# How to maintain CPT invariance in Kaon oscillations

by SteDolan
Tags: invariance, kaon, maintain, oscillations
 P: 14 Hey, I'm trying to get my head around neutral Kaon oscillations. As far as I understand it neutral Kaons can change between $K^0$ and $\overline{K^0}$ as they propagate. Going through the quantum mechanics of this implies that this oscillation must be facilitated by a mass difference between the $K^0$ and $\overline{K^0}$ in a $\cos(\Delta m t)$ term. But I thought a mass difference between any particle and its antiparticle implies CPT violation. As far as I know CPT is not known to be violated so my question is: How do Kaon Oscillations maintain CPT invariance? Thanks in advance for any help :D
 Mentor P: 14,723 The K0 and K0bar are not mass eigenstates.
 Sci Advisor P: 3,484 Consider the Hamiltonian across the K0 and K0-bar states. CPT says the masses of these states are equal: $$\left(\begin{array}{cc}M&0\\0&M\end{array}\right)$$ But the weak interaction adds a perturbation that turns K0 into K0-bar: $$\left(\begin{array}{cc}M&ε\\ε&M\end{array}\right)$$ As V50 says, K0 and K0-bar are no longer the eigenstates, and the originally degenerate mass eigenvalues have been split into two unequal eigenvalues.
P: 14

## How to maintain CPT invariance in Kaon oscillations

Thanks! That makes sense.

So I guess the $\Delta m$ refers to the difference in mass of the modified states coupling to the weak interaction (a superposition of K0 and K0-bar)?
 Mentor P: 9,730 No, those "particles" are the weak eigenstates. The mass eigenstates are called K1 and K2. Neglecting CP violation, K1 is the short-living KS and K2 is the "long"-living KL. CP violation introduces another, but small mixing in this system - KS has some small component of K2 and KL has a small component of K1.
 P: 14 Thanks. I think I understand that. Am I right in thinking that if there was no CP violation the mass eigenstates would have the same mass and therefore there would be no kaon oscillations?
 Mentor P: 9,730 Without CP violation, the mass eigenstates would be identical to K_S and K_L, but those would remain a mixture of the weak eigenstates.