# Does a symmetry operator $U$ and its generator $Q$ acting on a vacuum $|0\rangle$ both represent new degenerate vacuum?

After the spontaneous breakdown of a symmetry characterized by $$\hat{U}=e^{i\hat{Q}\theta},$$ the commutation relation $$[H,Q]=0$$ continues to hold. Let us consider two states after this symmetry breaking. One is $$\hat{U}|0\rangle$$ where $$|0\rangle$$ is a vacuum of the theory with zero energy. Using $$[\hat{H},\hat{Q}]=0$$, it can be easily established that the state $$\hat{U}|0\rangle$$ represents another vacuum of the theory with zero energy. Consider another state defined as $$\hat{Q}|0\rangle$$. Also by using $$[\hat{H},\hat{Q}]=0$$, it can the shown that it too has zero energy.

Does it mean $$\hat{Q}|0\rangle$$ represent another vacuum state like $$\hat{U}|0\rangle$$? Am I making a mistake?

Having the lowest possible energy (which we can take to be zero) is only a necessary condition, not a sufficient condition, for a state to qualify as a vacuum state. For a state $$|0\rangle$$ to qualify as a vacuum state, it must also have the cluster property. The cluster property is $$\langle 0|A(x)B(y)|0\rangle \rightarrow \langle 0|A(x)|0\rangle\, \langle 0|B(y)|0\rangle$$ as $$|x-y|\rightarrow\infty$$ with $$x-y$$ spacelike, for all local operators $$A(x)$$ and $$B(y)$$.

Section 19.1 in Weinberg, The Quantum Theory of Fields, volume 2, explains that among all of the zero-energy states, we can choose a basis $$|k\rangle$$ in which $$\langle j|A|k\rangle=0$$ for $$j\neq k$$ for all local operators $$A$$. (I'm considering the case of a spontaneously-broken discrete symmetry to avoid technical complications, but the same idea applies in case of a continuous symmetry.) Weinberg explains that each of these basis states has the cluster decomposition property, but other superpositions of these basis states do not. Only the basis states $$|k\rangle$$ qualify as vacuum states, even though arbitrary superpositions of them also have zero energy.

If $$|0\rangle$$ is one of the qualified vacuum states and $$\hat Q$$ is the generator of the continuous SSB symmetry, then $$|\theta\rangle \equiv\exp(i\hat Q\theta)|0\rangle$$ is another qualified vacuum state. In the continuous-symmetry case, we can't even regard the states $$|\theta\rangle$$ with different $$\theta$$ as belonging to the same separable Hilbert space, because a separable Hilbert space can't have a continuum of mutually orthonormal states. This is the "technical complication" I alluded to above. A related complication prevents $$\hat Q|0\rangle$$ from being a well-defined state-vector, because $$\hat Q$$ is supposed to be something like the derivative with respect to $$\theta$$, but states with different values of $$\theta$$ don't even belong to the same (seperable) Hilbert space.

Maybe these technical complications can be avoided by working in a finite spatial volume; I'm not sure. Strict SSB doesn't occur in a finite spatial volume, but we can study how SSB emerges in the infinite-volume limit.

Even though this vacuum-selection criterion (the cluster principle) is often portrayed as a separate principle, as I portrayed it here, it is really just an ideal special case of the principle that we could use to diagnose when an observable has been measured, in a model that includes the measurement equipment, etc, as part of the quantum system. In the SSB case, the observable that has been "measured" is an observable whose eigenstates are the basis states $$|k\rangle$$ that can't be connected to each other by local operators, so these are the vacuum states that we actually experience in the wake of this "measurement," even if we started with a superposition of them. From this perspective, SSB can occur for all practical purposes even in a finite-volume space, for the same reason ("decoherence") that we are practically unable to observe a superposition of measurement outcomes in the wake of a measurement. In practice, a ferromagnet doesn't have to be infinitely big to become spontaneously magnetized.

Related:

Why symmetry breaking?

What is spontaneous symmetry breaking in QUANTUM systems?

Why do we assume the spatial volume is infinite?

Appendix: another reference

In the context of spin systems (like the Ising model), section 23.3 ("Order Parameter and Cluster Properties") in Zinn-Justin's Quantum Field Theory and Critical Phenomena shows that requiring that vacuum states satisfy the cluster property uniquely selects the conventional SSB ground states and eliminates all other superpositions of them, even though they all have the same energy.