Studying the classical XY model (https://en.wikipedia.org/wiki/Classical_XY_model), I wish to compute the partition function: \begin{equation} Z=\int \mathrm{d}\mathbf{s}\; e^{-\beta H(\mathbf{s})} \end{equation} Where $H$ is: \begin{equation} H(\mathbf{s})=-\sum_{i \neq j} J_{i j} \mathbf{s}_{i} \cdot \mathbf{s}_{j}-\sum_{j} \mathbf{h}_{j} \cdot \mathbf{s}_{j} \end{equation}

The model assumes that $\mathbf{s}_{j}$ are unit vectors: $\mathbf{s}_{j}=\left(\cos \theta_{j}, \sin \theta_{j}\right)$.

From wikipedia and various articles, it seems that integrating over all possible angles $\theta_j$ is the way to go about it.

  • Why can't we uncouple the spins and then integrate directly over $(\mathbf{s}_i)_x\in [-1,1]$?
  • Wikipedia's article integrates over angles, why isn't there an extra factor $ -\sin(\theta_i)\mathrm{d} \theta_i$ in front of the exponential due to the change of variables? Since $ \mathrm{d} s_i=-\sin(\theta_i)\mathrm{d} \theta_i$. Wikipedia does not include this term:

$$Z=\int_{[-\pi, \pi]^{\Lambda}} \prod_{j \in \Lambda} d \theta_{j} e^{-\beta H(\mathbf{s})}$$


1 Answer 1


You can integrate over $\mathbf{s}$, but you also have to impose the constraint that $\|\boldsymbol{s}\|=1$, which does not seem to be presented in your partition function:$Z=\int \mathrm{d}\mathbf{s}\; e^{-\beta H(\mathbf{s})}$. For example, you can probably add the constraint using the Dirac delta function like $Z=\int \mathrm{d}\mathbf{s}\; \delta(\mathbf{s}^2-1) e^{-\beta H(\mathbf{s})}$, but I wouldn't bother to do that.

With the constraint, for the spin it ends up with only one degree of freedom which is the angle, so the integral over angle is correct.


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