= 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 $\widetilde G_0(0)=1/m^2$ and the <correlation length> both diverge. If the thermal scaling field is linear in $m^2$ with $\alpha$ finite and positive, then
$$
\xi=(\alpha m^2)^{-1/2}\propto|t|^{-1/2},\qquad \boxed{\nu=\tfrac12.}
$$
The 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,
$$
\int^{\Lambda}\frac{d^Dp}{(2\pi)^D}\frac1{(\alpha^{-1}p^2+m^2)^2}
\propto(m^2)^{D/2-2}
$$
as $m^2\downarrow0$. 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.
Back to article page
