Proof of traceless gamma matrices

  • #1

Main Question or Discussion Point

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

http://en.wikipedia.org/wiki/Gamma_matrices

(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]

thus

[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:

Answers and Replies

  • #2
Ah it seems like it was assumed that [tex]\mu \neq \nu[/tex].
 
  • #3
470
58
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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