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In chapter 19 of An Introduction to Quantum Field Theory by Peskin & Schroeder, they discuss spontaneous symmetry breaking (SSB) at low energies in massless (or nearly massless) QCD, given by $$\mathcal{L}_{\text{QCD}} = \bar{Q} (i\gamma^{\mu}D_{\mu}) Q - m\bar{Q}Q,$$ where $Q = \left( \begin{matrix} u \\ d \end{matrix}\right)$ is an $SU(2)$ quark doublet.

Due to quark-antiquark pairs being produced from the vacuum at low energies (the $q\bar{q}$ condensate), we have a non-zero vacuum expectation value (vev), which results in SSB of the global $SU(2)_A$ symmetry. For massless QCD, this is an exact symmetry so 3 goldstone bosons [one for each generator of $SU(2)$] appear. For massive QCD (where the quark masses are small), the mass term in $\mathcal{L}_{\text{QCD}}$ demotes this to an approximate symmetry, and so we obtain 3 massive pseudo-Goldstone bosons.

If we take the quark masses to be small (and so we have an approximate symmetry), they show (eq 19.92) that we have a partially conserved axial current (PCAC relation) from pion production from the vev, via $$ \langle 0 | \partial_{\mu} j^{\mu,a}_5 | \pi^b(p) \rangle = - p^2 f_{\pi} \delta^{ab} \quad \Rightarrow \quad \partial_{\mu} j^{\mu,a}_5 \propto m_{\pi}^2,$$ where $m_{\pi}$ is the mass of the pseudo-Goldstone bosons produced from SSB.

So if we have massive quarks in $\mathcal{L}_{\text{QCD}}$, then we have an approximate symmetry and obtain 3 massive pseudo-Goldstone bosons through SSB. If we have massless quarks in $\mathcal{L}_{\text{QCD}}$, then we have an exact symmetry and obtain 3 massless Goldstone bosons.

Now, here is the problem I am struggling to answer. In the text, they refer to these goldstones as being the three pions $\pi^+$, $\pi^-$, and $\pi^0$. From measurements, we know pions are massive particles. From the above reasoning, if the goldstones from SSB of QCD are in fact pions, there must be non-zero quark masses in the QCD Lagrangian. However, the usual Dirac mass terms violate the electroweak gauge symmetry $SU(2)_L \times U(1)_Y$, so it's not possible to have these mass terms appear in the Standard Model, and quarks must obtain their mass via the Higgs mechanism from SSB of the electroweak gauge group.

So, to preserve gauge invariance in the Standard Model $SU(3)_c \times SU(2)_L \times U(1)_Y$, we must then have that there are no quark mass terms and the global axial symmetry is exact, meaning our Goldstone bosons are massless and cannot be the same particles as pions. This seems to be a complete contradiction, and I really don't understand how we can have both massive pions that correspond to pseudo-goldstones and not break gauge invariance in the SM Lagrangian.

As a side question, I don't understand the logical jump made associating pions with the Goldstone fields in the first place. Pions are composite particles consisting of $u$ and $d$ quarks, so why should/can they even be candidates to Goldstone/pseudo-Goldstones bosons? It makes sense that they are related to the $q\bar{q}$ condensate in QCD due to the nature of their composition, but other than this I fail to see the connection.

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    $\begingroup$ I thought it is the gauge boson mass terms that would violate gauge invariance, not the quark mass terms. In QED, which is not broken, the photon is massless, but the charged particles have mass. Or do I misunderstand your question? $\endgroup$
    – flippiefanus
    Commented Jul 16, 2023 at 12:21
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    $\begingroup$ @flippiefanus Indeed, you might well. The weak bosons are hardly involved in his question. It is the quark masses which also violate gauge invariance, absent SSB, the second job of the Higgs, and the most important one, in the opinion of some (many?). $\endgroup$
    – Cosmas Zachos
    Commented Jul 16, 2023 at 13:27
  • $\begingroup$ @flippiefanus In the Chiral basis, $\psi = \left( \begin{matrix} \psi_L \\ \psi_R \end{matrix} \right)$, so the usual mass term would be $m\bar{\psi}\psi = m(\bar{\psi}_L \psi_R + \bar{\psi}_R \psi_L)$ which violates Electroweak gauge symmetry $SU(2)_L \times U(1)_Y$ since the LH and RH fields transform differently under the gauge group. $\endgroup$
    – Flynn Linton
    Commented Jul 22, 2023 at 6:05

2 Answers 2

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In a world without EW SSB, pions would, indeed, be perfect massless! Pion masses reflect two different SSBs.

There are two SSB scales involved in the SM: the electroweak spontaneous breaking of $SU(2)_L \times U(1)_Y$, with an order parameter of about 1/4 TeV; and the chiral spontaneous breaking of the three axial generators (not closing into an $SU(2)$, of course) with the quantum numbers of the three pions, with order parameter about 1/4–1/5 GeV, close to $\Lambda_{QCD}$; two completely different phenomena, understood very differently.

There is no sense in talking about pions above the higher scale; but below it, you do have effective quark masses, generated from the Yukawa couplings: the heart of the SM, regardless of how poorly they teach it to you.

So, below that upper scale, and around the lower one's χSSB that P&S detail, there are quark masses, for all practical purposes, and all they say makes eminent sense! Check that all statements are consistent and reasonable in all three regions demarcated by these two scales.

This includes your puzzlement about (pseudo)Goldstone bosons: the Goldstone property refers to the non-linear transformation law of the particles under the SSBroken symmetries. Pions, composite or not, are interpolated by operators transforming in the (Wigner-Weyl, or else) Goldstone mode: and our real world pions choose the latter.

This is why, as you are taught, it is very impractical to think of pions as ordinary hadron bound states, which they also are. Primarily, they are goldstons. In fact, there is a small energy region in-between $\Lambda_{QCD}$/confinement-scale and the above χSSB, where you may consider Goldstone degrees of freedom and constituent massive quarks, the chiral bag models, but angels fear to tread there... You really need not get interested in it.

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  • $\begingroup$ So below the Electroweak scale, does it make sense to say that the pion/Pseudo-Goldstone mass comes entirely from the QCD Lagrangian, or does the Yukawa interaction also contribute to the mass at this scale? In essence, I'm asking if it makes sense to discuss SSB of Electroweak below the electroweak scale. $\endgroup$
    – Flynn Linton
    Commented Jul 22, 2023 at 5:57
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    $\begingroup$ The latter; the pion mass comes from both SSBs. Below the EWSSB, only its consequences (quark masses, etc,...) are visible, and some hidden trees of it, not relevant here.... $\endgroup$
    – Cosmas Zachos
    Commented Jul 22, 2023 at 13:08
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As you say, $SU(2)_A$ is a symmetry of the Lagrangian before Electroweak symmetry breaking. But, what we call 'particles', (in particular pions) are always to be understood as excitations on top of the true vacuum, which is obtained after electroweak symmetry breaking. There is no contradiction here (but perhaps my words are not so helpful)

Also, $SU(2)_A$ is not actually a subgroup. $SU(2)_L \times SU(2)_R \to SU(2)_V$ is the breaking pattern, and $\mathfrak{su}(2)_A$ does indeed exist as the broken generators, but there is no $SU(2)_A$, $$(U, U^\dagger ) (V, V^\dagger) = (U V, U^\dagger V^\dagger) \neq (U V, (U, V)^\dagger)$$

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    $\begingroup$ Just to be clear, it's okay to have quark masses in the lagrangian since the Golstone bosons are interacting with the (already broken) $U(1)_{EM}$ vacuum, rather than the chiral $SU(2)_L \times U(1)_Y$ vacuum? $\endgroup$
    – Flynn Linton
    Commented Jul 22, 2023 at 5:53

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