2
$\begingroup$

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 $S_2=S_1^{T}$.

$\endgroup$
  • 5
    $\begingroup$ Maybe I'm missing something, but doesn't the fact $m_1 \neq m_2$ break the $SU(2)$ symmetry? $\endgroup$ – Michael Seifert May 30 at 20:39
  • 2
    $\begingroup$ 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. $\endgroup$ – Michael Seifert May 30 at 20:49
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for your reply but then the doublet still has some global internal symmetry a part from U(1) for each field? $\endgroup$ – Stefano Barone May 30 at 21:35
  • 1
    $\begingroup$ @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. $\endgroup$ – Cosmas Zachos May 30 at 22:51
  • $\begingroup$ @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? $\endgroup$ – Stefano Barone May 31 at 7:08
  • 1
    $\begingroup$ @StefanoBarone I added a clarification on the $\text{U}(1)$ in the answer. $\endgroup$ – Bob Knighton May 31 at 7:10
  • $\begingroup$ 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. $\endgroup$ – Michael Seifert May 31 at 12:29
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.