Is it true that gauge boson mix only around after the electroweak symmetry breaking, while mathematics do not explicitely state that?

Question

Is it true that gauge boson mix only around after the electroweak symmetry breaking, while mathematics do not explicitely state that ?

Indeed, in the mathematical description of the linear combination of $W_1$, $W_2$, $W_3$, $B$ to form $W^{\pm}$, $Z$, $\gamma$, we only need to have the covariant derivative operator in order to introduce $W_1$, $W_2$, $W_3$, $B$ : $D_{\mu}=\partial_{\mu}-f(W_1, W_2, W_3, B)$.

So we do not explicitely need that the field of the Higgs is after electroweak symmetry breaking. We only need the Higgs field ($\phi$) to exist, but it does exist also before the electroweak symmetry breaking. In particular $v$ (vacuum expectation value) does not appear in the formula of $W^{\pm}$, $Z$, $\gamma$

So what is the explanation?

See details of computation there Electroweak interaction: From $W^{1}_{\mu},W^{2}_{\mu},W^{3}_{\mu},B_{\mu}$ to $W^{\pm},Z_{\mu},A_{\mu}$

Possible duplicate: https://physics.stackexchange.com/q/545519/ — my2cts, Apr 21 '20 at 10:14
@my2cts : no : it is a different question, from myself as well. — Mathieu Krisztian, Apr 21 '20 at 10:17
The word is "explicitly". Indeed, v appears bimodally: it scales up and down all masses uniformly, and only its departure from zero matters to the details of the mixing. This is standard in several areas of physics touching upon masslessness. The explanation for what? Isn't it transparent in the formulas you are referring to and purportedly understand? — Cosmas Zachos, Apr 21 '20 at 14:39
@Cosmas Zachos : do you agree that the mathematics derivation used would drive to the exact same results in the case for the W+, W- and Z, if it was for the field of Higgs before the EWSB ? — Mathieu Krisztian, Apr 21 '20 at 15:04
If "before the EWSB" means v =0, then the Weinberg rotation is meaningless, and so is what you'd call a "mathematics derivation" vaguely alluded to. Explain how you diagonalize a null matrix. — Cosmas Zachos, Apr 21 '20 at 15:13
As with your other question, I think a lot of your confusion is due to sloppy usage of the word "exist". It's very unclear what it means for the Higgs field to start and stop "existing". — knzhou, May 05 '20 at 06:47

score 2 · Answer 1 · answered May 04 '20 at 13:12

2

Answer from an experimental physicist:

One has to keep in mind that the theory of physics has evolved within the universe and at the time we find ourselves in. This means that first there were observations and measurements, and then laboriously and slowly mathematical models evolved to fit the observations and measurements and, very important, to predict new observations and measurements.

One has to note that the photon was experimentally determined from measurements to be a particle of zero mass involved in electromagnetic interactions.

It helps to look at the history, of how three gauge bosons and a photon were proposed for an SU(2)xU(1) symmetry:

In 1964, Salam and Weinberg had the same idea, but predicted a massless photon and three massive gauge bosons with a manually broken symmetry. Later around 1967, while investigating spontaneous symmetry breaking, Weinberg found a set of symmetries predicting a massless, neutral gauge boson. Initially rejecting such a particle as useless, he later realized his symmetries produced the electroweak force, and he proceeded to predict rough masses for the W and Z bosons. Significantly, he suggested this new theory was renormalizable.

italics mine.

So the existence of a photon is in no doubt.

Identifying the photon with one of the four unbroken gauge bosons is not possible, as before symmetry breaking charge is a composite of weak hyperharge and weak isospin as defined when all gauge bosons are of mass zero.

The above spontaneous symmetry breaking makes the W3 and B bosons coalesce into two different physical bosons with different masses – the Z0 boson, and the photon (γ),

where $θ_W$ is the weak mixing angle.

Here is how charge is identified after symmetry breaking

The pattern of weak isospin, $T_3$, and weak hypercharge, $Y_W$, of the known elementary particles, showing the electric charge, Q, along the weak mixing angle. The neutral Higgs field (circled) breaks the electroweak symmetry and interacts with other particles to give them mass. Three components of the Higgs field become part of the massive W and Z bosons.

So even though zero mass gauge bosons exist before symmetry breaking it is not possible to identify one of them to the photon. It is combination according to the matrix above that ends up into being the experimentally observed photon.

As experiments at the LHC are nearing energies where the primordial quark gluon plasma can be studied, maybe in future higher energy colliders there will be some experimental evidence in the distributions of the change before and after symmetry breaking.

answered May 04 '20 at 13:12

anna v

233,453

Dear Anna. Thank you for your kind explanations. Please let me explain a detail point that it not clear to me.
The covariant derivative ($D_{\mu}$) has been introduced historically in order to make the Dirac Lagrangian stable in presence of a local gauge transformation, by introducing within a covariant derivative (that replaces the usual derivative) a counter term for the extra term that appears after local gauge transformation.
– Mathieu Krisztian May 04 '20 at 18:48
The same procedure was applied to the Yang-Mills theory with SU(2) transformation. The two allowed to introduced all together 4 gauge fields : B, W1, W2, W3.
The Higgs Lagrangian has a kinetic term $D^{\mu}\Phi D_{\mu}\Phi$, thus introduces also the terms with the B, W1, W2, W3 fields. After symmetry breaking, the vacuum expectation values $v$ terms appears, and we could finds naturally the masses for the fields $W^+$, $W^-$, $Z$ and no mass for $\gamma$.
– Mathieu Krisztian May 04 '20 at 18:48
With the Dirac and Yang-Mills theories alone, even with no Higgs, we always have the B, W1, W2, W3 fields, so we could already "define" $Z=f(W3, B)$, and $\gamma=f'(W3,B)$, and $W^+=f''(W1,W2)$, $W^+=f'''(W1,W2)$ with exactly the same formula as the ones after Higgs symmetry breaking. So I don't see why we would need the existence of the Higgs (appart to give mass) to make these linear combination to exist Maths don't seem to "need" the electroweak symmetry breaking to make these fields appear. – Mathieu Krisztian May 04 '20 at 18:48
In presence of the Higgs mechanism, the fact that the fields $W^+$, $W^-$, $Z$ have a mass in only due to the electroweak symmetry breaking, because by construction of symmetry breaking, the vaccum expectation value $v$ is non null. But I don't see again why the fields themselves $W^+$, $W^-$, $Z$, and the photon as well would not exist before electroweak symmetry breaking : they would just all have a null mass before the electroweak symmetry breaking when we just do the mathematics used for the electroweak symmetry breaking. This is a second example of my point raised. – Mathieu Krisztian May 04 '20 at 18:49
But apparently, particle physics community states that $W^+$, $W^-$, $Z$ and $\gamma$ exist (are defined) only after electroweak symmetry breaking. It is not clear to me why, since I have shown above that without Higgs, we could still define them with exactly the same formula as a function of $W1$, $W2$, $W3$, $B$, and that in presence of Higgs, but before electroweak symmetry breaking, we could again define them as well, with the only change that their mass before electroweak symmetry breaking would be 0. – Mathieu Krisztian May 04 '20 at 18:49
This was the meaning of my original question. Would you have an explanation ?
Thank you
– Mathieu Krisztian May 04 '20 at 18:49
Do not forget I see the matter as an experiment oriented physicist . Defining axiomatically is different than finding a theory that gives the same data results and predicts new observations. That is the time I quote above when Weinberg stumbled on breaking the symmetry by hand , i.e. have the four different mass gauge bosons in a theory. , axiomatically. Then he went to a mathematical stage where he could break the symmetry by setting the axioms at a higher energy state and "a spontaneous symmetry breaking mechanism" .became axiomatic, the gauge bosons absolutely derivable – anna v May 05 '20 at 04:33
from this axiom. Theories in physics, are considered better, the more the axioms are removed from data, i.e. putting things by hand.. Beauty of theory is a great attractor for physicists. That is why the holy grail of a Theory Of Everything (TOV) is sought. Spontaneous symmetry breaking , as the hat in the higgs potential, exists in classical physics too. In the format of SU(2)xU(1) it satisfied the desire to have the minimum of axioms in a compact theory (beauty). Now to the question whether what are measured particles at our present energies are also particles before symmetry – anna v May 05 '20 at 04:41
breaking. The formula I gave says that the Z and γ at energies above symmetry breaking are a quantum mechanical combination of $W_3$ and $Z$. in a fixed mathematically way. This means that there could be measurable differences in production mechanisms striclty governed by the mathematical setup, which does not define axiomatically the masses, but they come out as a result of the sponaneous breaking. As we have measured the three bosons that are the prediction of the model and have also found the Higgs , the model is validated at the moment, with fewer numbers put in by hand axiomatically, – anna v May 05 '20 at 04:46
(that is TOE not TOV). New theories in the future have to incorporate this symmetry breaking mechanism or show that it is emergent from the new mathematics, because the Standard model is a mathematical encapsulation of almost all the data up to now. – anna v May 05 '20 at 04:47
All right, thank you for your kind explanations. Let's wait for other possible answers of community in the remaining 6 days. – Mathieu Krisztian May 05 '20 at 07:58
Please do not think to accept my answer for the bonus. I have no need for the points, so unless you get a good theoretical answer from an expert, you do not have to give the bonus. I have been thinking on the way interactions would happen and if they could be checked experimentally, before electroweak breaking , and your question is relevant – anna v May 05 '20 at 08:58
Symmetry breaking and Electroweak theory is only the beginning. Scientists are finding ways to incorporate even the strong nuclear forces. In some cases even gravity is added which is a little hard to do considering the lack of quantum gravity. – MiltonTheMeme May 07 '20 at 18:21

score 0 · Answer 2 · answered May 07 '20 at 18:17

Lets start of with what particles we had before the symmetry breaking before the Higgs was powerful long, long time ago. This was the time when the so called graviton of the gravitational force and gluon of the strong force decided they no longer wanted peace with the other forces. They left and the electroweak force were having a happy time. Both of them were powerful. Electroweak had 4 enforcers of its force. W1, W2, W3, B. Then at this time the Higgs grew powerful as the hot universe kept expanding. More space for the same amount of radiation and bosons. Higgs already made a weak connection with the quarks which were starting to form the instant strong nuclear force left the other forces. The quarks were not the quarks we know today. So the W3 and B bosons started thinking of changing their identities. Then the Higgs knowing Electroweak plans let them create new particles. By the time the W3 and B were turning into the photon and the Z boson the Higgs came along. The Higgs added goldstone bosons to the Z boson as well as to the W bosons that were created due to the mixing of W1 and W2. The Higgs also gave the two W bosons charge. Now the Weak force split from Electromagnetism. The photon was forever destined to be alone. By now all the matter and antimatter particles were created. Now it was the Higgs bosons' job to use the weak force and send the remaining goldstone bosons at the fermions and antifermions. The fermions new the weak nuclear force was no longer working for them. It was working for the Higgs. At least the fermions found an ally. Electromagnetism and Strong force sometimes became friends with the fermions. By this time Higgs bosons found a way to use fermions allies to get rid of the antifermions so it could get even more powerful and try to prevent massless photons and gluons from stopping the Higgs. And now we get a Higgs that give us and all the fermions mass. Still the Higgs has some energy and could get even more powerful. Also the Higgs kept some mass for itself.

So all in all, the Higgs always existed before the Electroweak separation. It just was not as powerful as the other forces until it found a way to separate the weak force from electromagnetism. Then it found a way to give mass to all the fermions and the Weak force bosons and also the Higgs boson itself.

Also before symmetry breaking properties of electromagnetism such as electric charge and properties of weak force such as decay are going to be mixed. Also the Weak in the form of electroweak would have had an infinite range without the Higgs field. This is because without the Higgs the W and Z bosons would have zero mass allowing them to travel at the speed of light itself. — MiltonTheMeme, May 07 '20 at 18:24

Is it true that gauge boson mix only around after the electroweak symmetry breaking, while mathematics do not explicitely state that?

2 Answers2