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

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.