4

If I correctly understand this, the four gauge bosons that correspond to the electroweak force before symmetry breaking are the W1, W2, W3, and B. How come the W1, W2, W3, and B bosons are not listed in the Standard Model?

More clarification: I guess I am missing something from the Wiki article and other descriptions. For example, are the four bosons before symmetry breaking (i.e., W1, W2, W3, and B) only mathematical constructs that have never been observed in a high energy accelerator?

Qmechanic
  • 201,751
James Goetz
  • 541
  • 4
    I almost certain that any text or article discussing the three W bosons and B boson will also discuss that these aren't the particles observed in nature due to electroweak symmetry breaking. – Alfred Centauri Jun 25 '17 at 20:49
  • 1
    Yes Alfred, if I correctly recall, the temperature needed before symmetry breaking of the electroweak force requires a high energy accelerator. – James Goetz Jun 25 '17 at 22:52
  • Relevant https://physics.stackexchange.com/q/88270/226902 and https://physics.stackexchange.com/q/698793/226902 – Quillo Jan 13 '23 at 08:46

1 Answers1

6

They're listed under $W^+$, $W^-$, $Z$, and $\gamma$ (the photon). Those observed particles are constructed from linear combinations of $W^1$, $W^2$, $W^3$, and $B$ in the dynamical symmetry breaking of the Higgs mechanism, with $\gamma$ being the massless Nambu-Goldstone boson that results from the process (see comments for clarification, thanks @gj255). For specific details, go through the Wikipedia article on the Electroweak interaction, or any textbook that covers the standard model.

It's going to depend on what you mean by "observed". None of the weak theory bosons have been observed leaving a track in a detector. Instead, they're inferred as a "bump" in the interaction cross section scaling with energy, and as interaction vertices of other particle that have an invariant mass at the correct values. Because we are limited to detecting the $W^\pm$ and $Z$ bosons this way, there's no chance for observing the sort of superposition that the $W^{1-3}$ and $B$ bosons would be. If we could get the energy high enough for time dilation to cause one of these bosons to leave a track, it could, in principle, be possible to get an interaction to produce an observable superposition that would correspond to what you want to observe. Someone more familiar with the standard model Lagrangian would have to sit down to see whether the math allows this, given an appropriate superposition of input particles. Think of observing $K$ mesons that are produced in a superposition of two states that have different decay rates, just a whole lot more technically challenging to do at every level.

See also this question StackExchange recommends as related.

Sean E. Lake
  • 22,482
  • 2
    There are no Nambu-Goldstone bosons associated with electroweak symmetry breaking, since the electroweak symmetry is gauge. If we insisted on talking about them, the NG bosons are best thought of as the longitudinal polarisations of the three massive vector bosons. The photon is a vector particle, and not a NG boson. It was there as a gauge boson before EW symmetry-breaking, and it remains there after. – gj255 Jun 25 '17 at 21:08
  • I guess I am missing something from that Wiki article and other descriptions. For example, are the four bosons before symmetry breaking only mathematical constructs that have never been observed in a high energy accelerator? – James Goetz Jun 25 '17 at 23:01
  • "[the photon] was there as a gauge boson before EW symmetry-breaking" - @gj255 by 'there', do you mean there as a certain mixture of the neutral W and B bosons? Or do you mean something else? – Alfred Centauri Jun 25 '17 at 23:02
  • @JamesGoetz, why don't you add your comment above to your question. I think it would help clarify what it is your looking for. – Alfred Centauri Jun 25 '17 at 23:05
  • 2
    @AlfredCentauri I mean what you suspect. If we rotate the basis of electroweak gauge bosons before symmetry-breaking, to make the photon manifest, then we notice that nothing about the photon changes when the electroweak symmetry is broken. – gj255 Jun 25 '17 at 23:27
  • @gj255 Sorry that I might be missing something here. If the photon existed before EW symmetry breaking, then does this indicate that photons might have existed during the electroweak epoch in the early universe? – James Goetz Jun 26 '17 at 00:04
  • @JamesGoetz In a sense. Before EW symmetry-breaking there were four massless gauge bosons (plus the gluons), and one of those can be considered to be the photon. – gj255 Jun 26 '17 at 00:20