207
17.3. Spontaneously broken symmetry in condensed matter physics
There is still more to be learned from (17.46). If, in fact, ¯
v(|l|) ∼ 1/l
2 ,
then ω l → constant as |l| → 0, and the spectrum would not be phonon-like.
Indeed, if ¯
v(|l|) ∼ e
2 /l
2 , then ω l ∼ |e|(ρ/m)
1/2 for small |l|, which is just the
‘plasma frequency’ ω p . In particle physics terms, this would be analogous to
a dispersion relation of the form ω l ∼ (ω
2 + l
2 )
1/2 , which describes a particle
p
with mass ω p . Such a ¯
v is, of course, Colombic (the Fourier transform of
e
2 /|x|), indicating that in the case of such a long-range force the frequency
spectrum acquires a mass-gap. This will be the topic of chapter 19.
Having discussed the spectrum of quasiparticle excitations, let us now
concentrate on the ground state in this model. From (17.40), it is clear that
it is defined as the state |ground› B such that
α ˆ k |ground› B = 0
for all k = / 0;
(17.47)
i.e. as the state with no non-zero-momentum quasiparticles in it. This is a
†
complicated state in terms of the original ˆ
a k and ˆ
a operators, but we can
k
give a formal expression for it, as follows. Since the ˆ
α’s and ˆ
a’s are related by
a canonical transformation, there must exist a unitary operator U ˆ B such that
ˆ
U ˆ −1
U ˆ −1
ˆ
α ˆ k = U B a ˆ k B ,
a ˆ k =
α ˆ k U B .
(17.48)
B
Now we know that ˆ
a k |0› = 0. Hence it follows that
ˆ
α ˆ k U B |0› = 0,
(17.49)
and we can identify |ground› B with U ˆ B |0›. In problem 17.4, U ˆ B is evaluated for
ˆ
an H B consisting of a single k-mode only, in which case the operator effecting
the transformation analogous to (17.48) is U ˆ 1 = exp[θ(ˆ aa ˆ − a ˆ
† a ˆ
† )/2] where
θ replaces θ k in this case. This generalizes (in the form of products of such
operators) to the full H ˆ B case, but we shall not need the detailed result; an
analogous result for the BCS ground state is discussed more fully in section
17.7. The important point is the following. It is clear from expanding the
exponentials that U ˆ B creates a state in which the number of a-quanta (i.e. the
original bosons) is not fixed. Thus unlike the simple non-interacting ground
ˆ
state |N, 0› of (17.33), |ground› B = U B |0› does not have a fixed number of
particles in it: that is to say, it is not an eigenstate of the symmetry operator
ˆ
N , as anticipated in the comment following (17.36). This is just the situation
alluded to in the paragraph before equation (17.19), in our discussion of the
ferromagnet.
Consider now the expectation value of φ ˆ (x) in any state of definite particle
number – that is, in an eigenstate of the symmetry operator N ˆ . It is easy to
see that this must vanish (remember that φ ˆ destroys a boson, and so φ ˆ |N › is
proportional to |N − 1›, which is orthogonal to |N ›). On the other hand, this
is not true of φ ˆ B (x): for example, in the non-interacting ground state (17.33),
we have
1/2
‹N, 0|φ ˆ B (x)|N, 0› = ρ .
(17.50)
0
17.3. Spontaneously broken symmetry in condensed matter physics
There is still more to be learned from (17.46). If, in fact, ¯
v(|l|) ∼ 1/l
2 ,
then ω l → constant as |l| → 0, and the spectrum would not be phonon-like.
Indeed, if ¯
v(|l|) ∼ e
2 /l
2 , then ω l ∼ |e|(ρ/m)
1/2 for small |l|, which is just the
‘plasma frequency’ ω p . In particle physics terms, this would be analogous to
a dispersion relation of the form ω l ∼ (ω
2 + l
2 )
1/2 , which describes a particle
p
with mass ω p . Such a ¯
v is, of course, Colombic (the Fourier transform of
e
2 /|x|), indicating that in the case of such a long-range force the frequency
spectrum acquires a mass-gap. This will be the topic of chapter 19.
Having discussed the spectrum of quasiparticle excitations, let us now
concentrate on the ground state in this model. From (17.40), it is clear that
it is defined as the state |ground› B such that
α ˆ k |ground› B = 0
for all k = / 0;
(17.47)
i.e. as the state with no non-zero-momentum quasiparticles in it. This is a
†
complicated state in terms of the original ˆ
a k and ˆ
a operators, but we can
k
give a formal expression for it, as follows. Since the ˆ
α’s and ˆ
a’s are related by
a canonical transformation, there must exist a unitary operator U ˆ B such that
ˆ
U ˆ −1
U ˆ −1
ˆ
α ˆ k = U B a ˆ k B ,
a ˆ k =
α ˆ k U B .
(17.48)
B
Now we know that ˆ
a k |0› = 0. Hence it follows that
ˆ
α ˆ k U B |0› = 0,
(17.49)
and we can identify |ground› B with U ˆ B |0›. In problem 17.4, U ˆ B is evaluated for
ˆ
an H B consisting of a single k-mode only, in which case the operator effecting
the transformation analogous to (17.48) is U ˆ 1 = exp[θ(ˆ aa ˆ − a ˆ
† a ˆ
† )/2] where
θ replaces θ k in this case. This generalizes (in the form of products of such
operators) to the full H ˆ B case, but we shall not need the detailed result; an
analogous result for the BCS ground state is discussed more fully in section
17.7. The important point is the following. It is clear from expanding the
exponentials that U ˆ B creates a state in which the number of a-quanta (i.e. the
original bosons) is not fixed. Thus unlike the simple non-interacting ground
ˆ
state |N, 0› of (17.33), |ground› B = U B |0› does not have a fixed number of
particles in it: that is to say, it is not an eigenstate of the symmetry operator
ˆ
N , as anticipated in the comment following (17.36). This is just the situation
alluded to in the paragraph before equation (17.19), in our discussion of the
ferromagnet.
Consider now the expectation value of φ ˆ (x) in any state of definite particle
number – that is, in an eigenstate of the symmetry operator N ˆ . It is easy to
see that this must vanish (remember that φ ˆ destroys a boson, and so φ ˆ |N › is
proportional to |N − 1›, which is orthogonal to |N ›). On the other hand, this
is not true of φ ˆ B (x): for example, in the non-interacting ground state (17.33),
we have
1/2
‹N, 0|φ ˆ B (x)|N, 0› = ρ .
(17.50)
0