In section 8.2.1 of the book `Foundations of Perturbative QCD' by John Collins, there is a statement
Normal targets (nucleons and pions) are flavor eigenstates,... An exception would be DIS on a $K_L^0$ or $K_S^0$,...
What does he mean by a hadron being a flavor eigenstate or non-eigenstate here?
The flavour eigenstates (also known as strong eigenstates) have definite quark flavour. For neutral kaons, these are $K^0$, which is a $\overline{s}u$ state, and $\overline{K}^0$, which is a $s\overline{u}$ state. By contrast, the weak eigenstates $K^0_L$ and $K^0_S$ have definite lifetimes and are orthogonal linear combinations of $K^0$ and $\overline{K}^0$:
$$K^0_L = \frac{1}{\sqrt{1+|\eta|^2}}\left(K^0 + \eta\overline{K}^0\right),$$
$$K^0_S = \frac{1}{\sqrt{1+|\eta|^2}}\left(K^0 - \eta\overline{K}^0\right),$$
where $\eta$ is close to, but not exactly, 1. If $\eta=1$, then these would also be CP eigenstates.
