XY model free energy

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:

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

• What is it specifically about the expression you don't understand? In the limit $N\rightarrow\infty$, $\beta F=-\frac{ln \ Z}{N}=-ln \lambda$ Oct 22, 2021 at 6:39
• 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
– user275556
Oct 22, 2021 at 6:46
• 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)... Oct 22, 2021 at 6:52

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.
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 $$\lambda_0=\ln[2\pi I_0(\beta J)]$$
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.