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.


1 Answer 1


All these states, and so their linear combinations, are pseudoscalars, so, then, odd under P, and thus essentially P= –1 throughout.

This is a feature of Dirac equation solution bilinears. Pure C does give you a state of opposite strangeness, but not quite what you will use later on, when you will contrast weak decays of the standard linear combinations, where P and C are violated maximally, but CP is (mostly) preserved. Parity flips the chirality of the constituent quarks, but C does not.

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.

(Consider that only CP maps left-handed neutrinos to EW-flavored right-handed antineutrinos; C or P map to the sterile states, which is why the weak interactions violate them--maximally.)

You would have an analogous situation for the charged K brethren of these states, but, being of opposite charge, they would not interfere, as electromagnetic interactions would treat them very differently. So, for them, you never find yourself in a situation where the CP operation is needed instead of just plain C, I think (or do you?).

  • 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.

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