Why does isospin symmetry survive chiral symmetry breaking?

Question

In QCD there is a breaking of chiral symmetry such that, for the lightest quarks only, we have $$\langle \overline{u} u \rangle = \langle \overline{d} d \rangle = \mathrm{const.}$$ where the constant is zero in the symmetric phase and about (250 MeV)$^3$ in the broken phase.

I've never understood why it wouldn't be something more general, like e.g. $$\langle \overline{u} u \rangle = V_1 \\ \langle \overline{d} d \rangle = V_2 $$ which wouldn't correspond to a surviving isospin symmetry.

Is there some way of understanding from the QCD Lagrangian or path integral why $V_1=V_2$?

Answer 1

Chiral condensation underlying spontaneous chiral symmetry breaking is not manifest in the lagrangian or the functional integral (albeit deeply implicit in the latter): it is a nonperturbative property of the QCD vacuum.

Isospin breaking, by contrast, is largely due to the small quark mass differences in the lagrangian, possibly modified by the QCD vacuum—but it's basically a good (unbroken) symmetry of the vacuum, a just-so property only partially understood. These two types of symmetry logically connect in a tractable manner, below.

If you define the scalar bilinears underlying the chiral condensate, $$S=\bar q (x) q(x) ,\\ S^a =\bar q (x) \frac{\tau^a}{2}q(x), $$ you readily see that Isospin leaves $S$ invariant, but rotates $S^a$, $$[I^a,S^b(x)]= i\epsilon ^{abc} S^c(x)~~~~~~~\implies \\ S^a(x) = -\frac{i}{2}\epsilon^{abc} [I^b,S^c(x)].$$

The vacuum is isospin and translationally invariant (isospin doesn't break dynamically/spontaneously in QCD), so $I^a|0\rangle= 0$, whence $$\langle 0|S^a(x)|0\rangle = \langle 0|S^a(0)|0\rangle =0.$$ Focussing on a=3 directly implies $$ \langle \bar u u\rangle - \langle \bar d d \rangle =0. $$

• The takeaway is that isospin survives in the face of chiral condensation, as the two condensates must equal each other, whatever their value.

In sharp contrast to the arguments that chiral symmetry must break spontaneously, there isn't, can't be, an analog argument for isospin.

In real life, I is explicitly broken by the small quark mass difference, about 2 orders of magnitude smaller than the scale of the chiral condensate, so isospin is a good symmetry, and chiral symmetry dynamical breaking is largely unaffected by it.