elementary particles - Physics of K meson

$K^{0}$ meson consists of a $d$ quark and an $\bar{s}$ antiquark. Its antiparticle $\bar{K^{0}}$ consists of an $s$ quark and a $d$ antidown quark.

$$|K^{0}\rangle=|d\bar{s}\rangle$$ $$|\bar{K}^{0}\rangle=|\bar{d}s\rangle$$

There are two versions of $K^{0}$ meson. They have the same mass but different decay lifetimes.

K-zero-short: $$K_{S}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle+|\bar{K}^{0}\rangle\right]$$ with the observed lifetime $\tau_{S}=90$ ps.

K-zero-long: $$K_{L}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle-|\bar{K}^{0}\rangle\right]$$ with the observed lifetime $\tau_{L}=50$ ns.

What is the physical interpretation of the minus sign? I know that $\pm 1$ are the eigenvalues of $CP$ operator and $K_{S}^{0}$ and $K_{L}^{0}$ are its eigenstates. But what is the physical difference between $K_{S}^{0}$ and $K_{L}^{0}$ besides that they decay the way they do.