Question 5.9: Decomposition Into Vector and Axial Vector Couplings Prove r......
Decomposition Into Vector and Axial Vector Couplings
Prove relation (5.133),
\gamma_{\mu}\gamma_{\lambda}\gamma_{\nu}=(s_{\mu\lambda\nu\beta}+\mathrm{i}{\varepsilon_{\mu\lambda\nu\beta}}\gamma_{5})\gamma^{\beta} (5.133)
\gamma_{\mu}\gamma_{\alpha}\gamma_{\nu}=(s_{\mu\alpha\nu\beta}+\mathrm{i}\varepsilon_{\mu\alpha\nu\beta}\gamma_{5})\gamma^{\beta}\ . (1)
\gamma_{\mu}\gamma_{\alpha}\gamma_{\nu} anticommutes with \gamma_{5}. On the other hand, it must be a linear combination of the 16 matrices which span the Clifford space, namely 1,\gamma_{5},\gamma_{\mu},\gamma_{\mu} \gamma_{5}, and \sigma_{\mu\nu}. Obviously this implies \gamma_{\mu}\gamma_{\alpha}\gamma_{\nu}=a_{\mu\alpha\nu\beta}\gamma^{\beta}+b_{\mu\alpha\nu\beta}\gamma_{5}\gamma^{\beta}\;\;. (2)
Multiplying by \gamma_{\delta} and taking the trace we find with
\mathrm{tr}\{\gamma_{\mu}\gamma_{\alpha}\gamma_{\nu}\gamma_{5}\}=4(g_{\mu\alpha}g_{\nu\delta}+g_{\mu\delta}g_{\nu\alpha}-g_{\mu\nu}g_{\alpha\delta})=4a_{\mu\alpha\nu\delta} (3)
and
\mathrm{tr}\{\gamma^{\beta}\gamma_{\delta}\}=g_{\delta}^{\beta}\ ,\quad\mathrm{tr}\{\gamma_{5}\gamma^{\beta}\gamma^{\delta}\}=0that
g_{\mu\alpha}g_{\nu\delta}+g_{\mu\delta}g_{\nu\alpha}-g_{\mu\nu}g_{\alpha\delta}=a_{\mu\alpha\nu\delta}\ . (4)
The second coefficient is determined by multiplying (2) by \gamma_{\delta}\gamma_{5} and taking the trace
4b_{\mu\alpha\nu\delta}=4\mathrm{i}\varepsilon_{\mu\alpha\nu\delta}\;\;. (5)
Thus we have proven (1):
\gamma_{\mu}\gamma_{\alpha}\gamma_{\nu}=g_{\mu\alpha}\gamma_{\nu}+g_{\mathrm{\nu}\alpha}\gamma_{\mu}-g_{\mu\nu}\gamma_{\alpha}+\mathrm{i}\varepsilon_{\mu\alpha\nu\delta}\gamma_{5}\gamma^{\delta}~. (6)