Symmetry groups breaking for the lagrangian of two complex scalar fields

Question

Suppose we have a generic non-interacting Lagrangian of two complex scalar fields,
\begin{align} \mathcal{L} &= (\partial^\mu \Phi^\dagger)(\partial_\mu \Phi) - \Phi^\dagger\mathbb{M}^2\Phi \tag{1}\label{eq1}\\ &= \frac{1}{2}\partial^{\mu}\Phi_{i}\partial_{\mu}\Phi_{i}-\frac{m_i^{2}}{2}\Phi_{i}\Phi_{i},\quad i = 1,2,3,4. \tag{2}\label{eq2} \end{align} where \begin{align} \Phi = \begin{pmatrix}\phi_1 \\ \phi_2\end{pmatrix} && \Phi^\dagger = \begin{pmatrix}\phi_1^\dagger & \phi_2^\dagger\end{pmatrix} \end{align} The invariance of the kinetic term indicates that for the transformation $\Phi\rightarrow U\Phi$ and $\Phi^\dagger\rightarrow\Phi^\dagger U^\dagger$, $U^\dagger U = \mathbb{I}$. This indicates $U(2)$ symmetry for (1) and $SO(4)$ symmetry for (2). My question is how does the mass term break symmetries in this lagrangian in each case.

As suggested by this post, in the case where we have the same masses ($\mathbb{M}$ is proportional to $\mathbb{I}$), the $U(2)$ group survives (so is $SO(4)$). From Peskin and Schroeder question 2.2(d) (p.34), we can find the 4 conserved charges using the relation that $U(2)$ is locally isomorphic to $SU(2)\times U(1)$. If no symmetries are broken in this case, how can we see the two extra charges, which are clearly seen as generators of $SO(4)$? (An answer from the post linked below says the mechanism is $SO(4) \sim SU(2)×SU(2)$ but I don't quite understand this.)

Now if the two masses are not identical, this post suggests that $U(2)$ breaks into $U(1)\times U(1)$, whereas $SO(4)$ breaks into $U(1)\times U(1) \sim O(2) \times O(2)$. How can I understand if these two cases still have the same generators?

We have 3 generators for $(2)_$ and 3 for $(2)_$ , we still have the global $(1)$ and $(1)$ for each complex scalar field, is that right? So do we have 3+3+3 = 9 symmetry generators in total?

Answer 1

I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1). It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each component.

The resulting linear action generators in the formalism (2) of 4-vectors is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -( \alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 & \alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ could be absorbed in the normalization of $\vec\varphi$.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as your $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!