52
S. J. Blundell
g(ω)dω ∝ ω
1/2 dω
(2.36)
at low temperature where only small q and small ω are important. The spin waves
are quantized in the same way as lattice waves. The latter are termed phonons, and
so in the same way the former are termed magnons. They are bosons and have a spin
of one.
The number of magnon modes excited at temperature T , n magnon , is calculated by
integrating the magnon density of states over all frequencies after multiplying by the
Bose factor, [exp(ω/k B T ) − 1]
−1 , which must be included because magnons are
bosons. Thus the result is given by
n magnon =
∞
0
g(ω) dω
exp(ω/k B T ) − 1
,
(2.37)
which can be evaluated using the substitution x = ω/k B T . At low temperature,
where g(ω) ∝ ω
1/2 in three dimensions, this yields the result
n magnon =
k B T
3/2 ∞
0
x
1/2 dx
e x − 1
∝ T
3/2
.
(2.38)
Since each magnon mode which is thermally excited reduces the total magnetization
by one (because each magnon mode is a delocalized single reversed spin), then at low
temperature the reduction in the spontaneous magnetization from the T = 0 value is
given by
M(0) − M(T )
M(0)
∝ T
3/2
.
(2.39)
This result is known as the Bloch T
3/2 law. If one repeats this calculation in two
dimensions (rather than three), the integral diverges, showing that magnons spontaneously form at all non-zero temperatures, thereby destroying any magnetization.
The impossibility of spontaneous magnetization in two dimensions for the Heisenberg model is known as the Mermin–Wagner theorem [15–17] (see also [9]).
2.3.3 Three Spins
Let us return now to an apparently simpler system and consider three spins on the
corners of an equilateral triangle. We will put the exchange interaction to be negative
and thus the system is frustrated. If we put the first spin up, the next one down,
then we have a dilemma of how to arrange the third one because we cannot satisfy
the antiferromagnetic interactions on every bond. The solution has to be one of
compromise and in fact the ground state of the classical Heisenberg model on a
triangle is the so-called 120
◦ state shown in Fig. 2.7a.
S. J. Blundell
g(ω)dω ∝ ω
1/2 dω
(2.36)
at low temperature where only small q and small ω are important. The spin waves
are quantized in the same way as lattice waves. The latter are termed phonons, and
so in the same way the former are termed magnons. They are bosons and have a spin
of one.
The number of magnon modes excited at temperature T , n magnon , is calculated by
integrating the magnon density of states over all frequencies after multiplying by the
Bose factor, [exp(ω/k B T ) − 1]
−1 , which must be included because magnons are
bosons. Thus the result is given by
n magnon =
∞
0
g(ω) dω
exp(ω/k B T ) − 1
,
(2.37)
which can be evaluated using the substitution x = ω/k B T . At low temperature,
where g(ω) ∝ ω
1/2 in three dimensions, this yields the result
n magnon =
k B T
3/2 ∞
0
x
1/2 dx
e x − 1
∝ T
3/2
.
(2.38)
Since each magnon mode which is thermally excited reduces the total magnetization
by one (because each magnon mode is a delocalized single reversed spin), then at low
temperature the reduction in the spontaneous magnetization from the T = 0 value is
given by
M(0) − M(T )
M(0)
∝ T
3/2
.
(2.39)
This result is known as the Bloch T
3/2 law. If one repeats this calculation in two
dimensions (rather than three), the integral diverges, showing that magnons spontaneously form at all non-zero temperatures, thereby destroying any magnetization.
The impossibility of spontaneous magnetization in two dimensions for the Heisenberg model is known as the Mermin–Wagner theorem [15–17] (see also [9]).
2.3.3 Three Spins
Let us return now to an apparently simpler system and consider three spins on the
corners of an equilateral triangle. We will put the exchange interaction to be negative
and thus the system is frustrated. If we put the first spin up, the next one down,
then we have a dilemma of how to arrange the third one because we cannot satisfy
the antiferromagnetic interactions on every bond. The solution has to be one of
compromise and in fact the ground state of the classical Heisenberg model on a
triangle is the so-called 120
◦ state shown in Fig. 2.7a.