It is well known that the electroweak interaction can be studied as a $SU(2)_L \otimes U(1)_Y$ gauge theory. It is also known that the electromagnetic interaction is a $U(1)$ gauge theory, and it works perfectly on its own (by "it works" I mean on the paper, obviously there are many scattering processes that can't be explained via QED only, but one can study QED as a standalone theory). My question is: can the same be done for the weak interaction, with all three gauge bosons? Can it be described as a standalone gauge theory?

To clarify, I'm aware of the Fermi theory, but that's not what I'm interested in, because it isn't a gauge field theory.

  • $\begingroup$ I'm not sure what the question here is, exactly - what doubts do you have about the obvious choice of a SU(2) gauge theory as a standalone gauge theory? $\endgroup$
    – ACuriousMind
    Commented Oct 29, 2021 at 18:42
  • $\begingroup$ @ACuriousMind the only doubt that I have is that I've never seen it written down anywhere! Would it have an actual $Z$ boson, or just a $W^3$? Would it still be a $SU(2)_L$ group? $\endgroup$ Commented Oct 29, 2021 at 18:46
  • $\begingroup$ No, both the SU(2)$_L$ and the U(1) are broken. It is just a special combination of $T_3$ and Y which escapes breaking, but fails to commute with the charged generators, even though it commutes with the one coupling to the Z. It's the genius of the SM. $\endgroup$ Commented Oct 29, 2021 at 19:41

2 Answers 2


There is no gauge theory describing the weak interaction after electroweak breaking. The meaning of symmetry breaking, after all, is that the resulting effective theory no longer has all the symmetries of the theory prior to breaking.

There is no problem at all in writing down a standalone $\mathrm{SU}(2)$ gauge theory, and you're free to choose the representations of the particles, so it can also only act on particles of a certain chirality if you want. Such a gauge theory will have three massless gauge bosons, whether you call them "W and Z bosons" or something else is entirely up to you.

The reason you don't see such a theory discussed in the context of the weak interaction is that that's not what the weak interaction is: After electroweak breaking, the W and Z bosons are massive and the Ws are electrically charged - this isn't the particle content of a SU(2) gauge theory, so studying SU(2) gauge theories won't tell you anything useful about the weak interaction. In fact, since the bosons have become massive, they cannot any longer be described as the force carriers of any Yang-Mills gauge theory.

In contrast to this, the photon remains massless and uncharged after electroweak breaking, and so is well-described by a $\mathrm{U}(1)$ gauge theory.

  • $\begingroup$ Could the standalone $SU(2)$ gauge theory be written so that it couples two of its bosons to only left components, while the third couples to both L and R (as Z does)? In a gauge-invariant manner of course. $\endgroup$ Commented Oct 29, 2021 at 20:01
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    $\begingroup$ @MauroGiliberti No - what choice you have for every field (chiral or not) is the SU(2) representation it transforms in. Except for the trivial representation, there is no representation where one of the three generators does nothing - you always get coupling of all the gauge bosons, or coupling of none. They're just different components of the same gauge field, not individual fields you could treat independently from one another. $\endgroup$
    – ACuriousMind
    Commented Oct 29, 2021 at 20:11
  • $\begingroup$ @MauroGiliberti I didn't claim the SU(2) gauge theory would be "the weak interaction except for the masses and the charge". Adding masses alone inevitably destroys the gauge character of the theory - the resulting theory is wholly different, it's not a slight modification or anything. There's plenty of other things that will differ between the SU(2) gauge theories and "the weak interaction" (note that the mass adds an additional physical polarization to the boson, for instance), and trying to enumerate a complete list of such differences strikes me as pointless. $\endgroup$
    – ACuriousMind
    Commented Oct 29, 2021 at 20:19
  • $\begingroup$ So the answer to my question is "No, you can write a SU(2) theory but that would be nothing like the weak interaction", right? There is a fundamental difference between the electromagnetic U(1) (or the color SU(3)) and the SU(2). $\endgroup$ Commented Oct 29, 2021 at 20:28

No; it's tricky. Recall, QED runs on the single generator of the four in the SM ($\tau_3+{\mathbb I}/2$) that escapes SSBreaking by leaving the Higgs doublet v.e.v. invariant upon exponentiation, even though it does not commute with all three of them.

So, singling out QED from the SM is the world before 1967 and is easy because the photon is massless, in contrast to the Ws and the Z, so massive that they damp themselves out at low energies; this is what makes the Weak interactions weak.

You could go to higher energies, around 85GeV and study the WI subtractively, by taking out QED, but this is a bit contrived, to the extent there is no surviving group for it which is unbroken, unlike in QED. All generators of the WI are SSBroken, and fail to commute with QED, so it's not a standalone theory.

In practice however, writing down the "after SSBreaking" Lagrangian is what one in fact does, focussing on the charged and neutral weak currents and their chiral peculiarities, and suitably renormalizing them, etc... The cynosure or signature in these studies is focussing on the three quasi-Goldstone-boson components of the Higgs multiplet, but I don't have a dramatic, cogent illustration of how to do this.


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