We have that the neutral pion has quark composition given by:
$$\pi^0=\frac{u\bar{u}-d\bar{d}}{\sqrt{2}}$$
I want to justify this. This can be deduced by reasoning with isospin. We know that:
$$|\pi^0\rangle=|1,0\rangle, \quad |u\rangle=|1/2,1/2\rangle, \quad |d\rangle=|1/2,-1/2\rangle$$
so it follows that:
$$|\pi^0\rangle=\frac{|u\rangle|d\rangle+|d\rangle|u\rangle}{\sqrt{2}}$$
From this, we want to show that:
$$|\pi^0\rangle=\frac{|u\rangle|\bar{u}\rangle-|d\rangle|\bar{d}\rangle}{\sqrt{2}}$$
from which follows the quark composition of $\pi^0$. So let's define:
$$|\bar{u}\rangle \equiv |d\rangle=|1/2,-1/2\rangle$$
We now define the isospin $|\bar{d}\rangle$ of $\bar{d}$ reasoning as follows. Let's consider the unitary matrix:
$$U=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
This matrix represents a unitary operator $U$ in the representation where:
$$|u\rangle=(1,0), \quad |d\rangle=(0,1)$$
These are to be understood as column vectors. We obviously have:
$$U|u\rangle=|d\rangle$$
So it must be also that:
$$U|\bar{u}\rangle=|\bar{d}\rangle$$
since a transformation that changes $u$ with $d$ also has to change $\bar{u}$ with $\bar{d}$. But we have that:
$$U|\bar{u}\rangle=U|d\rangle=-|u\rangle$$
So we impose that:
$$|\bar{d}\rangle \equiv -|u\rangle=-|1/2,1/2\rangle$$
It follows that:
$$|\pi^0\rangle=\frac{|u\rangle|d\rangle+|d\rangle|u\rangle}{\sqrt{2}}=\frac{|u\rangle|\bar{u}\rangle-|d\rangle|\bar{d}\rangle}{\sqrt{2}}$$
since $|\bar{u}\rangle=|d\rangle$ and $|\bar{d}\rangle=-|u\rangle$. Then finally:
$$\pi^0=\frac{u\bar{u}-d\bar{d}}{\sqrt{2}}$$
My question: I'm pretty sure that my reasonement is correct. I only have a doubt in the section where I state that from $U|u\rangle=|d\rangle$ must follow that $U|\bar{u}\rangle=|\bar{d}\rangle$ (I stress this part in bold in my question). Is this assumption correct?
