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In classical field theories, it is with no difficulty to imagine a system to have a continuum of ground states, but how can this be in the quantum case?
Suppose a continuous symmetry with charge $Q$ is spontaneously broken, that would means $Q|0\rangle\ne0$, and hence the symmetry transformation transforms continuously $|0\rangle$ into anther vacuum, but how can a separable Hilbert space have a continuum of vacuums deferent from each other?

I saw somewhere that says the quantum states are built upon one vacuum, and others simply doesn't belong to it, what does this mean? and then how could $Q$ be a well defined operator which acting on a state (the vacuum) actually gives a state (another "vacuum") out of the space considered?

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up vote 4 down vote accepted

My understanding is that as you state, these states do not belong to the same hilbert space but are formally connected by the symmetry transformation. The essence is that the degeneracy of the vaccum is expressed in the notion that there are multiple equally well suited states (from separate hilbert spaces) from which one can build up excited states. Nature chooses one of these equivalent hilbert spaces.

It is also helpful to note that the charge operator is in fact not well-defined for infinite volumes and $Q|0\rangle\neq 0$, and in this infinite volume limit all vacua are orthogonal to each other, and hence would be part of an uncountable orthonormal basis, which prohibits them to be part of a signle separable hilbert space.

Some more information can be found here:

Spontaneous Symmetry Breaking and Nambu-Goldstone Bosons in Quantum Many-Body Systems (arXiv)

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