Pseudoscalar particle decay

Question

Suppose I want to calculate amplitude of pseudoscalar particle decay into electron + positron. Interaction Hamiltonian is given by (ignoring the positive and real constants) $\mathcal{H} = \bar{\psi} \gamma_5 \psi A(x) $, where A is pseudoscalar field. So I want to calculate the amplitude.

I think it will be: $\mathcal{M}=\bar{u}(k,\alpha) \gamma_5 v(k, \beta)$. Is that correct? When I square this amplitude and sum over polarizations I get: $\mathcal{M^2} = \sum_{pol} \bar{u}(p,\alpha) \gamma_5 v(k, \beta) \bar{v} \gamma_5 u(p,\beta) = \bar{u} \gamma_5 (\gamma^{\mu}k_{\mu} - m) \gamma_5 u = -\bar{u} (\gamma^{\mu}k_{\mu})u -m \bar{u} u=...=-Tr(\gamma^{\mu}\gamma^{\nu} l_{\nu} p_{\mu} + m^2)=-4l\cdot p - 4m^2=-(2M^2-4m^2+4m^2)=-2M^2$ After all identities with gamma matrices I get that squared amplitude is negative and it equals to $\mathcal{M^2} = -2M^2$, where M is the mass of A particle. Am I doing something wrong? Is $\bar{u} \gamma_{\mu}k^{\mu} u=(\gamma^{\nu}p_{\nu}+m) \gamma_{\mu} k^{\mu}$ ?

Answer 1

Hey I had the same problem, but the key is that you have to take care about the following thing, if you write $$ \mathcal{M}^\dagger = v^\dagger \gamma^5 (\gamma^0)^\dagger u = - v^\dagger \gamma^0 \gamma^5 u = -\bar{v} \gamma^5 u $$ so taking the conjugate of a spinor don't give you automatically the "barred version" The same actually happens when you take the conjugate of some vector coupling like the photon, here you get a minus sign by the conjugate of $\gamma^i$ ($(\gamma^i)^\dagger = - \gamma^i$) and an extra minus sign for interchanging $\gamma^i$ with $\gamma^0$, so this subtlety is hidden (For the zero-component, of course, you won't have to interchange the matrices but here $(\gamma^0)^\dagger = + \gamma^0$)