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.

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.
Let's understand this from the point of view that pions are pseudogoldstone bosons. Suppose for case of $SU_{L}(2)\times SU_{R}(2)$ we have bilinear form $$\tag 0 L_{qq} = \bar{q}_{i}q_{i}, \quad i = u, d$$ As we know, below $\Lambda_{QCD}$ scale spontaneously breaking symmetry group down to diagonal group $SU_{V}(2)$ arises. By using the usual technique we may extract goldstone degress of freedom from quark fields, $$q_{i} \equiv (U\tilde{q})_{i}, \quad U \equiv \text{exp}\left[ i\gamma_{5}\frac{\epsilon_{a}t_{a}}{f_{\pi}}\right],$$ where $t_{a}$ are Pauli matrices and $\epsilon_{a}$ are real-valued coordinate dependent parameters, ans then replace bilinear forms to VEVs: $$\tag 1 \bar{\tilde{q}}_{i}\tilde{q}_{j} \to V\delta_{ij}, \quad \bar{\tilde{q}}_{i}\gamma_{5}\tilde{q}_{j} \to 0$$ Note that $\epsilon_{a}t_{a}$ may be parametrized, by using explicit form of Pauli matrices, in a form $$\tag 2 \epsilon_{a}t_{a} = \begin{pmatrix} \frac{\pi^{0}}{\sqrt{2}} & \pi^{-} \\ \pi^{+} & -\frac{\pi^{0}}{\sqrt{2}}\end{pmatrix},$$ where $\pi^{\pm} \equiv \epsilon_{1} \pm i\epsilon_{2}$. As we see, we explicitly obtain one neutral degree of freedom, $\pi^{0}$, and two charged, $\pi^{\pm}$. Now let's calculate one-pion transition amplitude from $(0)$, $$\langle 0 | \bar{q}_{i}q_{i}|\pi^{0}\rangle,$$ by using $(1)$ and $(2)$. We immediately obtain that $$\langle 0 | \bar{q}_{i}q_{i}|\pi^{0}\rangle \equiv \frac{i}{\sqrt{2}f_{\pi}}\langle 0|\bar{\tilde{q}}_{i}t_{3}^{ij}\tilde{q}_{j}\pi^{0}|\pi^{0}\rangle \simeq \frac{i}{\sqrt{2}f_{\pi}}\langle 0|\bar{\tilde{u}}\tilde{u}-\bar{\tilde{d}}d|0\rangle,$$ which immediately gives statement that the pion is combination of $uu, dd$ with the minus sign because of it is the parametrization of goldstone degree of freedom in case of broken $SU_{L}(2)\times SU_{R}(2)$ group, not $U_{L}(2)\times U_{R}(2)$. The combination with the "plus" sign corresponds to parametrization of $U(1)$ group. As you know, this is $\eta$ meson for $SU_{L}(2)\times SU_{R}(2)$ group and $\eta{'}$ meson for $SU_{L}(3)\times SU_{R}(3)$ group. Of course, these combinations appear in nature.
• @HenryDeith : the underlying reason is that phenomenologically all of the QCD's staff (the isospin symmetry, vector current conservation, PCAC, Goldberg-Treiman relation) at low energies can be explained by the spontaneous breaking of the $G\simeq SU_{L}(3)\times SU_{R}(3)$ approximate global symmetry down to $SU_{V}(3)$. In terms of the fundamental quark fields, this means nothing but appearance of non-zero VEV of the quark bilinear $\bar{q}q$: it breaks the axial part of $G$, leaving the vectorial part unbroken. – Name YYY Aug 22 '17 at 19:47