Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
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)