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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?

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  • $\begingroup$ Your latter, real triplet Higgs model, is called the Georgi Glashow model. Convince yourself as taught that the v.e.v. can be chosen in one of the 3 components, and so the SU(2)~SO(3) breaks down to a U(1)~O(2), so one generator survives. Such a model lacks neutral currents, so it is useless for the electroweak interactions. $\endgroup$ Apr 12, 2023 at 14:57
  • $\begingroup$ G-G SU(2) model. $\endgroup$ Apr 14, 2023 at 15:00

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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.

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  • $\begingroup$ What if $\delta_Q \hat O(x) |0\rangle \neq 0$? Isn't that sufficient to say that $Q$ is (or generates) a spontaneously broken symmetry? $\endgroup$
    – Myridium
    Apr 12, 2023 at 4:07
  • $\begingroup$ We would like an indicator of SSB that distinguishes a broken symmetry and unbroken symmetry, but clearly your indicator is not a good one because it does not have to be zero, even if it is unbroken symmetry. (i.e. $\hat Q_i \hat O(x)|0\rangle$ is not necessarily zero) $\endgroup$
    – Siam
    Apr 12, 2023 at 6:22
  • $\begingroup$ I thought SSB was when the vacuum state is not invariant under the transformation? What if $\langle 0| \delta_Q \hat O(x)|0\rangle = 0$, but $\langle v| \delta_Q \hat O(x)|0\rangle \neq 0$? $\endgroup$
    – Myridium
    Apr 12, 2023 at 18:39
  • $\begingroup$ If we are considering ordinary symmetry, then it is expected that some field (in general, it is composite operator) has a vacuum expectation value from the shape of the potential if SSB happens. On the other hand, if there is this vacuum expectation value, can we say that spontaneous symmetry breaking is taking place? That is what the above discussion complements; it states that if $\langle 0|\delta_Q O| 0\rangle\neq 0$ for some operator, then SSB is taking place. $\endgroup$
    – Siam
    Apr 13, 2023 at 6:07
  • $\begingroup$ Ah I see, thank you. $\endgroup$
    – Myridium
    Apr 13, 2023 at 17:47

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