# Why is $|\bar{K}^0\rangle=\mathscr{CP}|K^0\rangle$ and not $|\bar{K}^0\rangle=\mathscr{C}|K^0\rangle$?

If the charge conjugation operator $\mathscr{C}$ changes a particle state into the corresponding anti-particle state then we must write $|\bar{K}^0\rangle=\mathscr{C}|K^0\rangle$. But instead, we write $|\bar{K}^0\rangle=\mathscr{CP}|K^0\rangle$ where $\mathscr{P}$ is the parity operator. $|K^0\rangle$ and $|\bar{K}^0\rangle$ stands for the neutral Kaon states. Please help me understand this.

Consider the states in question, $\bar{K}^0\sim \bar{d} i\gamma_5 s$ versus $K^0\sim \bar{s}i\gamma_5 d$. The operation that completely transforms them to each other with the role of s completely supplanted by that of d is CP. In a world with only 2 generations, weak charged currents would couple to these quarks identically (In that world, the weak interactions would preserve CP, all the while breaking C and P identically. Only the strong interactions preserve C and P separately.), as the left-handed s supplanted the left-handed d, instead of the right-handed one, had one skipped operation by P.
• A small nitpick on the question's unwarranted assertion that $|\bar{K}^0\rangle=C|K^0\rangle$. Not quite! All you need is $C^2 |K^0\rangle= |K^0\rangle$, so $-|\bar{K}^0\rangle=C|K^0\rangle$ would also do, provided $C|\bar{K}^0\rangle=-|K^0\rangle$. This amounts to $C(|K^0\rangle+|\bar{K}^0\rangle)=-(|K^0\rangle+|\bar{K}^0\rangle)$, so then CP-even, and the orthogonal combination C-even but CP-odd. So, then, the effective term of the common weak decay $K_s\to \pi^0 \pi^0$ in the effective action is $K_S \pi^0\pi^0$. Since everybody has P= -1, and C for the neutral πs is +, the term violates P and C, but preserves CP, like a good weak effective vertex.