# Path integral XY model

I am given the Hamiltonian for the XY model of spins

$$H[\theta] = \frac{K}{2} \int |\nabla \theta|^2dxdy,$$

where $$K = J/k_B T$$ with $$J$$ the interaction energy between neighbouring spins.

I calculated that $$\beta H[\theta] = \frac{K}{2L^2} \sum_{\mathbf{q}} \theta_{\mathbf{q}}\theta_{-\mathbf{q}} \mathbf{q}^2$$, where $$\theta_{-\mathbf{q}} = \theta_{\mathbf{q}}^\star$$ since $$\theta$$ is a real function.

From this I have to calculate $$\langle e^{i \theta(\mathbf{0})} \rangle$$.

For this I used the partition function

$$Z = \prod_{\mathbf{q}} \int d\theta_{\mathbf{q}} e^{-\frac{K}{2L^2} \mathbf{q}^2 |\theta_{\mathbf{q}}|^2} = \prod_{\mathbf{q}}\sqrt{\frac{2\pi L^2}{K\mathbf{q}^2}}.$$

And similarly,

$$\prod_{\mathbf{q}} \int d\theta_{\mathbf{q}} e^{-\frac{K}{2L^2}|\theta_{\mathbf{q}}|^2 \mathbf{q}^2 + iL^{-2} \theta_{\mathbf{q}}} = \prod_{\mathbf{q}} \sqrt{\frac{2\pi L^2}{K\mathbf{q}^2}} e^{-\frac{1}{2L^2}\frac{1}{K\mathbf{q}^2}} ,$$

where $$\theta(\mathbf{0}) = \frac{1}{L^2} \sum_{\mathbf{q}} \theta_{\mathbf{q}}$$ has been used. Therefore we find

$$\langle e^{i \theta(\mathbf{0})} \rangle = \exp \Big(-\frac{1}{2KL^2} \sum_{\mathbf{q}} \mathbf{q}^{-2}\Big).$$

This is exactly the result I had to prove, however I do not understand why I can evaluate the integrals in the way I did. For example in the first integral we just pretend as if $$\theta_{\mathbf{q}}$$ is real and $$|\theta_{\mathbf{q}}|^2$$ is just $$\theta_{\mathbf{q}}^2$$. Why can we do such things?

Due to the relation $$\theta_{-\mathbf{q}} = \theta_{\mathbf{q}}^{\ast}$$, one needs to be careful about integrating over distinct degrees of freedom. One possible choice is to restrict to $$q_i > 0$$ for one of the components of $$\mathbf{q}$$, and then to integrate over both real and imaginary parts of $$\theta_{\mathbf{q}}$$.
Then you can write $$\theta(\mathbf{x} = 0) = \frac{1}{L} \sum_{\mathbf{q}} \theta_{\mathbf{q}} = \frac{1}{L} \sum_{q_i>0} \left( \theta_{\mathbf{q}} + \theta^{\ast}_{\mathbf{q}} \right) = \frac{2}{L} \sum_{q_i>0} \mathrm{Re}[\theta_{\mathbf{q}}],$$ Similarly, you can write $$|\theta_{\mathbf{q}}|^2 = \mathrm{Re}[\theta_{\mathbf{q}}]^2 + \mathrm{Im}[\theta_{\mathbf{q}}]^2$$. So for your case, the partition function reads $$Z = \prod_{q_i > 0} \int d\theta_{\mathbf{q}} e^{-\frac{K}{L^2} \mathbf{q}^2 \left[ \mathrm{Re}[\theta_{\mathbf{q}}]^2 + \mathrm{Im}[\theta_{\mathbf{q}}]^2 \right]},$$ while the numerator of the correlation function can be written $$\prod_{q_i > 0} \int d\theta_{\mathbf{q}} e^{-\frac{K}{L^2}\left[ \mathrm{Re}[\theta_{\mathbf{q}}]^2 + \mathrm{Im}[\theta_{\mathbf{q}}]^2 \right] \mathbf{q}^2 + 2iL^{-2} \mathrm{Re}[\theta_{\mathbf{q}}]}.$$ So the integration measure is really over all real and imaginary parts of $$\theta_{\mathbf{q}}$$, that is, $$d\mathrm{Re}[\theta_{\mathbf{q}}]d\mathrm{Im}[\theta_{\mathbf{q}}]$$. If you do this integral, you'll find $$\langle e^{i \theta(\mathbf{0})} \rangle = \exp \Big(-\frac{1}{KL^2} \sum_{q_i > 0} \mathbf{q}^{-2}\Big) = \exp \Big(-\frac{1}{2KL^2} \sum_{\mathbf{q}} \mathbf{q}^{-2}\Big).$$
Why did your calculation work out when you just assumed $$\theta_{\mathbf{q}}$$ was real? Because the integration over the imaginary part of $$\theta_{\mathbf{q}}$$ just factored out of the above integrals, so it canceled out in the ratio. But I claim that your partition function is actually wrong (there should not be a 2 inside the square root).
• One final question. We can we just discard the case $\mathbf{q} = \mathbf{0}$? Oct 13, 2019 at 12:34
• The $\mathbf{q} = 0$ mode is problematic in gapless free theories, so the usual "solution" is just to discard it. This problem goes away if you include an infinitesimal amount of interactions, so I usually think of this as being a symptom of working in a pathological limit. Oct 13, 2019 at 18:28