Quark condensate and VEV of $\pi^0$

Question

In David Tong's lectures on the Standard Model I saw that there is a quark condensate, which is just a Vacuum Expectation Value (VEV) of the $\bar{q}_{Li}\, q_{Ri}$ operator, $$ \left< \bar{q}_{Li}\, q_{Ri} \right> = - A \delta_{ij} \, . \tag{3.47} $$

Now, this leads me to think that there might also be a VEV for the $\pi^0$, which is composed of the linear combination $$ \pi^0 \equiv \dfrac{u \bar{u} - d \bar{d}}{\sqrt{2}} \, . $$

I don't think there are any reason why this would be forbidden, as it seems like the vacuum "can give you the needed energy", and it doesn't seem like we are violating any of the QCD symmetries (like Isospin!).

So, am I right? And, if so, how can I compute the VEV of the 0-pion $$ \left<\pi^0 \right> \equiv \left<\dfrac{u \bar{u} - d \bar{d}}{\sqrt{2}} \right>\, ? $$

Answer 1

Nonono! You've made a bad hash of standard notation.

The interpolating operator for the neutral pion you wrote is far too schematic/complacent, and lacks an all-important $\gamma^5$. You might consider $(\bar u \gamma^5 u - \bar d \gamma^5 d)/\sqrt 2$.

QCD, of course, supports a scalar condensate of quarks triggering spontaneous chiral symmetry breaking, but not a pseudo-scalar one, as the QCD vacuum does not break parity! Specifically, $\langle \Omega| \pi^0|\Omega\rangle= \langle \Omega|\Pi^\dagger \pi^0 \Pi|\Omega\rangle=-\langle \Omega| \pi^0|\Omega\rangle$ so that $\langle \pi^0\rangle=0$.

It is more complicated than that, covered in good QFT books. The neutral pion is the pseudogoldstone boson of the SSBroken axial charge $\int d^3x~~\bar q \gamma^5 \gamma^0 \tau^3 q$, so it pops into and out of the roiling QCD vacuum, a much longer story.