Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 69 / 2

The renormalization group compares descriptions at successively longer length scales. First eliminate short-wavelength degrees of freedom, then rescale coordinates and fields to restore the reference cutoff. This changes the couplings, including interactions and additive free-energy terms generated during elimination. A renormalization-group fixed point is a scale-invariant description. Linearized directions growing under coarse graining are relevant, shrinking directions are irrelevant, and zero-eigenvalue directions are marginal and require higher-order analysis.

A diverging correlation length means that many coarse-graining steps can approach a renormalization-group fixed point before a relevant perturbation becomes large. Systems with different microscopic details but the same renormalization-group fixed point and relevant directions share universal critical behavior. For the quadratic model below, Fourier modes decouple, so eliminating a high-momentum shell produces a Gaussian determinant but does not couple the remaining modes. The anisotropic gradient kernel requires different length rescalings in the parallel and transverse directions. This is an anisotropic renormalization group, not an isotropic rescaling of all $d$ coordinates.

ChooseFor a real scalar field $\tilde{m}(-q)=\tilde{m}(q{)}^{\ast }$. The parallel derivative becomes $i{q}_{\parallel }$, and the transverse Laplacian becomes $-q_{\perp }^2$. The latter is squared in the energy, producing $q_{\perp }^4$, not $q_{\perp }^2$. ThusThe zero-momentum source is understood in a finite box or as a uniform-field distribution before taking the thermodynamic limit. With an orthonormal box transform the source is $-h\sqrt{V}{m}_0$. The positive coefficients and $t>0$ give a normalizable Gaussian field theory, with mean uniform field $\overline{m}=h/t$ and connected Fourier covariance $[K{q}_{\parallel }^2+L{q}_{\perp }^4+t{]}^{-1}$.

Let $b,c>1$, rescale $q_{\parallel }^{\prime }=b{q}_{\parallel }$, ${\mathbf{q}}_{\perp }^{\prime }=c{\mathbf{q}}_{\perp }$, and set $B=b{c}^{d-1}$. Use the Fourier-amplitude conventionThe momentum measure contributes $B^{-1}$, and substitution into the quadratic statistical Hamiltonian givesIn the uniform source the momentum delta function cancels the Jacobian determinant of the measure, explaining why $h^{\prime }$ does not have an additional $B^{-1}$. In a full momentum-shell renormalization group transformation integrate out the fast modes first. Their uncoupled Gaussian integrals contribute an additive free-energy density; the displayed transformations describe the slow-mode statistical Hamiltonian after coordinate and field rescaling.

The field convention must be stated. In real space, $x_{\parallel }^{\prime }=x_{\parallel }/b$ and ${\mathbf{x}}_{\perp }^{\prime }={\mathbf{x}}_{\perp }/c$, and Fourier inversion gives $m^{\prime }(x^{\prime })=Bm(x)/z$. Thus if instead the symbol $m^{\prime }=m/z_x$ refers to real-space amplitudes, $z=B{z}_x$. Then the same transformation reads $K^{\prime }=B{z}_x^{2}K/b^2$, $L^{\prime }=B{z}_x^{2}L/c^4$, $t^{\prime }=B{z}_x^{2}t$, $h^{\prime }=B{z}_{x}h$. These are equivalent normalizations, not different critical exponents.

Keeping both gradient coefficients fixed requires $z^2=B{b}^2=B{c}^4$. HenceEquivalently the real-space field normalization is $z_x=z/B=b^{(3-d)/4}$. The transformed thermal and source couplings areThis is the Gaussian fixed point at $t=h=0$, with anisotropic effective volume exponent $D_{\mathrm{eff}}=(d+1)/2$. The stiffnesses remain positive parameters labeling the normalization of coordinates and fields. A strictly positive $t$ is a relevant perturbation away from the renormalization-group fixed point, not itself a critical renormalization-group fixed point.

The two correlation lengths also follow directly by balancing the kernel: $K{q}_{\parallel }^2\sim t$ gives ${\xi }_{\parallel }\sim (K/t{)}^{1/2}$, while $L{q}_{\perp }^4\sim t$ gives ${\xi }_{\perp }\sim (L/t{)}^{1/4}$. Their different exponents are consistent with the required anisotropic scaling. An omitted scalar quartic interaction has scaling eigenvalue $4-D_{\mathrm{eff}}=(7-d)/2$, so the quadratic renormalization-group fixed point alone does not establish the interacting smectic's critical behavior below physical dimension seven.

Let $f=-V^{-1}\log \mathcal{Z}$ be the dimensionless free-energy density. The transformed volume is $V^{\prime }=V/B$. Including the normalization from eliminated modes giveswhere $g_b$ is an additive shell contribution. After subtraction of regular backgrounds, the leading singular part obeys the scaling hypothesis for critical phenomenaChoose $b=t^{-1/2}$ for the high-temperature side $t>0$. Then the nominal homogeneous form isThe volume exponent is $D_{\mathrm{eff}}$, so inserting $d$ in the isotropic hyperscaling formula would give the wrong answer.

There are two qualifications to interpreting this as the total free energy. Completing the Gaussian square gives the source contribution exactly as $-h^2/(2t)$, and the zero-field determinant contributesRescaling $q_{\parallel }\sim t^{1/2}$ and $q_{\perp }\sim t^{1/4}$ gives the power $t^{(d+1)/4}$, while $2\Delta -1=(d+1)/4$ makes the source term consistent with it. Cutoff-dependent analytic terms need not obey the homogeneous law. Furthermore Gaussian free-energy logarithms at integer singular powers replace the pure zero-field power by $t^n\log t$ when $(d+1)/4=n$ is a positive integer, including $d=3,7$. This is the usual logarithmic qualification of the nominal exponents. The quadratic model also has no stable negative-$t$ phase unless a stabilizing interaction is added.