Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 303 / 3 / b / Solution

When $u_0=0$, the field theory is a Gaussian field theory, so integrating out short-wavelength modes can be done exactly rather than merely at a saddle. Introduce an ultraviolet momentum cutoff $\Lambda$ and a scale factor $b>1$, then split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into $|q|<\Lambda /b$ and $\Lambda /b<|q|<\Lambda$. Orthogonality of Fourier modes eliminates cross terms in the quadratic Hamiltonian. A constant source couples only to the zero-momentum mode and therefore does not couple to ${\varphi }_{>}$.

Before rescaling, the Gaussian functional integral factorizes:where $H_{<}$ has the original $r_0$ and $h$, restricted to low momenta. Up to a field-independent normalization, the eliminated modes contributeThe determinant affects the additive free energy, but generates neither a mass correction nor a source correction, since no interaction mixes low and high modes. This remains true beyond the Landau approximation: it is an exact integration of a Gaussian field theory, not neglect of the high-mode fluctuations.

To restore the original cutoff, use $x=b{x}^{\prime }$, $q^{\prime }=bq$, and choose the field rescalingSubstitution into the low-mode Hamiltonian gives the Gaussian momentum-shell scalingThe gradient coefficient is unchanged: the volume factor is $b^D$, the two derivatives contribute $b^{-2}$, and the two fields contribute $b^{2-D}$. The mass term gains $b^2$, while the source term gains $b^{(D+2)/2}$. ThusFor a nonconstant source the corresponding formula is $h^{\prime }(x^{\prime })=b^{(D+2)/2}h(b{x}^{\prime })$, after restricting its coupling to retained modes. The constant-source case avoids that extra source filtering.

Writing $b=e^{\ell }$, the exact Gaussian renormalization-group flow hasBoth perturbations are relevant at the massless zero-source Gaussian fixed point, with $y_t=2$ and $y_h=(D+2)/2$. There is no anomalous field rescaling here, so $\eta =0$ and the correlation-length critical exponent is again $\nu =1/y_t=1/2$. The coefficient changes arise from rescaling, not from an interaction-induced shift in $r_0$ during shell integration.

A stable real Gaussian functional integral requires $r_0>0$, or an infrared prescription that treats the zero mode at the massless point. With $u_0=0$ and $r_0<0$, the field energy is unbounded below; the Gaussian field theory alone cannot describe a stable ordered phase. The scaling laws are therefore interpreted about the Gaussian fixed point from the stable side, with finite-volume zero-mode regularization when needed.