CP-Violation in the $K_0$-System

Without CP-symmetry, we do not have (1.8), but still have (1.4) and (1.7). The main consequence of this is that - with our phase convention - $M_{12}$ and $\Gamma_{12}$ are no longer real, but get a phase. The CP-conserving matrix (1.9) has to be generalized to

$$H = \left( \begin{array}{ccc} H_{11} & H_{12}(1+2\epsilon) \\ H_{12}(1-2\epsilon) & H_{11} \\ \end{array} \right)$$ (1.19)

where $H_{12}$ contains only the real parts of $M_{12}$ and $\Gamma_{12}$, while $\epsilon$ collects both phases like this^1.7:

$$\epsilon := \frac{1}{2} \frac{\frac{1}{2}Im \Gamma_{12} + i Im M_{12}}{Re M_{12} - \frac{i}{2}Re \Gamma_{12}}$$ (1.20)

Assuming a small CP-violation, $\epsilon$ is also small, which motivates the form (1.19). Neglecting all terms $O(\epsilon^2)$, this CP-violating matrix gives the same eigenvalues^1.8

$$H_{S,L} = H_{11} \pm H_{12}$$ (1.21)

and therefore also the same masses and lifetimes. However, the eigenstates change - at $O(\epsilon)$ - to

$$\vert K_{S,L}\rangle = \frac{1}{\sqrt{2}}((1+\epsilon)\vert K_0\rangle \pm (1-\epsilon)\vert\overline{K}_0\rangle )$$ (1.22)

which of course can also be expressed in $\vert K_{1}\rangle$ and $\vert K_{2}\rangle$:

$$\vert K_{S}\rangle = \vert K_1\rangle +\epsilon\vert K_2\rangle$$ (1.23)

$$\vert K_{L}\rangle = \vert K_2\rangle +\epsilon\vert K_1\rangle$$ (1.24)

This can be interpreted as such: in the presence of small CP-violation, the kaon eigenstates are almost CP-eigenstates, but with a small admixture of the opposite CP-eigenstate of the order $\epsilon$. Masses and lifetimes are not affected in first order.