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Spontaneous symmetry breaking

  1. Mar 11, 2013 #1
    Is there a reason why we have to expand a field ψ about the true vacuum |Ω>? Can't we just do field theory about ψ=0 instead of about ψ=<Ω|ψ|Ω>?

    Also, I'm a bit confused about other fields. For the E&M potential, under the true vacuum, wouldn't we need to expand about A=<Ω|A|Ω> instead of A=0?

    Also, how do we find the true vacuum anyhow? The way that it seems to be done is to take the derivative of the potential V, and set it equal to zero. The potential V usually has a negative mass or something shifting the true vaccum away from ψ=0. Is this an approximation? Don't we have to take the derivative of the effective potential Veff instead of V?

    And how can one show that two derivatives of Veff gives the mass generated? I thought to get the mass generated, you have to shift the fields, and identify the coefficients of terms proportional to the square of the field. But a paper I'm reading claims you can just take two derivatives of Veff to get the mass generated?
  2. jcsd
  3. Mar 12, 2013 #2


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    Yes, you have to use V_eff. "The way it seems to be done..." Are you referring to some specific book? There are presentations of varying level of precision.
  4. Mar 12, 2013 #3
    Most textbooks just state [tex]V(\phi)=-\mu^2\phi^2+λ\phi^4 [/tex] as the beginning potential which leads to spontaneous symmetry breaking.

    V is not the effective potential, but the potential appearing in the Lagrangian, which is just the tree-level effective potential.

    Something feels rotten about just stating that the potential is [tex]V(\phi)=-\mu^2\phi^2+\lambda \phi^4 [/tex]. Spontaneous symmetry breaking depends crucially on the sign of μ2, but how do we know the sign of μ2? Aren't these couplings supposed to depend on scale? Is there a scale where the coupling goes back to being negative?

    I can see how λ must always be positive at any scale or else your theory doesn't have a ground state at all. But I'm not sure how we know the sign of μ2 by just looking at the bare Lagrangian.
  5. Mar 28, 2013 #4


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    I just now saw your reply.
    Zee, Quantum field theory in a nutshell explains that the effective potential has to be used to determine whether a symmetry is broken or not.
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