# Feynman slash identity

by nicksauce
Tags: feynman, identity, slash
 Sci Advisor HW Helper P: 1,275 1. The problem statement, all variables and given/known data I am trying to prove that $$\displaystyle{\not} a \displaystyle{\not} b + \displaystyle{\not} b \displaystyle{\not} a = 2a\cdot b$$ using the relation $$\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}$$ 2. Relevant equations 3. The attempt at a solution If I work backwards, $$2a\cdot b = 2a_{\mu} g^{\mu \nu} b_{\nu} = a_{\mu}(\gamma^{\mu}\gamma^{\nu})b_{\nu} + a_{\mu}(\gamma^{\nu}\gamma^{\mu})b_{\nu}$$ The first term is $$\displaystyle{\not} a \displaystyle{\not} b$$ but the second term doesn't seem to look like $$\displaystyle{\not} b \displaystyle{\not} a$$. Am I missing something here?
 Sci Advisor P: 1,588 You're missing the fact that $a_\mu$ and $b_\nu$ are ordinary numbers, and so commute with everything.