 Solution

Fix the uniform angular mode, for example by fixing the mean phase in a large box, and introduce a microscopic ultraviolet cutoff $a$. The quadratic Goldstone-mode effective free energy becomes, under a Fourier transform,To evaluate the Gaussian functional integral, add a source $j$. Completing the square in every nonzero Fourier transform mode givesTaking two source derivatives yields the correlation functionThe cutoff and the treatment of the uniform mode regulate this expression. At separations much larger than $a$, its nonconstant part is the Green function satisfying $-\bar{K}{\nabla }^{2}G={\delta }^{(d)}(x)$. Rotation invariance and the radial Laplacian give $(r^{d-1}{G}^{\prime }{)}^{\prime }=0$ away from the origin. Integrating the Green function equation over a small sphere, whose unit-sphere area is $S_d=2{\pi }^{d/2}/\Gamma (d/2)$, fixes the flux:Thus for $d\ne 2$ the required long-distance correlation function isAt $d=2$, the finite separation-dependent limit isThe constant absorbs the divergent constant term in the $d\to 2$ limit. In low dimension the individual $G(r)$ has no regulator-independent infinite-volume value; its differences do. This is why the phase-difference variance, rather than an absolute phase variance, is the useful observable.