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The Dirac Equation

  1. Jan 3, 2008 #1
    [SOLVED] The Dirac Equation

    I'm trying to understand the following property of the Dirac equation:

    [tex](i \gamma^{\mu}\partial_{\mu} - m)\Psi(x) = 0[/tex]

    Acting twice with [tex](i \gamma^{\mu}\partial_{\mu} - m)[/tex]:

    [tex](i \gamma^{\mu}\partial_{\mu} - m)^{2} \Psi(x) = 0[/tex]

    [tex] = [ - \gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} - 2im\gamma^{\mu}\partial_{\mu} + m^{2}]\Psi = 0[/tex]

    But then somehow the book jumps to this step:

    [tex] = [ 1/2 \left{\{\gamma^{\mu}, \gamma^{\nu}\right}\} \partial_{\mu}\partial_{\nu} + m^{2}]\Psi = 0[/tex]

    And I have no idea how it got there! I understand the { } denote an anticommutator, but I just can't see how the factor of 1/2 has appeared, where the minus has gone and where the middle term has gone. Help!!!
     
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  3. Jan 3, 2008 #2

    Dick

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    [tex]\gamma^\mu \gamma^\nu \partial_\mu \partial _\nu=\gamma^\nu \gamma^\mu \partial_\mu \partial _\nu[/tex]. Add those and divide by two, to get the anticommutator. Let the middle term act on [tex]\Psi[/tex] and use the dirac equation to see where it has gone. The minus will become an overall minus once you do all of this.
     
  4. Jan 3, 2008 #3
    Ah, I understand now why the commutator arises, thanks Dick. However, I'm still confused about the middle term business. If I understand you right, you're saying that:

    [tex](-2im\gamma^{\mu}\partial_{\mu})\Psi[/tex]

    is analagous to:

    [tex](i\gamma^{\mu}\partial_{\mu} - m)\Psi = 0[/tex]

    and hence should disappear? I don't quite understand that part...
     
  5. Jan 3, 2008 #4

    Dick

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    It doesn't disappear. It's equal to -2m^2*psi. Do you see why?
     
  6. Jan 3, 2008 #5
    ^ Ah! Got it, [tex]i\gamma^{\mu}\partial_{\mu} = m[/tex] from the original eq. Yes, I see now! Thanks so much, Dick!
     
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