In the $\mu \rightarrow e+\gamma$ calculation in Cheng and Li "Gauge theory of elementary particle physics" p.422 they have

$$ T=A\bar{u}_e(p-q)(1+\gamma_5)i\sigma_{\lambda\nu}q^\nu\epsilon^\lambda u_\mu(p)= A\bar{u}_e(p-q)(1+\gamma_5)(2p \cdot\epsilon -m_\mu\gamma \cdot \epsilon)u_\mu(p). $$

I struggle to reproduce this. What would be the right way to do this? I need some sort of vector like structure in between the two spinors and I know that

$ \sigma_{\lambda\nu}= \frac{i}{2}[ \gamma_\lambda, \gamma_\nu] $ And I also think that when I use the Gordon decomposition for vector and pseudo vector (since I already wrote them down asuming I get the correct form) I will find terms $ \epsilon^\lambda q_\lambda=0$

Any hints how to reproduce their result?


I found the answer myself. Actually it wasn't that difficult. Just take the Gordon decomposition for a vector-like bilinear and a pseudo-vector-like bilinear. Adjust them with the corresponding momenta and note that here you don't have the $2m$ but $(m_e+m_\mu)$ since there are two different particles.

The adjusted identities are: $$\bar{u}_j(p-q) i \sigma_{\lambda\nu}q^\nu u_i(p) = \bar{u}_j(p-q) \left[(2p+q)_\lambda - m_i \gamma_\lambda\right] u_i(p)$$ $$\bar{u}_j(p-q) \gamma_5 i \sigma_{\lambda\nu}q^\nu u_i(p) =\bar{u}_j(p-q)\left[-q_\lambda\gamma_5 + 2 i \sigma_{\lambda\nu}p^\nu\gamma_5 -m_i\gamma_\lambda\gamma_5\right]u_i(p)$$

Adding both and using that $2i\sigma_{\lambda\nu}p^\nu\epsilon^\lambda=2(p\cdot \epsilon-\gamma^\alpha p_\alpha\gamma^\beta \epsilon_\beta$ We get the correct result.


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