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Dirac equation with gamma_5 mass term?

  1. Oct 22, 2008 #1
    During my research a while ago, I have unexpectedly derived a "modified Dirac equation" with a [tex]\gamma_{5}[/tex] mass term.

    [tex]
    (\gamma^{\mu}\partial_{\mu}+\gamma^{5}m)\psi(x)=0
    [/tex]

    I was quite surprised, and went about asking a few people. The answer I got is this equation is not new and has been studied by Sakurai in the context of parity violation. Can anyone lead me to some relevant sources? Of course any comment is helpful.
     
  2. jcsd
  3. Oct 23, 2008 #2

    samalkhaiat

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  4. Oct 23, 2008 #3

    Avodyne

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    Your equation is actually equivalent to the Dirac equation. To see this, define a new field [itex]\Psi(x)[/itex] via [itex]\psi(x)=\exp(i\alpha\gamma_5)\Psi(x)[/itex], where [itex]\alpha[/itex] is a real parameter (specified later). Then, multiply your equation by [itex]i\exp(i\alpha\gamma_5)[/itex]. Now we have
    [tex]ie^{i\alpha\gamma_5}\gamma^{\mu}e^{i\alpha\gamma_5}\partial_{\mu}\Psi+ie^{i\alpha\gamma_5}\gamma_{5} e^{i\alpha\gamma_5}m\Psi=0.[/tex]
    Since [itex]\gamma_5[/itex] anitcommutes with [itex]\gamma^\mu[/itex], we have [tex]e^{i\alpha\gamma_5}\gamma^{\mu}=\gamma^\mu e^{-i\alpha\gamma_5}[/tex], while we also have [tex]e^{i\alpha\gamma_5}\gamma_5=\gamma_5 e^{i\alpha\gamma_5}[/tex]. This allows us to simplify the equation to
    [tex]i\gamma^{\mu}\partial_{\mu}\Psi+i\gamma_{5} e^{2i\alpha\gamma_5}m\Psi=0.[/tex]
    Since [itex]\gamma_5^2=1[/itex], we also have
    [tex]e^{2i\alpha\gamma_5}=\cos(2\alpha)+i\sin(2\alpha)\gamma_5.[/tex]
    If we now choose [itex]\alpha=\pi/4[/itex], we get
    [tex]e^{2i\alpha\gamma_5}=i\gamma_5.[/tex]
    For this choice of [itex]\alpha[/itex], the equation becomes
    [tex]i\gamma^{\mu}\partial_{\mu}\Psi-m\Psi=0,[/tex]
    which is the usual Dirac equation.
     
  5. Oct 23, 2008 #4
    Thanks Avodyne, this is very nice indeed. I recall seeing such a gauge transformation involving [tex]\gamma_{5}[/tex] somewhere.

    Are there any physical significance to such a transformation?
     
  6. Oct 24, 2008 #5

    Haelfix

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    Not really, its just a straight forward field redefinition, consistent with the gauge symmetries of the system..

    Is it a useful gauge to see anything nontrivial about the system? I don't know, its not apparent to me.
     
  7. Oct 24, 2008 #6

    Avodyne

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    This transformation is a "chiral rotation". It has no physical significance in QED. In QCD, it would change the vacuum angle. This is discussed in a number of books, e.g., Srednicki.
     
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