17.3. Spontaneously broken symmetry in condensed matter physics
201
the axis n ˆ = (0, sin α, cos α). Thus the first ground state is
( ) (
)
(
)
1
1
1
χ 0 =
. . .
(N products)
(17.19)
0
0
0
1
2
N
while the second is (cf (4.31), (4.32))
(
)
(
)
(α)
cos α/2
cos α/2
χ 0 =
. . .
.
(17.20)
i sin α/2
i sin α/2
1
N
The scalar product of (17.19) and (17.20) is (cos α/2)
N , which goes to zero as
N → ∞. Thus any two such ‘rotated ground states’ are indeed orthogonal in
the infinite volume (or infinite number of degrees of freedom) limit.
We may also enquire about the excited states built on one such ground
ˆ
state, say the one with S z eigenvalue N/2. Suppose for simplicity that
the magnet is one-dimensional (but the spins have all three components).
ˆ
ˆ
Consider the state χ n = S n χ 0 where S n is the spin lowering operator
−
−
ˆ
ˆ
S n = (S nx − ˆ
iS ny ) at site n, such that
−
(
)
( )
1
0
ˆ
S n
=
;
(17.21)
−
0
1
n
n
ˆ
so S n χ 0 differs from the ground state χ
−
0 by having the spin at site n flipped.
ˆ
The action of H S on χ n can be found by writing
∑
∑ 1
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
S i · S j =
(S i S + S S ) + S S
(17.22)
2
− j+
j− i+
iz jz
i=j
i=j
ˆ
ˆ
(remembering that spins on different sites commute), where S i+ = S ix +
ˆ
ˆ
iS iy . Since all S i+ operators give zero on a spin ‘up’ state, the only non-zero
contributions from the first (bracketed) term in (17.22) come from terms in
ˆ
ˆ
which either S i+ or S j+ act on the ‘down’ spin at n, so as to restore it to ‘up’.
ˆ
ˆ
The ‘partner’ operator S i (or S j ) then simply lowers the spin at i (or j),
−
−
leading to the result
∑ 1
∑
ˆ ˆ
ˆ ˆ
(S i S j+ + S j S i+ )χ n =
χ
2
−
−
i .
(17.23)
i=j
i=n
ˆ
Thus the state χ n is not an eigenstate of H S . However, a little more work
shows that the superpostitions
1 ∑
χ ˜ q = √
e
iqna χ n
(17.24)
N n
are eigenstates; here q is one of the discretized wavenumbers produced by
appropriate boundary conditions, as is usual in one-dimensional ‘chain’ problems. The states (17.24) represent spin waves, and they have the important
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/
/
/
201
the axis n ˆ = (0, sin α, cos α). Thus the first ground state is
( ) (
)
(
)
1
1
1
χ 0 =
. . .
(N products)
(17.19)
0
0
0
1
2
N
while the second is (cf (4.31), (4.32))
(
)
(
)
(α)
cos α/2
cos α/2
χ 0 =
. . .
.
(17.20)
i sin α/2
i sin α/2
1
N
The scalar product of (17.19) and (17.20) is (cos α/2)
N , which goes to zero as
N → ∞. Thus any two such ‘rotated ground states’ are indeed orthogonal in
the infinite volume (or infinite number of degrees of freedom) limit.
We may also enquire about the excited states built on one such ground
ˆ
state, say the one with S z eigenvalue N/2. Suppose for simplicity that
the magnet is one-dimensional (but the spins have all three components).
ˆ
ˆ
Consider the state χ n = S n χ 0 where S n is the spin lowering operator
−
−
ˆ
ˆ
S n = (S nx − ˆ
iS ny ) at site n, such that
−
(
)
( )
1
0
ˆ
S n
=
;
(17.21)
−
0
1
n
n
ˆ
so S n χ 0 differs from the ground state χ
−
0 by having the spin at site n flipped.
ˆ
The action of H S on χ n can be found by writing
∑
∑ 1
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
S i · S j =
(S i S + S S ) + S S
(17.22)
2
− j+
j− i+
iz jz
i=j
i=j
ˆ
ˆ
(remembering that spins on different sites commute), where S i+ = S ix +
ˆ
ˆ
iS iy . Since all S i+ operators give zero on a spin ‘up’ state, the only non-zero
contributions from the first (bracketed) term in (17.22) come from terms in
ˆ
ˆ
which either S i+ or S j+ act on the ‘down’ spin at n, so as to restore it to ‘up’.
ˆ
ˆ
The ‘partner’ operator S i (or S j ) then simply lowers the spin at i (or j),
−
−
leading to the result
∑ 1
∑
ˆ ˆ
ˆ ˆ
(S i S j+ + S j S i+ )χ n =
χ
2
−
−
i .
(17.23)
i=j
i=n
ˆ
Thus the state χ n is not an eigenstate of H S . However, a little more work
shows that the superpostitions
1 ∑
χ ˜ q = √
e
iqna χ n
(17.24)
N n
are eigenstates; here q is one of the discretized wavenumbers produced by
appropriate boundary conditions, as is usual in one-dimensional ‘chain’ problems. The states (17.24) represent spin waves, and they have the important
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/
/
/