Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 41 / 3 / Solution

The ultraviolet cutoff specifies which fluctuations remain explicit. Changing it integrates out some modes; to preserve long-distance predictions, their effect must be absorbed into the mass, interactions, and other couplings. Thus bare coefficients depend on $\Lambda$. The long-distance inverse connected correlation function has the form $\tilde{\Gamma }(p)\simeq Z^{-1}(p^2+{\xi }^{-2})$. After fixing the gradient normalization, the infrared mass is an inverse correlation length. With the question's unrescaled gradient coefficient, ${\xi }^{-2}=\alpha m^2$ at Gaussian level, so proportionality rather than literal equality is appropriate until that normalization is chosen.

As a concrete momentum-shell renormalization group step, split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into modes below $\Lambda /b$ and in the shell $\Lambda /b<|p|<\Lambda$, and defineThen rescale coordinates and the field to restore the cutoff and gradient coefficient. A local derivative expansion of $H_{\mathrm{eff}}$ gives new quadratic, quartic, and higher even couplings. If the system has short-range interactions, the field varies slowly, the effective potential is analytic, and remaining fluctuations are small, retaining the leading gradient and lowest stabilizing powers and minimizing the resulting functional gives Landau-Ginzburg theory. Suppressing the omitted fluctuations is the substantive Landau approximation; it is justified sufficiently far from the narrow fluctuation region or above the appropriate upper critical dimension.

The truncated two-point function in this question is the one-particle-irreducible two-point vertex, $\tilde{\Gamma }(p)=\tilde{G}(p{)}^{-1}$. It is the Hessian of the Legendre effective action, not merely the connected two-point function. In perturbation theory,Here ${\tilde{G}}_0$ is the Gaussian propagator about the chosen reference mass, $\delta m^2$ is the mass counterterm or bare-to-reference mass shift, and $\Sigma$ is the interaction self-energy, computed from proper diagrams. Fix the gradient normalization so ${\tilde{G}}_0(p)=(p^2+M^2{)}^{-1}$, and choose the infrared reference mass $M^2=m^2(0,T)$. Then $\delta m^2=m^2(\Lambda ,T)-M^2$ at this order. The mass renormalization condition is $\tilde{\Gamma }(0)=M^2$.

At one loop the only two-point diagram is the quartic tadpole diagram. There are $4\cdot 3$ choices for attaching its two external lines, leaving the other pair contracted; dividing by $4!$ gives its factor $1/2$. It is independent of external momentum, soThe renormalization condition therefore gives the self-consistent one-loop equationUsing $M$ inside the loop is a self-consistent resummation; replacing it by the bare mass gives the same formal first-order expansion away from infrared problems.

Subtract the equation at $T_c$, where $M=0$, and absorb regular temperature dependence of $g$ into the thermal coefficient. Let $\tau =m^2(\Lambda ,T)-m^2(\Lambda ,T_c)\propto T-T_c$ and $I(M)={\int }_{|q|<\Lambda }{d}^{D}q/(2\pi {)}^D(q^2+M^2{)}^{-1}$. ThenFor $D>4$, the integral divided by $M^2$ tends to a finite cutoff-dependent constant, since its small-momentum radial integrand is $q^{D-5}$. Thus $\tau$ is proportional to $M^2$, consistent with the Landau prediction. For $2<D<4$, set $q=Mz$: the correction instead scales as $M^{D-2}$ and dominates $M^2$. At $D=4$ it is proportional to $M^2\log (\Lambda /M)$. Hence a pure linear Landau mass law without logarithmic modification is consistent only aboveFor $D\le 2$, the massless integral itself is infrared divergent, so the subtraction already needs additional care. The sub-four-dimensional self-consistent approximation locates the breakdown of mean-field behavior; it is not an exact calculation of interacting critical exponents.

At a tricritical point the quartic coupling is tuned away and the leading stabilizing interaction is $g_6{\varphi }^6$. At the Gaussian fixed point, the field has engineering dimension $(D-2)/2$, so the sextic coefficient rescales as $g_6^{\prime }=b^{D-3(D-2)}{g}_6=b^{6-2D}{g}_6$. It changes from irrelevant to relevant atThis is the sextic case of the upper critical dimension of an even scalar interaction; marginality at three dimensions allows logarithmic corrections.