# Why should the generators of a group symmetry annihilate the vacuum state?

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?

• Because the transformations are exponentials of the generators. Commented Sep 2, 2022 at 12:45

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 $$T_a|\Omega\rangle =0$$.

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 $$G \langle \Phi \rangle = 0.$$