hilbert space - Quark composition of the neutral pion

I wonder why the neutral pi meson is

$$| \pi^0\rangle = \frac{1}{\sqrt{2}}\left(\vert u\overline {u}\rangle - \vert d \overline{d} \rangle \right)$$

and not

$$| \pi^0\rangle = \frac{1}{\sqrt{2}}\left(\vert u\overline {u}\rangle + \vert d \overline{d} \rangle \right) .$$

Pions are a pair of quark and antiquark which both are isospin doublets. Since pions are isospin triplets $\pi^0$ should have plus in the quarks decomposition (?). I would expect minus in the isospin singlet state which, unfortunately, does not seem to be realized in the nature.

Answer

The reason the signs are flipped from what you expect has to do with the fact that the antiquark transforms in the opposite way under isospin rotations. If the ordinary quark doublet is a column vector

$$q=(u, d)^T$$ and transforms under rotations as $$q\rightarrow U(R) q$$ the antiquark doublet is a row vector $$\bar{q}=(\bar{u}, \bar{d})\rightarrow \bar{q} U(R)^\dagger.$$

But $SU(2)$ has a special property called being "pseudoreal" so we can write the antiquarks as a column vector that transforms normally like

$$(-\bar{d}, \bar{u})^T\rightarrow U(R)(-\bar{d}, \bar{u})^T$$ This is related to the Pauli matrix $\sigma_2$ being like a charge conjugation operator if you are familiar with that.

To do the addition of isospin in the ordinary way we need both quark and antiquark in the same representation, so the singlet $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$ is in this case

$$u\bar{u}-d(-\bar{d})$$ so we pick up a plus sign.