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youngurlee
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Spontaneous symmetry breaking: the vacuum be infinitly degenerate?
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 [itex]Q[/itex] is spontaneously broken, that would means [itex]Q|0\rangle\ne 0[/itex], and hence the symmetry transformation transforms continuously [itex]|0\rangle[/itex] 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 [itex]Q[/itex] be a well defined operator which acting on a state (the vacuum) actually gives a state (another "vacuum") out of the space considered?
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 [itex]Q[/itex] is spontaneously broken, that would means [itex]Q|0\rangle\ne 0[/itex], and hence the symmetry transformation transforms continuously [itex]|0\rangle[/itex] 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 [itex]Q[/itex] 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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