I know that an adjoint field $\Phi$ spontaneously breaks an SU(N) symmetry when it gets a VEV in the diagonal form
$$ \langle\Phi\rangle = diag(v_1, v_1,...,v_1, v_2,.., v_2,...) $$ where the number of $v_1$ is $n_1$, $v_2$ is $n_2$, etc.
The unbroken symmetry is SU($n_1$)$\times$SU($n_2$)$\times$U(1). The last U(1) comes from the combination of all the unbroken subgroup generators, i.e. proportional to $\langle\Phi\rangle$ itself. I have a couple of questions
- How to relate this to the fact the unbroken symmetry generator should vanish the vacuum, $Q|0\rangle=0$? I don't see, say, the SU($n_1$) generator vanish the diagonal VEV matrix.
- What if in the VEV matrix, there are some entries which only have one copy. For example, consider an SU(5) adjoint with a VEV $$ \langle\Phi\rangle \propto diag(-1,-1,-1,-1,4) $$ Does the last entry (which only has one copy) generate an unbroken U(1)? But apparently the whole $\langle\Phi\rangle$ generates an unbroken U(1). So I'm not sure whether the unbroken symmetry group is SU(4)$\times$U(1)$\times$U(1) or SU(4)$\times$U(1)?
- The most extremal situation is that all the entries in VEV only have one copy. Let's say the SU(2) adjoint with a VEV $$ \langle\Phi\rangle \propto diag(-1,1) $$ If the two elements separately generate an unbroken U(1), then $\langle\Phi\rangle$ itself apparently does not generate an independent U(1), otherwise originally SU(2) have three generators whereas after SSB there are still three generators in three U(1)'s...But it's also possible that the two elements do not generate unbroken U(1) while $\langle\Phi\rangle$ does. So, in this case, is the unbroken symmetry U(1)$\times$U(1) or U(1)?