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Physical interpretation of V and A couplings

  1. Mar 16, 2009 #1
    Hi,

    I'm trying to remind myself of exactly what, physically, is the difference between V and A couplings. Now, a vector coupling is of the form [tex]\bar{\psi}\gamma^\mu\psi[/tex], and axial coupling of the form [tex]\bar{\psi}\gamma^\mu\gamma^5\psi[/tex]. Thinking in terms of a chiral fermion expanded as:

    [tex]
    f = \left[\left(\frac{1-\gamma^5}{2}\right) + \left(\frac{1+\gamma^5}{2}\right)\right]\psi
    [/tex]

    and where [tex]\bar{f} = \gamma^\dagger\gamma^0[/tex], I assume the difference between the V and A couplings has to do with how the L and R projection operators commute through either [tex]\gamma^\mu[/tex] or [tex]\gamma^\mu\gamma^5[/tex] from the 'coupling' term, and the [tex]\gamma^0[/tex] from the conjugate field term in a given Lagrangian.

    However, as [tex][\gamma^5,\gamma^5] = 0[/tex], I don't see how a difference in the chiral treatment between V and A couplings can arise. Am I barking up completely the wrong tree? Any insight welcomed!
     
  2. jcsd
  3. Mar 16, 2009 #2

    Vanadium 50

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    You can always redefine what you mean by "V" and "A", as purely vector or purely axial currents conserve parity. It's the mixed terms that generate parity violation.
     
  4. Mar 16, 2009 #3
    So, a mixed term is something like [tex]\bar{\psi}\gamma^\mu\left(1-\gamma^5\right)\psi[/tex]? How can it be shown that this term violates parity?

    Cheers.
     
  5. Mar 16, 2009 #4

    clem

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    Gamma 5 is a pseudoscalar. Adding a pseudoscalar to a scalar violates parity.
    When used in beta decay, the V-A interaction leads to negative helicity for outgoing leptons. This violates parity.
     
  6. Mar 16, 2009 #5
    Sure, but I'm interested in showing why this is the case mathematically...
     
  7. Mar 17, 2009 #6

    Vanadium 50

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    Given [itex]P|\psi> = (-1)^n|\psi>[/itex]

    Then [itex]P(a + b\gamma^5)|\psi> = aP|\psi> + bP \gamma^5|\psi> [/itex]

    Which is [itex]\left( (-1)^na + (-1)^{n+1}b \right) |\psi>[/itex], so is only an eigenvalue of parity if [itex]a=0[/itex] or [itex]b=0[/itex].
     
  8. Mar 17, 2009 #7
    Ah yes, very elegant. Thanks!
     
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