I am learning about Electron Capture.

$$p+e\to n+\nu_e$$

My question is whether the $W^+$ boson or the $W^-$ boson is involved in this transfer. I can consider this two ways:

1) Since the electron is losing a negative charge in this situation, this would indicate that this should be mediated by the $W^-$ boson.

2) Since inside the proton (uud), there is a change from an up quark $\left(+\frac{2}{3}\right)$ to a down quark $\left(-\frac{1}{3}\right)$ in the neutron (udd), we are losing a positive charge of $+1$, suggesting that this reaction should be mediated by the $W^+$ boson to 'carry' this charge away.

My question is which boson is the correct boson that is involved in this transfer?

## Best Answer

In time-ordered diagrams, you would have to deal with both:

1) $e^- \rightarrow \nu_e+W^-$ followed by $W^- + u \rightarrow d$

2) $u \rightarrow d + W^+$ followed by $W^+ + e^-\rightarrow \nu_e$

but those aren't covariant. In the modern language of Feynman diagrams as $W$ boson is exchanged. There is no time ordering to the vertices (Feynman diagrams are manifestly covariant). Four momentum is conserved at all vertices, so if you look at scattering processes, the square for momentum of the exchanged boson (called $Q^2$) is less than zero: it is space-like.