Why are the lightest mesons (Goldstone bosons) pseudoscalar? I get that the lightest mesons are generically the (approximate) Goldstone bosons of spontaneous chiral symmetry breaking in the quark sector. I understand that they are spin-0, but why are they necessarily pseudoscalars (odd-parity)? You can see it in this PDG listing of known $\bar q q$-mesons — all the lightest entries are $J^{\pi}=0^-$.

Here is what I know. The generators of axial flavor transformations is given in spin-space (Weyl basis):
$$T_a^A=\begin{pmatrix}T_a & 0 \\ 0 & -T_a\end{pmatrix}=\gamma^5 T_a$$
where $T_a$ is the $N_f\times N_f$ flavor generator. The CCWZ construction tells us that the correct way to incorporate these Goldstone bosons is via:
$$U(x)=\exp \left(\frac{i}{f_\pi}\Pi^a T_a^A\right)$$
where under chiral rotations we have $U\rightarrow R U L^\dagger$ with $R,L\in SU(N_f)$. But how does this transform under parity transformations, or any other discrete transformations for that matter? Anyway, I know that $\bar\psi \gamma^5\psi$ is a pseudoscalar quantity, and therefore $\bar\psi T_a^A \Pi^a\psi$ is a scalar so long as $\Pi^a$ is a pseudoscalar.
How do I piece all these together to show that indeed $\Pi^a$ must be a pseudoscalar? What am I missing?
 A: First, recall from your favorite QFT text that $P\psi(t,x)P=\gamma^0\psi(t,-x)$.
The SSBroken generators you wrote are odd under parity, by dint of the $\gamma^5$, effectively
$$
P T_a^A P = -T^A_a ~;
$$
and ditto for the charge you make out of them, integration being invariant under parity,
$$
Q^A_a=\int dx^3 ~~ J_{0~~a}^A (x)  = i\int dx^3 ~~ \bar \psi \gamma^0 T_a \gamma^5 \psi, \\
PQ^A_a P = -Q^A_a.
$$


*

*But the SSBroken currents are always linear  in the gradient of the goldstons  (essentially by definition of the nonlinear realization involved: $\langle \Pi_a(p)| J_{\mu~~b}^A(x)|0\rangle\sim f_\pi \delta_{ab} e^{ip\cdot x} p_\mu$),
$$  \bbox[yellow]{
J_{0~~a}^A (x) \sim f_\pi \partial_0 \Pi^A_a (x)+ ...} \\
P~J_{0~~a}^A (x)~P =i \bar \psi (-x)\gamma^0 \gamma^0T_a \gamma^5 \gamma^0 \psi (-x) = - J_{0~~a}^A (-x) ,
$$
So, likewise for the goldstons: they are parity odd. 
You already saw that, above, for their interpolating field operators, $$ \bar \psi (x)  T_a P\gamma^5P \psi (x)=- \bar \psi (-x)  T_a \gamma^5 \psi(-x). $$
