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How to trace over spinor indices?

  1. Nov 21, 2014 #1
    I would like to take the trace over spinorial indices of the following expression:

    [tex](\gamma_{\mu}\gamma^{0})_{\alpha}^{\beta}=(\gamma_{\mu})_{\alpha}^{\gamma}(\gamma^{0})_{\gamma}^{\beta}[/tex].

    How do I go about doing this? I reckon I could expand the trace out (let's say I want to do this in 4D) and use a particular representation of the algebra of the gammas, but is there a representation-independent way of doing it?

    Also, it confuses me that in the equation above, if one traces over it, it becomes [itex](\gamma_{\mu})_{\alpha}^{\beta}(\gamma^{0})_{\beta}^{\alpha}[/itex], where the beta index is summed over properly according to the spinor northwest-southeast convention, while the alphas aren't. Is this a problem?

    I guess my main question is, how do you take a trace over spinor indices?

    Thanks
     
  2. jcsd
  3. Nov 21, 2014 #2

    Orodruin

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    You can start from the anti-commutation relation ##\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu\nu}## and use the cyclic property of the trace.
     
  4. Nov 21, 2014 #3
    So you mean:

    [tex]Tr(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu})=2Tr(g^{\mu\nu})[/tex]

    and the LHS can be expanded as

    [tex]Tr(\gamma^{\mu}\gamma^{\nu})+Tr(\gamma^{\nu}\gamma^{\mu})=Tr(\gamma^{\mu}\gamma^{\nu})+Tr(\gamma^{\mu}\gamma^{\nu})=2Tr(\gamma^{\mu}\gamma^{\nu})[/tex]

    So [itex]Tr(\gamma^{\mu}\gamma^{\nu})=Tr(g^{\mu\nu})=4I[/itex]
     
  5. Nov 21, 2014 #4

    Orodruin

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    Exactly, apart from the last step. The last step is really a trace of the identity matrix multiplied by the metric (##\mu## and ##\nu## are Lorentz indices, not spinor indices) and hence ##tr(g^{\mu\nu}) = g^{\mu\nu} tr(\mathbb 1) = 4g^{\mu\nu}##.
     
  6. Nov 21, 2014 #5
    Of course, you are right. Thanks!
     
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