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Gamma matrices

  • Thread starter WarnK
  • Start date
  • #1
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Homework Statement


Show that [tex]tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = 0[/tex]


Homework Equations


(anti-)commutation rules for the gammas, trace is cyclic


The Attempt at a Solution


I can do
[tex] tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = -tr(\gamma^{\mu}\gamma^{5}\gamma^{\nu}) = - tr(\gamma^{\nu}\gamma^{\mu}\gamma^{5}) [/tex]
and so on, but I don't see how that helps me. Any suggestions?
 

Answers and Replies

  • #2
Avodyne
Science Advisor
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You've used the first of your "relevant equations", now use the second.
 
  • #3
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If I continue like this
[tex] tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = - tr(\gamma^{\nu}\gamma^{\mu}\gamma^{5}) = - tr( (2\eta^{\nu \mu} - \gamma^{\mu}\gamma^{\nu} )\gamma^{5}) = tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) - tr(2\eta^{\nu \mu} \gamma^{5}) [/tex]
which I guess means that [tex]tr(\gamma^{5})=0[/tex]
 
  • #4
Avodyne
Science Advisor
1,396
86
Oops, I didn't notice that you've already gotten as far as you need to. You've shown that your original expression is equal to minus itself. What is the only number with this property?
 
  • #5
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But, I have [tex]\gamma^{\mu}[/tex] and [tex]\gamma^{\nu}[/tex] in a different order, and the entire point is that they don't commute?
 
Last edited:
  • #6
Avodyne
Science Advisor
1,396
86
Oops, sorry, I screwed up again.

To do this one, I think you need to insert (gamma^rho)^2 (which is either one or minus one, depending on your conventions and on whether or not rho=0) into the trace,
with rho not equal to mu or nu, and then move one of the gamma^rho's around by commutation.
 

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