Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 303 (Statistical Field Theory)

The equilibrium magnetization is a global minimum of the Landau free energy. Its candidates are the stationary pointsThus $m=0$, or $x=m^2\ge 0$ is a nonnegative root ofOne retains candidates with nonnegative curvature $f^{\prime \prime }(m)$ and compares their free energies; the candidate with the smallest value is the equilibrium state. Equal global minima describe phase coexistence.

At the first-order phase transition, the central minimum $m=0$ and two nonzero minima $m=\pm m_0$ are degenerate, with barriers between them. Writing $x=m_0^2>0$, stationarity and equal free energies giveTheir difference gives $a_{4}x+2a_6{x}^2=0$, soThe magnitude of the magnetization therefore jumps from zero toThe two possible signs are related by the model's spin-reversal symmetry.

At the tricritical point, $a_4=0$ and the zero-field Landau free energy is $f=a_2{m}^2+a_6{m}^6$. Below $T_c$, minimization givesSince $a_2\sim T-T_c$, the order-parameter critical exponent isAt this minimum, $f_{\mathrm{sing}}=a_2{m}^2+a_6{m}^6$ is proportional to $-|T-T_c{|}^{3/2}$. Comparing this with $f_{\mathrm{sing}}\sim |T-T_c{|}^{2-\alpha }$ gives the heat-capacity critical exponentAfter adding the magnetic contribution $-Bm$, the inverse zero-field magnetic susceptibility is the curvature $f^{\prime \prime }(m)$. Above $T_c$ it is $2a_2$, while below $T_c$ it is $-8a_2$, so $\chi \sim |T-T_c{|}^{-1}$ and the magnetic-susceptibility critical exponent isFinally, exactly at $T_c$ the equation of state is $B=6a_6{m}^5$. Hence $m\sim B^{1/5}$ and the critical-isotherm exponent is

The mean-field approximation suppresses correlated order-parameter fluctuations. Near a continuous phase transition, the correlation length diverges and long-wavelength fluctuations become increasingly important. The Ginzburg criterion therefore fails in sufficiently low dimension, and the interacting renormalization-group fixed point changes the mean-field exponents. For the tricritical $m^6$ theory the upper critical dimension is three: below $d=3$ the exponents are generally non-mean-field, while at $d=3$ one expects logarithmic corrections to mean-field scaling.

This is the Blume–Capel model. In the mean-field approximation, write $m=⟨{\sigma }_i⟩$ and replaceBecause every site has coordination number $q$, the resulting energy isThe one-site partition function is thereforeWith $\kappa =e^{-\beta g}$, the mean-field partition function is $Z_{\mathrm{MF}}=e^{-\beta NJq{m}^2/2}{Z}_1^N$. Taking $F=-T\log Z_{\mathrm{MF}}$ gives

Set $B=0$, $h=\beta Jqm$, and $A=1+2\kappa$. The required power series is obtained fromThe quadratic Landau coefficient isIts vanishing gives the line of continuous transitionsThe quartic coefficient isAt a tricritical point, both $a_2$ and $a_4$ vanish. Since $A=1+2\kappa$, the condition $a_4=0$ gives $A=6\kappa$ and hence $\kappa =1/4$. Substitution into the critical line yieldsThe sextic coefficient is positive there, so the sixth-order term stabilizes the free energy. Equivalently, $g_{\mathrm{tri}}=T_{\mathrm{tri}}\log 4$.

A momentum-shell renormalization group transformation consists of three steps. First split $\varphi ={\varphi }_{-}+{\varphi }_{+}$ into slow modes with $|k|<\Lambda /\zeta$ and fast modes with $\Lambda /\zeta <|k|<\Lambda$, and integrate out ${\varphi }_{+}$ to obtain a Wilsonian effective action for ${\varphi }_{-}$. Next rescale momenta by $k^{\prime }=\zeta k$, equivalently coordinates by $x^{\prime }=x/\zeta$, to restore the cutoff to $\Lambda$. Finally rescale the field to restore the chosen normalization of the gradient term. Repeating these operations produces a renormalization-group flow of every permitted mass and coupling.

The free energy is dimensionless, while $d^{d}x$ has engineering dimension $-d$ and each derivative has dimension one. Requiringto be dimensionless gives $2+2[\varphi ]-d=0$. Thus the engineering dimension of the field is

A scalar field of scaling dimension ${\Delta }_{\varphi }$ has a scale-invariant two-point correlation function proportional to $|x{|}^{-2{\Delta }_{\varphi }}$. Comparison with the stated form givesThe difference $\eta /2$ from the engineering value is the field's anomalous dimension, generated by fluctuations and interactions at a non-Gaussian renormalization-group fixed point.

Ignoring interactions, every field has dimension $(d-2)/2$, and every Laplacian contributes two derivatives. The operator ${\varphi }^n({\nabla }^2\varphi {)}^m$ therefore has dimensionBecause the integrated interaction $\int d^{d}xg\mathcal{O}$ is dimensionless, the coupling has scaling dimensionIt is a relevant coupling when ${\Delta }_g>0$, a marginal coupling when ${\Delta }_g=0$, and an irrelevant coupling when ${\Delta }_g<0$.

Expand the quintic interaction after the slow-fast split. Its term with three slow fields and two fast fields isThe first term of the cumulant expansion therefore containsSince the coincident fast-mode propagator isthe lowest-order correction isIts Feynman diagram is one five-valent vertex with three external slow-field legs and the remaining two legs contracted into a tadpole diagram.

One cubic vertex cannot leave four external legs. Two cubic vertices can be joined by one contracted fast-field line, leaving four slow-field legs, so the first correction to ${\lambda }_0$ involving ${\alpha }_0$ isIt arises from the second term of the cumulant expansion.

A correction with three external legs can be made from one cubic and one quartic vertex by contracting four of their seven fields into internal lines. Consequently the first correction to ${\alpha }_0$ involving ${\lambda }_0$ isA purely quartic interaction cannot generate an odd interaction because its ${\mathbb{Z}}_2$ symmetry forbids odd powers of $\varphi$.

The constraint $\mathbf{n}\cdot \mathbf{n}=1$ makes $\mathbf{n}$ dimensionless. Since $d^{d}x({\partial }_i{n}_A{)}^2$ has engineering dimension $2-d$, dimensional consistency of the O(N) nonlinear sigma model gives

Choose the positive local branch of the constraint,The chain rule givesand thereforeSubstitution into the original free energy gives exactly

For small $\mathbf{\pi }$, expand $(1-{\mathbf{\pi }}^2{)}^{-1}=1+O({\mathbf{\pi }}^2)$ and retain the quartic interaction $(\mathbf{\pi }\cdot {\partial }_i\mathbf{\pi }{)}^2/(2g_0)$. Split $\mathbf{\pi }={\mathbf{\pi }}^{-}+{\mathbf{\pi }}^{+}$ and contract the fast fields in the shell $\Lambda /\zeta <|q|<\Lambda$. To first order inthe contraction adds $I_d$ to the inverse coefficient of the slow-field kinetic term. The prescribed wave-function renormalizationthen multiplies that coefficient by $A^2$. HenceFinally the coordinate rescaling contributes ${\zeta }^{d-2}$, and thus

Let $s=\log \zeta$. For an infinitesimal momentum shell,Differentiating the inverse-coupling relation and using $d(1/g)/ds=-g^{-2}dg/ds$ gives the one-loop beta functionFor $d=2+ϵ$, ${\Omega }_{d-1}/(2\pi {)}^d=1/(2\pi )+O(ϵ)$, so

The renormalization-group fixed points are the zeros of the beta function:For $N>2$ and $ϵ>0$, the positive fixed point $g_c^{\ast }$ separates the low-temperature ordered flow toward $g=0$ from the high-temperature strong-coupling flow. At $ϵ=0$ the two perturbative fixed points merge at zero.

Identifying $g$ with temperature, write $g=g_c^{\ast }+\delta g$ near the nonzero critical fixed point. The derivative of the beta function there isThus the temperature-like perturbation has renormalization-group eigenvalue $y_t=ϵ$. Since the correlation-length critical exponent satisfies $\nu =1/y_t$,The Gaussian fixed point $g=0$ is the stable ordered-phase fixed point for $d>2$, rather than the finite-temperature transition. In exactly two dimensions the flow instead gives an essential, exponential correlation-length divergence, corresponding formally to $\nu =\infty$ rather than a finite power-law exponent.