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?

  • 5
    $\begingroup$ Because the transformations are exponentials of the generators. $\endgroup$
    – Meng Cheng
    Commented Sep 2, 2022 at 12:45

2 Answers 2


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.