Was photon defined according to Standard Model theory before electroweak symmetry breaking ? Or were there only $W_3$ and $B$ ?

Indeed, in Electroweak to Electro/Weak Bosons?

Aman pawar answers "no" to the question :

"does that mean the W3 and B bosons are created for an instant and then mix to form a photon in the heart of a star?"

His answer is : "No!. Only photons are created directly. The reason is that W3 and B fileds do not exists in this stage and therefore we can't have W3 and B quanta. Only A and Z fields exists."

So he seems to mean that $W_3$ and $B$ have never existed in the history of the universe, and that the photon has existed since the beginning. If this is the case, then the $Z$ could not be created from the $W_3$ and $B$ if they don't exist...

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    $\begingroup$ Have you defined a workable definition of "exist"? Write down the observables and specify what you have in mind. Follow the math and give good names. The rest is theology. $\endgroup$ Commented Apr 21, 2020 at 14:07
  • $\begingroup$ @Cosmas Zachos : exist was meaning "defined by the Standard Model". It is not a theology question but a question for the Standard Model $\endgroup$ Commented Apr 21, 2020 at 15:01
  • $\begingroup$ @Cosmas Zachos :sure this is why I had already replaced it once I had replied to you. $\endgroup$ Commented Apr 21, 2020 at 16:07
  • $\begingroup$ Well, a subjective definition is always thinkable, provided it makes sense. What you actually do with it is what matters. What do you propose doing with it? $\endgroup$ Commented Apr 21, 2020 at 16:11
  • $\begingroup$ @Cosmas Zachos : I'm not sure what you mean : I am asking the question in order to understand this point. My_ question is not subjective at all since I ask about the Standard Model. It is a theory. What is "subjective" in my question. As I said, I have corrected it. $\endgroup$ Commented Apr 21, 2020 at 17:13

2 Answers 2


The Higgs potential, depending on the medium involved (early universe versus our present world, described by "before" and "after" SSB), has different vacuum properties, leading to two dramatically different phases of the standard model.

The "before" phase is unbroken (linear realization of the symmetry): all four gauge bosons are massless, and there is no meaningful mixing between $W^3$ and $B$. At tree level, these two bosons simply "do not know about each other". So it would be meaningless to be talking about a photon or a Z in any consequential way. You may define any and all linear combinations of them, but that would be hardly meaningful or useful.

In the SSB "after" phase, the symmetry is realized nonlinearly, and the $W^\pm$ bosons "develop" masses, while $W^3$ and $B$ possess messy bilinear terms, $(\frac{g'}{g}B_\mu - W^3_\mu)^2$. It is customary to study these combinations of formal field operators by diagonalizing such quadratic forms (mass matrices) to their eigenstate bases: a massless photon and a massive Z, so the fields that propagate.

At this stage, nature has no acknowledgement or knowledge of $W^3$ and $B$: these are practically non-existent particles, and correspond to abstract operators serving to aid your intuition on group theory. The effective lagrangian of the SM does not have them, and only the propagating photon and Z make any sense to talk about.

The takeaway is that it is unmeaning to be talking about photons and Zs in the symmetric phase, just as senseless to be talking about $W^3$ and $B$s in our world, the SSB phase, except perhaps as group-theoretic mnemonics. A good text on the SM always takes care to write down the effective lagrangian of the actual physical fields involved, which is why the PDG wastes no time on $W^3$ and $B$s.

Edits in response to questions

1) 3) At tree level, you see no mixing on p 16 of these notes. It is all in your head. In fact, gauge invariance protects these gauge bosons from picking up a mass through fermion interactions, and the SU(2)xU(1) structure means just that. There is no mixing: they sit apart, but they are not physical. You must be misreading p 16 if you understand it as mixing. In practice, after Higgs-mixing, you get the physical messier structure of the PDG booklet!

2) The basis change of two degenerate fields $W^{1,2}\to W^{\pm}$ is an option you have to make their interactions more transparent vis-a-vis electric charge, hardly definable before SSB, which is effected through the Higgs mechanism, not the Higgs particle. Charge is only defined by the above mixing, the subject of the original question.

4) Look at the before breaking (symmetric) Weinberg-Salam lagrangian. The $W^3$ and $B$s do not interact, as they live in completely different group manifolds; just look. They don't know about each other. That's what the SU(2)×U(1) group Cartesian product actually means! Historically, we knew very very much that the W and Z had masses! That's precisely what made their interactions weak! The SM was invented to give them masses, which appeared almost impossible at the time.

5) The Weinberg-Salam model was cooked up in 1967 and used the Higgs mechanism, not the Higgs particle (largely irrelevant to SSB) to give the W a mass and inevitably predict the massive Z: it picked up the slack in keeping the photon massless... a price to pay. Discovery of the neutral currents confirmed the SM, by validating a hitherto pointless Z (which the Georgi-Glashow model lacked!). You misunderstand the structure of Yukawa couplings and half the point of the SM: it is designed to allow chiral weak interactions all the while generating fermion mass terms. The only way to achieve that is by coupling to a massless photon and a massive Z, and and the only way to get those is SSBreaking an SU(2)×U(1) theory, the "hidden brilliant math" behind the physical answers. (Had the Higgs not been discovered, the SM could work fine, with a Higgs mechanism driven by a "Higgsless" dynamical mechanism: the subject of a truckload of brilliant models.)

It's all in your SM book, I'm sure.

  • $\begingroup$ Dear Cosmas. Thank you for your explanation. But I don't understand why you say that defining a mixing of $W^3$ and $B$ would be meaningful before electroweak symmetry breaking. For example, one could consider the case of the neutral current Lagrangian, in absence of Higgs mechanism, the neutral current Lagrangian (before any electroweak symmetric breaking) for example would introduce naturally the mixing of $W^3$ and $B$, so why would it be not meaningfull to have done the mixing before or in absence of electroweak symmetry breaking ? $\endgroup$ Commented May 4, 2020 at 21:21
  • $\begingroup$ And why would we change $W_1$ and $W_2$ "by hand" to $W^+$ and $W^-$ by a trick of $W^{\pm}~W_1 \mp i W_2$ ? I don't see relationship with the Higgs for that. $\endgroup$ Commented May 4, 2020 at 21:23
  • $\begingroup$ See proof page 16 of physics.lbl.gov/shapiro/Physics226/lecture19.pdf that one does not need Higgs to introduce the mixing of B and W3. $\endgroup$ Commented May 4, 2020 at 21:25
  • $\begingroup$ sorry I don't understand. Your point is very interesting though. Why could we not write neutral current before electroweak symmetric breaking. It means that there is interaction between particle. This is meaning something. Of course, without electroweak symmetric breaking, the gauge bosons would have no mass, but I don't see why the lagrangian would mean nothing. Why does it mean nothing ? Also, historically, we didn't know that W and Z had masses. $\endgroup$ Commented May 4, 2020 at 21:33
  • $\begingroup$ The neutral current was found in 1973, 50 years before we knew that Higgs boson was really existing. Why did physicists search for neutral current while they even don't know that Higgs was existing if they would believe that neutral lagrangian means nothing ? $\endgroup$ Commented May 4, 2020 at 21:33

Your question is essentially about classical field theory. In fact, the core issue can be explained even within classical mechanics -- all the field theory language is just a distraction.

Consider two independent masses on separate springs, with spring constant $k$ and mass $m$. The equations of motion for the masses are $$m \ddot{x}_1 = - k x_1, \quad m \ddot{x}_2 = - k x_2.$$ Therefore, the solutions are $$x_1(t) = A_1 \cos(\omega t), \quad x_2(t) = A_2 \cos(\omega t), \quad \omega = \sqrt{\frac{k}{m}}.$$ The two masses oscillate independently. So when we quantize the system, we get two independent quantum harmonic oscillators $x_1$ and $x_2$, and the "particles" (i.e. the excitations about the ground state) involve oscillations of either mass individually.

Now suppose we connect the two masses by another spring, so that the equations of motion become coupled, $$m \ddot{x}_1 = - k (x_1 - x_2), \quad m \ddot{x}_2 = - k (x_2 - x_1).$$ The solutions will look different. For example, energy in one mass will end up being transferred to the second mass, and back. To understand what is going on, it's better to work in terms of normal modes. We notice that if we define $u = x_1 + x_2$, and $v = x_1 - x_2$, then $$m \ddot{u} = 0, \quad m \ddot{v} = - 2 k v.$$ Therefore, the solutions are $$u(t) = A_1 t, \quad v(t) = A_2 \cos(\sqrt{2} \omega t).$$ We can then quantize $u$ and $v$ independently. Note that $u$ happens to have zero "spring constant".

The notation of my example is as simple as possible, but all the physics relevant to your question is precisely the same. The degrees of freedom $x_1$ and $x_2$ are analogous to the $W$ and $B$ degrees of freedom before electroweak symmetry breaking, which is analogous to connecting the extra spring. When we connect the extra spring, the classical equations of motion change, which means the elementary excitations look different. The useful variables become $u$ and $v$, which are analogous to $A$ and $Z$.

Was $v$ defined before connecting the spring? Or were there only $x_1$ and $x_2$?

Indeed, you could have defined $v = x_1 + x_2$ before connecting the spring. It just wouldn't have been a very useful quantity to consider. It's useful after connecting the spring because it is the coordinate of a normal mode.

No!. Only $v$ is created directly. The reason is that $x_1$ and $x_2$ fileds do not exists in this stage and therefore we can't have $x_1$ and $x_2$ quanta. Only $u$ and $v$ fields exists.

It doesn't make sense to say some variables exist and some don't, since they're related by a trivial change of variable. What is true is that before attaching the extra spring, it is especially easy to see the result of quantization when we work in terms of $x_1$ and $x_2$. And afterwards, it is easiest to do the quantization in terms of $u$ and $v$. Neither "exists" more in some metaphysical sense.

$x_1$ and $x_2$ have never existed in the history of the universe, and $u$ has existed since the beginning. If this is the case, then the $v$ could not be created from $x_1$ and $x_2$ if they don't exist...

I think you can fill in the rest of the analogy here!

  • $\begingroup$ thank you. It is a bit different to the subject, but useful information $\endgroup$ Commented May 5, 2020 at 12:11

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