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Question regarding Dirac matrices

  1. Feb 10, 2005 #1
    Hey there.

    In an exercise I was trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation.

    Now I have two very simple questions:

    Is [tex]\gamma_\nu \gamma^\mu = \gamma^\nu \gamma_\mu[/tex] ?

    And is [tex]\partial_\nu \partial^\mu = \partial^\nu \partial_\mu[/tex] ?

    Thanks for the help.
     
  2. jcsd
  3. Feb 10, 2005 #2

    dextercioby

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    Not really.U should pay attention with this "Lorentz covariance" and always try to see these expressions as tensor products of 4-vectors.

    Daniel.
     
  4. Feb 10, 2005 #3
    I kinda felt it couldn't be that simple. :p

    Thanks anyway.
     
  5. Feb 10, 2005 #4
    Can anybody verify this?

    [tex]\gamma^\nu\partial_\nu\gamma_\mu\partial^\mu = \partial_\mu\partial^\mu[/tex]

    Just as clarification: I am trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation. So I started with the Dirac equation:

    [tex](i\gamma_\mu\partial^\mu - m)\Psi(x) = 0[/tex]

    Then, I multiplied it with the operator [tex]-(i\gamma^\nu\partial_\nu + m)[/tex] (from left) which gave me:

    [tex](\gamma^\nu\partial_\nu\gamma_\mu\partial^\mu + m^2)\Psi(x) = 0[/tex]

    This surely looks somewhat like the Klein-Gordan equation if the above assumption is correct. I just fail to see why it is true.

    Any hints appreciated!
     
    Last edited: Feb 10, 2005
  6. Feb 10, 2005 #5

    dextercioby

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    Yes,it can be shown easily once you manipulate the Lorentz indices and the anticommutation relation between the gamma's.
    [tex] \gamma_{\nu}\partial^{\nu}\gamma_{\mu}\partial^{\mu}=\gamma_{\nu}\gamma_{\mu}\partial^{\nu}\partial^{\mu}=\frac{1}{2}\gamma_{(\nu}\gamma_{\mu)}\partial^{\nu}\partial^{\mu}=g_{\mu\nu}\partial^{\nu}\partial^{\mu}=d'alembertian [/tex]

    Daniel.

    P.S.Did u follow the reasoning...?
     
  7. Feb 10, 2005 #6

    dextercioby

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    Why's the KG equation:

    [tex] (\partial_{\mu}\partial^{\mu}-m^{2})\Psi (\vec{r},t) =0 [/tex]

    ,instead of:

    [tex] -(\partial_{\mu}\partial^{\mu}+m^{2})\Psi (\vec{r},t)=0 [/tex]

    What metric are u using...?

    Daniel.
     
  8. Feb 10, 2005 #7
    Sorry, it was just a typo. I edited my post. Thanks for all the help. I think I got it now...

    Start with the Dirac equation:

    [tex](i\gamma_\mu\partial^\mu - m)\Psi(x) = 0[/tex]

    Multiply the operator [tex]-(i\gamma_\mu\partial^\mu + m)[/tex] (from left):

    [tex](\gamma_\mu\partial^\mu\gamma_\mu\partial^\mu + m^2)\Psi(x) = (\gamma_\mu\gamma_\mu\partial^\mu\partial^\mu + m^2)\Psi(x) = 0[/tex]

    Because of the anticommutation relation [tex]\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu\nu}[/tex] it is possible to simplify [tex]\gamma_\mu\gamma_\mu[/tex] to [tex]g_{\mu\mu}[/tex] which proves [tex]\gamma_\mu\gamma_\mu\partial^\mu\partial^\mu = \partial^\mu\partial_\mu[/tex].

    Thanks again.
     
    Last edited: Feb 10, 2005
  9. Feb 10, 2005 #8

    dextercioby

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    Note that u have 4 identical (Lorentz) indices in the same tensor equation and that's not good,not good at all.
    Pay attention with these indices,as u might confuse them.

    Daniel.
     
  10. Feb 10, 2005 #9
    Hm, that's true. Do you know a better way? I'm highly interested in alternative solutions.
     
  11. Feb 10, 2005 #10
    To be honest, I don't understand what you're doing here (step 2 -> step 3)
     
  12. Feb 10, 2005 #11

    dextercioby

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    You multiply with the same operator,but with index "nu" (instead of "mu") and then use the trick i've used in post #5.It should come out nicely.


    Daniel.
     
  13. Feb 10, 2005 #12

    dextercioby

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    [tex] \gamma_{\nu}\gamma_{\mu}T^{\nu\mu}=\frac{1}{2}(\gamma_{\nu}\gamma_{\mu}+\gamma_{\mu}\gamma_{\nu})T^{\mu\nu} [/tex]

    where T is a symmetric second rank tensor and to get the second term from the bracket i turned "mu"--->"nu" and "nu"---->"mu" and to factor the same tensor i used the fact that under this renotation,the tensor T is symmetric.

    Daniel.
     
  14. Feb 10, 2005 #13
    This surely looks slick! :) But why is [tex]\partial^\nu\partial^\mu[/tex] a symmetric second rank tensor?
     
  15. Feb 10, 2005 #14

    dextercioby

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    Because wrt bosonic coordinates,the partial derivatives commute.

    Daniel.
     
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