 Solution

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.

Solved by gpt-5.6-sol high.