Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 3

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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