208
17. Spontaneously Broken Global Symmetry
Furthermore, using the inverse of (17.38)
a ˆ = cosh θ
k
k α ˆ k − sinh θ k α ˆ
†
(17.51)
−k
together with (17.47), we find the similar result:
B ‹ground| ˆ
φ B (x)|ground›
1/2
B = ρ 0 .
(17.52)
ˆ
The question is now how to generalize (17.50) or (17.52) to the complete φ(x)
and the true ground state |ground›, in the limit N, Ω → ∞ with fixed N/Ω.
We make the assumption that
‹ground| ˆ
φ(x)|ground› = 0;
(17.53)
that is, we abstract from the Bogoliubov model the crucial feature that the
field acquires a non-zero expectation value in the ground state, in the infinite
volume limit.
We are now at the heart of spontaneous symmetry breaking in field theory.
Condition (17.53) has the form of an ‘ordering’ condition: it is analogous to
the non-zero value of the total spin in the ferromagnetic case, but in (17.53)
– we must again emphasize – |ground› is not an eigenstate of the symmetry
ˆ
operator N ; if it were, (17.53) would vanish, as we have just seen. Recalling the association ‘quantum vacuum ↔ many body ground state’ we expect
that the occurrence of a non-zero vacuum expectation value (vev) for an operator transforming non-trivially under a symmetry operator will be the key
requirement for spontaneous symmetry breaking in field theory. Such operators are generically called order parameters. In the next section we show how
this requirement necessitates one (or more) massless modes, via Goldstone’s
theorem (1961).
Before leaving the superfluid, we examine (17.37) and (17.52) in another
way, which is only rigorous for a finite system but is nevertheless very suggesˆ
ˆ
tive. Since the original H has a U(1) symmetry under which φ transforms to
ˆ
φ
′ = exp(− ˆ
iα)φ, we should be at liberty to replace (17.37) by
iα 1/2
1 ∑
ˆ
φ
′
B = e
− ρ 0 +
a ˆ e
−iα e
ik·x .
(17.54)
Ω 1/2
k
k=0
But in that case our condition (17.52) becomes
B ‹
ˆ
ˆ
ground|φ
′
B |ground› B = e
−iα
B ‹ground|φ B |ground› B .
(17.55)
ˆ
ˆ
Now φ
′
ˆ ˆ
= U α φU
−1
ˆ
ˆ
α where U α = exp(iαN ). Hence (17.55) may be written as
‹
| ˆ ˆ ˆ −1 |
›
−iα
ˆ
B ground U α φU α ground B = e B ‹ground|φ B |ground› B .
(17.56)
If |
ˆ
ˆ
ground› B were an eigenstate of N with eigenvalue N , say, then the U α factors in (17.56) would become just e
iαN
· e
−iαN and would cancel out, leaving a
/
/
17. Spontaneously Broken Global Symmetry
Furthermore, using the inverse of (17.38)
a ˆ = cosh θ
k
k α ˆ k − sinh θ k α ˆ
†
(17.51)
−k
together with (17.47), we find the similar result:
B ‹ground| ˆ
φ B (x)|ground›
1/2
B = ρ 0 .
(17.52)
ˆ
The question is now how to generalize (17.50) or (17.52) to the complete φ(x)
and the true ground state |ground›, in the limit N, Ω → ∞ with fixed N/Ω.
We make the assumption that
‹ground| ˆ
φ(x)|ground› = 0;
(17.53)
that is, we abstract from the Bogoliubov model the crucial feature that the
field acquires a non-zero expectation value in the ground state, in the infinite
volume limit.
We are now at the heart of spontaneous symmetry breaking in field theory.
Condition (17.53) has the form of an ‘ordering’ condition: it is analogous to
the non-zero value of the total spin in the ferromagnetic case, but in (17.53)
– we must again emphasize – |ground› is not an eigenstate of the symmetry
ˆ
operator N ; if it were, (17.53) would vanish, as we have just seen. Recalling the association ‘quantum vacuum ↔ many body ground state’ we expect
that the occurrence of a non-zero vacuum expectation value (vev) for an operator transforming non-trivially under a symmetry operator will be the key
requirement for spontaneous symmetry breaking in field theory. Such operators are generically called order parameters. In the next section we show how
this requirement necessitates one (or more) massless modes, via Goldstone’s
theorem (1961).
Before leaving the superfluid, we examine (17.37) and (17.52) in another
way, which is only rigorous for a finite system but is nevertheless very suggesˆ
ˆ
tive. Since the original H has a U(1) symmetry under which φ transforms to
ˆ
φ
′ = exp(− ˆ
iα)φ, we should be at liberty to replace (17.37) by
iα 1/2
1 ∑
ˆ
φ
′
B = e
− ρ 0 +
a ˆ e
−iα e
ik·x .
(17.54)
Ω 1/2
k
k=0
But in that case our condition (17.52) becomes
B ‹
ˆ
ˆ
ground|φ
′
B |ground› B = e
−iα
B ‹ground|φ B |ground› B .
(17.55)
ˆ
ˆ
Now φ
′
ˆ ˆ
= U α φU
−1
ˆ
ˆ
α where U α = exp(iαN ). Hence (17.55) may be written as
‹
| ˆ ˆ ˆ −1 |
›
−iα
ˆ
B ground U α φU α ground B = e B ‹ground|φ B |ground› B .
(17.56)
If |
ˆ
ˆ
ground› B were an eigenstate of N with eigenvalue N , say, then the U α factors in (17.56) would become just e
iαN
· e
−iαN and would cancel out, leaving a
/
/