# Correlation Function of One-Dimensional XY Model

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?

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.