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What are the weak isospins (T3 values) of various hadrons, including the proton, neutron, mesons, hyperons and other hadrons? How is the weak isospin calculated for any hadron?

Published sources provide T3 only for fundamental fermions, that is, quarks and leptons. In the fundamental bosonic sector, the photon's T3 is (0, 1), the gluon's is 0, the Higgs boson's is -1/2, the Z boson's is 0 and the charged weak bosons' is ±1. No such information appears for composite particles.

One could calculate this using Q=T3+YW/2. However, the weak hypercharge (YW) values for hadrons are also not available.

Supposedly, it is possible that the weak isospin of all hadrons is 0, since the weak interaction does not operate on the hadron as such, only on its constituent quarks. Is this the case?

Thanks in advance.

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  • $\begingroup$ you could use this en.wikipedia.org/wiki/Eightfold_Way_(physics) to get the quantum numbers for the calculation $\endgroup$
    – anna v
    Commented Jan 19, 2017 at 11:09
  • $\begingroup$ Well, you came close: Since weak isospin and hypercharge are spontaneously broken, they are mostly ill-defined (Fabry-Picasso theorem) and are pointless to even "fake", although see 296618. Very cautiously, you could go through the pointless, but possibly instructive, exercise of coupling the Ws and Z's to the quarks inside hadrons, and inferring the corresponding W/Z-hadron couplings in an effective lagrangian, and thereby assign Weak I to the hadrons as a mnemonic of the W/Z couplings. It might be like climbing Mt Olympus in flipflops... $\endgroup$
    – Cosmas Zachos
    Commented Jan 19, 2017 at 20:11
  • $\begingroup$ As @anna_v points out, looking at the valence quarks of hadrons and ensuring their weakly interacting valence quarks are available, you might write effective vertices, e.g. for $\pi^- \to \mu +\bar{\nu}_\mu$, so $G_F \cos\theta_c \partial^\mu \pi^+ ~\bar{\nu}_\mu \gamma_\mu P_L \mu ~f_\pi$. So $Q_3(\pi^-)=-1, ~ Q_3(\mu)=-1/2=Q_3(\bar{\nu})$, etc, but these are broken symmetries, so I almost feel like inviting you to play with a malfunctioning gun... one wrong move and you get nonsense. $\endgroup$
    – Cosmas Zachos
    Commented Jan 19, 2017 at 20:50
  • $\begingroup$ Finally, Y is a canard, since it is a linear combination of charge, Q, a good quantum number, and $T_3$ the spontaneous broken one you insist on using past expiration. So you write the effective lagrangian term for quark and lepton currents, $2\sqrt{2}G_F(J_\mu^+ J_\mu ^- ~+~ J_\mu^{n.c.} J_\mu^{n.c.}) $ and rummage through it for pieces overlapping hadrons, and sum up the weak isospin numbers of the engulfing hadron. By why bother? if you have the underlying interaction, you have already found your vertex--you don't need an ill-defined quantum number to restrict it!! $\endgroup$
    – Cosmas Zachos
    Commented Jan 20, 2017 at 17:12

2 Answers 2

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This is an older question, but for reference, I think it's worth pointing out that just because the case realised in nature is that quarks confine below the electroweak scale (and therefore the hadrons are not in irreps of $SU(2)_L$), doesn't mean we couldn't explore hadronic $SU(2)_L \times U(1)_Y$ as a learning tool.

For example, in the simplest case of Dynamical Electroweak Symmetry Breaking, where we remove the Standard Model Higgs and attempt to give mass to $W^{\pm}$ and $Z$ using only the quarks. Here the strong QCD gauge coupling condenses pairs of quarks $(q_L)_i (q_R)_i$, giving them a vacuum expectation value. We can then introduce a field $\Sigma$ that parameterises these pairs as a sigma model, which could be linear or non-linear depending on the observables we're interested in.

For the simplest breaking case of the up and down quark condensing, we have $$ SU(2)_L \times SU(2)_R \rightarrow SU(2)_V,\\ \implies \Sigma_{LSM} \sim (\sigma, \pi_1, \pi_2, \pi_3) \rightarrow \Sigma_{NLSM} \sim (\pi^0, \pi^+, \pi^-) $$ (abusing sigma model notation a little).

These pions couple to the gauge bosons, and therefore have perfectly well-defined hypercharges. Shifting to the broken vacuum, the pions become the masses of the gauge bosons and then the hypercharge isn't well-defined.

One could certainly add quarks to the model. For example, adding the strange quark gives a type of Eightfold Way ($SU(3)_L \times SU(3)_R \rightarrow SU(3)_V$), except that the hypercharge of this is the genuine weak hypercharge. One can add all of the quarks in this way, such that we have $35$ mesons with well-defined hypercharges, and a $W^\pm$ mass of around $50$ MeV.

Unfortunately, this doesn't match nature, and we need the Higgs field. But we couldn't know this without experimentalists.

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First of all, You should note that left quark doublets form the fundamental representation $2$ of the weak $SU_{L}(2)$, while right quarks are $SU_{L}(2)$ singlets. The quark bounded states formed below the QCD confinement scale consist of massive non-chiral quarks. This means that they aren't irreducible representations of the electroweak symmetry group (except, of course, its EM sub-group, which is abelian), and thereofore don't have well-defined isospin. At most they can be represented as the direct sum of the irreducible representations of the $SU_{L}(2)$ group with given weak isospins.

Second, these bounded states even can't be expanded on irreps! Really, they are formed from three kinds of quarks, one of which (precisely, the $s$-quark) doesn't have its doublet pair (the $c$-quark)! This is not the problem from the theoretical point of view, since below the Higgs mechanism scale even interaction mediators (namely, the photon, $Z$- and $W_{\pm}$ bosons) don't belong to the irrep below the Higgs mechanism scale.

In the result, when discussing weak processes involving quark bounded states (mesons, nucleons, hyperons etc) people typically prefer describe them microscopically, by assuming the weak mediated scattering of one of quarks located inside the chiral bag.

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