1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ One final question. We can we just discard the case $\mathbf{q} = \mathbf{0}$? $\endgroup$ Oct 13, 2019 at 12:34
  • $\begingroup$ 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. $\endgroup$ Oct 13, 2019 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.