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Why can't there be spontaneous symmetry breaking in finite volume?

  1. Dec 22, 2008 #1
    Hello everyone, I'm a bit confused by something I've read as have been unable to find resources to clarify it. Here is the statement that confused me, from Binetruy's Supersymmetry textbook (p.26):

    It is well-known that [in the case of ordinary continuous symmetries] no possibility of spontaneous symmetry breaking in a finite volume. Through superposition, the ground state can always be made invariant under the symmetry because mixing between states is always possible in finite volume. It is only in the infinite volume limit that one may define unmixed ground states in spontaneously broken symmetry.​

    Why can there be no spontaneous symmetry breaking in finite volume? Why can one only take superpositions of ground states in finite volume?

    Thanks,
    JB
     
  2. jcsd
  3. Dec 22, 2008 #2
    In a finite volume, there's a nonzero probability of quantum tunneling from one ground state to the other. It does not make sense to talk about a specific random ground state because it's not special in the long run.

    Consider a simple potential [tex]\phi^4 - 2 \phi^2[/tex]. There are two ground states [tex]\phi = +/-1[/tex]. On very general grounds, you can expect the probability for the system to tunnel from one ground state to the other in a given time to decrease exponentially with the height of the energy barrier, which is [tex]~V[/tex].

    In the limit of infinite volume, the same mechanism also keeps you from going into infinitesimally close ground states, e.g. in "mexican hat" potential.

    There's a couple of pages on this subject in the second volume of Weinberg.
     
  4. Dec 23, 2008 #3
    Thanks for the reply, Hamster. This brings up a follow-up question that I've had in the back of my mind for some time.

    My understanding is that for infinite volume/continuous space the probability of quantum tunneling goes to zero because one has an infinite product of decaying exponentials (one for each point of spacetime). Thus the vacuum state of a theory does not tunnel. However, is it possible for the vacuum to be a superposition of the two minima at each point in spacetime?

    Suppose that one has a double well potential for some Higgs field. I.e. there are two minima at each point in spacetime, [itex]| \uparrow \rangle [/itex] and [itex]| \downarrow \rangle [/itex]. Then we could write a state as:

    [itex]|\uparrow\oplus \uparrow\oplus\downarrow\oplus\cdots\oplus\uparrow\oplus\downarrow\rangle[/itex]

    My question is that why can't we have a vacuum state that looks like

    [itex]2^{-n/2}|(\uparrow+\downarrow)\oplus (\uparrow+\downarrow)\oplus\cdots\oplus(\uparrow+\downarrow)\rangle[/itex]?

    (n represents the number of 'lattice sites,' this formally goes to infinity) Would this be a valid vacuum state?




    Would it be correct that one should instead change basis to

    [itex]|\pm\rangle = 2^{-1/2}(|\uparrow\rangle\pm|\downarrow\rangle)[/itex]

    And then the product states look like

    [itex]|+\oplus + \oplus - \oplus\cdots\oplus+\rangle[/itex].

    Then this goes back to a problem without superpositions. So what does this mean? If there are well defined states in the [itex]|\uparrow\rangle,|\downarrow\rangle[/itex] basis, are we then free to pick a vacuum state where each point in spacetime has a vacuum that is a superposition?
     
  5. Dec 23, 2008 #4
    I'm not sure what you mean by "having a vacuum at each point in spacetime". A vacuum is a ground energy state of the entire system.

    In your example (double well potential), there are only two global vacua that are also eigenstates of Higgs. One where every point in space is [tex]\uparrow[/tex] and the other where every point in space is [tex]\downarrow[/tex]. We could call them [tex]|0_{\uparrow}\rangle[/tex] and [tex]|0_{\downarrow}\rangle[/tex]. Their superposition is also a vacuum, but it's not an eigenstate.

    You could in principle work with a superposition, but it is more convenient to work with individual states. The reason is that any physically relevant state (a state that can be obtained from any vacuum with a nonzero probability) would be a superposition of a state in the vicinity of [tex]|0_{\uparrow}\rangle[/tex] and a state in the vicinity of [tex]|0_{\downarrow}\rangle[/tex]. You could work with either of these separately in the original basis.

    Try to look at it from the many-worlds perspective. During symmetry breaking, your universe "branches" into a superposition of a universe with [tex]|0_{\uparrow}\rangle[/tex] vacuum and a universe with [tex]|0_{\downarrow}\rangle[/tex] vacuum. These two proceed to evolve completely unaware of each other. Eventually they develop intelligent life (us). We measure the expectation value of Higgs and see that it's [tex]\uparrow[/tex]. Our parallel brothers do the same and see [tex]\downarrow[/tex].

    Does any of this have anything to do with your question?
     
  6. Jan 4, 2009 #5
    Hi Hamster -- I think that answers my question exactly. So even if the vacuum were set to be a superposition of the two states, a measurement by an observer in the universe will yield a definite state.

    Would it be correct to say that the choice of vacuum (up or down) is an observable that commutes with the Hamiltonian?
     
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