# Why the four gauge bosons that correspond to the $SU(2)\times U(1)$ electroweak force before symmetry breaking are not listed in the Standard Model?

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?

• 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
• 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

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.