Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 3

When all masses are equal, the free energy depends on the real vector $\mathbf{\varphi }$ only through inner products, so its symmetry is the orthogonal group $O(N)$. The mean-field potential isFor ${\mu }_0^2>0$, its unique minimum is $\mathbf{\varphi }=0$, which preserves $O(N)$. For ${\mu }_0^2<0$, the minima form the sphereChoosing one minimum leaves the subgroup $O(N-1)$ that fixes its direction, so spontaneous symmetry breaking gives $\boxed{O(N)\to O(N-1)}$.

The leading mass corrections are one-vertex tadpole diagrams. In index notation, one contraction closes a freely summed component loop and is proportional to $N$; the two exchange contractions force the internal component to equal the external one and are not proportional to $N$. Including their multiplicities givesThus the $N$ in $N+2$ comes from the closed index loop, while the $2$ comes from the two same-component contractions.

The leading quartic corrections contain two quartic vertices joined by two internal propagators: the three familiar exchange channels distribute the four external legs in the $s$, $t$, and $u$ pairings. One index contraction contains a freely summed closed component loop and is proportional to $N$; the remaining contractions have indices fixed by the external legs. Their sum givesThe $N$ term is the closed-index-loop topology and the eight is the combined multiplicity of the other contractions.

Under $x^{\prime }=x/\zeta$ and the Gaussian field rescaling ${\varphi }^{\prime }(x^{\prime })={\zeta }^{(d-2)/2}\varphi (x)$, a mass squared has engineering dimension two and a quartic coupling has dimension $4-d$. Therefore

LetFor a radial integrand $h(q)$, differentiating the thin shell at $\zeta =e^s$ givesReplacing the bare parameters by running ones after each infinitesimal step yields the beta functions

For $d=4-ϵ$, define the dimensionless couplings $r={\mu }^2/{\Lambda }^2$ and $u=\tilde{g}={\Lambda }^{-ϵ}g$. Since $K_4=1/(8{\pi }^2)$, the leading epsilon expansion of the flow isThere is a Gaussian fixed point $(r_{\ast },u_{\ast })=(0,0)$. Its thermal eigenvalue is $y_t=2$, so its correlation-length critical exponent is $\boxed{{\nu }_{\mathrm{G}}=1/2}$.

The interacting Wilson-Fisher fixed point isLinearizing the renormalization-group flow gives the thermal eigenvalueTaking its reciprocal gives

Write $r_a={\mu }_{0,a}^2$. The stationary equations forareFor $r_1,r_2>0$, the disordered minimum is $(0,0)$. If $r_1<0$ and $r_1<r_2$, the first component orders with ${\varphi }_1^2=-r_1/(4g_0)$ and ${\varphi }_2=0$. If $r_2<0$ and $r_2<r_1$, the second component orders analogously.

The positive $r_2$ half of $r_1=0$ and the positive $r_1$ half of $r_2=0$ are continuous-transition lines. Along the negative diagonal $r_1=r_2<0$, the potential has an enhanced $O(2)$ symmetry and a circle of minima. Crossing that diagonal exchanges the two ordered axes and makes derivatives of the minimum free energy jump, so it is a first-order phase transition line. The origin, where the two continuous lines meet the first-order line, is a bicritical point.

Define the two shell integralsFor an external ${\varphi }_1$, the ${\varphi }_1^4$ part of the interaction gives the scalar tadpole coefficient $12g_0{I}_1$, while $2g_0{\varphi }_1^2{\varphi }_2^2$ gives $4g_0{I}_2$. Interchanging the components givesWhen the masses agree, $I_1=I_2$ and both formulas reduce to $4(N+2)g_0{I}_1$ with $N=2$.

No. Even when the two masses differ, the free energy retains an independent Z2 symmetry for each component: ${\varphi }_1\mapsto -{\varphi }_1$ and ${\varphi }_2\mapsto -{\varphi }_2$. The bilinear ${\varphi }_1{\varphi }_2$ is odd under either transformation, so integrating out modes cannot generate it.