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In the continuum limit the Hamiltonian of the classical XY model is given by, ignoring the inessential constant: $$H=\int d\vec{r}\ (\nabla\theta)^2$$ and the x-component of the order parameter is given by $$\langle S_x\rangle=\langle \cos\theta(\vec{r}) \rangle=\langle \cos\theta(0) \rangle=\frac{\int D\theta \cos\theta(0)e^{-\beta H}}{Z}=\operatorname{Re}\frac{\int D\theta e^{-\beta H+i\theta(0)}}{Z}$$ Going to Fourier space: $$\theta(0)=\int \frac{d\vec{k}}{(2\pi)^d}\theta(\vec{k})$$ $$\int d\vec{r}\ (\nabla\theta)^2=\int \frac{d\vec{k}}{(2\pi)^d}k^2\theta(\vec{k})\theta(-\vec{k})=\int \frac{d\vec{k}}{(2\pi)^d}k^2\theta(\vec{k})\theta^*(\vec{k})$$ We have, using $\theta(\vec{k})=\theta_R+i\theta_I$: \begin{equation} \begin{aligned} \int D\theta e^{-\beta H+i\theta(0)}&=\int D\theta_R D\theta_Ie^{-\int\frac{d\vec{k}}{(2\pi)^d}(\beta k^2\theta^2_R-i\theta_R)+(\beta k^2\theta^2_I+\theta_I)}\\ &=\int D\theta_R D\theta_Ie^{-\int\frac{d\vec{k}}{(2\pi)^d}\beta k^2(\theta_R-\frac{i}{2\beta k^2})^2+1/(4\beta k^2)+\beta k^2(\theta_I+\frac{1}{2\beta k^2})^2-1/(4\beta k^2)}\\ &=\int D\theta_R D\theta_Ie^{-\int\frac{d\vec{k}}{(2\pi)^d}\beta k^2(\theta_R-\frac{i}{2\beta k^2})^2+\beta k^2(\theta_I+\frac{1}{2\beta k^2})^2}\\ &=\int D\theta_R D\theta_Ie^{-\int\frac{d\vec{k}}{(2\pi)^d}\beta k^2(\theta_R)^2+\beta k^2(\theta_I)^2}\\ &=Z \end{aligned} \end{equation} which is not correct. The above steps basically involve completing the square in Fourier space followed by shifting the functional variables. The $1/(4\beta k^2)$ shouldn't have been canceled out, but I don't see any sign problem here. Note that the $k$ integration is only over half of the $k$-space, because both $\theta_R$ and $\theta_I$ are only independently defined over half of the $k$-space due to the reality of $\theta(x)$. What I have been trying to do is to fill the gap between Eq.(19) and (20) in this XY model notes.

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Thanks to Shane for drawing my attention to the integration measure. In my original calculation, I directly used $\theta_R$ and $\theta_I$ as the two independent variables, which is not exactly right here. The reason is the following:

Due to the reality of $\theta(x)$, we have $\theta^*(\vec{k})=\theta(-\vec{k})$, which means $\theta_R(\vec{k})-i\theta_I(\vec{k})=\theta_R(-\vec{k})+i\theta_I(-\vec{k})$, which again means $\theta_R$ is symmetric and $\theta_I$ is anti-symmetric. Using the anti-symmetry of $\theta_I$, the $\int d\vec{k}\theta_I$ part of my above integration readily vanishes, i.e. $$\int D\theta e^{-\beta H+i\theta(0)}=\int D\theta_R D\theta_Ie^{-\int\frac{d\vec{k}}{(2\pi)^d}(\beta k^2\theta^2_R-i\theta_R)+(\beta k^2\theta^2_I)}$$ and this is why there would be no completing of square for the $\theta_I$ part and extra term $1/(4\beta k^2)$ remains. In the end will have $$\int D\theta e^{-\beta H+i\theta(0)}=Z\mathrm{exp}\left(-\int \frac{d\vec{k}}{(2\pi)^d}1/(4\beta k^2)\right)$$ where the exponent $\sim -\int^{\pi/a}_{\pi/L}dk k^{d-3}$.

For $d=1,2$, the exponent $\to -\infty$ when system size $L\to \infty$, meaning $S_x\to 0$. Whereas, for $d>2$ the order parameter takes a non-zero value.

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So I think there is an issue with pulling the $Re$ of $Re(e^{i\theta})$ out of the path integral because the measure $\mathcal{D}$ has a bunch of factors of i in it after Fourier transforming. I don't know for sure but I think this messes with your ability to do: $\theta(-k)=\theta(k)^*$(i.e. assume the field is real).

I might be completely wrong though but my reason for believing this is completing the square as follows gets the right result:

$ \int_{\infty}^{\infty} \frac{d\vec{k}}{(2\pi)^d} \beta k^2\theta(\vec{k})\theta(-\vec{k}) +i\theta(\vec{k})= $

$ 2\int_{0}^{\infty} \frac{d\vec{k}}{(2\pi)^d} \beta k^2\theta_1(\vec{k})\theta_2(\vec{k}) +i\theta_1(\vec{k})+i\theta_2(\vec{k})= $

$ \int_{0}^{\infty} \frac{d\vec{k}}{(2\pi)^d} 2\beta k^2(\theta_1(\vec{k})+i/2\beta k^2)(\theta_2(\vec{k})+i/2\beta k^2) +1/\beta k^2 $

With $\theta_1(k)=\theta(k)$ and $\theta_2(k)=\theta(-k)$. Assuming you can shift $\theta_1$ and $\theta_2$ independently. Things now become:

$ 2\int_{0}^{\infty} \frac{d\vec{k}}{(2\pi)^d} \beta k^2\theta_1(\vec{k})\theta_2(\vec{k}) +1/4\beta k^2 $

You can now do the integration on the second term and pull it out of the path integral. The remain first term is the same as the $-\beta H$ so the remaining path integral becomes the the partition function $Z$ after functional integration.

You can do the rotation: $\theta_1=\theta_+ +\theta_-$, $\theta_2=\theta_+ -\theta_-$ To diagonalize and do the functional integration.

Note to preform the shift independently, the following constraint must be relaxed $\theta_1=\theta_2^*$. This why I think there is something funny about the complex analysis. It might be related to symmetry breaking but I don't see that clearly.

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  • $\begingroup$ Pulling the Re out at the very beginning isn't really a problem, because the $\theta(x)$ is a real field. Now, a bunch of questions regarding your answer: 1. How are your $\theta_1, \theta_2$ defined? 2. Why is there a summation over $i=1,2$? 3. In your final expression, the $\theta_1$ part and the $\theta_2$ part factorize, instead of splitting into a sum, making the functional integral hard to solve if your measure is $d\theta_1 d\theta_2$. $\endgroup$ – M. Zeng Aug 26 '18 at 19:17
  • $\begingroup$ Did I clarify your questions with my edits? $\endgroup$ – Shane P Kelly Aug 26 '18 at 21:41
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    $\begingroup$ Thanks for the clarification and for drawing my attention to the integration measure. Even though it's not the problem of taking the real part, the problem is indeed due the integration measure. I fixed the problem based on my original calculation, edited above. In your calculation you are right in the sense that $\theta_1$ and $\theta_2$ can be treated as independent variables, but only on half of the $k$-space. This point is extremely crucial especially for my approach. If this is corrected, your approach will also give the exact right answer. $\endgroup$ – M. Zeng Aug 26 '18 at 23:20
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I don't know what the author of the notes has in mind, but in any dimension, and in the absence of any direction breaking field, we must have $\langle \cos\theta \rangle=0$ because all directions of the order parameter are equally likely. To get spontanous symmetry breaking you need to impose a small symmetry breaking perturbation that makes $\langle \cos\theta \rangle$ non-zero, and then take the perturbation to zero to see if $\langle \cos\theta \rangle$ has a non-zero limit.

I suspect that the author has ignored the effect of the singular ${\bf k}=0$ mode in doing his momentum space calculation.

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  • $\begingroup$ I think the asymmetry is introduced when the original hamiltonian takes the form $cos(\theta_i-\theta_j)$. The $k=0$ singularity is important and is in fact included in the calculation by taking the thermodynamic limit $L\to \infty$ so that $k_{min}=\pi/L\to 0$. $\endgroup$ – M. Zeng Aug 26 '18 at 17:37
  • $\begingroup$ @M. Zeng Nope. The original hamiltonian is invariant under $\theta_i\to \theta_i+a$ with $a$ independent of $i$. So if you have a configuration in the Path integral pointing in one direction there is one with the same weight in another direction. Thus all components of ${\bf S}_i$ have zero average. To get a non zero anwer you must do something like fixing a single boundary spin in some direction. Long range order on the other hand just requires that $<{\bf S}_i {\bf S_j}>$ go to a non-zero constant as site $i$ and $j$ get far apart. $\endgroup$ – mike stone Aug 26 '18 at 21:08
  • $\begingroup$ You had a good point, but what about Mermin-Wagner theorem? I managed to fix my calculation, presented in the edit above, which I believe reproduced the proof of the theorem in the XY model case. In the proof, the only thing matters is the dimensionality. $\endgroup$ – M. Zeng Aug 27 '18 at 0:26
  • $\begingroup$ @M.Zeng Mermin Wagner only says that no long range order (LRO) is possible in two-d and under. It says nothing about when LRO can occur. The issue here is rather what we need to do to the Path integral to satisfy the Cluster Decomposition property $<S_iS_j>\to <S_i><S_j>$. as $i,j$ get far apart. Without (infinesimal) explict symmetry breaking the path integral/partition function does not give Cluster Decomposition. $\endgroup$ – mike stone Aug 27 '18 at 12:55
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    $\begingroup$ The cluster decomposition property is the statement that $<S_iS_j>\to <S_i><S_j>$ at large distance. The path integral without infinitesimal expllicit symmetry breaking can have $<S_iS_j>\to const$ (and hence LRO) but at the same time have $<S>= 0$. Thus the PI does not necessarily satisfy the cluster decomposition property. $\endgroup$ – mike stone Aug 28 '18 at 12:14

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