What are the weak hypercharge and weak isospin quantum numbers of the helicity states of the $W^\pm$ and $Z^0$ bosons?

The W boson is a spin-1 massless particle. Consequently it has three helicity states, negative (left-handed), positive (right-handed), and neutral. As discussed in this paper, these states restrict which orientations of quarks they can interact with:

The emitted $b$ quark can be regarded as massless compared to the top quark, and hence expected to be predominantly of negative helicity (left-handed), meaning that its spin points opposite to its line of flight. The emitted $W$ boson, being a massive spin-1 particle, can assume any of three helicities: one longitudinal ($W_0$) and two transverse states ($W_−$, left-handed and $W_+$, right-handed)

Int.J.Mod.Phys.A24:2899-3037,(2009), Marc-André Pleier, Review of Properties of the Top Quark from Measurements at the Tevatron

The helicity states of a particle like the electron act very much differently, and they carry different weak hypercharge and/or weak isospin quantum numbers. So really this is two questions. First, are weak hypercharge and weak isospin good quantum numbers for the helicity states of the W and Z bosons? And if they are good states, what are the quantum numbers?

  • $\begingroup$ You mean probably massive not massless $\endgroup$
    – ohneVal
    Dec 17, 2020 at 8:39

1 Answer 1


Hypercharge and weak isospin are obviously broken generators of $SU(2)\times U(1)$, so they're not conserved which also means that the mass eigenstates are not eigenstates (or representations) of those generators. W-bosons and Z-bosons are states where the breaking of these symmetries is most manifest, so of course, in general, they don't have well-defined values of those quantities.

The fundamental fields $W^i_\mu$ and $B_\mu$ carry vanishing hypercharge $Y=0$ because $U(1)_Y$ is an Abelian group so its messengers are neutral. However, their helicity components, especially the longitudinal ones, if they're moving quickly, are mixing with components of the Higgs boson (or several Higgs bosons) that carries a nonzero hypercharge $Y$. So especially for the longitudinal component, the value of $Y$ is a superposition of two possibilities.

The third component of the isospin suffers from the very same uncertainty because the sum of the isospin and the hypercharge is the electric charge - which is well-defined and conserved. One could discuss that the $T_3=-1,0,+1$ triplet (the W-bosons and the $W_3$ combination of the photon and the Z-boson) and the $T_3=0$ singlet states (the $B_0$ combination of the photon and the Z-boson) are getting mixed with the $T_3=\pm 1/2$ doublet of the Higgs field but the conclusion - and the degree of mixing of states with different eigenvalues - would be equivalent to the hypercharge discussion.

(At loop level, the states of the W-bosons and Z-bosons may receive additional contributions from multi-particle states with even different values of the hypercharge and isospin. If $T_3$ may change by $1/2$ by a simple processes, a sequence of processes in a loop diagram may change it by anything.)

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    $\begingroup$ Oh this is so precisely what I needed and so obvious now that you say it. I'll +1 it when I get my votes back tomorrow. Thank you thank you thank you. $\endgroup$ Feb 18, 2011 at 7:37
  • $\begingroup$ It was a pleasure, Carl, and don't get too worried about the points, my life (and even reputation here) don't depend on it. ;-) $\endgroup$ Feb 18, 2011 at 9:46
  • $\begingroup$ It's just a reminder to me. By the way, last night I found my copy of "The Standard Model: A Primer" by Burgess and Moore, and it helped a lot. Meanwhile, my new paper is moving along. I just had the realization that I can probably trim off the "diquark" solutions by restricting myself to the solutions that have a non-zero volume in their basin of attraction. I.e., if you start with a random Hopf algebra element, and iterate it by squaring, the results are all quarks and leptons. Some of the quarks and leptons are missing but the mass interaction makes them. But it skips the diquark solutions. $\endgroup$ Feb 18, 2011 at 21:38
  • $\begingroup$ @carl, when did you seriously decide to become an amateur physicist? $\endgroup$ Feb 19, 2011 at 4:13

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