Electroweak to Electro/Weak Bosons?

Question

I apologise in advance, this is something I just can't seem to get my head around. So it's my understanding that before the electroweak force split, there were four bosons - $W^1$, $W^2$, $W^3$ and $B$ - and that they mix or interact to form photons $\gamma$, $W^+$/$W^-$ and $Z$ bosons. So, in our current universe with the split weak nuclear and electromagnetic forces, what happens to $W^1$, $W^2$, $W^3$ and $B$? If photons are mixes of $W^3$ and $B$, does that mean the $W^3$ and $B$ bosons are created for an instant and then mix to form a photon in the heart of a star? Or can photons be made without W3 and B?

Answer 1

This strictly depends of the scale.

At the electroweak scale $\Lambda_{EW}$ symmetry breaking occurs.

At $E>\Lambda_{EW}$ no SSB occurs and 4 massless fields propagates:

$W^1 ; W^2 ; W^3 ; B$

Those 4 fields have each 2 polarizations as you should know for massless vector bosons.

Then your energy goes under such a scale, the Higgs mass, which depends on the energy of your process, triggers spontaneous symmetry breaking.

At $E < \Lambda_{EW}$ then you have simmetry breaking. 3 Goldstone modes are created and a new symmetry is now present. You have the same 4 fields above to propagate and the new 3 Goldstone modes to propagate.

There is a catch. Your 4 bosons can no longer propagate freely due to the new symmetry established after SSB. They are propagated only in a specific combination:

$W^1$ and $W^2$ propagate ONLY in the combinations $W^{\pm} = W^1 \mp iW^2$

$W^3$ and $B$ propagate only in yhe combinations $A = cos(\theta_W)B + sin(\theta_W)W^3$ and $Z = -sin(\theta_W)B + cos(\theta_W)W^3$.

Those are still massless tho. The $W^{\pm}$ and $Z$ boson gain mass because they are propagated togheter with the 3 Goldstone bosons, each of those provide the longitudinal polarization needed for a massive vector field.

So for example photons are the simultaneous propagation of $W^3$ and $B$. Inside a star you generate $W^3$ and $B$ bosons which are constrained by the new symmetry to propagate only in a way we label photons. Photons are not the real fields. The electroweak bosons are the real fields, but when a SSB occurs they are constrained to propagate togheter in such a way that we interpret as photons.

The same is valid for the 3 weak force bosons.

Answer 2

I had the same question about a month ago when i tried to interprete the mathematics of electroweak SSB(sponteneous symmetry breaking). I investigated the obvious resolutions: either [1]maths is givining wrong interpretations or [2]It is giving correct interpretations.

Since Standard Model of EW interaction is so profound, I assumed [2] is correct.

So, what is the interpretation then if we analyze the math?

Consider this (understand in terms of fields, which are more fundamental than particles):

[a]Before SSB, above electroweak unification energy($E_{ew}$), we had 4 fields.

[b]At $E_{ew}$, these 4 fields start interacting with each other and mix in different combinations.

[c]After SSB, we have 4 new fields.

{$E_{ew}$ is about 246 GeV(approximately $10^{15}$ K, a temperature exceeded until shortly after the Big Bang)}

i.e., the real world is energy dependent. The cases [a], [b] and [c] do not exist together. The fields in [c] are completely new and its like resultant fields of combinations. And Our current stage of the universe falls under [c], therefore only [c] fields exist.

"what happens to $W^1$, $W^2$, $W^3$ and $B$?" These fields do not exist in this stage of the universe.

"does that mean the $W^3$ and $B$ bosons are created for an instant and then mix to form a photon in the heart of a star?"

No!. Only photons are created directly. The reason is that $W^3$ and $B$ fields do not exist in this stage and therefore we can't have $W^3$ and $B$ quanta. Only $A$ and $Z$ fields exist.