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Gamma matrix traceless proof

  1. Oct 18, 2014 #1
    I'm reading through some lecture notes and there is a proof that the gamma matrices are traceless that I've never seen before (I've seen the "identity 0" on wikipedia proof) and I can't work out some of the steps:

    \begin{align*}
    2\eta_{\mu\nu}Tr(\gamma_\lambda) &= Tr(\{\gamma_{\mu},\gamma_{\nu}\}\gamma_\lambda)
    \\ &= Tr(\gamma_\mu\gamma_\nu\gamma_\lambda + \gamma_\nu\gamma_\mu\gamma_\lambda)
    \\ &= Tr(\gamma_\mu\gamma_\nu\gamma_\lambda + \gamma_\mu\gamma_\lambda\gamma_\nu)
    \\ &= Tr(\gamma_\mu\{\gamma_{\nu},\gamma_{\lambda}\})
    \\ &= 2\eta_{\nu\lambda}Tr(\gamma_\mu)
    \\ \mu = \nu \neq \lambda \implies Tr(\gamma_\lambda) = 0
    \end{align*}

    In particular I don't understand the very first equality and the very last (I assume the same method is being used in both) but I understand the rest using the trace and anti commutator properties. I understand that the 2 eta factor is equal to the anti commutator but I don't see how this allows you to pull it inside of the trace, I tried to work it out explicitly using a generic 4x4 matrix for the gammas but I can't get it.

    Thank you in advance for any help
     
  2. jcsd
  3. Oct 18, 2014 #2

    Orodruin

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    The identity used is ##\{\gamma_\mu,\gamma_\nu\} = 2\eta_{\mu\nu}##. The ##2\eta_{\mu\nu}## has first been moved into the trace which is allowed since it is a number.
     
  4. Oct 18, 2014 #3
    I think I must have misunderstood something, I thought eta was a matrix?

    eta=diag(-1,1,1,1)
     
  5. Oct 18, 2014 #4

    Orodruin

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    It is the metric tensor, the indices belonging to it are not the ones the trace is over.
     
  6. Oct 18, 2014 #5

    If it's a tensor not a scalar why can we put it inside the trace? Sorry if I'm missing something obvious. I thought the indices on the eta referred to the elements in eta and the trace is the sum of the diagonal elements and so the trace would also apply to the eta?
     
  7. Oct 18, 2014 #6

    Orodruin

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    The trace is over the gamma matrices. The indices in this particular case are not summation indices and so each set of indices refer to a particular relation.
     
  8. Oct 18, 2014 #7
    I really appreciate your help, thanks for your time but I'm still struggling to understand. I understand that the indices aren't being summed over, rather that the sum of the diagonal part of the gamma matrix is the trace. I'm still not understanding why the eta matrix can be moved inside the trace as this changes it from a matrix to just a scalar once the trace is taken.
     
  9. Oct 18, 2014 #8

    Orodruin

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    For given μ and ν, the component of the metric is just a number that you can move into the trace. The anti-commutation relation between the gamma matrices allows you to exchange this number for an anti-commutator of two matrices (in the anti-commutation relation, the metric should really be multiplied by an identity matrix in the gamma matrix space). Remember that ##\gamma^\mu## for a fixed μ is a matrix, this is the matrix space you are going to take the trace in. The trace of ##\gamma^\mu## is simply ##\sum_a \gamma^\mu_{aa}## so the trace has nothing to do with the Lorentz indices.
     
  10. Oct 18, 2014 #9
    Thank you very much, this really clears it up for me, you've been very helpful. So just to check, on the lhs we are thinking of eta not as a whole matrix, but are instead just considering the specific components given by mu and nu?
     
  11. Oct 18, 2014 #10

    Orodruin

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    Yes. In general, in high-energy physics you will have a lot of different indices floating around, often suppressed for readability. It is important to know which indices are intended with traces, matrix multiplications, and other operations.
     
  12. Oct 18, 2014 #11
    Brilliant, thank you very much.
     
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