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I've read in a book that when I have a quantum field theory with a symmetry under a group of transformations generated by a basis of generators, these generators should annihilate the vacuum state $\langle \Phi \rangle$ in order that the symmetry be respected by the theory.

I can't understand why I need to use the generators and not a generic element of the group. For example, in the SSB of the $SU(2)_L\times U(1)_Y$, I see that the generators $T_i$ and $Y$ don't annihilate the vacuum state anymore after symmetry breaking, in equations $T_i\langle \Phi \rangle\neq 0$.

I would rather expect $U(\theta)\langle \Phi \rangle\neq 0$, instead, with $U(\theta) \in SU(2)$. Why do I need the generator and not the transformation operator itself?

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    $\begingroup$ Because the transformations are exponentials of the generators. $\endgroup$
    – Meng Cheng
    Commented Sep 2, 2022 at 12:45

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The generators of a symmetry annihilating the vacuum state is equivalent to the vacuum state being invariant under the symmetry. Let me expand on that. Let $G$ be a simply connected Lie group and $\mathfrak{g}$ its Lie algebra. If $G$ is a symmetry of the quantum system, its Hilbert space ${\cal H}$ carries a unitary representation of $G$ (were $G$ not simply connected we would have to replace $G$ by its universal cover here). In other words, we have unitary transformations $U(g)$ for every $g\in G$ respecting the composition law of $G$: $$U(g)U(g')=U(gg')\tag{1}.$$

Near the identity of the group we can expand $U(g)$ in terms of generators $$U(g)=\mathbf{1}+i\theta^a T_a+O(\theta^2)\tag{2}$$ and one may check that the $T_a$ are hermitian operators which obey the commutation relations of $\mathfrak{g}$. The higher order terms can then be written in terms of the $T_a$. For some groups, all elements of the group turn out to be exponentials of Lie algebra elements, and then (2) resums to an exponential $\exp(i\theta^a T_a)$ in the Hilbert space. For other groups, the exponentials of group elements only recover a neighborhood of the identity. Even then it is still true that the terms are determined by the $T_a$ and the proof can be found in Appendix B to Weinberg's The Quantum Theory of Fields Chapter 2.

In any case, let $|\Omega\rangle$ be the vacuum state. Applying (1) to $|\Omega\rangle$ we find $$U(g)|\Omega\rangle = |\Omega\rangle+i\theta^a T_a|\Omega\rangle+O(\theta^2)\tag{3}.$$

The key thing is that in order for $|\Omega\rangle$ to be symmetric, $U(g)|\Omega\rangle=|\Omega\rangle$ for all $g\in G$, you must have $T_a|\Omega\rangle=0$.

In particular, imagine you have a set of operator-valued fields $\Phi_i$ which transform in several representations of $G$. Let $F(\Phi_i)$ be some function of these fields which combines them to form a scalar. If we evaluate the mean value $\langle \Omega|F(\Phi_i)|\Omega\rangle$ we may transform it as follows:

$$\langle \Omega|F(\Phi_i)|\Omega\rangle = \langle \Omega| U(g)^\dagger U(g)F(\Phi_i)U(g)^\dagger U(g)|\Omega\rangle = \langle \Omega |U(g)^\dagger F(\Phi_i) U(g)|\Omega\rangle$$

where in the last equality we used the hypothesis that $F(\Phi_i)$ makes a scalar out of the fields. We see that the mean value will be a scalar, and hence invariant, if and only if $|\Omega\rangle$ is invariant, which in turn happens if and only if $T_a|\Omega\rangle =0$.

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The unique ground state is invariant under a group transformation of an unbroken symmetry: it is a non-degenerate singlet, $$U(\theta)\langle \Phi \rangle =\langle \Phi \rangle. $$

Typically, the (Lie) group transformation is of the form $U(\theta)= e^{\theta G}\sim {\mathbb 1} + \theta G + O(\theta^2)$, for G the relevant generator (direction). It is then evident that you need $$ G \langle \Phi \rangle = 0. $$

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