= Solution
Put $\rho_i=|\psi_i|^2$. The uniform potential is
$$
V=\mu_1^2\rho_1+\mu_2^2\rho_2+g(\rho_1+\rho_2)^2.
$$
If 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 with
$$
|\psi_1|^2+|\psi_2|^2=-\frac{\mu^2}{2g}
$$
is 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.
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