Appendix C Trace identities

As we have seen in Section 3, we often will have to calculate traces of γ\gamma matrices. These can be calculated independently of the representation as 4×44\times 4 matrices using just (18). In this appendix, we will list number of identities and then prove some of them

tr⁢[γμ]\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\Big{]} =0,\displaystyle=0\,, (229a)
tr⁢[γ5]\displaystyle{\rm tr}\Big{[}\gamma^{5}\Big{]} =0,\displaystyle=0\,, (229b)
tr⁢[γμ1⁢⋯⁢γμn⏟odd]\displaystyle{\rm tr}\Big{[}\underbrace{\gamma^{\mu_{1}}\cdots\gamma^{\mu_{n}}% }_{\text{odd}}\Big{]} =0,\displaystyle=0\,, (229c)
tr⁢[γ5⁢γμ1⁢⋯⁢γμn⏟odd]\displaystyle{\rm tr}\Big{[}\gamma^{5}\underbrace{\gamma^{\mu_{1}}\cdots\gamma% ^{\mu_{n}}}_{\text{odd}}\Big{]} =0,\displaystyle=0\,, (229d)
tr⁢[γμ⁢γν]\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\Big{]} =4⁢gμ⁢ν,\displaystyle=4g^{\mu\nu}\,, (229e)
tr⁢[γ5⁢γμ⁢γν]\displaystyle{\rm tr}\Big{[}\gamma^{5}\gamma^{\mu}\gamma^{\nu}\Big{]} =0,\displaystyle=0\,, (229f)
tr⁢[γμ⁢γν⁢γρ⁢γσ]\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{% \sigma}\Big{]} =4⁢gμ⁢ν⁢gρ⁢σ−4⁢gμ⁢ρ⁢gν⁢σ+4⁢gμ⁢σ⁢gν⁢ρ.\displaystyle=4g^{\mu\nu}g^{\rho\sigma}-4g^{\mu\rho}g^{\nu\sigma}+4g^{\mu% \sigma}g^{\nu\rho}\,. (229g)

To proof these, let us start with γν⁢γν=4\gamma^{\nu}\gamma_{\nu}=4 (cf. tutorial questions) and write

tr⁢[γμ]=14⁢tr⁢[γμ⁢γν⁢γν]=14⁢tr⁢[(2⁢gμ⁢ν−γν⁢γμ)⁢γν]=12⁢tr⁢[γμ]−14⁢tr⁢[γν⁢γμ⁢γν],\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\Big{]}=\frac{1}{4}{\rm tr}\Big{[}% \gamma^{\mu}\gamma^{\nu}\gamma_{\nu}\Big{]}=\frac{1}{4}{\rm tr}\Big{[}(2g^{\mu% \nu}-\gamma^{\nu}\gamma^{\mu})\gamma_{\nu}\Big{]}=\frac{1}{2}{\rm tr}\Big{[}% \gamma^{\mu}\Big{]}-\frac{1}{4}{\rm tr}\Big{[}\gamma^{\nu}\gamma^{\mu}\gamma_{% \nu}\Big{]}\,, (230)

where we have used the anti-commutator. Next, we write

tr⁢[γμ]=−12⁢tr⁢[γν⁢γμ⁢γν]=−12⁢tr⁢[γν⁢γν⁢γμ]=−2⁢t⁢r⁢[γμ],\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\Big{]}=-\frac{1}{2}{\rm tr}\Big{[}% \gamma^{\nu}\gamma^{\mu}\gamma_{\nu}\Big{]}=-\frac{1}{2}{\rm tr}\Big{[}\gamma_% {\nu}\gamma^{\nu}\gamma^{\mu}\Big{]}=-2{\rm tr}\Big{[}\gamma^{\mu}\Big{]}\,, (231)

where we have used the cyclicity of the trace. Therefore, the trace must be zero.

Next, let us prove (229e)

tr⁢[γμ⁢γν]=∗12⁢(tr⁢[γμ⁢γν]+tr⁢[γν⁢γμ])=12⁢tr⁢[{γμ,γν}]=†gμ⁢ν⁢tr⁢[1]=4⁢gμ⁢ν.\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\Big{]}\stackrel{{% \scriptstyle*}}{{=}}\frac{1}{2}\bigg{(}{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}% \Big{]}+{\rm tr}\Big{[}\gamma^{\nu}\gamma^{\mu}\Big{]}\bigg{)}=\frac{1}{2}{\rm tr% }\Big{[}\big{\{}\gamma^{\mu},\gamma^{\nu}\big{\}}\Big{]}\stackrel{{% \scriptstyle{\dagger}}}{{=}}g^{\mu\nu}{\rm tr}\Big{[}1\Big{]}=4g^{\mu\nu}\,. (232)

Here we have used the cyclicity at ∗* and the anti-commutator (18) at †{\dagger}.

For (229g), we follow the same procedure

tr⁢[γμ⁢γν⁢γρ⁢γσ]\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{% \sigma}\Big{]} =∗2⁢gρ⁢σ⁢tr⁢[γμ⁢γν]−tr⁢[γμ⁢γν⁢γσ⁢γρ]=(⁢229e⁢)8⁢gρ⁢σ⁢gμ⁢ν−tr⁢[γμ⁢γν⁢γσ⁢γρ]\displaystyle\stackrel{{\scriptstyle*}}{{=}}2g^{\rho\sigma}{\rm tr}\Big{[}% \gamma^{\mu}\gamma^{\nu}\Big{]}-{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\gamma^% {\sigma}\gamma^{\rho}\Big{]}\stackrel{{\scriptstyle\eqref{eq:trace:2}}}{{=}}8g% ^{\rho\sigma}g^{\mu\nu}-{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}% \gamma^{\rho}\Big{]}
=†8⁢gρ⁢σ⁢gμ⁢ν−2⁢gσ⁢ν⁢tr⁢[γμ⁢γρ]+tr⁢[γμ⁢γσ⁢γν⁢γρ]=(⁢229e⁢)8⁢gρ⁢σ⁢gμ⁢ν−8⁢gσ⁢ν⁢gμ⁢ρ+tr⁢[γμ⁢γσ⁢γν⁢γρ]\displaystyle\stackrel{{\scriptstyle{\dagger}}}{{=}}8g^{\rho\sigma}g^{\mu\nu}-% 2g^{\sigma\nu}{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\rho}\Big{]}+{\rm tr}\Big{[}% \gamma^{\mu}\gamma^{\sigma}\gamma^{\nu}\gamma^{\rho}\Big{]}\stackrel{{% \scriptstyle\eqref{eq:trace:2}}}{{=}}8g^{\rho\sigma}g^{\mu\nu}-8g^{\sigma\nu}g% ^{\mu\rho}+{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\sigma}\gamma^{\nu}\gamma^{\rho}% \Big{]}
=#8⁢gρ⁢σ⁢gμ⁢ν−8⁢gσ⁢ν⁢gμ⁢ρ+2⁢gμ⁢σ⁢tr⁢[γν⁢γρ]−tr⁢[γσ⁢γμ⁢γν⁢γρ]\displaystyle\stackrel{{\scriptstyle\#}}{{=}}8g^{\rho\sigma}g^{\mu\nu}-8g^{% \sigma\nu}g^{\mu\rho}+2g^{\mu\sigma}{\rm tr}\Big{[}\gamma^{\nu}\gamma^{\rho}% \Big{]}-{\rm tr}\Big{[}\gamma^{\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}% \Big{]}
=(⁢229e⁢)8⁢gρ⁢σ⁢gμ⁢ν−8⁢gσ⁢ν⁢gμ⁢ρ+8⁢gμ⁢σ⁢gν⁢ρ−tr⁢[γσ⁢γμ⁢γν⁢γρ].\displaystyle\stackrel{{\scriptstyle\eqref{eq:trace:2}}}{{=}}8g^{\rho\sigma}g^% {\mu\nu}-8g^{\sigma\nu}g^{\mu\rho}+8g^{\mu\sigma}g^{\nu\rho}-{\rm tr}\Big{[}% \gamma^{\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\Big{]}\,. (233)

Here we have used the anticommutator to swap γρ\gamma^{\rho} and γσ\gamma^{\sigma} at ∗*, γν\gamma^{\nu} and γσ\gamma^{\sigma} at †{\dagger} and γμ\gamma^{\mu} and γσ\gamma^{\sigma} at #\#. Finally, we can use the cyclicity to bring γσ\gamma^{\sigma} back to the end and which just results in the l.h.s. again. Therefore,

tr⁢[γμ⁢γν⁢γρ⁢γσ]\displaystyle{\rm tr}\Big{[}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{% \sigma}\Big{]} =4⁢gρ⁢σ⁢gμ⁢ν−4⁢gσ⁢ν⁢gμ⁢ρ+4⁢gμ⁢σ⁢gν⁢ρ.\displaystyle=4g^{\rho\sigma}g^{\mu\nu}-4g^{\sigma\nu}g^{\mu\rho}+4g^{\mu% \sigma}g^{\nu\rho}\,. (234)