Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 (Statistical Field Theory)

Write $t=T-T_c$. Analyticity gives ${\alpha }_2=at+O(t^2)$ with $a>0$ and ${\alpha }_4(T_c)=u>0$. For $t<0$, minimization gives $m^2=-{\alpha }_2/{\alpha }_4\sim -at/u$, so $\beta =1/2$. At the minima,Two temperature derivatives give a finite heat-capacity jump, hence $\alpha =0$. This is the ordinary mean-field transition of Landau theory.

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At the tricritical point, let ${\alpha }_2=at+O(t^2)$, ${\alpha }_4=bt+O(t^2)$ and ${\alpha }_6(T_c)=u>0$. The nonzero stationarity equation isThe middle term is subleading, so $m^4\sim -at/u$ and $\beta =1/4$. Substitution gives $F_{\min }\asymp -|t{|}^{3/2}$, so the heat capacity behaves as $|t{|}^{-1/2}$ and $\alpha =1/2$.

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Fourier expansion in volume $V$ diagonalizes the quadratic Landau-Ginzburg theory:whereEach independent real mode contributes a Gaussian integral proportional to $(T/K{)}^{1/2}$. Taking $-T\log Z$, dividing by $V$, and replacing the sum by an integral gives, up to field-independent conventions,

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For one mode put $f=-\frac{T}{2}\log (CT/K)$. Since $c=-T{\partial }_T^{2}f$,ThusAt $(T_c,0)$, $K_T=({\mu }^2{)}^{\prime }(T_c)\ne 0$ because ${\mu }^2$ has a simple zero, so $A(T_c,0)\ne 0$.

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Near the ordinary critical point, $K\sim at+\gamma k^2$. Rescaling $k=t^{1/2}q$ gives the Gaussian fluctuation correction near a critical pointThus ${\alpha }_{\mathrm{fl}}=2-d/2$. It becomes as important as the mean-field value $\alpha =0$ at $d=4$, so the ordinary upper critical dimension is $d_c=4$.

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The quadratic kernel still has $K\sim at+\gamma k^2$, so ${\alpha }_{\mathrm{fl}}=2-d/2$. The tricritical mean-field exponent is $\alpha =1/2$. Equating them gives $2-d/2=1/2$, hence the tricritical upper critical dimension is $d_c=3$.

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A Wilsonian renormalization group step:

The resulting effective free energy defines the transformed couplings and their flow.

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At the Gaussian fixed point, set $a_d=12{\Omega }_{d-1}{\Lambda }^{d-2}/(2\pi {)}^d$. The linearized flow isThe mass direction, corresponding to ${\varphi }^2$, has coupling dimension $y_t=2$. For $d\ne 2$, the second eigendirection isand has $y_g=4-d$; it is the tadpole-subtracted mixture of ${\varphi }^4$ and ${\varphi }^2$. The corresponding operator dimensions are $d-2$ and $2d-4$.

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For $d<4$, $y_g=4-d>0$, so the quartic coupling is a relevant direction of a fixed point. Generic Ising-like systems therefore flow away from the Gaussian point toward the interacting Wilson-Fisher fixed point, which controls the Ising universality class below four dimensions.

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Put $r={\mu }^2/{\Lambda }^2$ and $\tilde{g}={\Lambda }^{d-4}g$. At $d=4-ϵ$, ${\Omega }_3/(2\pi {)}^4=1/(8{\pi }^2)$ givesThe nonzero fixed point iswhich is the stated Wilson-Fisher fixed point.

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Linearization at the fixed point givesThe relevant eigendirection is $(\delta r,\delta \tilde{g})=(1,O({ϵ}^2))$. An irrelevant eigendirection isThus the mass-like combination is relevant and the quartic-like combination is irrelevant; neither is marginal for $ϵ>0$.

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The coefficient $A$ comes from one sextic vertex with two external slow legs and two fast tadpole loops. The coefficient $B$ comes from one sextic vertex with four external slow legs and one fast tadpole. The coefficient $C$ comes from one quartic and one sextic vertex joined by two fast propagators, leaving six external slow legs.

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Choosing two slow legs in $h({\varphi }_{<}+{\varphi }_{>}{)}^6$ and contracting four fast legs gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)(3)h{I}_1^2{\varphi }_{<}^2$. Since the mass term is ${\mu }^2{\varphi }_{<}^2/2$,Choosing four slow legs and contracting the remaining pair gives $B=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$. For the connected quartic-sextic cumulant, the factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{2}\right)2!=180$ and the cumulant has a minus sign, soAs a check, two quartic vertices give $-\frac{1}{2}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=-36$, matching the supplied coefficient.

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The canonical eigenvalue of $h$ isThus ${\beta }_h=(-2+O(ϵ))h+O(g^3)+O(h^2)$. At a fixed point this forces $h_{\ast }=O(g_{\ast }^3)=O({ϵ}^3)$. Its feedback into the mass and quartic flows begins only at order ${ϵ}^3$, so the order-$ϵ$ fixed point from part (b)(iii) is unaffected.

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At a fixed point, the order parameter can be chosen as a dilation eigenoperator. Therefore, for $x^{\prime }=x/\zeta$,Interactions shift ${\Delta }_{\varphi }$ from $(d-2)/2$ to $(d-2+\eta )/2$.

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If $t^{\prime }={\zeta }^{{\Delta }_t}t$, lengths shrink by $\zeta$, so $\xi (t^{\prime })={\zeta }^{-1}\xi (t)$. Substituting $\xi (t)\sim t^{-\nu }$ gives ${\zeta }^{-\nu {\Delta }_t}={\zeta }^{-1}$ and hence

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The source term transforms asTherefore

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Choose $\zeta =t^{-\nu }$. The singular free-energy density scales as $t^{d\nu }$, giving the hyperscaling relation $\alpha =2-d\nu$. The order parameter givesThe susceptibility has exponentAt $t=0$, choosing $\zeta =B^{-1/{\Delta }_B}$ gives

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For $d>4$, the quartic coupling has $y_g=4-d<0$, but setting it to zero removes the stabilizing term. It is a dangerously irrelevant coupling. At $t=0$, minimizing $g{\varphi }^4-B\varphi$ gives $\varphi \sim (B/g{)}^{1/3}$ and $\delta =3$. This is consistent with Gaussian scaling becauseFor $d<4$, $g$ flows to a nonzero Wilson-Fisher fixed point, so no dangerously irrelevant zero-coupling limit spoils the ordinary scaling relations.

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Put ${\rho }_i=|{\psi }_i{|}^2$. The uniform potential isIf both masses are positive, ${\rho }_1={\rho }_2=0$. If ${\mu }_1^2<\min (0,{\mu }_2^2)$, only ${\psi }_1$ condenses, with ${\rho }_1=-{\mu }_1^2/(2g)$; symmetrically, only ${\psi }_2$ condenses when ${\mu }_2^2<\min (0,{\mu }_1^2)$. The positive coordinate axes are continuous-transition lines. On ${\mu }_1^2={\mu }_2^2={\mu }^2<0$, every pair withis a minimum; crossing this diagonal exchanges the two condensates, so it is a coexistence line ending at the origin.

Away from the diagonal the symmetry is $U(1{)}_1\times U(1{)}_2$. It is unbroken in the normal phase. In either one-condensate phase one factor is broken and the other remains, giving one Goldstone boson. On the diagonal the symmetry is enhanced to $U(2)$. For positive equal mass it is unbroken; for negative equal mass it breaks as $U(2)\to U(1)$ and gives three Goldstone bosons. At the origin $U(2)$ is unbroken although both quadratic modes are critical.

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