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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 $$ T_{a} ‎\langle 0|\phi|0\rangle ‎\neq‎ 0‎, \quad T_{a}|0\rangle ‎\neq‎ 0 $$

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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.)

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