How can a virtual $W$ boson turn into an electron and electron antineutrino?

Question

This is specifically with regard to beta decay. My current understanding is that in one type of beta decay, a neutron turns into a proton, an electron, and an electron antineutrino. In order to change the down quark in the neutron into an up quark, a W minus boson is emitted. Is this $W$ boson virtual or not? And if it is virtual how does this virtual particle (created from a fluctuation defined by the uncertainty principle) last long enough and even turn into two particles such as an electron and electron antineutrino?

Answer 1

Your question is a result of "just winging it" with quantum field theory. The idea that a $W$ boson with mass $M_W$ is created for a short period of time by borrowing energy via the HUP is very popular on YouTube pop-sci videos, but in my opinion, the ratio:

$$ R = \frac{\rm Physical\ Insight}{\rm Confusion} \ll 1$$

So that whole idea is from old fashioned perturbation theory (OFPT), and involves time ordered diagrams where on-shell particle steal energy (there are no off-shell particles, it's just that energy isn't conserved a vertices).

Of course, every time beta decay is discussed, half the process is ignored, since you also need to include the diagram in-which there is a spontaneous vacuum poppage:

$$|0\rangle \rightarrow W^+e^-\bar \nu_e $$

followed by:

$$ W^++d \rightarrow u $$

So in modern relativistic quantum field theory (in which I was not an expert--though my 2 graduate profs were Preskill and Feynman, and I passed), energy and momentum are conserved at all vertices, and Feynman diagrams (he just call them diagrams in his class, Gell-Mann?, he used his own name whenever possible), and there is no time-ordering. A $W^{\pm}$ is exchanged between the quark and lepton sides, and the Feynman diagram is a Lorentz invariant combination of both the OFPT diagrams.

The necessitates particles be off shell, where the shell is:

$$ E^2 + p^2 = M_W^2 $$

The amplitude to be off-shell is suppressed far from resonance, see: Breit-Wigner distribution...it looks just like a forced damped harmonic oscillator.

The other part of the amplitude comes from the $(u, d)$ and $(\nu_e, e^-)$ being weak isospin doublets, and the $W^{\pm}$ acting as a ladder operator, much in the same way you couple two atomic states with a $J^{\pm}$.

From there: Fermi's Golden Rule and a phase space calculation and you're done.

Answer 2

Yes, the W boson in beta decay is virtual (off-shell). It is not necessary for the W boson to "last enough" for it to decay. It can decay into an electron and an antineutrino because the corresponding quantum fields are coupled.

Also, the uncertainty principle plays an important role in understanding some features of the weak force (for instance why it is "weak"), but it has nothing to do with the fact that the W boson can decay into an electron and antineutrino, as far as I know.

This is certainly not a straightforward topic in the Standard Model, but there is plenty of literature on this subject; I would recommend looking for it.