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Physical interpretation of Lorentz invariant fermion field product?

  1. Nov 24, 2009 #1
    Hey all!
    Just a very short question: May I interpret the Lorenz invariant quantity

    [tex]\bar\psi\psi[/tex]

    as being the probability density of a fermion field? Thanks!
    Blue2script
     
  2. jcsd
  3. Nov 24, 2009 #2

    Demystifier

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    No, because it is not positive.
     
  4. Nov 24, 2009 #3
    Hmm.. right. But then, what is the physical interpretation of the product above? And what is the probability density of a fermion field?
     
  5. Nov 24, 2009 #4
    as far as i know there is no propability interpretation of fermion fields only charge density.
     
  6. Nov 24, 2009 #5
    the interpretation of [tex]\bar\psi\psi[/tex] is that it transforms under lorentz boosts like a scalar.
     
  7. Nov 25, 2009 #6

    Demystifier

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    Fermion field is an operator, so it does not have a probabilistic interpretation. However, one should distinguish the field operator from the wave function which is a c-number function representing a quantum state. For a 1-particle wave function [tex]\psi[/tex], the probability density is
    [tex]\psi^{\dagger}\psi[/tex]
     
  8. Nov 25, 2009 #7
    you mean [tex]\psi^{*}\psi[/tex] because [tex]\psi[/tex] is a scalar ? but eitherway its okay to write it with a dagger.
     
  9. Nov 26, 2009 #8

    Demystifier

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    I mean dagger because psi a spinor, i.e., a 4-component wave function.
     
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