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 [itex]K^0[/itex] and [itex]\overline{K^0}[/itex] as they propagate. Going through the quantum mechanics of this implies that this oscillation must be facilitated by a mass difference between the [itex]K^0[/itex] and [itex]\overline{K^0}[/itex] in a [itex]\cos(\Delta m t)[/itex] 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:
[tex]\left(\begin{array}{cc}M&0\\0&M\end{array}\right)[/tex]
But the weak interaction adds a perturbation that turns K⁰ into K⁰-bar:
[tex]\left(\begin{array}{cc}M&ε\\ε&M\end{array}\right)[/tex]
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 [itex]\Delta m[/itex] 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

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.