1549380323-Statistical Mechanics Theory and Molecular Simulation

628 Critical phenomena

E(μj,μk) =J

∑n

i=1

σi,jσi,k

E(μj) =J

∑n

i=1

σi,jσi+1,j+h

∑

i,j

σi,j (16.7.4)

in terms of which the Hamiltonian becomes

H=−

∑n

j=1

[E(μj,μj+1) +E(μj)], (16.7.5)

and the partition function can be expressed as

∆(N,h,T) =

∑

μ 1

∑

μ 2

···

∑

μn

exp







β

∑n

j=1

[E(μj,μj+1) +E(μj)]







. (16.7.6)

Although eqn. (16.7.6) now resembles the partition function of a one-dimensional Ising
model, each sum overμjnow represents 2nterms. We can, nevertheless, define a 2n× 2 n
transfer matrix P with elements

〈μ|P|μ′〉= exp

{

β

[

E(μ,μ′) +

1

2

(E(μ) +E(μ′))

]}

, (16.7.7)

so that the partition function becomes

∆(N,h,T) =

∑

μ 1

∑

μ 2

···

∑

μn

〈μ 1 |P|μ 2 〉〈μ 2 |P|μ 3 〉···〈μn|P|μ 1 〉= Tr [Pn]. (16.7.8)

The partition function can now be computed from the 2neigenvalues of P as

∆(N,h,T) =λn 1 +λn 2 +···λn 2 n. (16.7.9)

As in the one-dimensional case, however, asN → ∞,n→ ∞, and the contribu-
tion from the largest eigenvalue will dominate. Thus, to a very good approximation,
∆(N,h,T)≈λnmax, and the problem of computing the partition function becomes one
of simply finding the largest eigenvalue of P.
A detailed mathematical discussion of how the largest eigenvalue of Pcan be found
is given by Huang (1963), which we will not replicate here. We simply quote the final
result for the Gibbs free energy per spin at zero field in the thermodynamic limit:

g(0,T) =−kTln [2cosh(2βJ)]−

kT

2 π

∫π

0

dφln

1

2

(

1 +

√

1 −K^2 sin^2 φ

)

(16.7.10)

(Onsager, 1944; Kaufmann, 1949), whereK= 2/[cosh(2βJ)coth(2βJ)]. The integral is
the result of taking the thermodynamic limit. In his 1952 paper, C. N.Yang obtained