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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?

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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.
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  • $\begingroup$ I was taught that $U(1)×SU(2)$ was not the same symmetry group as (not isometric to) $U(2)$ because the former has copies of the center of $U(2)$. Why or why not is this a problem for what you're saying? $\endgroup$ – Craig Feb 2 at 19:04
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    $\begingroup$ I think you are right, perhaps this is just bad terminology from the physicists. I always said $\mathrm{U}(n) =\mathrm{U}(1)\times \mathrm{SU}(n)$, but now that you make me think about it, the first has center $\mathrm{U}(1)$ and the second has center $\mathrm{U}(1)\times \mathbb{Z}_n$. So one should mod out that $\mathbb{Z}_n$ factor. That's basically saying that describing a matrix $m\in \mathrm{U}(n)$ as a product of $(e^{i\varphi},m'\in\mathrm{SU}(n))$ has a redundancy. For example for $\mathrm{U}(2)$ the matrix $m=-\mathbb{1}$ is either $(e^{i\pi},\mathbb{1})$ or $(1,-\mathbb{1})$ $\endgroup$ – MannyC Feb 2 at 22:43
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    $\begingroup$ Then the correct statement is that the symmetry group is $\mathrm{U}(2)$, forget about $\mathrm{SU}(2)\times \mathrm{U}(1)$. (Somebody correct me if I'm saying nonsense). $\endgroup$ – MannyC Feb 2 at 22:45
  • $\begingroup$ Okay good this does make sense with what I've been taught. Would these two very similar symmetry groups ever lead to ovservationally different physics? Either as a symmetry group or as a gauge symmetry? Perhaps I should post this as a question of it's own. $\endgroup$ – Craig Feb 3 at 0:39
  • $\begingroup$ FWIW, the mass matrix $\mathbb{M}^2$ must be Hermitian and hence diagonalizable, say with eigenvalues $m_1^2$ and $m_2^2$. The symmetry group $G:=\{U\in U(2)| [U,\mathbb{M}^2]=0\}$ is the commutant of $\mathbb{M}^2$. 1. Case with same eigenvalues: $G=U(2)$. 2. Case with different eigenvalues: $G\cong U(1)^2$. $\endgroup$ – Qmechanic Feb 7 at 3:46

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