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Goldstone boson state, conserved current

  1. Apr 23, 2013 #1
    I don't understand these sentences from Peskin:

    "At long wavelength (*why long wavelength?*), the Goldstone bosons become infinitesimal symmetry rotations of the vacuum, ##Q^a|0 \rangle##, where ##Q^a## is the global charge associated with ##J^{\mu a}##. Thus, the operators ##J^{\mu a}## have the correct quantum numbers to create Goldstone boson state. In general, there will be a current ##J^{\mu a}## that can create or destroy this boson; we can parametrize the corresponding matrix element as $$\langle 0| J^{\mu a} | \pi_k(p) \rangle = -i p^\mu F^a{}_k e^{-ip \cdot x}$$, where ##p^{\mu}## is the on-shell momentum of the boson and ##F^a{}_k## is a matrix of constants. The elements ##F^a{}_k## vanish when ##a## deontes a generator of an unbroken symmetry.""

    I don't understand every single sentence here. Can anybody explain. It is from Peskin&Schroeder 20.1 formula 20.46. Thanks.
     
  2. jcsd
  3. Apr 24, 2013 #2

    DrDu

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    Take for example a ferromagnet. The broken symmetry is specified by the orientation of the magnetization axis. The Goldstone boson is the magnon, an oscillation of the magnetization axis.
    A long wavelength magnon corresponds to an (at least locally) uniform rotation of the magnetization axis. There is some charge (here e.g. the total spin in the z-direction## Q= S_z=\sum_i s_{zi}=\int J^0 dV ##) which induces this rotation.
    The treatment is quite sloppy as the operator ##Q^a## can be shown not to exist. However you can work with an operator ##Q^a_V=\int_V J^{0 a} dV## defined for a finite region V.
     
  4. Apr 24, 2013 #3
    But in ferromagnetic, I so understand, SO(3) is broken into SO(2) so rotation around axis leaves the particles massless and has nothing to do with massless boson emerged because of the symmetry breaking. Am I wrong?

    How about the second sentence?
     
  5. Apr 25, 2013 #4

    DrDu

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    The appearance of Goldstone Bosons has nothing to to with whether the symmetry is completely broken or only broken down to a sub-group.
    If you want to simplify the model consider a ferromagnet in two dimensions (or an anisotropic three dimensional one).
     
  6. Apr 25, 2013 #5
    how about the second and most importantly the third sentence?
     
  7. Apr 25, 2013 #6

    DrDu

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    The second sentence is a rather weak statement. For a general operator we expect non-vanishing matrix elements between almost any states, including the vacuum and states with one Goldstone boson. Furthermore we know explicitly that some integral aka Q of J over V generates long wavelength GBs. So J itself certainly also has non-vanishing matrix elements between the vacuum and 1-GB states.
    Again, I find analogies with non-relativistic systems helpful: E.g. rotation of all spins in a ferromagnet changes one broken symmetry state into another. Rotation of just one spin corresponds to the excitation of a superposition magnons of all wavelengths. You could also consider a crystal where translation symmetry is broken. A translation of the whole crystal transforms one broken symmetry state into another one. Translation of just one atom yields a superposition of acoustic lattice vibrations (phonons) of all wavelengths.


    The third sentence is the most general expression possible with all the indices and being Poincare covariant (or only covariant with respect to the translation group).
    I am sorry if this is still a bit hand-waving, but I am not so familiar with relativistic field theory.
     
  8. Apr 27, 2013 #7
    I understand these: the lagrangian has a symmetry with generators ##\hat{T}^a## and parameters say ##\alpha_a##, this leads to conserved current operators ##\hat{J}^{\mu}_a##; on the other hand we have symmetry breaking, which , according to my understanding, is that you fix the minimum of the potential and make an expansion around it and a massless boson alongside a massive one emerges. Question: how precisely mathematically is ##\hat{J}^{\mu}_a## related to operator of goldstone boson creation ##\hat{a}^\dagger##?
     
  9. Apr 28, 2013 #8

    DrDu

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    In general such a simple relation only exists in the long wavelength limit. E.g. you won't be able to derive a complete phonon spectrum just from the broken translational symmetry.
    In chapter 19.2 of "The quantum theory of fields" by Stephen Weinberg, vol. 2 might contain the answer you are looking for.
     
  10. Apr 28, 2013 #9
    I have to read previous chapters to understand it. :/
     
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