Chemistry Reference
In-Depth Information
any spin and its two neighbours is given by
2
Js
2
cos
W
=−
θ
(7.27)
For a sufficiently long loop,
θ
is small, so we can approximate cos
θ
as
2
2
−
θ
/
=
−
(π/
)
/
1
2
1
N
2. The total energy of the chain, 2
NW
,isthen
given by
4
JNs
2
1
2
2
Js
2
2
−
(π/
N
)
π
4
JNs
2
2
NW
=−
=−
+
(7.28)
2
N
and the increase in energy over the ground state, where all spins are par-
allel, is only 2
Js
2
2
. Clearly, mean field
theory fails badly in its estimate of the energy of this simple excitation.
The classical unit of spin excitation involves flipping just one spin. (In
quantummechanics, we reduce the total angular momentum by one unit.)
We see that we have effectively flipped
N
spins in order to produce the
state illustrated in fig. 7.8. The net energy associated with each individual
spin flip,
E
flip
is therefore
π
/
N
, which goes to zero as
N
→∞
2
Js
2
N
2
2
Js
2
2
π
π
E
flip
=
=
(7.29)
N
2
N
We can describe the state in fig. 7.8 as being formed by combining
N
'spin
waves', each of wavelength
2
Na
, where
a
is the nearest neighbour
separation. Each spin wave then has wavevector
k
, with
k
λ
=
=
2
π/(
2
Na
)
=
π/
N
in eq. (7.29), we, therefore,
estimate that the energy
E
k
of the basic quantum unit of spin excitation,
referred to as a
magnon
, depends quadratically on the wavevector
k
,as
Na
,sothat
ka
=
π/
N
. Substituting for
π/
2
Js
2
2
2
Js
2
a
2
k
2
E
k
=
ω
≈
(
ka
)
=
(
)
(7.30)
k
We have made a couple of gross assumptions in the above analysis. First,
we presumed that the excitation energies add linearly, so that the total
energy of
N
spin waves (eq. (7.28)) is just
N
times the energy of a single
spin wave (hence eq. (7.29)). We then further presumed that the result for
a classical spin excitation in eq. (7.29) remains true when we quantise the
spins, and so derived an expression as to how the energy of a quantum of
spin excitation varies with wavevector
k
(eq. (7.30)).
It is perhaps surprising to find that the conclusion of eq. (7.30) is in fact
true; that is, that the energy
E
k
of a magnon depends on wavevector
k
as
E
k
k
2
. As a consequence, even at the lowest temperatures, as
T
∼
→
0,
there will always be some long wavelength excitations within an energy
kT
of the ground state. By contrast, mean field theory overestimates the
