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Feynman slash identity

  1. Nov 23, 2009 #1

    nicksauce

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    1. The problem statement, all variables and given/known data
    I am trying to prove that [tex]\displaystyle{\not} a \displaystyle{\not} b + \displaystyle{\not} b \displaystyle{\not} a = 2a\cdot b[/tex] using the relation [tex]\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    If I work backwards,
    [tex]
    2a\cdot b = 2a_{\mu} g^{\mu \nu} b_{\nu}
    = a_{\mu}(\gamma^{\mu}\gamma^{\nu})b_{\nu} + a_{\mu}(\gamma^{\nu}\gamma^{\mu})b_{\nu}[/tex]

    The first term is [tex]\displaystyle{\not} a \displaystyle{\not} b[/tex] but the second term doesn't seem to look like [tex]\displaystyle{\not} b \displaystyle{\not} a[/tex]. Am I missing something here?
     
  2. jcsd
  3. Nov 23, 2009 #2

    Ben Niehoff

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    You're missing the fact that [itex]a_\mu[/itex] and [itex]b_\nu[/itex] are ordinary numbers, and so commute with everything.
     
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