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Pion as a goldstone boson

  1. Oct 7, 2011 #1
    I'm reading Zee book on quantum field theory.
    He wants to explain that pion is the goldstone boson arising from the spontaneous simmetry breaking of the chiral symmetry.
    So he
    start with the weak decay
    [itex]\pi^{-} \rightarrow \bar{\nu} + e^{-} [/itex]
    and write this equation
    [itex] <0|J^{\mu}_{5}|k>=f k^{\mu}[/itex] (1),
    where [itex] k [/itex] is the momentum of the pion.

    Then, of course, if you act with [itex] k_{\mu} [/itex] on the left, you get
    [itex] k_{\mu }<0|J^{\mu}_{5}|k>=f m _{\pi} [/itex] (2)
    and we see that if we consider the pion as a massless spin 0 particle, it is a good candidate for a goldstone boson associated with the spontaneous breaking of the chiral symmetry, because it then follows that
    [itex] \partial_{\mu} J_{5}^{\mu} = 0 [/itex]

    My questions are:
    -why in (1) he wrote only the axial current and he doesn't write something like
    [itex] <0|J^{\mu}-J^{\mu}_{5}|k> [/itex]
    ?
    -why the right hand side of (1) is just a vector while the left hand side is a pseudovector?

    Thank you!
     
    Last edited: Oct 7, 2011
  2. jcsd
  3. Oct 7, 2011 #2
    The answer to my second question may be that under spatial riflession [itex] \pi [/itex] is a pseudo scalar and thus, the first member is a vector too. And so follows the answer to the first question: the expectation value of the vector part of the current vanishes because it's a pseudovector and it's impossible to form a pesudo vector with the only vector k.
     
    Last edited: Oct 7, 2011
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