How can we determine which subgroup remains unbroken after spontaneous symmetry breaking for $SU(2)\times U(1)$ symmetry?

Question

Consider an $SU(2)$ doublet of bosons $\Phi = (\phi^+, \phi^0)^T$, where the complex scalar field $\phi^+$ destroys positively charged particles and creates negatively charged ones, and the complex scalar field $\phi^0$ destroys neutral particles and creates neutral anti-particles. The theory has $SU(2)\times U(1)$ symmetry, with Lagrangian $$ \mathcal L = (\partial_\mu\Phi^\dagger)(\partial^\mu\Phi)+\mu^2\Phi^\dagger\Phi-\frac{\lambda}{4}(\Phi^\dagger\Phi)^2 $$ I saw another example with $SU(2)$ triplet of real scalar fields $\Phi = (\phi_1, \phi_2, \phi_3)^T$, with symmetry $SU(2)\times U(1)_Y, Y=0$ and it has a similar scalar potential, $$ V(\Phi) = -\frac{m^2}{2}\Phi^\dagger\Phi-\lambda(\Phi^T\Phi)^2 $$ Are there physical interpretations of those scalar fields $\phi_1, \phi_2, \phi_3$ like those in the doublet case (so we can adjust the notation for $V(\Phi)$? Should we expect to obtain the same Lagrangian density as the doublet case? Also, how can we determine which subgroup remains unbroken after spontaneous symmetry breaking?

Answer 1

If some of symmetries of theory are spontaneously broken, we expect some charge operator will satisfy $\hat Q|0\rangle\neq 0$. However, because of the existence of NG mode, $\hat Q|0\rangle$ has a divergent norm so the condition above is ill-defined statement. Instead of this statement, usually we use a following statement as an indicator of the spontaneous symmetry broken (SSB): $$^\exists \hat O(x)\ s.t. [\hat Q, \hat O(x)]=-i\delta_Q\hat O(x)\ \mathrm{and}\ \langle 0| \delta_Q\hat O(x) |0\rangle\neq 0.$$

In your example of the Higgs model, $O(x)$ corresponds to $\Phi(x)$. In general, unbroken symmetry ${H}$, which is sub group of spontaneously broken symmetry $ G$, is characterized by the vacuum expectation above: $\langle 0| \delta_Q\hat O(x) |0\rangle$.

I mean, if $H$ is unbroken, its charge operator and a vacuum of theory must satisfy $\langle 0|[\hat Q, \hat O(x)]|0\rangle=0,$ for arbitrary $\hat O(x)$. In the other hand, from the discussion above, some of charge operators of $G$ is not satisfies this relation. Thus, we can say that if we use a label $a$ for generators of $G-H$ and a label $i$ for generators of $H$ , $\hat Q_a$ and $\hat Q_i$ must satisfy the following relation: $$\langle 0| \delta_{Q_i}\hat O(x) |0\rangle =0,$$ $$\langle 0| \delta_{Q_a}\hat O(x) |0\rangle \neq0.$$

This is how we know the unbroken symmetry. In principle, what we should do is to observe $\langle 0| \delta_Q\hat O_i(x) |0\rangle$ and to find specific generators of $G$ which satisfies $\langle 0| \delta_Q\hat O_i(x) |0\rangle =0 .$

In your example, you can check $\langle 0|\Phi(x)|0\rangle$ is invariant only $U(1)_{\mathrm{em}}$. To show this, one should notice that we can choose the following Ansatz indefinitely: $$\langle 0|\Phi(x)|0\rangle = (0,v)^{\mathsf T}.$$

Roughly speaking, elements of $SU(2)\times U(1)_Y$ are the unitary $2\times 2$ matrices. Thus, all that needs to be done is to let this $2\times 2$ matrix act on the vacuum expectation value above to find the condition that satisfies $\langle 0|\delta_i\Phi(x)|0\rangle=0$.

This is essentially all that needs to be done, but its looking becomes a bit more complicated to try to develop a general theory. For more details: see Weinberg’s QFT.