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Dirac Matrix Property? Possible Book mistake?

  1. Jul 16, 2012 #1
    Dirac Matrix Property? Possible Book mistake? Derive KG from Dirac

    I got a copy of QFT demystified and on pg. 89 he says he can write [itex]\gamma_{\nu} \gamma^{\mu} = g_{\nu \sigma} \gamma^{\sigma} \gamma^{\mu} = g_{\nu \sigma} \frac{1}{2} (\gamma^{\sigma} \gamma^{\mu} + \gamma^{\mu} \gamma^{\sigma}) [/itex]

    and i am trying to figure out why this is because the only reason I could see why it's true is if [itex]\gamma^{\mu} \gamma^{\nu} = \gamma^{\nu} \gamma^{\mu}[/itex] which for the love of my brain I can't figure out why that would be true, I'm pretty sure it's not. Is this a book mistake. For reference what he is doing is deriving the KG equation starting from Dirac.

    on another note, regardless of the answer what i am actually looking for is a derivation of the kg equation starting from dirac, or perhaps the other way around. if someone can point me to that, that is a fine answer as well.
    Last edited: Jul 16, 2012
  2. jcsd
  3. Jul 16, 2012 #2
    Just guessing here, but isn't [itex]g_{\mu \sigma}[/itex] symmetric, so any antisymmetric terms would cancel?
  4. Jul 16, 2012 #3
    i don't see what you mean
  5. Jul 16, 2012 #4
    Yeah, I think what I was thinking isn't relevant after all. If anything, I had thought the gammas anticommute.
  6. Jul 16, 2012 #5
    Re: Dirac Matrix Property? Possible Book mistake? Derive KG from Dirac

    To prove the formula, you just need to sum over the dummy indices in the right-hand side: remember that the metric tensor [itex]g_{\nu \sigma} [/itex] lowers indices.
  7. Jul 16, 2012 #6


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    I had a look at that book, he's being careless in writing that equation. The important thing is that the term he wants to simplify can be written as ##\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu##. The derivatives here are symmetric in ##\mu\nu##, so we want to compute the symmetric part of

    $$ \gamma^\mu \gamma^\nu = \frac{1}{2} \{ \gamma^\mu, \gamma^\nu\} + \frac{1}{2} [\gamma^\mu ,\gamma^\nu].$$

    The first term is symmetric, while the second, commutator, part is antisymmetric. The antisymmetric part vanishes when we sum against ##\partial_\mu \partial_\nu##.

    The formula in your OP does not hold in general, only in a sum against a symmetric object.

    As for deriving the KG equation from the Dirac eq, the method in this book is fine as long as you realize the sloppiness. Usually, we just note that, from

    $$ (i\gamma^\mu \partial_\mu - m) \psi =0,$$

    we can just compute

    $$ 0 = (i\gamma^\mu \partial_\mu + m) (i\gamma^\mu \partial_\mu - m) \psi = - (\partial^\mu\partial_\mu + m^2 ) \psi,$$

    which is the KG equation. This is entirely equivalent to the derivation given in the book.
  8. Jul 16, 2012 #7
    I am sorry, I was not careful enough: indeed, only the first equality is correct.
  9. Jul 16, 2012 #8
    okie doke, thanks frank.
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