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For the Gaussian functional integral with dimensionless phase stiffness $K$ and a microscopic cutoff $a$, the Gaussian phase averaging identity gives the displayed correlation function. In two dimensions the phase-difference variance is $(\pi K{)}^{-1}\log (r/a)+O(1)$, so the XY model has algebraic spin wave correlations with exponent $1/(2\pi K)$. In one dimension the variance is $r/K$, giving exponential decay. Above two dimensions the variance has a finite large-distance limit within the low-temperature spin wave approximation. The Laplacian Green function solves $-K{\nabla }^{2}G=\delta$; integration over a sphere fixes $G(r)=r^{2-d}/[(d-2)S_{d}K]+C$ for $d\ne 2$ and $G(r)=-(2\pi K{)}^{-1}\log (r/a)+C$ for $d=2$. A fixed uniform phase and ultraviolet regularization are required to define $G$ itself; phase differences eliminate the constant.

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.

The dot product of the two magnetizations is ${\bar{m}}^2\cos [\theta (x)-\theta (0)]$. For the Gaussian spin wave field, the Gaussian phase averaging identity $⟨e^{iX}⟩=e^{-⟨X^2⟩/2}$ for a centered Gaussian variable givesKeeping this exponential is essential: expanding the cosine to quadratic order would fail when phase differences grow with separation. Equivalently, the phase-difference variance isFor $d>2$, the integral defining the absolute phase variance is finite at small momentum after fixing the uniform mode; its microscopic ultraviolet cutoff still matters. Its nonconstant Green function decays as $r^{2-d}$, so $D(r)$ tends to a finite constant $D_{\infty }$ andIn two dimensions, the logarithmic Green function gives $D(r)=(\pi \bar{K}{)}^{-1}\log (r/a)+O(1)$. ConsequentlyFor $1\le d<2$, $D(r)$ grows as $2r^{2-d}/[(2-d)S_d\bar{K}]$, up to bounded microscopic terms, and the correlation function again vanishes. In particular, in one dimension $S_1=2$ gives $D(r)=r/\bar{K}$ and exponential decay ${\bar{m}}^2{e}^{-r/(2\bar{K})}$. Therefore there is no conventional long-range order for $d\le 2$ at positive temperature in this short-range continuous-symmetry model. In two dimensions, algebraic decay can still support quasi-long-range order.

These infrared spin-wave correlations of the XY model explain the mechanism behind the Mermin-Wagner theorem. They are a test of the low-temperature spin wave approximation, not a claim that every $d>2$ XY model is ordered at every temperature. Large local fluctuations and defects destroy the ordered phase at sufficiently high temperature. Also, the assumed small angle is a local smooth-field approximation; phase differences at arbitrarily large separation need not remain small.

When $m_1=m_2={\mu }^2$, the free energy has continuous symmetry $O(2)$. Its ordered minima satisfyContinuous phase fluctuations make the lower critical dimension $d_{\mathrm{l}}=2$, in agreement with the Mermin-Wagner theorem. In two dimensions write the complex order parameter as $\psi ={\rho }_0{e}^{i\theta }$. Neglecting the massive amplitude mode gives the Goldstone-mode effective free energyThe phase-difference variance isand thereforewithThus spin waves replace true long-range order by quasi-long-range order.

A vortex-antivortex pair of separation $R$ has the logarithmic energy $E_{\mathrm{pair}}\simeq 2\pi {\rho }_0^2\log (R/a)$. The number of pair separations below $R$ grows as $R^2$, giving the coarse entropy $S_{\mathrm{pair}}\simeq 2\log (R/a)$. Thusand widely separated pairs become favorable at ${\rho }_0^2\simeq 1/\pi$. Substituting the mean-field stiffness givessuggesting a Berezinskii–Kosterlitz–Thouless transition. The numerical location is only a bare-stiffness estimate: vortex-core fluctuations renormalize the stiffness near the transition.

For the $O(2)$ model writeAt long distances the massive radial field $\rho$ can be neglected, leaving the Goldstone-mode effective free energyThereforeThe mode is massless, so its correlation length is infinite. For $d>2$ its large-distance Green function isFor $d=2$ it is $-(2\pi \gamma v^2{)}^{-1}\log (r/a)$ up to an infrared-dependent constant, and in $d=1$ it is $-|r|/(2\gamma v^2)$ up to such a constant.

The invariant diagnostic is the phase-difference varianceIt diverges linearly in $d=1$ and logarithmically in $d=2$, destroying true long-range order, but approaches a finite infrared limit for $d=3,4$. Thus the continuous-symmetry ordered phase exists for $d=3,4$ and not for $d=1,2$, in agreement with the Mermin-Wagner theorem. The lower critical dimension is $d_{\mathrm{lc}}=2$; the two-dimensional $O(2)$ model can instead show quasi-long-range order below a BKT transition.

Quasi-long-range order means that correlations decay algebraically rather than approaching a nonzero constant. It occurs in the low-temperature phase of the two-dimensional XY model, where spin waves destroy true long-range order but bound vortex pairs preserve power-law correlations.

For a dimensionless Gaussian action $F/(k_{B}T)=(K_R/2)\int d^{2}r|\nabla \theta {|}^2$, the phase-difference variance at $a\ll r\ll L$ is $\log (r/a)/(\pi K_R)$ up to bounded cutoff terms. Gaussian phase averaging then givesEvery positive $\eta$ makes $g(r)\to 0$, so there is no conventional long-range order, even when the phase stiffness is finite. This is how an accumulation of gentle, cheap twists destroys global alignment without making every local twist free.