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

In Cheng and Li p. 421 they consider the transition

$$T(\mu \rightarrow e\gamma) = \epsilon^\lambda<e|J^{em}_\lambda|\mu>$$

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?


1 Answer 1


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 $\sigma_{\lambda\nu}$.


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