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Statistical Mechanics - Physics at Oregon State University

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9.5. EXACT SOLUTION FOR THE ISING CHAIN. 205<br />

the value of the square is always one. But the susceptibility diverges even in<br />

mean-field theory! This means th<strong>at</strong> the spin-correl<strong>at</strong>ion in mean field theory for<br />

large distances is non-zero. There seems to be a contradiction here, but th<strong>at</strong> is<br />

not true. The only assumption we make in mean field theory is th<strong>at</strong> Γi = 0 for<br />

nearest-neighbors! For further neighbors we do not make any assumptions. We<br />

are not able to calcul<strong>at</strong>e the correl<strong>at</strong>ion function <strong>at</strong> larger distances. Similar,<br />

in the Bethe approxim<strong>at</strong>ion we can only calcul<strong>at</strong>e the spin correl<strong>at</strong>ion function<br />

inside the cluster. In this case the condition involving h ′ is not readily expressed<br />

in terms of the spin-correl<strong>at</strong>ion function.<br />

9.5 Exact solution for the Ising chain.<br />

The Ising model in one dimension can be solved exactly. The same st<strong>at</strong>ement<br />

is true in two dimensions, but the solution is much more difficult. We will not<br />

do th<strong>at</strong> here. First consider a chain of N <strong>at</strong>oms without an external field. The<br />

energy of a configur<strong>at</strong>ion {σ1, · · · , σN} is given by<br />

The partition function is<br />

H(σ1, · · · , σN) = −J<br />

Z(T, N) = <br />

σ1,···,σN<br />

N−1 <br />

i=1<br />

which can be calcul<strong>at</strong>ed by starting <strong>at</strong> the end via<br />

σiσi+1<br />

(9.65)<br />

e βJ i σiσi+1 (9.66)<br />

Z(T, N) = <br />

e βJσ1σ2 · · · <br />

e βJσN−1σN (9.67)<br />

σ1<br />

σ2<br />

Since the last factor is the only place where σN plays a role, the sum over σN<br />

can be performed giving<br />

σN<br />

<br />

e βJσN−1σN = 2 cosh(βJ) (9.68)<br />

σN<br />

Note th<strong>at</strong> σN−1 drops out since it only takes the values ±1 and since the hyperbolic<br />

cosine is symmetric! Therefore the partition function is<br />

Z(T, N) = 2 cosh(βJ) <br />

e βJσ1σ2 · · · <br />

e βJσN−1σN (9.69)<br />

This process is now repe<strong>at</strong>ed and leads to<br />

σ1<br />

σ2<br />

Z(T, N) = 2 N−1 cosh N−1 (βJ) <br />

and the free energy <strong>at</strong> h = 0 is<br />

σ1<br />

σN−1<br />

= 2 N cosh N−1 (βJ) (9.70)

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