# How to maintain CPT invariance in Kaon oscillations

1. Feb 17, 2013

### SteDolan

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

2. Feb 17, 2013

Staff Emeritus
The K0 and K0bar are not mass eigenstates.

3. Feb 17, 2013

### Bill_K

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.

4. Feb 17, 2013

### SteDolan

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)?

5. Feb 17, 2013

### Staff: Mentor

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.

6. Feb 17, 2013

### SteDolan

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?

7. Feb 17, 2013

### Staff: Mentor

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.

8. Feb 17, 2013

### Bill_K

Isn't it true that, regardless of CP violation, eigenstates have definite mass values and lifetimes. So KS and KL must be by definition the eigenstates, even though they won't be exact eigenstates of CP.

9. Feb 17, 2013

### Staff: Mentor

Wait, I messed it up.
$K^0 = d\bar{s}$ (and its antiparticle) is a strong eigenstate.
Without CP violation, $K_1 = K_S = \frac{1}{\sqrt{2}} (d\bar{s} + \bar{d}s$ and $K_2 = K_L = \frac{1}{\sqrt{2}} (d\bar{s} - \bar{d}s$ are weak and mass eigenstates. The first one has CP=+1, the latter one has CP=-1. This explains their lifetime difference, $K_L$ has to decay to three pions while $K_S$ can decay to two.

With CP violation, but conserved CPT,
$K_S = \frac{1}{1+\epsilon^2}(K_1 + \epsilon K_2)$ and $K_L = \frac{1}{1+\epsilon^2}(K_2 + \epsilon K_1)$
$K_S$ and $K_L$ are weak (and mass?) eigenstates, giving them their different lifetime. The small CP violation allows (rare) decays of a $K_L$ to two pions and decays of a $K_S$ to three pions.
The real part of ϵ is about 1.6 permille.