# Partition function for 1D XY model [closed]

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_, unsymJun 26 '17 at 22:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – By Symmetry, Jon Custer, Yashas, Mo_, unsym
If this question can be reworded to fit the rules in the help center, please edit the question.

• As in the Ising case, the largest eigenvalue dominates the sum. Find it, and you are done. – Adam Jun 19 '17 at 15:14
• To find the eigenvalues I have to first diagonalize the matrix and I was trying to do that but could not. – Swarnadeep Seth Jun 19 '17 at 15:18
• 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). – Adam Jun 19 '17 at 15:52
• (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.) – Yvan Velenik Jun 19 '17 at 16:49

## 1 Answer

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.