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Let's have pure QCD. I know that after spontaneous symmetry breaking quark bilinear form are replaced by their averaged values: $$ \bar{q}_{i}q_{j} \to \langle \bar{q}_{i}q_{j}\rangle \approx \Lambda_{QCD}^3, \quad \bar{q}_{i}\gamma_{5}q_{j} \to \langle \bar{q}_{i}\gamma_{5}q_{j}\rangle \approx 0 $$

What can be said about VEVs of $\partial_{\mu}\bar{q}_{i}\gamma^{\mu}\gamma_{5}q_{i}$, $$ \int d^4x d^4y\langle 0|T\left(\partial^{x}_{\mu}\bar{q}_{i}\gamma_{\mu}\gamma_{5}q_{i}(x))(\partial^{y}_{\nu}\bar{q}_{i}\gamma^{\nu}\gamma_{5}q_{i}(y))\right)|0\rangle? $$

An edit. It seems that the second correlator is zero in momentum space for $k \to 0$, since no massless states couples to correlator $\Pi^{\mu \nu}(k) \equiv \int d^{4}x e^{ikx}\langle 0|T(J^{\mu}_{5}(x)J^{\nu}_{5}(0))|0\rangle$ in QCD.

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By the chiral anomaly equation $$ \partial^\mu \bar{q}_f\gamma_\mu\gamma_5 q_f = \frac{N_f}{16\pi^2} \tilde{G}^a_{\alpha\beta}G^{a\,\alpha\beta} $$ this correlator is proportional to the topological susceptibility $$ \chi_{top}= \frac{1}{V}\frac{1}{(16\pi^2)^2}\int d^4x \int d^4 y \; \langle T\, \tilde{G}^a_{\alpha\beta}G^{a\,\alpha\beta}(x) \tilde{G}^b_{\gamma\delta}G^{b\,\gamma\delta}(y) \rangle $$ The topological susceptibility is zero if one of the quarks is massless, but it is non-zero in general, and of $O(\Lambda^4_{QCD})$ in pure gauge theory or the large $N_c$ limit.

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