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If

$$ Z = tr\left[ \Pi_{i=1}^N \int_0^{2\pi} d\theta_i \exp{(\beta J \cos{(\theta_i - \theta_{i+1})})}\right] $$

Then why does the eigenvalue problem:

$$ \exp{(\beta J \cos{(\theta - \theta^\prime)})} = \sum_{n=-\infty}^\infty I_n(\beta J)e^{in(\theta - \theta^\prime)} $$

where

$$ I_n(\beta J) = \int_0^{2\pi} \dfrac{d\phi}{2\pi} e^{\beta J \cos{\phi}}\cos{n\phi}? $$

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  • $\begingroup$ This is just a Fourier series. $\endgroup$
    – d_b
    Oct 21, 2021 at 0:04
  • $\begingroup$ I know that it's an inverse Fourier transform but I'm not seeing how to get the cos(n phi) term. $\endgroup$
    – Dizzy
    Oct 21, 2021 at 2:46

1 Answer 1

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We can write a function on the circle in a Fourier series as \begin{align} f(\theta) = \sum_{n=-\infty}^{\infty} c_ne^{in\theta}. \end{align} The Fourier coefficients are \begin{align} c_n &= \frac{1}{2\pi}\int_{0}^{2\pi} f(\theta)e^{-in\theta}\,d\theta\\ &=\frac{1}{2\pi}\int_{0}^{2\pi} f(\theta)\left(\cos n\theta - i \sin n\theta\right)\,d\theta \end{align} If $f$ is even, $f(-\theta) = f(\theta)$, then \begin{align} \int_0^{2\pi} f(\theta) \sin n\theta\,d\theta = 0. \end{align} (In case this isn't obvious, translate the integrand to make the limits symmetric about zero, then since $\sin$ is odd and $f$ is even, the integral vanishes.) So \begin{align} c_n = \boxed{\frac{1}{2\pi}\int_0^{2\pi}f(\theta)\cos n \theta\,d\theta,} \end{align} which is just what you have with $f(\theta) = e^{\beta J \cos \theta}$.

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