It is well known in QCD that the light quark condensates are a unit matrix in flavor space:

$\langle \bar{u}u\rangle=\langle \bar{d}d\rangle=\langle \bar{s}s\rangle\sim\Lambda_{\rm QCD}^3$.

Thus, we observe that the quark condensates are aligned with the Yukawa couplings to the Higgs: In the basis in which the quark masses are diagonal, the condensates are diagonal as well.

Now let us consider a hypothetical world in which we have no experimental knowledge about QCD and thus no information about the chiral symmetry breaking pattern, the meson spectrum, etc.

Would it a priori be possible to have misaligned quark condensates, such as $\langle \bar{u}c\rangle\neq 0$?

Or is there any (perhaps obvious) theoretical reason why these condensates have to vanish or have to be strongly suppressed in Nature, even though the formation of these condensates is completely unrelated to the Yukawa Higgs sector?


1 Answer 1


The Yukawa couplings start life as an awful off-diagonal mess, and, upon diagonalization of the induced mass matrix they define flavor, as the resulting eigenstates. From this point on, we may ignore the weak interactions, keeping the small quark mass terms that ensued.

Since the strong interactions couple equally/indifferently to these flavors, they have no way to misalign them, so the flavor eigenstates persist, and flavor is conserved there. So the strong interactions do recognize that residual part of the Yukawa sector, contrary to your assertion. Gluons pair-create such flavorless quark-antiquark pairs: associated production. The same then holds for soft effects involving the sea and chiral symmetry breaking condensation.

Only if two quark masses were degenerate, then, yes, the quark-antiquark combinations produced by gluons would segue into each other, in search of a degeneracy-breaking option (like the weak interaction we have been ignoring all along? So Cabbibo mixing would undo itself, and, yes, we'd be working with Weak eigenstates instead of flavor ones!). So you would not be able to define a different u and a c, in meaningful contrast to a rotation of that basis.

The basic pattern of $SU(3)_L\times SU(3)_R\rightarrow SU(3)_V$, would, of course, persist; in fact, the flavor-vector group (eightfold way) on the right would work even better, given the degeneracy. (Recall it is explicitly broken by these very masses).

Since however, these (current) quark masses are so much smaller than the chiral condensation scale Λ, all the patterns and interesting differences of the pseudoscalar meson spectrum are pegged to these tiny masses... this is the reason behind "Dashen's formula" relating the square of these mesons' masses to linear functions of the quark masses, e.g., $m_\pi^2=-\langle\bar{u} u \rangle 2(m_u+m_d)/f_\pi^2 $.

Edit: terminology clarification as per comment/question of the OP.

Mass terms, so the mass basis, define flavor. The mass operator commutes with flavor operators. The so called "flavor basis" of some texts is expressly designed to confuse: it consists of WI eigenstates, not flavor operator eigenstates. So a charged weak gauge boson coupling term changes flavor; flavor basis does not mean it does not: Flavor changing charged currents. Mass terms do not change flavor, thus they define it. That is why the SI preserve flavor but the WI violate it.

Historical footnote. I found out where the absurd term "flavor basis" originated. It has never been used in the quark sector, as it is liable to confuse, as here. It is used in the neutrino sector, where $\nu_1,\nu_2, \nu_3$ constitute the mass basis, and the "convenience mixtures" $\nu_e,\nu_\mu,\nu_\tau$ the "flavor basis", as the latter are only meaningful in a weak coupling to a charged lepton flavor, e,μ,τ, and there is no other way to probe their identity. But let me stress that using the term "flavor basis" for quarks is asking for trouble, as right here.

  • $\begingroup$ What do you mean with "they define flavor, as the resulting eigenstates"? The Yukawa couplings are off-diagonal in the flavor basis, and after diagonalization of the mass matrix, they get diagonal in the mass basis (obtained by multiplying the original flavor eigenstates by unitary matrices $U_i$). Thus, the small quark masses are diagonal in the mass basis and the QCD condensates are diagonal in the flavor basis. Why isn't there a mismatch? Because the QCD couplings commute with $U_i$? $\endgroup$
    – Thomas
    Jun 13, 2017 at 14:25
  • $\begingroup$ And why should this argument change if we consider two degenerate quark masses? I don't understand your comparison to weak interactions in this context. In weak interactions, the reason of the mismatch is the mixing of $u$ and $d$, so we get nonzero $U^\dagger_u U_d$ rotations in the quark currents that couple to the $W$ boson field. $\endgroup$
    – Thomas
    Jun 13, 2017 at 14:57
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    $\begingroup$ Concerning your second comment question, the WI Cabbibo-mix the d and the s, or, equivalently, the u and the c, that is they choose to couple to rotated states, not the mass eigenstates. But if there are no mass eigenstates, a Cabbibo rotation is free: we might as well use the weak eigenstates as the mass ones! $\endgroup$ Jun 13, 2017 at 19:05
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    $\begingroup$ Arguably, 253751#253751 liable to shed light. $\endgroup$ Jun 13, 2017 at 22:26
  • $\begingroup$ One final question comes to my mind: According to your argumentation, how can gravity mediate flavor-violating processes if it couples equally to all flavors? Because gravity generically violates all global symmetries? $\endgroup$
    – Thomas
    Jun 14, 2017 at 11:33

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