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How to maintain CPT invariance in Kaon oscillations 
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#1
Feb1713, 08:29 AM

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 [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 


#3
Feb1713, 09:06 AM

Sci Advisor
Thanks
P: 4,160

Consider the Hamiltonian across the K^{0} and K^{0}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 K^{0} into K^{0}bar: [tex]\left(\begin{array}{cc}M&ε\\ε&M\end{array}\right)[/tex] As V^{50} says, K^{0} and K^{0}bar are no longer the eigenstates, and the originally degenerate mass eigenvalues have been split into two unequal eigenvalues. 


#4
Feb1713, 10:31 AM

P: 14

How to maintain CPT invariance in Kaon oscillations
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 K^{0} and K^{0}bar)? 


#5
Feb1713, 10:41 AM

Mentor
P: 11,890

No, those "particles" are the weak eigenstates. The mass eigenstates are called K_{1} and K_{2}. Neglecting CP violation, K_{1} is the shortliving K_{S} and K_{2} is the "long"living K_{L}.
CP violation introduces another, but small mixing in this system  K_{S} has some small component of K_{2} and K_{L} has a small component of K_{1}. 


#6
Feb1713, 10:55 AM

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?



#7
Feb1713, 11:20 AM

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P: 11,890

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
Feb1713, 11:36 AM

Sci Advisor
Thanks
P: 4,160




#9
Feb1713, 03:22 PM

Mentor
P: 11,890

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. 


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