 Solution

Assume $g>0$ for stability. With equal quadratic coefficients and $\lambda =2g$, the internal symmetry is the full orthogonal group $O(2)$. In two dimensions at positive temperature, with short-range interactions, the Mermin-Wagner theorem forbids spontaneous symmetry breaking of this continuous symmetry. Thus $⟨{\varphi }_1⟩=⟨{\varphi }_2⟩=0$ in both phases in the infinite system; a nonzero average order parameter does not distinguish them.

Instead measure the correlation functionThe low-temperature phase has quasi-long-range order and algebraic correlations; the high-temperature phase has a finite correlation length and exponential correlations. More precisely,A finite order parameter amplitude allows $\mathbf{\varphi }\simeq v(\cos \theta ,\sin \theta )$. Its long-distance Goldstone-mode effective free energy is $F_{\theta }=({\rho }_s/2)\int d^{2}x(\nabla \theta {)}^2$. Restoring the Boltzmann constant and using the renormalized phase stiffness, the thermal phase fluctuations giveThe Berezinskii–Kosterlitz–Thouless transition separates vortex-antivortex binding from unbound vortex defects. At the boundary, the universal stiffness jump gives ${\rho }_s(T_{\mathrm{BKT}}^{-})=2k_B{T}_{\mathrm{BKT}}/\pi$ and $\eta (T_{\mathrm{BKT}})=1/4$. The low-temperature phase stiffness is nonzero; it vanishes at long distances in the disordered phase. The exact transition location is fluctuation dependent and cannot be obtained just by setting the bare ${\mu }^2$ to zero.