Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 303 (Statistical Field Theory)

Write $a={\alpha }_2(T)$ and $u={\alpha }_{2n}>0$. At high temperature $a>0$, so $f(m)=a{m}^2+u{m}^{2n}$ has one minimum at $m=0$. At low temperature $a<0$, the origin is a local maximum and two symmetry-related minima appear. The stationary equation isso the equilibrium magnetization isandBecause $a(T)\sim T-T_c$, the two nonzero minima approach zero continuously as $T\uparrow T_c$. The order parameter is continuous while its response becomes singular, so this is a continuous phase transition.

The ordered solution givesso the order-parameter critical exponent isAt $T=T_c$, adding a magnetic term $-hm$ gives $h=2nu{m}^{2n-1}$, henceAbove $T_c$, the zero-field susceptibility is $\chi =(2a{)}^{-1}$. Below $T_c$, the curvature at a minimum is $-4(n-1)a$, so both sides giveAt the ordered minimum, $a=-nu{m}^{2n-2}$ andSince $f_{\mathrm{sing}}\sim |T-T_c{|}^{2-\alpha }$, the heat-capacity critical exponent isThese values obey the Widom scaling relation and Rushbrooke scaling relation at mean-field level.

The Landau-Ginzburg theory promotes the order parameter to a field and penalizes spatial gradients:with stiffness $\kappa >0$. Further even powers and higher-derivative terms may be included when required by accuracy.

Mean field givesThe stated Gaussian fluctuation term isAs $T\to T_c$, mean field is more singular precisely whenorThus the upper critical dimension for the $m^{2n}$ interaction is

Let $z=2d$ be the coordination number of the hypercubic lattice. The mean interaction energy per site is $(Jz/2)m_e{m}_o=Jd,m_e{m}_o$. DefineThe two sublattices each contain half the sites, so the mean-field free energy per site isThe entropy terms are the binary-spin mixing entropies and the positive $J$ favors opposite sublattice magnetizations.

Since $I^{\prime }(m)=\mathrm{artanh}m$, stationarity givesThereforewhere

For $h=g=0$, the expansion $I(m)=-\log 2+m^2/2+O(m^4)$ givesWith $m_{\pm }=(m_e\pm m_o)/2$, one has $m_e{m}_o=m_{+}^2-m_{-}^2$ and $m_e^2+m_o^2=2(m_{+}^2+m_{-}^2)$. HenceThe staggered magnetization $m_{-}$ becomes unstable atproducing the antiferromagnetic transition. The coefficient of the uniform magnetization $m_{+}$ remains positive for every $T>0$, so there is no $m_{+}$ transition.

A momentum-shell renormalization group step has three parts:

1. Split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into slow modes $|k|<\Lambda /\zeta$ and fast modes $\Lambda /\zeta <|k|<\Lambda$, then integrate out ${\varphi }_{>}$ using a cumulant expansion to obtain a Wilsonian effective action for ${\varphi }_{<}$.
2. Rescale momenta by $k^{\prime }=\zeta k$, equivalently coordinates by $x^{\prime }=x/\zeta$, restoring the cutoff from $\Lambda /\zeta$ to $\Lambda$.
3. Rescale the field by ${\varphi }^{\prime }(x^{\prime })={\zeta }^{(d-2)/2}{\varphi }_{<}(x)$ at the Gaussian fixed point, restoring the normalized kinetic term.

The resulting functional has the original form and cutoff but new masses and couplings. Iteration produces a renormalization-group flow.

Requiring $\int d^{d}x(\nabla \varphi {)}^2$ to be dimensionless gives the engineering dimensionsandThus ${\mu }_0$ is relevant. The cubic coupling is relevant, marginal, or irrelevant for $d<6$, $d=6$, or $d>6$; the quartic coupling has the same classification around $d=4$; and the sextic coupling has it around $d=3$.

LetAt order $g_0^2$, the connected six-point topology consists of two quartic vertices joined by one fast line. The second cumulant givesFor a sharp shell and external slow momenta approaching zero, the connecting momentum cannot lie in the fast shell, so this nonlocal tree topology gives no local zero-momentum ${\varphi }_{<}^6$ coupling.

At order $g_0^3$, the local connected contribution is the triangle of three quartic vertices, with two external slow legs at each vertex and one fast propagator along each edge. The factor isConsequently, at zero external momentum,Ignoring contributions involving ${\lambda }_0,{\gamma }_0$ and anomalous field rescaling, the running sextic coupling is thereforeThe diagrams are respectively the one-line two-vertex six-point tree and the three-vertex triangular one-loop six-point diagram; only the latter contributes to the local shell coupling at vanishing external momenta.

The leading mass correction is the one-vertex tadpole diagram: two legs of the quartic $g_0({\mathbf{\varphi }}^2{)}^2$ vertex are external and the remaining two form one fast loop. One contraction has a freely summed component index and contributes $N$, while two exchange contractions have their index fixed by the external component. The result is proportional to $N+2$.

The leading quartic correction is the one-loop bubble diagram with two quartic vertices joined by two fast propagators. The four external legs can be distributed in the three exchange channels. One family of index contractions contains a freely summed closed component loop and contributes $N$; the other contractions contribute eight, giving the factor $N+8$.

Define the shell integralsThe index multiplicities described in part a and the canonical rescaling giveFor $N=1$, these reduce to the stated Ising coefficients $12$ and $36$.

PutFor $\zeta =e^s$, differentiating a thin radial shell at $s=0$ givesReplacing bare parameters by running parameters after each infinitesimal step yields the beta functions

For $d=4-ϵ$, define dimensionless variablesTo leading order, with $K_4=1/(8{\pi }^2)$,The Gaussian fixed point is $(r_{\ast },u_{\ast })=(0,0)$. Its thermal eigenvalue is $y_t=2$, soThe interacting Wilson-Fisher fixed point isLinearizing the flow givesSince the correlation-length critical exponent is $\nu =1/y_t$,

Corrections quadratic in the cubic-anisotropy coupling use two ${\lambda }_0{\sum }_a{\varphi }_a^4$ vertices joined by two fast propagators. Because all four legs at each anisotropic vertex carry the same component index, the internal propagators force the two vertices to have that same index. The resulting one-loop bubble renormalizes the anisotropic tensor ${\sum }_a{\varphi }_a^4$, hence $\lambda$, but does not generate the distinct-index structure in $({\mathbf{\varphi }}^2{)}^2$ needed for a correction to $g$. There is no freely summed closed component index, so the diagram is not proportional to $N$.

At order ${\lambda }_0{g}_0$, use one anisotropic quartic vertex and one $O(N)$-invariant quartic vertex joined by two fast propagators. With four equal external component labels, the bubble renormalizes $\lambda$; with external labels in two equal pairs, it also renormalizes $g$. The anisotropic vertex fixes the component index carried by the internal lines, so none of these mixed diagrams contains an independent component loop. Consequently no correction proportional to $N$ occurs at order ${\lambda }_0^2$ or ${\lambda }_0{g}_0$.