This is a follow-up question to this question.

How can I compute which generators remain unbroken when a linear combination of Higgs fields $a \Phi_1+ b\Phi_2$ get a vev?

If I compute the unbroken generators as I would do for a single Higgs field, my result is always that all generators get broken, because no generator annihilates both Higgs fields at the same time. For example

$$ T_1 (a \Phi_1+ b\Phi_2) = a \underbrace{ T_1 \Phi_1}_{=0}+ b \underbrace{ T_1 \Phi_2}_{\neq 0} \neq 0 \rightarrow \ T_1 \text{ is broken} $$


1 Answer 1


Follow the goldstons! I would expect any useful textbook on symmetry breaking covering the Goldstone theorem at any depth tells you exactly how to find the broken generators, and thus the surviving ones, as well. Sometimes, however, texts fall in love with mathematical abstraction and their message is obscured for students.

Here is the seat-of-the-pants procedure, which, if you master, you may abstract in mathematese to (mostly) confuse your friends-- life is too short to find an excuse for it. I am providing it here, as your odd formal metaphor above may miss a crucial point of the linear algebra involved. But I will break all symmetries, in this case SO(3). The potential is important. You may enlarge the group to higher N for surviving symmetries, but the point is to properly track the broken ones!

Consider a double SO(3) σ-model with two Higgs real triplets, so in the vector irrep, $\Phi_1$ and $\Phi_2$. Take the potential, with malice aforethought, to be $$ V\propto (\vec{\Phi_1} ^2 -a^2)^2 + (\vec{\Phi_2} ^2 -b^2)^2 + \lambda (\vec{\Phi_1}\cdot \vec{\Phi_2} )^2. $$ The function of the last term, with strictly positive λ is to discourage alignment of the v.e.v.s of the two triplets, so they can be chosen or rotated to be, w.l.o.g., for convenience, $$ \langle \vec{\Phi_1}\rangle = \left( \begin{array}{c} a \\ 0 \\ 0 \end{array} \right) \qquad \langle \vec{\Phi_2}\rangle= \left( \begin{array}{c} 0 \\ b\\ 0 \end{array} \right) $$.

Now, the central point of SSB, presumably emphasized in your texts, is the vanishing or not of the rotations of the Higgses evaluated at the vacuum, so, then, for the infinitesimal angles θ labelled according to the corresponding generators $L_x,L_y,L_z$ they attach to, $$ \langle \delta\vec{\Phi_1}\rangle = \left( \begin{array}{c} 0 \\ \theta_z a\\ -\theta_y a \end{array} \right) \qquad \langle \delta\vec{\Phi_2}\rangle= \left( \begin{array}{c} -\theta_z b \\ 0\\ \theta_x b \end{array} \right) $$.

Non-vanishing of these transform v.e.v.s signals Goldstone modes of the broken generators corresponding to the angles present on the r.h.side. That is to say the 1st Higgs breaks $L_y,L_z$,
and the second $L_x,L_z$, but which is the goldston φ of the broken $L_z$? Note the (un-normalized) combination $\langle \delta\varpi\rangle=\langle \delta (b\Phi^y_1 +a\Phi^x_2 )\rangle=0$, so it is not the goldston of the broken $L_z$. Instead, its orthogonal combination, $\phi=(a\Phi^y_1 -b\Phi^x_2 )$ is, so that $\langle \delta (a\Phi^y_1 -b\Phi^x_2)\rangle = \theta_z (a^2+b^2)$. Further note, of course, $\langle \phi\rangle=0$. (Naturally, $\langle \varpi\rangle=0$, as well.)

It is also instructive to compute the vacuum value of the second variation of the potential, the non diagonal 6x6 mass matrix, $\langle \delta^2 V/\delta \phi_i \delta \phi_j\rangle $, to notice that, there are 3 zero and three non vanishing eigenvalues: beyond the two original, evident goldstons noted above, the third null eigenvector is precisely the new goldston, (0,a,0,-b,0,0)=φ in this notation. The takeaway message is that it is the variation of the linear combinations in the vacuum that has to be analyzed, and not the related v.e.v.-acquiring combination, directly.


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