing vacuum expectation value for a field, in a theory
with a continuous symmetry, appears to lead inevitably to the necessity of
having a massless particle, or particles, in the theory. This is the Goldstone
result.
The superfluid provided us with an explicit model exhibiting the crucial
non-zero expectation value ‹ground|φ ˆ |ground› = / 0, in which the now expected
massless mode emerged dynamically. We now discuss a simpler, relativistic
model, in which the symmetry breaking is brought about more ‘by hand’ –
that is, by choosing a parameter in the Lagrangian appropriately. Although
in a sense less ‘dynamical’ than the Bogoliubov superfluid (or the BCS superconductor, to be discussed shortly) this Goldstone model does provide a
very simple example of the phenomenon of spontaneous symmetry breaking
in field theory.
17.5 Spontaneously broken global U(1) symmetry: the
Goldstone model
We consider, following Goldstone (1961), a complex scalar field φ ˆ as in section 7.1, with
φ ˆ = √
1 (φ ˆ 1 − iφ ˆ 2 ),
φ ˆ † = √
1 (φ ˆ 1 + iφ ˆ 2 ),
(17.68)
2
2
described by the Lagrangian
L ˆ G = (∂ μ φ ˆ † )(∂
μ φ ˆ ) − V ˆ (φ ˆ ).
(17.69)
We begin by considering the ‘normal’ case in which the potential has the form
ˆ
ˆ
1 λ(φ ˆ † φ ˆ )
2 + μ
2 φ ˆ † φ ˆ
(17.70)
V = V S ≡ 4
with μ
2 , λ > 0. The Hamiltonian density is then
H ˆ G = φ ˆ ˙
†
φ ˆ ˙ + ∇φ ˆ † · ∇φ ˆ + V ˆ (φ ˆ ).
(17.71)
But this expression is independent of x 0 . Massive states |n› will produce
±iMnx0
explicit x 0 -dependent factors e
(p n0 → M n as the δ-function constrains
p = 0), hence the matrix elements of ˆ j 0 between |0› and such a massive state
n
must vanish, and such states contribute zero to (17.67). Equally, if we take
|n› = |0›, (17.67) vanishes identically. But it has been assumed to be not zero.
Hence some state or states must exist among |n› such that ‹0|j 0 |n› / = 0 and
yet (17.67) is independent of x 0 . The only possibility is states whose energy
p n0 goes to zero as their 3-momentum does (from δ
3 (p )). Such states are,
n
of course, massless; they are called generically Goldstone modes. Thus the
existence of a non-vanishing vacuum expectation value for a field, in a theory
with a continuous symmetry, appears to lead inevitably to the necessity of
having a massless particle, or particles, in the theory. This is the Goldstone
result.
The superfluid provided us with an explicit model exhibiting the crucial
non-zero expectation value ‹ground|φ ˆ |ground› = / 0, in which the now expected
massless mode emerged dynamically. We now discuss a simpler, relativistic
model, in which the symmetry breaking is brought about more ‘by hand’ –
that is, by choosing a parameter in the Lagrangian appropriately. Although
in a sense less ‘dynamical’ than the Bogoliubov superfluid (or the BCS superconductor, to be discussed shortly) this Goldstone model does provide a
very simple example of the phenomenon of spontaneous symmetry breaking
in field theory.
17.5 Spontaneously broken global U(1) symmetry: the
Goldstone model
We consider, following Goldstone (1961), a complex scalar field φ ˆ as in section 7.1, with
φ ˆ = √
1 (φ ˆ 1 − iφ ˆ 2 ),
φ ˆ † = √
1 (φ ˆ 1 + iφ ˆ 2 ),
(17.68)
2
2
described by the Lagrangian
L ˆ G = (∂ μ φ ˆ † )(∂
μ φ ˆ ) − V ˆ (φ ˆ ).
(17.69)
We begin by considering the ‘normal’ case in which the potential has the form
ˆ
ˆ
1 λ(φ ˆ † φ ˆ )
2 + μ
2 φ ˆ † φ ˆ
(17.70)
V = V S ≡ 4
with μ
2 , λ > 0. The Hamiltonian density is then
H ˆ G = φ ˆ ˙
†
φ ˆ ˙ + ∇φ ˆ † · ∇φ ˆ + V ˆ (φ ˆ ).
(17.71)