# Lorentz decomposition of electromagnetic current Cheng&Li $\mu\rightarrow e\gamma$ p.421

I've been struggeling with this for a while, should be very easy probably....

In Cheng and Li p. 421 they consider the transition

$$T(\mu \rightarrow e\gamma) = \epsilon^\lambda$$

with $$\epsilon^\lambda$$ the photon polarization. To simplify the further calculations they argue that the em current is vector-like thus it can be generally written as:

$$T(\mu \rightarrow e\gamma) = \epsilon^\lambda = \bar{u}_e(p-q)\left[iq^\nu\sigma_{\lambda\nu}(A+B\gamma_5) + \gamma_\lambda(C+D\gamma_5)+q_\lambda(E+F\gamma_5)\right]u_\mu(p)$$

With A,...,F invariant amplitudes. Using the em gauge condition $$\partial^\lambda J^{em}_\lambda = 0$$ will give:

$$-m_e(C+D\gamma_5)+m_\mu(C-D\gamma_5) + q^2(E+F\gamma_5) = 0$$

That gives with $$q^2= 0$$ That $$C=D=0$$ and only A and B survive. Im struggling to reproduce that. the C,D term is easy. For E,F I would argue that such a term can't exist right from the beginning because $$\epsilon^\lambda q_\lambda = 0$$ anyways. But the A,B terms gives me trouble. The derivative should be zero independently of A and B. So here is what I did: $$\partial^\lambda\left[\bar{u}_e(p-q)(iq^\nu\sigma_{\lambda\nu}(A+B\gamma_5))u_\mu(p)\right]$$ Using the definition of $$\sigma$$ and the product rule as well as the Dirac equation for spinor and adjoint spinor gives:

$$\frac{i}{2}\left[i\partial^\lambda\bar{u}_e(p-q)\gamma_\lambda q^\nu\gamma_\nu(A+B\gamma_5)u_\mu(p)+\bar{u}_e(p-q)q^\nu\gamma_\lambda \gamma_\nu(A+B\gamma_5)i\partial^\lambda u_\mu(p)-i\partial^\lambda\bar{u}_e(p-q)q^\nu\gamma_\nu\gamma_\lambda(A+B\gamma_5)u_\mu(p) - \bar{u}_e(p-q)q^\nu\gamma_\nu(A-B\gamma_5)i\partial^\lambda \gamma_\lambda u_\mu(p)\right]$$

Now using the anticommutator for the gamma matrices and seeing that the two terms $$\propto 2\eta_{\lambda\nu}$$ cancel due to the different signs will give:

$$-i\left[m_e\bar{u}_e(p-q)q^\nu \gamma_\nu(A+B\gamma_5)u_\mu(p) + m_\mu\bar{u}_e(p-q)q^\nu \gamma_\nu (A-B\gamma_5) u_\mu(p) \right]$$

So I don't see how this could be 0. first there is that minus sign in front of the second $$B$$ coming from the anti commutator of $$\gamma_\mu$$ and $$\gamma_5$$ and more important the different masses as well as spinors for different particles. In QED, when calculating the electron vertex function one has a similar situation with $$\bar{u}(p')q^\nu\gamma_\nu u(p) = 0$$ which is easy to show since here you have the same particle and you can just use the dirac equation for spinor and adjoint spinor by adding (p-p)= 1 Can anyone please help me with this calculation?

The term with $$C$$ and $$D$$ can indeed be ignored because $$\epsilon\cdot q =0$$. That $$A=B$$ follows from the fact that we are considering $$m_e=0$$ and hence a left-handed electron. So we expect a $$u_{e,L} = (1-\gamma^5)u_e$$ or $$\bar{u}_e(1+\gamma^5)$$. You can move the $$1+\gamma^5$$ to the left of the $$\sigma_{\lambda\nu}$$.