How to maintain CPT invariance in Kaon oscillations

[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

 

[Vanadium 50]

The K0 and K0bar are not mass eigenstates.

 

[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.

 

[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)?

 

[mfb]

No, those "particles" are the weak eigenstates. The mass eigenstates are called K₁ and K₂. Neglecting CP violation, K₁ is the short-living K_S and K₂ is the "long"-living K_L.
CP violation introduces another, but small mixing in this system - K_S has some small component of K₂ and K_L has a small component of K₁.

 

[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?

 

[mfb]

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.

 

[Bill_K]

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.

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

 

[mfb]

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.