# Applying ladder operators to Isospin states for quarks

I'm having trouble finding the isospin representation of the meson isospin triplet $|\rho\rangle$.

Starting from the $|\rho^+\rangle$ meson $|u^\uparrow\bar d^\uparrow\rangle$ (where the arrows signify the 3rd component of the normal spin), I should get the $|\rho^0\rangle$ state by applying the $I_-$ operator.

I assume $|\rho^+\rangle$ is (with $|I,I_3\rangle$ as the notation) the $|1,+1\rangle$ state, but if I then apply the ladder operator I always get $\frac{1}{\sqrt{2}}(|u^\uparrow\bar u^\uparrow\rangle+|d^\uparrow\bar d^\uparrow\rangle)$ while it should be $\frac{1}{\sqrt{2}}(|u^\uparrow\bar u^\uparrow\rangle-|d^\uparrow\bar d^\uparrow\rangle)$.

What am I doing wrong? Is my assumption about the Isospin state corresponding to the positively charged meson wrong?

The antiquarks are in the "antifundamental representation." If a quark in the fundamental representation transforms as $q_i\rightarrow U_{ij}q_j$, an antiquark transforms as $\bar{q}_i\rightarrow \bar{q}_jU^{-1}_{ji}=U^*_{ij}\bar{q}_j.$
In terms of the generators $I$, $$U^*=\exp[{i\alpha_a I_a}]^*=\exp[{i\alpha_a (-I^*_a)}]$$ So when acting on the antiquark you should use $\bar{I}=-I^*$. (I'll use the bar on the generator to distinguish the representations)
So $\bar{I}_-\bar{d}=(-I_x^*-i(-I_y^*))\bar{d}=-I_+\bar{d}=-\bar{u},$ that's where the sign difference comes from.