The Hamiltonian for 1D XY Model for $N$ Ising spins is written as, $$H=\sum_{i=1}^{N} \vec{S_i} . \vec{S_{i+1}} =\sum_{i=1}^{N} cos(\theta_i-\theta_{i+1})$$ Here we will implement periodic boundary condition as $\vec{S_{N+1}}=\vec{S_{1}}$. Using the Transfer Matrix method we can arrive at the partition function as, $$ \begin{align} Z&=\int_{0}^{2\pi} \Big(\prod_{i=1}^{N} d\theta_{i}\Big) \langle \theta_1 |T|\theta_2 \rangle \langle \theta_2 |T|\theta_3 \rangle\dots \langle \theta_N |T|\theta_1 \rangle\\ &=\int_{0}^{2\pi}d\theta_{1} \langle \theta_1 |T^N|\theta_1 \rangle =Tr(T^N) \end{align} $$ And we identify $\langle \theta_i |T|\theta_{i+1} \rangle =e^{\beta J cos(\theta_i-\theta_{i+1})}$.
Now my question is how to find out the partition function $Z(\beta J)$, given that $f(\theta_i) \sim e^{in\theta_i}$ diagonalizes the matrix? And what is the integral form of the matrix $T$?


closed as off-topic by By Symmetry, Jon Custer, Yashas, Mo_, unsym Jun 26 '17 at 22:08

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  • $\begingroup$ As in the Ising case, the largest eigenvalue dominates the sum. Find it, and you are done. $\endgroup$ – Adam Jun 19 '17 at 15:14
  • $\begingroup$ To find the eigenvalues I have to first diagonalize the matrix and I was trying to do that but could not. $\endgroup$ – Swarnadeep Seth Jun 19 '17 at 15:18
  • 1
    $\begingroup$ You already know the eigenvectors, so you just need to compute $\int d\theta e^{\beta J \cos(\theta_1-\theta)}e^{i n \theta}=\lambda_ne^{i n \theta_1}$ to find $\lambda_n$ (spoiler alert, it is a Bessel function). $\endgroup$ – Adam Jun 19 '17 at 15:52
  • $\begingroup$ (Alternatively, you can avoid the transfer matrix completely, by using free boundary condition and integrating one spin at a time; this is actually a slightly simpler approach.) $\endgroup$ – Yvan Velenik Jun 19 '17 at 16:49

The eigenvalue corresponding to $e^{in\theta}$ is a multiple of $I_n(\beta J)$ where $I_n$ is the modified Bessel function. Just use the definition of this function as an integral to see that this is so.


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