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For the XY model, we have $$ Z = \int_0^{2\pi} \prod_{i=1}^N d\theta_i \exp(\beta J \sum_{i=1}^N \cos(\theta_i - \theta_{i+1}))$$

and eigenvectors $\vec{v}(\theta)=e^{in\theta} $ and eigenvalues $\lambda_n = 2 \pi I_n(\beta J)$ where $I_n(z) = \int_0^{2\pi} \dfrac{d\phi}{2\pi} e^{z\cos \phi}\cos n\phi$.

Why is the free energy of the XY model for $ N \rightarrow \infty $ given by:

$$ F = -\lim_{N\rightarrow\infty}\dfrac{1}{\beta N} \ln Z = -\dfrac{1}{\beta}\ln[2\pi I_0(\beta J)]~? $$

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  • $\begingroup$ What is it specifically about the expression you don't understand? In the limit $N\rightarrow\infty$, $\beta F=-\frac{ln \ Z}{N}=-ln \lambda$ $\endgroup$
    – joseph h
    Oct 22, 2021 at 6:39
  • $\begingroup$ note that $I_0(x) > I_n(x)$ for $n>0$, which means that in the thermodynamic limit it is the only one that remains $\endgroup$
    – user275556
    Oct 22, 2021 at 6:46
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    $\begingroup$ Note also that you can compute the free energy by a trivial change of variables that factorizes the partition function (using free b.c., which in any case does not matter in the thermodynamic limit)... $\endgroup$ Oct 22, 2021 at 6:52

3 Answers 3

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An important relation for the Bessel functions is $$ e^{zcos\theta}=I_0(z)+2\sum_{n=1}^{+\infty}I_n(z)\cos(n\theta) $$ Substituting this into the partition function and carrying out the integration should produce the desired result.

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Note that in the limit $N\rightarrow\infty$ the larger eigenvalue (as pointed out in the comment by yyy $I_0\gt I_n$) dominates meaning $$\beta F=-\frac{ln\ Z}{N}=-\ln \lambda_0$$ where $\lambda_0=\ln[2\pi I_0(\beta J)]$

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Just to complement the other answers, here is the most elementary derivation.

Simply change variables to $\phi_i = \theta_i - \theta_{i-1}$, for $i=1, \dots, n$ (with the convention that $\theta_0=0$). Since the associated Jacobian is equal to $1$, one obtains \begin{align} Z_n &= \int_0^{2\pi}\mathrm{d}\phi_1\cdots\int_0^{2\pi}\mathrm{d}\phi_n \exp\Bigl( \beta J \sum_{i=1}^n \cos\phi_i \Bigr)\\ &= \prod_{i=1}^n \int_0^{2\pi} \mathrm{d}\phi_i \exp( \beta J \cos\phi_i ) = (2\pi I_0(\beta J))^n. \end{align} This then immediately yields $$ F = -\lim_{n\to\infty} \frac1{\beta n} \ln Z_n = -\frac1\beta \ln \bigl(2\pi I_0(\beta J)\bigr) $$

Note that, in the derivation above, I have used the following boundary condition: $\theta_0=0$ on the left and free on the right. Of course, the choice of boundary condition does not affect the free energy density in the thermodynamic limit.

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