What happens during a weak interaction?

For e.g. during $\beta^-$ decay a $W^-$ boson is emitted changing an up quark to a down quark. This seems very weird to me as it looks like that up quark is not interacting with some other particle but simply decays suddenly by emitting a boson. Is this process something like an electron falling into a lower energy state by emitting photons? Or is there something more happening. And what are some other ways that weak interactions take place? And I have no clue what the $Z$ boson does. Why and how do particles interact with $Z$ bosons?

• You can think of it as three quantum fields interacting at a point. An excitation in the up-quark field is replaced by excitations in the down-quark field and the W-boson field. It is not too different to differential equations like Maxwell's, except that there is quantum uncertainty on top of the field dynamics. Sep 15, 2018 at 10:36
• @MitchellPorter So is it something like that there is a certain probability of this configuration happening and so it happens? How is it similar to Maxwell’s — is it just like how an excitation in the electric field produces an excitation in the magnetic field?
– Foon
Sep 15, 2018 at 10:50
• A Z boson is the neutral carrier of the weak force. Sep 15, 2018 at 14:36

1 Answer

The Universe is permeated by a bunch of quantum fields. These are fermionic (like the electron field, or the quark field) or bosonic, like the electromagnetic field $A^{\mu}$ (the "photon"), or the three weak interaction gauge fields $W^{\pm} Z^0$ (the "$W^+, W^+, Z^0$ bosons").

$A^{\mu}$ is in the context of the quantum field theory of electromagnetism, called QED. To get back to the "classical" concepts of electric and magnetic fields, $A^{\mu}$ is related to the Faraday tensor which contains $\mathbf{E}$ and $\mathbf{B}$.

There is always a finite non-zero chance of interaction between the various fields. As you know from quantum mechanics, the actual outcome of the interaction is probabilistic. Meaning you have to wait an "average" of say time $\tau$ for an interaction to take place. What this means in practice, is that given a large amount of interacting particles $N$, by time $\tau$ a number of particles $N/e$ have interacted. If this time is short, the interaction is said to be strong - otherwise, it's weak. You have to wait, on average, a long time to see something happening.

A weak interaction is just a type of interaction.
An electrically charged particle sees the electromagnetic field and can interact with it, by emitting/absorbing photons$^{\dagger}$ which then may be absorbed/emitted by other charged particles, causing attraction/repulsion between unlike and like charged particles.

Analogously, some particles (every particle in existance, actually) have another charge called weak isospin. This allows them to "see" and interact with the weak fields, $W^{\pm}$ and $Z^{0}$. They can emit/absorb$^{\dagger}$ these bosons, which can then decay into other particles - like you have described in the beta decay.

The $\pm$ and $0$ refer to the electrically charged of the weak bosons. You will interact with the $W$ or the $Z$ according to electric charge conservation.
In your case, you start with a neutron (charge 0) and end up with a proton, so you need the $W^-$ to ensure the total charge is still zero.

For example, in a $e^- e^+ \rightarrow \mu^- \mu^+$ process, you could have the electron & positron annihilate into a either a photon or a $Z^0$ boson, which then decays into the muon pair.

$\dagger$: There would be so many technicalities to add to this question. From the connection between weak isospin, weak hypercharge and electrical charge, to the reason as two why there are $3$ weak fields and just one EM one.

Also, the picture that I have (mea culpa) employed is that of virtual particles, i.e. charges emit photons which then are absorbed by other particles, thereby mediating the interaction.
This is, technically wrong, though understaning of this requires understanding of pertubration theory and QFT.

• Great answer! As for "There is always a finite chance of interaction between the various fields", I would add for OP's sake that, here, the word "finite" is not defined as in mathematics (i.e. "smaller than infinity") but in the usual way physicists employ it, namely as "non-zero".
– balu
Sep 15, 2018 at 15:04
• Noted. I corrected it. Sep 15, 2018 at 15:09
• @SuperCiocia Thank you for the great answer! Yes, I know a little bit about the incorrectness of the virtual photon picture. So, instead, is it okay to think that when two electrons are closer together there is a larger disturbance (and a chance of disturbance?) in the electromagnetic field between them, and that disturbance is reabsorbed by both causing them to repel (at least in the simplest scenario)?
– Foon
Sep 15, 2018 at 15:18
• The correct picture is that QED is a long range interaction, so charged particle are never "free", but always interact. You would have to diagonalise the full QED hamiltonian to find the real eigenstates of the system, which would show a smooth change from $ee$ to $\mu \mu$ (for instance), without the need to introduce the photon "at short range". This is not feasible, since there is not analytic solution. So instead, you consider the EM field as a perturbation that only acts in a small region of space - hence the vertices in the Feynman diagrams. Sep 15, 2018 at 15:21
• Also, for most things you don't even have to quantise the EM field. You can just work with a quantum fermionc field, and assume a classical field $A^{\mu}$ with a continuum of modes. This is enough, for example, to recover Maxwell's equations from the Lagrangian formalism of the Standard Model. You need to quantise the EM field if you want to see emission/absorption of single photons, i.e. of single modes of the field. Sep 15, 2018 at 15:25