How can we have a quark condensate without a quark potential?
Yes, you are. The Goldstone theorem is 95% of SSB. SSB is a statement of the realization of a symmetry; the relevant currents are still conserved, but the vacuum, the ground state (the lowest energy state/s of the theory) is not invariant under the symmetries in question. Such symmetries shift you from one ground state to another. The symmetry is realized nonlinearly (in the Nambu-Goldstone mode), and it is hard to see it directly, but the relevant action is still invariant under such transformations.
Explicit breaking means no conserved currents, no invariant action, no degenerate ground states.
(Actually, in chiral SSB systems, you may also have a small explicit breaking on top of it, so your currents are not fully conserved--they are "partially conserved: PCAC.)
If you wish to model the SSB with a potential in an effective theory, then the SSB is more tractable, and Landau-Ginsburg utilized such in superconductivity, with an intricate Cooper-pair collective structure for the ground state; further, Gell-Mann and Levy in the σ model, which captures the salient features of QCD χSSB, in lieu of the dynamical condensation mechanism that seems to grate on you, etc. (Their flavor group is that of only the two lightest nucleons/quarks, $SU(2)\times SU(2)$, but conceptually the same as here.)
In the weak interactions, the jury is still out, for lack of information, but technicolorists could argue with you endlessly about such... They think of Weinberg's model as the analog of the σ-model for the EW interactions.
Before QCD was discovered, in 1960, the σ-model envisioned, prophetically, correctly (!), all the properties you ask for, and is the mainstay in any good QFT course. Itzykson & Zuber, Cheng & Li, T D Lee, M Schwartz, or P&S cover it extensively. Mainstay is not a breezy exaggeration. Really.
It has a quartic potential involving the σ and three πs (you may trivially extend it to 8 pseudo scalars, as, of course, one has), the σ picks up the vev. at the bottom of the potential, (the prototype for the Higgs), and, sure enough, this vev generates nucleon masses through the Yukawa couplings. More importantly, the current algebra works out like a charm, including the explicit breaking overlay on top of the SSB.
With the advent of QCD and lattice simulations of its nonperturbative strong coupling dynamics, one could confirm any and all predictions of the σ-model, using subtler tools, and confirmed the nontrivial v.e.v. s of scalar operators, quark field bilinear now, with the quantum numbers of the σ, like the one you write down, etc... They further confirmed the Dashen formulas for the masses of the pseudoscalar pseudogoldstons just right.
The relevant WP section explains the basics--yes, the Goldstone construction is SSB, all right.
Edit in response to comment question : Yes but.
The σ-model, of course, summarizes the group multiplet behavior of the quark bilinear interpolating fields, $$ \vec{\pi} \leftrightarrow \bar{q}\gamma_5\vec{\tau} q, \qquad \sigma \leftrightarrow \bar{q} q, $$ so that the three SSB axial transformations rotate them the same, $$ \delta \vec{\pi}= \vec{\theta} \sigma, \qquad \delta \sigma = \vec{\theta} \times \vec{\pi} ,\\ \delta (\bar{q}\gamma_5\vec{\tau} q)= \vec{\theta} \bar{q} q, \qquad \delta (\bar{q} q) = \vec{\theta} \times \bar{q}\gamma_5\vec{\tau} q . $$
Now, while the gimmick fiat potential of the σ-model $\lambda (\sigma^2+\vec{\pi}^2 -f_\pi^2)^2$ constrains the σ vev to $\langle \sigma\rangle=f_\pi$ and $\langle \vec{\pi}\rangle=0$, so $ \langle\delta \vec{\pi}\rangle= \vec{\theta} f_\pi, \quad \langle\delta \sigma\rangle = 0$, so the pions are goldstones and the sigma is massive, there simply is no such thing as a meaningful "potential" for the composite interpolating fields.
The vev is enforced by the aggressively nonperturbative roiling glue QCD vacuum, (Thacker has done impressive work probing it), so that $\langle \bar{q} q\rangle=-(250MeV)^3$, and $\langle \bar{q}\gamma_5 \vec{\tau} q\rangle=0$, so that $$ \langle \delta (\bar{q}\gamma_5\vec{\tau} q)\rangle= -\vec{\theta} (250MeV)^3 , \quad \delta (\bar{q} q) = 0 . $$
The group theory is the same, which is what SSB is all about, but no usable "potential" is there to enforce the vevs. This is the reason such effective lagrangian models are still around and kicking: to humanize group theory.
I'll address part of the second question first.
Question 2: Ground state of what? There is no potential. What potential are we minimizing...?
"Ground state" (aka vacuum state in relativistic QFT) simply means the state that minimizes the expectation value of the Hamiltonian. No separation into "kinetic" and "potential" terms is needed for this, and even when such a separation makes sense, the ground/vacuum state is still the one that minimizes the expectation value of the whole Hamiltonian, including the kinetic terms.
Question: How does this qualify as spontaneous symmetry breaking?
Chiral symmetry is explicitly broken by the quark mass terms in QCD, but the symmetry is not restored when the mass terms are omitted, so the symmetry is also spontaneously broken.
For a scalar-field illustration of explicit-and-spontaneous breaking, let $\phi$ is a complex scalar field with real part $\phi_r$ and consider a potential of the form $$ V(\phi)= \beta\phi_r + \mu|\phi|^2+\lambda|\phi|^4 \tag{1} $$ where the parameter $\beta$ is very close to zero, and $\mu<0$ and $\lambda>0$. If $\beta$ were exactly zero, we'd have SSB (if $\mu$ is sufficiently negative)${}^{[1]}$, because the energy cost of oscillations in the phase of $\phi$ goes to zero in the long-wavelength (low-momentum) limit; the Hamiltonian is independent of the phase when $\beta=0$. If $\beta$ is non-zero but close to zero, then the symmetry is explicitly broken. Heuristically, instead of a circular valley (parameterized by the phase of $\phi$) with the same depth everywhere around the circle, now we have a circular valley that is tilted, so one point in the valley is lower than all the others. But we still have a circular valley, and the energy cost of oscillations along the valley is still relatively small in the long-wavelength limit, so we still get almost-Goldstone-bosons that are almost massless. In this sense, the symmetry is both explicitly and spontaneously broken.
In QCD, with no quark mass terms, baryons would still be massive, but pions would be massless — they would be exactly Goldstone bosons. Heuristically, they correspond to oscillations of the phase $\theta$ in $\exp(\theta\gamma^5)\psi$, and they're massless because the energy cost of those oscillations goes to zero in the long-wavelength (low-momentum) limit, because the Hamiltonian is independent of $\theta$ if $\theta$ is spatially uniform. Quark mass terms break the symmetry explicitly, so that the Hamiltonian depends on $\theta$ even when $\theta$ is spatially uniform, but as long as the quark mass $m$ is small, the original Goldstone-boson picture is still approximately valid. Pions are no longer exactly massless, but their masses are still much smaller than the baryon masses — much more so than the classical valence-quark picture would suggest.
... whose VEV we are expanding about?
I'm not sure I understand this part of the question, but I'll try this: In the context of chiral symmetry breaking, the quantity $\langle 0|\bar \psi \psi|0\rangle$ (the "condensate") plays the role that $\langle 0|\phi|0\rangle$ would play in the more familiar complex-scalar-field case. The condensate is not invariant under $$ \psi\to \exp(\theta\gamma^5)\psi \tag{2} $$ unless it is zero, so a non-zero condensate is a sign that the symmetry is broken (either explicitly or spontaneously); it's an order parameter. Applying the transformation (2) to the condensate just changes the relative phase of the $LR$ and $RL$ terms. In the case of purely spontaneous symmetry breaking, this corresponds to a different choice of vacuum state from the "valley" of equally-good vacuum states, all having the same minimum energy. This is all analogous to the case of a complex scalar field with order parameter $\langle 0|\phi|0\rangle$, where the transformation $\phi\to\exp(i\theta)\phi$ changes the relative phase of $\langle 0|\phi_r|0\rangle$ and $\langle 0|\phi_i|0\rangle$.
Some references:
There's a whole book about "spontaneous symmetry breaking without a potential," also called dynamical symmetry breaking:
- Miransky (1994), Dynamical Symmetry Breaking In Quantum Field Theories, https://www.worldscientific.com/worldscibooks/10.1142/2170
That book attempts to give some insight into why chiral symmetry is spontaneously broken in QCD, which my answer does not do. That's a relatively difficult non-perturbative issue that, as far as I know, is still not completely understood. The paper
- Coleman and Witten (1980), "Chiral Symmetry Breakdown in Large N Chromodynamics," https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.45.100
shows that if QCD is confining, then chiral symmetry is spontaneously broken, at least in the large-number-of-colors limit. However, confinement is not a necessary condition for chiral SSB. The book
- Greensite (2011), An Introduction to the Confinement Problem (https://www.springer.com/us/book/9783642143816)
says this on page 77:
Chiral symmetry breaking does not even require gauge fields. The breaking can occur in theories with other types of forces between fermions, as in the Nambu Jona-Lasinio model [ref], where the effect is due to four-fermion interactions.
Page 79 in the same book says
The evidence [from lattice simulations] is that center vortices are not only responsible for the confining force, but also, by themselves, can be responsible for the spontaneous breaking of chiral symmetry.
The point is that chiral SSB is a relatively difficult non-perturbative issue, one that does not (yet) have such a clear heuristic picture as the familiar scalar-field version of SSB. However, according to the review paper
- Rajagopal and Wilczek (2000), "The Condensed Matter Physics of QCD," https://arxiv.org/abs/hep-ph/0011333
the problem is more tractable in QCD at high density (still at low temperature). I haven't studied it enough yet to understand it myself, but here's a teaser from the intro section:
The resulting predictions regarding the low-energy spectrum and dynamics are striking. Color symmetry and chiral symmetry are spontaneously broken. ... Altogether, one finds an uncanny resemblance between the properties one computes at asymptotic densities, directly from the microscopic Lagrangian, and the properties one expects to hold at low density, based on the known phenomenology of hadrons. In particular, the traditional `mysteries' of confinement and chiral symmetry breaking are fully embodied in a controlled, fully microscopic, weak-coupling (but nonperturbative!) calculation, that accurately describes a directly physical, intrinsically interesting regime.
Footnote:
${}^{[1]}$ In models with a scalar field where the Hamiltonian has the form kinetic + potential, people often use a heuristic classical picture to introduce the idea of spontaneous symmetry breaking, as though the scalar field were an ordinary real variable. That heuristic picture has merit as a starting point for perturbation theory (which is $\sim$99% of the content in most QFT textbooks), but a non-perturbative perspective reveals limitations of the heuristic picture. For example, in a real-scalar-field model with potential $V(\phi)=\mu\phi^2+\lambda\phi^4$, where $\mu$ and $\lambda$ are the "bare" coefficients at the cutoff scale, just making $\mu$ negative is not sufficient to spontaneously break the $\phi\to-\phi$ symmetry. The transition between the SSB and non-SSB phases occurs at a finite negative value of $\mu$ (dependent on both $\lambda$ and on the cutoff), not at zero. This can be deduced intuitively as explained here: https://physics.stackexchange.com/a/435461, and it is confirmed in numerical studies. Both of those approaches (intuitive and numerical) are non-perturbative.