Feynman slash identity


by nicksauce
Tags: feynman, identity, slash
nicksauce
nicksauce is offline
#1
Nov23-09, 03:49 PM
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1. The problem statement, all variables and given/known data
I am trying to prove that [tex]\displaystyle{\not} a \displaystyle{\not} b + \displaystyle{\not} b \displaystyle{\not} a = 2a\cdot b[/tex] using the relation [tex]\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}[/tex]


2. Relevant equations



3. The attempt at a solution
If I work backwards,
[tex]
2a\cdot b = 2a_{\mu} g^{\mu \nu} b_{\nu}
= a_{\mu}(\gamma^{\mu}\gamma^{\nu})b_{\nu} + a_{\mu}(\gamma^{\nu}\gamma^{\mu})b_{\nu}[/tex]

The first term is [tex]\displaystyle{\not} a \displaystyle{\not} b[/tex] but the second term doesn't seem to look like [tex]\displaystyle{\not} b \displaystyle{\not} a[/tex]. Am I missing something here?
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Ben Niehoff
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#2
Nov23-09, 04:31 PM
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P: 1,563
You're missing the fact that [itex]a_\mu[/itex] and [itex]b_\nu[/itex] are ordinary numbers, and so commute with everything.


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