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From the Harvard lecture notes XY model: particle-vortex duality by Subir Sachdev, the path-integral of 1D XY-model is given by $$\mathcal{Z}=\int\mathcal{D}\theta\exp{\left\{-\frac{K}{2}\int \!dx~(\frac{d\theta}{dx})^{2}\right\}}.\tag{4}$$ Introducing a complex order parameter $$\psi=e^{i\theta},\tag{3}$$ the correlation function is given by $$\left\langle\psi(x)\psi^{\ast}(0)\right\rangle=\exp{\left(-\frac{1}{K}\int\!\frac{dk}{2\pi}\frac{1-\cos(kx)}{k^{2}}\right)}.\tag{5}$$ My question is how I should perform the path-integral to obtain the above correlation function?

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2 Answers 2

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To answer this question we will first find the correlator: $$\langle e^{i\alpha \theta(x)} e^{-i\alpha \theta(0)}\rangle=\frac{1}{\mathcal{Z}}\int \mathcal{D}\theta \exp\left\{-\int dx\frac{K}{2} \left( \frac{d\theta}{dx}\right)^2+i\alpha \theta(x)-i\alpha\theta(0) \right\}$$ Take the Fourier Transform: $$\langle e^{i\alpha \theta(x)} e^{i\alpha' \theta(0)}\rangle=\frac{1}{\mathcal{Z}}\int \mathcal{D}\theta \exp\left\{\int \frac{d^D k}{(2\pi)^D}\left(-\frac{K}{2} k^2 \theta(k)\theta(-k)+i\alpha \theta(k)e^{ikx}-i\alpha\theta(k) \right)\right\}$$ $$\langle e^{i\alpha \theta(x)} e^{i\alpha' \theta(0)}\rangle=\frac{1}{\mathcal{Z}}\int \mathcal{D}\theta \exp\left\{\int \frac{d^D k}{(2\pi)^D}\left(-\frac{K}{2} k^2 \theta(k)\theta(-k)+2\alpha \theta(k)e^{ikx/2} \sin \left(\frac{kx}{2}\right)\right)\right\}$$ $$=\frac{1}{\mathcal{Z}}\int \mathcal{D}\theta \exp\left\{\int \frac{d^D k}{(2\pi)^D}\left(-\frac{K}{2} k^2 \theta(k)\theta(-k)+\alpha \theta(k)e^{ikx/2} \sin \left(\frac{kx}{2}\right)+\alpha \theta(-k)e^{-ikx/2} \sin \left(-\frac{kx}{2}\right)\right)\right\}$$ Complete the square: $$\langle e^{i\alpha \theta(x)} e^{i\alpha' \theta(0)}\rangle=\frac{1}{\mathcal{Z}}\int \mathcal{D}\theta \exp\left\{\int \frac{d^D k}{(2\pi)^D}\left(-\frac{K}{2} k^2 \left(\theta(k)+\frac{2}{K k^2}\alpha e^{-ikx/2} \sin\left( -\frac{kx}{2}\right) \right) \left(\theta(-k)+\frac{2}{K k^2}\alpha e^{ikx/2} \sin\left( \frac{kx}{2}\right) \right)-\frac{2\alpha^2}{Kk^2} \sin^2\left(\frac{kx}{2}\right)\right)\right\}$$ Redefining the fields so that: $$\theta(k)\rightarrow \theta(k)+\frac{2}{K k^2}\alpha e^{-ikx/2} \sin\left( -\frac{kx}{2}\right)$$ we get $$\langle e^{i\alpha \theta(x)} e^{i\alpha' \theta(0)}\rangle=\frac{1}{\mathcal{Z}}\int \mathcal{D}\theta \exp\left\{\int \frac{d^D k}{(2\pi)^D}\left(-\frac{K}{2} k^2 \theta(k) \theta(-k)-\frac{2\alpha^2}{Kk^2} \sin^2\left(\frac{kx}{2}\right)\right)\right\}$$ $$=\exp\left\{-\int \frac{d^D k}{(2\pi)^D} \frac{2\alpha^2}{Kk^2} \sin^2\left(\frac{kx}{2}\right)\right\}\frac{\mathcal{Z}}{\mathcal{Z}}$$ $$=\exp\left\{-\int \frac{d^D k}{(2\pi)^D} \frac{\alpha^2}{Kk^2} (1-\cos\left(kx\right))\right\}$$ which setting $D=1$ and $\alpha=1$ gives you your answer. (you could set $\alpha=1$ initially - I didn't as I thought it may help in the derivation. p.s. sorry for the long equations.

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    $\begingroup$ Why is the step where you redefine the field legitimate? $\endgroup$
    – Milarepa
    Commented Apr 29, 2020 at 19:22
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It seems that we can also use a trick in Xiao-Gang Wen's book (Quantum field theory of many body systems, page 93).

Now $\mathcal{L}=\frac{K}{2\pi}(\partial_x\theta)^2$, then the correlation function (in imaginary time) is $$\langle e^{i\theta(x_1)}e^{-i\theta(0)}\rangle=\frac{\int D\theta(x)e^{-\int dx\frac{K}{2\pi}(\partial_x\theta)^2+\int dx f(x)\theta(x)}}{\int D\theta(x)e^{-\int dx\frac{K}{2\pi}(\partial_x\theta)^2}}=e^{\frac{1}{2}\int dxdy\,f(x)G(x-y)f(y)},$$ where $f(x)=\delta(x-x_1)-\delta(x)$, and $G(x-y)=(-\frac{K}{\pi}\partial_x^2)^{-1}$. $$\frac{1}{2}\int dxdy\,f(x)G(x-y)f(y)=G(0)-G(x_1)=-\int\frac{dk}{2\pi}\frac{\pi}{K}\frac{1-e^{ikx}}{k^2},$$ and $\int dke^{ikx}/k^2$ can be further simplified to $\int dk \cos(kx)/k^2$. And also, this method can be generalized to D dimension, which is the same as Quantum spaghettification, but here we do not need the re-definition procedure.

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  • $\begingroup$ Thank you very much. $\endgroup$
    – Valac
    Commented Nov 23, 2022 at 18:32

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