Physics Asked by A quarky name on November 18, 2020
I was trying to prove the identity $overline{displaystyle{not}{a}displaystyle{not}{b}dots displaystyle{not}{p}} = displaystyle{not}{p}dots displaystyle{not}{b}displaystyle{not}{a}$. On simplifying the LHS I end up with $ overline{ displaystyle{not}{p}^{}} dots overline{ displaystyle{not}{b}^{}} overline{ displaystyle{not}{a}^{}} $.
I’m wondering if $ overline{ displaystyle{not} p} = gamma^0 displaystyle{not} p^{dagger} gamma^0 = displaystyle{not} p$ as this would let me complete the final step.
Usually, gamma-matrices are chosen in such a way that they satisfy the hermiticity condition $gamma^{mudagger}=gamma^0gamma^mugamma^0$, and that is what you need, but this choice is not mandatory.
Answered by akhmeteli on November 18, 2020
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