Equation (28.33) in Matthew Schwartz's QFT text book $$ \mathcal{M}(\pi^+ \rightarrow \mu^+ \nu_\mu) = \frac{G_F}{\sqrt{2}}F_\pi p^\mu \bar{\psi}_{\nu_\mu}\gamma^\mu(1-\gamma^5)\psi_\mu \tag{28.33} $$ where $F_\pi$ is pion decay constant. But, how can QCD pions interact with SM neutrinos and leptons?

Axial current associated with axial part of chiral symmetry excites vacuum to generate QCD pions $$\langle0|J^{a5}_\mu|\pi^a\rangle=i\frac{f_\pi p_\mu}{\sqrt{2}}.\tag{28.30'}$$ Since chiral symmetry is among 3 flavors of quarks, this axial current consists only of quarks. But Matthew Schartz's textbook seems to identify axial current of leptons with axial currents arising from QCD chiral symmetry by introducing Equation (28.32) $$ \mathcal{L}_{4F} = \frac{G_F}{\sqrt{2}}J_\mu^LJ_\mu^L \tag{28.31} $$ where $$ J_\mu^L = \bar{\psi}_u\gamma^\mu(1-\gamma^5)\psi_d + \bar{\psi}_{\nu_\mu}\gamma^\mu\gamma^5\psi_\mu+\cdots \tag{28.32} $$ where $\psi_\mu, \psi_{\mu_\nu}$ refers to muon and muon neutrino fields.

If subgroup of chiral symmetry is gauged with electroweak gauge bosons corresponding to electroweak symmetry $SU(2)_L\times U(1)_Y$ and is spontaneously broken by QCD vacuum, then 3 pions that are associated Goldstone bosons should be eaten by gauge bosons and become unphysical. Does that pions are physical mean electroweak symmetry $SU(2)_L\times U(1)_Y$ is not broken by QCD vacuum and therefore these are not subgroup of chiral symmetry?


1 Answer 1


This is a cluster of questions, with a very good one in the end, which is, in fact, the very question that inspired Susskind to introduce technicolor for EW SSB in the late 70s. I'll be very schematic to get the ideas across, rather than catching every factor of $\sqrt 2$ and such.

Both the strong and the EW interactions work on conserved currents and gauge fields. In turn, the currents consist of quarks (and also leptons for EW), but, through the magic of confinement and chiral symmetry breaking, the very same currents couple characteristically to their proper pseudoscalar mesons (with the same quantum numbers), such as the pions, for the sake of argument here. In an effective low-energy lagrangian for such mesons, the very same currents consist of mesons, instead of quarks, but are, naturally, equivalent.

So take the conserved hadronic axial current proper to the positive pion, for specificity, $$ J_{\mu}^{+5} \sim \bar{\psi}_d \gamma_\mu \gamma_5 \psi_u \sim f_\pi \partial_\mu ~\pi^+ +..., $$ whose vacuum excitation matrix element you acknowledge above, (28.30').

Through ferocious nonperturbative workings, the gluons of QCD have spontaneously broken the chiral symmetry, i.e., even though this current is still conserved, it is linear in the field and pops its excitation in and out of the vacuum, with strength $f_\pi\sim 92$MeV, a strong/hadronic scale. You might consider terms like $\partial^\mu \pi^- J^{-+5}_\mu/f_\pi $ in the corresponding effective lagrangian. Farewell strong interactions, for now.

The EW interactions, through the Higgs mechanism, also SSBreak some symmetries, with characteristic currents, among others, (28.32), $$ J^{+L}_\mu= \bar{\psi}_d\gamma_\mu(1-\gamma^5)\psi_u + \bar{\psi}_{\mu}\gamma^\mu(1-\gamma^5)\psi_{\nu_\mu}+\cdots $$ The gauge symmetry of these interactions dictate/require a coupling of the form $$ gW^{-\mu}(J^{L+}_\mu + v \partial_\mu h^+ +...) $$ in the lagrangian, where $h^+$ is the Higgs doublet goldston to be promptly eaten by the $W^+$. Here, "for now", we emphasize it has the quantum numbers of a spontaneously broken EW generator, and the corresponding gauge field, $W^+$. Here the numbers matter: in relative terms, $v\sim 246$GeV is enormous.

Since the mass of the obese Ws is large, the relevant effective lagrangian devolves to the (28.31) you wrote.

The key point: even though different, the quark part of the EW current $J^{+L}_\mu$ overlaps that of the purely quark axial $J^{+5}_\mu$. So taking a matrix element of (28.31) between the hadronic vacuum and a charged pion will produce a $\bar{\nu} \mu^+$ as in (28.33) you are curious about.

The good question: back to the couplings of the W, it actually couples to the $\pi^+$ as well as the Higgs-doubet goldston, $h^+$, $$ gW^{-\mu}( v \partial_\mu h^+ +f_\pi\partial_\mu \pi^+ +...) $$ Why wouldn't it eat it too?

It actually does, but... The state the W ate is mostly Higgs: this is the combination $$ \frac{v}{\sqrt{v^2+f^2_\pi}}\left (h^+ +\frac{f_\pi}{v}\pi^+\right ), $$ so it is "piony" by less than half per mil. And of course, the orthogonal intact state cavorting about and impersonating a pion is effectively pure pion.

Nevertheless, you now probably saw the charm of the idea to model-builders... If you started speculating about hypothetical strong interactions at a much higher scale, comparable to v, could you then..., then... ah, nevermind, that should be a sequel question...

There are untold subtleties in the story, needless to say, that have led to lots of wrong papers by grownups.

  • $\begingroup$ Thank you for a clear answer. Can you explain the last comments that you mentioned as a sequel question? Do you refer to technicolor model? $\endgroup$
    – Liberty
    Commented Sep 30, 2017 at 1:27
  • $\begingroup$ Yes, indeed... . $\endgroup$ Commented Sep 30, 2017 at 2:46
  • $\begingroup$ I guess it depends on what we mean by "eaten". We say the three Higgs Goldstones are eaten, because we could redefine the SU(2) scaler field and set those Goldstones to zero, and the two Ws and Z become massive. In the case of pions, although they couple to Ws and Z, I don't see how to set the pions to zero and to make Ws or Z massive at the same time. $\endgroup$ Commented Nov 13, 2022 at 17:41
  • $\begingroup$ You may mix pions and Higgs goldstons, as described, and gauge-absorb one linear combination, while leaving the other one as a propagating physical field. "Eating", in this sense, is a narrowly and strictly defined operation, not a fantasy procedure, as you might be misconstruing it as... $\endgroup$ Commented Nov 13, 2022 at 18:45
  • $\begingroup$ ... so a propagating pion which is less than 0.05% higgs is not a big deal: do the calculation! $\endgroup$ Commented Nov 13, 2022 at 20:34

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