In his chapter on the electroweak unification, Griffiths introduces the weak hypercharge explaining that his attempt at introducing full weak isospin symmetry with the Pauli spin matrices failed because the neutral current corresponding to the third isospin matrix does not contain right-handed fields (the actual weak neutral current does).

This seems to imply that it is impossible to construct a realistic $SU(2)$ model for the weak interaction with the associated conserved weak isospin and therefore it makes sense to introduce the $SU(2)\times U(1)$ and the associated conserved charge (weak hypercharge).

However, the weak interaction IS described by the $SU(2)$ group and it already has its three generators corresponding to the three gauge bosons (two charged and one neutral).

If the above notion is true, then the statement that the weak isospin is the conserved charge under $SU(2)$ is wrong (after all, according to Griffiths "we could contemplate a full “weak isospin” symmetry, if only there were a third weak current").

I am really struggling to understand what he means here. Does the existence of a "right-less" neutral current spoil the $SU(2)$ model of the weak interaction? Is this yet another reason (among the absence of massive gauge bosons) we need the electroweak unification?

  • $\begingroup$ Glashow understood that weak mixing is necessary; is this your question? Your text details weak mixing to a "t". $\endgroup$ Commented Apr 21, 2023 at 22:17
  • $\begingroup$ Does the answer address your confusion? $\endgroup$ Commented Apr 22, 2023 at 11:08

1 Answer 1


Linked answers; ditto, and also. Glashow's neat argument in 1961 had the power of simplicity/minimality, a full dozen years before the experimental discovery of neutral currents in 1973. Griffiths tries to make things simple for you by already assuming the experimental results of 1973, namely the existence of the neutral current which contains both L-chiral and R-chiral pieces, as you summarize. Glashow, by contrast, had to guess this structure by pure logic and economy of modeling!

He observed the commutator of the known weak charge currents (actually their charges) is L-chiral, $T_3$, which could therefore not be the electric current/charge Q, vectorlike: an equal blend of both L and R-chiral. So the most economical extension of this L-chiral weak isospin SU(2) is to just embed it and electromagnetism into a minimal structure involving an independent weak hypercharge U(1) commuting with the SU(2); and here comes the daring: unlike what most theorists driven by beauty were expecting, the hypercharge would need to have capricious and arbitrary chirality, dictated by consistency constraints, a tight structure, and supremely counterintuitive! (Feynman called it "cockeyed", a gut sentiment shared by most students...)

Both $T_3$ and Y would be neutral, and trace-orthogonal to each other, Tr$T_3 Y=0$, but could contain $Q=T_3+Y/2$, vectorlike, generating a very different U(1) indeed. The trace-orthogonal combination, call it $N= T_3-Y/2$, would thus also be neutral, connected to the charge of an undiscovered neutral current$^\natural$, coupling to an undiscovered gauge boson (triumphantly discovered a dozen years later). Tr$QN=0$.

By above, the chiralities of $N=2T_3-Q$ and its linked neutral current are freaky: a piece L= (V-A)/2 from the SU(2) generators and a piece V from electromagnetism!! Quite daring. A prediction of a neutral current, twisted brother to the EM current, both of them a Weinberg angle rotation of hypercharge and $T_3$. So you definitely do have a full weak SU(2) symmetry before spontaneous breaking. "Unification" might confuse you as too ambitious. "Entwining" might be more intuitive.

Both your text and WP, as well as the PDG reviews, detail which combinations of the two neutral gauge bosons couple to which currents/charges.

  • In the fantasist theorist's limit where the mixing ("Weinberg's" !!??!!) angle goes to zero, the coupling of the charged currents ($g=e/\!\sin\theta_W$) becomes hugely larger than the coupling to photons, and so, for notional purposes, the hypercharge and EM U(1)s serve the same role, while the weak interaction retreats to the SU(2) and the neutral current becomes pure L. This is not our real world, as per experiment. Glashow had the good sense to stay away from that limit!

$^\natural$ Geeky: you won't find this in footnotes in books. The above charge N is a special limit for $\theta_W= \pi/4$, that is $g=g'$, the limit where SU(2) couples with the same strength as U(1); however, in real life, the U(1) couples less strongly than SU(2), so the effective charge coupling to the Z boson is, instead, $2T_3 -2\sin^2\theta_W ~ Q$, as summarized in texts, including yours. For the above even-handed limit, this charge reduces to N.

  • $\begingroup$ What I take from this answer (and the answers you have linked) is that there is no "complete" $SU(2)$ theory of weak interaction - $SU(2)\times U(1)$ is required. Complete in the sense that it also takes into consideration the electric charges of the weak current. $\endgroup$
    – Hrach
    Commented Apr 23, 2023 at 7:26
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    $\begingroup$ The actual non-vanishing of the Weinberg angle is an experimental fact, albeit a hardly surprising one… $\endgroup$ Commented Apr 23, 2023 at 11:15
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    $\begingroup$ I see. I think the vanishing of the Weinberg angle illustrates your point. Thanks. $\endgroup$
    – Hrach
    Commented Apr 23, 2023 at 14:41
  • $\begingroup$ Sphaleron physics utilizes that unrealistic fantasist's limit of $\theta_W\to 0$ where weak and EM are unentwined. $\endgroup$ Commented Apr 24, 2023 at 16:11

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