= Solution
The $O(2)$ rotation symmetry requires the quadratic dependence on $(\phi_2,\phi_3)$ to be proportional to $\phi_2^2+\phi_3^2$, so
$$
\mu_2^2=\mu_3^2.
$$
The quartic terms involving only that pair must be proportional to $(\phi_2^2+\phi_3^2)^2$, and the terms coupling it to $\phi_1$ must be proportional to $\phi_1^2(\phi_2^2+\phi_3^2)$. With the convention that the symmetric double sum counts off-diagonal terms twice, this gives
$$
g_{22}=g_{33}=g_{23},
\qquad
g_{12}=g_{13},
$$
while $\mu_1^2$, $g_{11}$, and the common mixed coupling are unrestricted. The stated $\mathbb Z_2$ symmetry imposes no further relation because every term is already even in $\phi_1$.
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