0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $\sigma_{\lambda\nu}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.