Statistical Mechanics - Physics at Oregon State University

9.2. INTEGRATION OVER THE COUPLING CONSTANT. 195
Note that in this case we could also have obtained the same result by calculating
the entropy from the specific heat
S =
T
0
dT ′
T ′
∞
dU
= NkBJ
dT ′
β
β ′ dβ ′
cosh 2 β ′ J
(9.22)
This integral is harder to evaluate, however. But it can be done, especially since
we already know the answer!
We can also analyze the Ising model in some more detail. The energy of a
configuration is
E{σ1, σ2, · · ·} = −J
σiσj − h
(9.23)
The interaction term is large, of course, and the division we made before is not
optimal. We would like to subtract some average value of the spins and write
E{σ1, σ2, · · ·} = −J
(σi − µ)(σj − µ) − (Jµq + h)
2 1
σi + Jµ Nq (9.24)
2
and we define this for a variable coupling constant via
Eλ{σ1, σ2, · · ·} = −λ
(σi − µ)(σj − µ) − (Jµq + h)
2 1
σi + Jµ Nq (9.25)
2
It would be natural to define µ as the average magnetization. But that value depends
on the value of λ, which makes the λ dependence of the Hamiltonian quite
complicated. Hence we need to choose µ as some kind of average magnetization,
and an appropriate value of the coupling constant.
The reference system has energy eigenvalues given by
and the partition function is
which gives
and the free energy is
E0{σ1, σ2, · · ·} = −(Jµq + h)
2 1
σi + Jµ Nq (9.26)
2
Z0(T, h, N) = e
1 −βJµ2 2 Nq
i
σ1,σ2,···
1 −βJµ2
Z0(T, h, N) = e 2 Nq [2 cosh β(Jµq + h)] N
i
i
i
σi
e β(Jµq+h) i σi (9.27)
(9.28)
G0(T, h, N) = Jµ 2 1
2 Nq − NkBT log(2 cosh β(Jµq + h)) (9.29)
The interaction term is now