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Proof of traceless gamma matrices

  1. Sep 14, 2012 #1
    Hi I'm trying to figure out the proof of why the gamma matrices are traceless. I found a proof at wikipedia under 'trace identities' here


    (it's the 0'th identity)
    and from the clifford algebra relation

    [tex] \{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu \nu}[/tex]

    one get that

    [tex] \frac{\gamma^\mu \gamma^\mu}{\eta^{\mu \mu}} = I[/tex]


    [tex] Tr(\gamma^\nu) = \frac{1}{\eta^{\mu \mu}} Tr(\gamma^\nu \gamma^\mu \gamma^\mu) = \frac{1}{\eta^{\mu \mu}} Tr(\{\gamma^\nu, \gamma^\mu\} \gamma^\mu - \gamma^\mu \gamma^\nu \gamma^\mu)[/tex]

    and here it seems like they set

    [tex]\frac{1}{\eta^{\mu \mu}} Tr(\{\gamma^\nu, \gamma^\mu\} \gamma^\mu) = \frac{1}{\eta^{\mu \mu}} Tr(2 \eta^{\nu \mu} \gamma^\mu ) = 0.[/tex]

    But why is this true?
    Last edited: Sep 14, 2012
  2. jcsd
  3. Sep 14, 2012 #2
    Ah it seems like it was assumed that [tex]\mu \neq \nu[/tex].
  4. Sep 15, 2012 #3
    If you assume $$\mu\neq \nu$$ then you have $$\gamma^{\nu}\gamma^{\mu}=-\gamma^{\mu}\gamma^{\nu}$$ and so you have (using the invariance of the trace under cyclic permutations of the matrices):

    $$Tr(\gamma^{\nu})=\frac{1}{\eta^{\mu\mu}}Tr( \gamma^{\nu} \gamma^\mu \gamma^\mu)=\frac{1}{\eta^{\mu\mu}}Tr( \gamma^\mu \gamma^\nu \gamma^\mu)$$

    You can now anticommutate and thus obtain:

    $$-\frac{1}{\eta^{\mu\mu}}Tr(\gamma^\nu \gamma^\mu \gamma^\mu)=-Tr(\gamma^\nu) \Rightarrow Tr(\gamma^\nu)=-Tr(\gamma^\nu)=0$$
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