Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 43 / 3 / 5 / Solution

At $m^2=0$ and $h=0$ the model is at a Gaussian fixed point. Its connected correlation function has the power-law critical form $r^{-(D-2)}$ rather than a finite exponential decay length. Approaching from $m^2>0$, the uniform susceptibility ${\tilde{G}}_0(0)=1/m^2$ and the correlation length both diverge. If the thermal scaling field is linear in $m^2$ with $\alpha$ finite and positive, thenThe same result follows from the mass multiplier $b^2$: its logarithmic thermal eigenvalue is two, so $\nu =1/2$.

The determinant also shows a thermodynamic singularity. Two derivatives of the free-energy density with respect to $m^2$ give, up to sign and a nonzero normalization,as $m^2\downarrow 0$. Rescaling $p=\sqrt{\alpha m^2}q$ establishes the power, and the dimensionless integral is ultraviolet-convergent for $D<4$. Integrating back gives a nonanalytic free-energy term proportional to $(m^2{)}^{D/2}$ after regular backgrounds are removed. This is critical behavior of the quadratic model.

There is a stability qualification: a pure Gaussian potential with $m^2<0$ is unbounded below in the uniform mode, so it does not by itself provide a stable ordered phase on the other side. The ordered-phase obstruction in a Gaussian scalar model is removed by adding a positive quartic interaction. In $2<D<4$ that interaction is relevant, so the resulting generic scalar critical point need not retain the Gaussian exponents. Thus the calculation proves the Gaussian critical singularity and $\nu =1/2$, without claiming that the unmodified quadratic integral defines both phases of a physical ordered transition.