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I'm considering a Lagrangian of two complex scalar field: $$\mathcal{L}=\partial_{\mu}\phi_1^{*}\partial^{\mu}\phi_1-m_1^2\phi_1^{*}\phi_1+\partial_{\mu}\phi_2^{*}\partial^{\mu}\phi_2-m_2^2\phi_2^{*}\phi_2$$ It can be written in a doublet:

$$\Phi_1 =\begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} \,\,\,\,\,\,\,\,\,\,\,\,\, \Phi_1^{\dagger} = \begin{pmatrix} \phi_1^\dagger & \phi_2^\dagger \end{pmatrix} \,\,\,\,\,\,\,\,\,\,\,\,\, M = \begin{pmatrix} m_1 & 0 \\ 0 & m_2 \end{pmatrix} $$

$$\mathcal{L}=\partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi - \Phi^{\dagger}M\Phi$$ It has an internal global symmetry $SU(2)$: $$ \begin{cases} \Phi^{'}= e^{\frac{i}{2}\vec\alpha \cdot \vec\sigma} \Phi\\ \Phi^{'\dagger} = \Phi^{\dagger}e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} \end{cases} $$ I'd like to check it explicitly but I'm stuck: $$\mathcal{L^{'}}=\partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi - \Phi^{\dagger}M\Phi$$

$$\mathcal{L^{'}}=\begin{pmatrix} \partial_{\mu}\phi_1^{*} & \partial_{\mu}\phi_2^{*} \end{pmatrix} e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} e^{+\frac{i}{2}\vec\alpha \cdot \vec\sigma} \begin{pmatrix} \partial^{\mu}\phi_1 \\ \partial^{\mu}\phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1^{*} & \phi_2^{*} \end{pmatrix} e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} \begin{pmatrix} m_1 & 0 \\ 0 & m_2 \end{pmatrix} e^{+\frac{i}{2}\vec\alpha \cdot \vec\sigma} \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix}$$ However: $$ S_1=e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} \begin{pmatrix} m_1 & 0 \\ 0 & m_2 \end{pmatrix} = \begin{pmatrix} m_1e^{-\frac{i\alpha_3}{2}} & m_2e^{-\frac{i(\alpha_1-i\alpha_2)}{2}} \\ m_1 e^{-\frac{i(\alpha_1+i\alpha_2)}{2}}& m_2e^{\frac{i\alpha_3}{2}} \end{pmatrix}$$ $$ S_2=\begin{pmatrix} m_1 & 0 \\ 0 & m_2 \end{pmatrix} e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} =\begin{pmatrix} m_1e^{-\frac{i\alpha_3}{2}} & m_1 e^{-\frac{i(\alpha_1-i\alpha_2)}{2}} \\ m_2 e^{-\frac{i(\alpha_1+i\alpha_2)}{2}} & m_2e^{\frac{i\alpha_3}{2}} \end{pmatrix}$$ So $S_2=S_1^{T}$.

Stefano Barone
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    Maybe I'm missing something, but doesn't the fact $m_1 \neq m_2$ break the $SU(2)$ symmetry? – Michael Seifert May 30 '19 at 20:39
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    Also, I'm pretty sure that you're not exponentiating your matrices correctly; it looks like you've found the exponential of $-\frac{i}{2} \vec{\alpha} \cdot \vec{\sigma}$ by taking the exponential of each entry. But this is incorrect. – Michael Seifert May 30 '19 at 20:49

2 Answers2

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The $\text{SU}(2)$ symmetry of this theory is only preserved when $m_1=m_2\equiv m$, in which case $M=m\textbf{1}$. Otherwise, the internal $\text{SU}(2)$ symmetry is broken down to $\text{U}(1)\times\text{U}(1)$, one for each scalar field. Something similar to this happens in QCD, in which the up and down quarks have nearly the same mass, and the approximate $\text{SU}(2)$ isospin symmetry is useful for classifying low-mass mesons and baryons.

Bob Knighton
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  • Thanks for your reply but then the doublet still has some global internal symmetry a part from U(1) for each field? – Stefano Barone May 30 '19 at 21:35
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    @StefanoBarone Of course, there are two commuting U(1)s, one for each scalar field. You could make life easier for you if you eschewed complex notation and wrote down the 4 real degrees of freedom; then, ask yourself: How does the mass matrix break the original SO(4) symmetry explicitly? Take $m_1=0$ for simplicity. – Cosmas Zachos May 30 '19 at 22:51
  • @CosmasZachos Thanks for your reply, do you know some references where these calculations about symmetries for multiplet for both complex fields and fermions are performed in pedantic detail? – Stefano Barone May 31 '19 at 07:08
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    @StefanoBarone I added a clarification on the $\text{U}(1)$ in the answer. – Bob Knighton May 31 '19 at 07:10
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    If I'm not mistaken, the symmetry of the original Lagrangian is $U(2) = U(1) \times SU(2)$, which is then broken to $U(1) \times U(1)$ when $m_1 \neq n_2$. In other words, one of the $U(1)$ factors that remains after the symmetry is broken was "outside" of the original $SU(2)$ subgroup that the OP was concerned with. – Michael Seifert May 31 '19 at 12:29
  • @StefanoBarone The thesis paper of Longhitano referenced in my geeky footnote on the custodial group question should cover it, but any remedial group theory review in a serviceable QFT course should do. Do you want me to do the explicit SO(4) ~ SU(2) x SU(2) breaking with 4 scalars here ? – Cosmas Zachos May 31 '19 at 15:15
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Ignore all normalizations, since this is an issue of symmetry, and consider real numbers $a,b,c,d$, s.t. $$ \phi_1=a+ib, \qquad \phi_2=c+id , $$ and define $m_1=m_2 + \Delta$.

Then, for the real 4-vector $\vec \varphi\equiv (a,b,c,d)^T$, you have $$\mathcal{L}=\partial_{\mu} \vec \varphi \cdot \partial^{\mu} \vec \varphi -m_2 \vec \varphi \cdot \vec \varphi -\Delta (a^2+b^2). $$ It is manifest that, for $\Delta=0$, this expression is SO(4) ~ SU(2)×SU(2) invariant. (One of these two SU(2) s is the SU(2) you display, but the other one is less easy to see in your language, and corresponds to the "right custodial" SU(2) of the SM, a global approximate symmetry of it. Through SO(4), you can appreciate there are 6 transformations mixing your 4 real scalars.)

But the introduction of the explicit perturbation $\Delta$ breaks this SO(4) into $U(1)\times U(1) \sim O(2) \times O(2)$, the two O(2) s rotating the doublets (a,b) and (c,d), independently, respectively. The other 4 generators of SO(4) are explicitly broken, since you cannot preserve the last, perturbation term, any other way--try it.

Cosmas Zachos
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