# SSB In Electroweak Theory

I know that when vacuum state does not remain invariant under generators of lagrangian group symmetry, we have SSB in our theory: $$U_{a} | 0 \rangle‎ = |0\rangle‎ \Rightarrow T_{a} | 0\rangle‎ = 0 \qquad U_{a} \in SU(2) \times U(1)$$ in A Pedagogical Review of Electroweak Symmetry Breaking Scan by Gautam Bhattacharyya, I found that we can show SSB in electroweak theory by acting generators on scalar field vacuum expectation value (scalar vev) : $$T_{a}\langle0|\phi|0\rangle‎$$ the problem is I can not figure out what relation between these two can be according to QM $$T_{a} ‎\langle 0|\phi|0\rangle ‎\neq‎ 0‎, \quad T_{a}|0\rangle ‎\neq‎ 0$$

You need to think carefully about what space your quantities act on. $$\phi$$ is a quantum field (operator) acting on Hilbert space. However, by itself, $$T^a$$ is just an element of a Lie algebra in some representation. It acts on the appropriate representation space, not on Hilbert space.
Therefore, the notation $$T^a \lvert 0 \rangle$$ is sloppy. If you go back to Noether's theorem and see how the symmetry generators are defined, the symmetry generator $$Q^a$$, as an operator on Hilbert space, is actually given by
$$Q^a = \int d^3 \; \dot{\phi}^\dagger_i(x) T^a_{ij} \phi_j(x) + \mathrm{h.c.}$$
for a multi-component complex field $$\phi_i(x)$$ and $$T^a_{ij}$$ in the appropriate matrix representation. This corresponds to the infinitesimal symmetry transformation
$$\phi_i(x) \to \phi_i(x) + i \epsilon T^a_{ij} \phi_j(x), \qquad \epsilon \ll 1.$$
(I have assumed here a set of quite simple Lagrangians with kinetic term $$\partial_t \phi^\dagger \partial_t \phi$$, so that the canonical momentum for $$\phi$$ is $$\dot{\phi}^\dagger = \partial_t \phi^\dagger$$, but that is not necessary for this answer.)