Symmetry group of two complex scalar fields with different masses Which is the symmetry group of the following Lagrangian:
$$
\mathcal{L} = (\partial^\mu \phi_1^\dagger)(\partial_\mu \phi_1) + (\partial^\mu \phi_2^\dagger)(\partial_\mu \phi_2) - m_1^2\phi_1^\dagger\phi_1 - m_2^2\phi_2^\dagger\phi_2
$$
where $\phi_1$ and $\phi_2$ are two complex scalar fields with, in general, different masses $m_1 \neq m_2$?
$\mathcal{L}$ is manifestly invariant under the two $U(1)$ transformations of $\phi_1 \rightarrow \phi_1e^{iq_1\alpha}$ and $\phi_2 \rightarrow \phi_2e^{iq_2\beta}$, independently of each other, since I have no term like $\phi_1^\dagger\phi_2$ that would have required a $U(1)$ transformation of both fields to be left invariant. Does that mean that I have an $U(1)_{\phi_1} \times U(1)_{\phi_2}$ group of symmetries?
Moreover, we can write $\mathcal{L}$ in a more compact form defining:
\begin{align}
\Phi = \begin{pmatrix}\phi_1 \\ \phi_2\end{pmatrix} && \Phi^\dagger = \begin{pmatrix}\phi_1^\dagger & \phi_2^\dagger\end{pmatrix} && \mathbb{M^2} = \begin{pmatrix} m_1^2 & 0 \\ 0 & m_2^2\end{pmatrix} \\
\end{align}
$$
\mathcal{L} = (\partial^\mu \Phi^\dagger)(\partial_\mu \Phi) - \Phi^\dagger\mathbb{M}^2\Phi
$$
In this case I would say that $\mathcal{L}$ has only a $U(1)$ symmetry $$\Phi \rightarrow \Phi e^{iq\alpha} = \begin{pmatrix}\phi_1e^{iq\alpha} \\ \phi_2e^{iq\alpha}\end{pmatrix}$$ which transforms both the fields at the same time, losing in a certain sense, the independence of the two fields.
Finally, how is this argument related to the case $m_1 = m_2$, where I should expect an $SU(2) \times U(1)$ symmetry group?
 A: The way you wrote it as a matrix product is perfectly good for finding symmetries. In that notation, suppose first that the masses are absent. Then we have a symmetry
$$
\Phi \to U \Phi\,,\qquad \Phi^\dagger \to \Phi^\dagger U^\dagger\,.
$$
The invariance of the kinetic term imposes $U^\dagger U = \mathbb{1}$, which gives you an $\mathrm{U}(2)$ symmetry group (or equivalently $\mathrm{U}(1)\times\mathrm{SU}(2)$). If you have masses you have to further satisfy
$$
U^\dagger\mathbb{M}^2U =\mathbb{M}^2\,.\tag{1}\label{eq1}
$$


*

*If the mass matrix is generic (i.e. you have a mixing $\phi_1^\dagger\phi_2$) then this equation will only have the trivial solution $U = e^{iq\alpha} \mathbb{1}$, that is, a single $\mathrm{U}(1)$

*If you have a diagonal mass matrix, then $U$ can be any diagonal matrix, so
$$
U=\left(\matrix{e^{iq_1\alpha}&0\\0&e^{i q_2 \beta}}\right)\in \mathrm{U}(1)\times\mathrm{U}(1)\,.
$$

*If finally $\mathbb{M}^2$ is proportional to the identity (i.e. equal masses), equation \eqref{eq1} is the same as the one we had before: $U^\dagger U = \mathbb{1}$, therefore the entire $\mathrm{U}(2)$ group survives.

