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Spontaneous symmetry breaking: How can the vacuum be infinitly degener

  1. Mar 11, 2013 #1
    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?
     
    Last edited: Mar 11, 2013
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  3. Mar 11, 2013 #2

    Physics Monkey

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    To understand these statements it is best to start with a finite system. Thus consider a chain of [itex] N [/itex] spin 1/2 particles. Take the Hamiltonian to be [tex] H = - J \sum_i \vec{S}_i \cdot \vec{S}_{i+1}. [/tex]
    The Hamiltonian is symmetric with respect to global spin rotations generated by [tex] \sum_i \vec{S}_i . [/tex]
    However, the ground state of the Hamiltonian spontaneously breaks this symmetry because the Hamiltonian favors all spins to point in the same direction. Indeed, the state
    [tex]
    | \uparrow_1 ... \uparrow_N \rangle
    [/tex]
    is an exact ground state as is
    [tex]
    | \downarrow_1 ... \downarrow_N \rangle.
    [/tex]
    In fact, the ground state subspace is the [itex] N+1 [/itex] dimensional space of total spin [itex] N/2 [/itex].

    What can one say about this space? Well, for finite [itex] N [/itex] it is finite dimensional, however, as [itex] N \rightarrow \infty [/itex] it approaches a continuous collection of degenerate ground states. To show this, consider the coherent state labelled by spin direction [itex] \hat{n} [/itex]. Then in the limit [itex] N \rightarrow \infty [/itex] all the states
    [tex]
    |\hat{n},N\rangle = \otimes_{i=1}^N |\hat{n}_i \rangle
    [/tex]
    are orthogonal. Furthermore, for finite [itex] N [/itex] the overlap between coherent states in different directions is exponentially small, i.e. goes like [itex] e^{-c N}[/itex].

    Thus for physical systems consisting of a finite number of degrees spontaneous symmetry breaking, as defined in your original post, is an approximate asymptotic statement about the ground state manifold for large systems.

    Note, however, that for any finite N everything is perfectly well defined and there are no subtleties about different spaces for different ground states, etc. This is why the finite N approach is in my opinion simpler and more physical.

    Hope this helps.
     
  4. Mar 12, 2013 #3

    DrDu

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    The point is exactly that these vacua don't live in the same Hilbert space and that the operator Q is not defined.
     
  5. Mar 12, 2013 #4

    Bill_K

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    I believe this type of Hilbert Space is called a direct integral (as opposed to a direct sum)
     
  6. Mar 12, 2013 #5
    Physics Monkey's answer is completely correct. In the infinite limit there is no probability of tunneling between vacua, so they live separately. This question is explained in Maggiore "Modern Introduction in QFT" in the chapter on SSB.
     
    Last edited: Mar 12, 2013
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