1549380323-Statistical Mechanics Theory and Molecular Simulation

One-dimensional Ising model 625

Sinceσandσ′can be only 1 or -1, P is a 2×2 matrix with elements given by

〈 1 |P| 1 〉= eβ(J+h)

〈− 1 |P|− 1 〉= eβ(J−h)

〈 1 |P|− 1 〉=〈− 1 |P| 1 〉= e−βJ. (16.6.5)

Written as a matrix, P appears as

P =

(

eβ(J+h) e−βJ

e−βJ eβ(J−h)

)

. (16.6.6)

In terms of P, the partition function can be expressed as

∆(N,h,T) =

∑

σ 1

···

∑

σN

〈σ 1 |P|σ 2 〉〈σ 2 |P|σ 3 〉···〈σN− 1 |P|σN〉〈σN|P|σ 1 〉. (16.6.7)

Because the partition function is now a matrix product ofNfactors of P, each sand-
wiched between spin states with a spin in common, P is known as thetransfer matrix.
Using the completeness of the spin eigenvectors, each factor of the form

∑

σk|σk〉〈σk|
appearing in eqn. (16.6.7) is an identity operatorI, and the sum overNspins can be
collapsed to a sum over just one spinσ 1 :

∆(N,h,T) =

∑

σ 1

〈σ 1 |PN|σ 1 〉

= Tr

(

PN

)

. (16.6.8)

Interestingly, in deriving eqn. (16.6.8), we performed the oppositeset of operations
used in eqns. (12.2.9) and (12.2.11) to derive the Feynman path integral. In the latter,
an operator product was expanded by theintroductionof the identity between factors
of the operator.
The simplest way to calculate the trace is to diagonalize P, which yields two eigen-
valuesλ 1 andλ 2 , in terms of which the trace is simplyλN 1 +λN 2. The eigenvalues of
P are solutions of det(P−λI) = 0, which gives the eigenvalues

λ= eβJ

[

cosh(βh)±

√

sinh^2 (βh) + e−^4 βJ

]

. (16.6.9)

We denote these as valuesλ±(instead ofλ 1 , 2 ), whereλ±corresponds to the choice of
+/- in eqn. (16.6.9). Thus, the partition function becomes

∆(N,h,T) = Tr

[

PN

]

=λN++λN−. (16.6.10)

Although eqn. (16.6.10) is exact, sinceλ+> λ−, it follows that forN→∞,λN+≫λN−
so that the partition function is accurately approximated using thesingle eigenvalue
λ+. Thus, ∆(N,h,T)≈λN+, and the free energy per spin is simply