Register to reply

How to maintain CPT invariance in Kaon oscillations

by SteDolan
Tags: invariance, kaon, maintain, oscillations
Share this thread:
SteDolan
#1
Feb17-13, 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
Phys.Org News Partner Physics news on Phys.org
A new, tunable device for spintronics
Watching the structure of glass under pressure
New imaging technique shows how cocaine shuts down blood flow in mouse brains
Vanadium 50
#2
Feb17-13, 08:47 AM
Mentor
Vanadium 50's Avatar
P: 16,356
The K0 and K0bar are not mass eigenstates.
Bill_K
#3
Feb17-13, 09:06 AM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
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
#4
Feb17-13, 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 K0 and K0-bar)?
mfb
#5
Feb17-13, 10:41 AM
Mentor
P: 11,890
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.
SteDolan
#6
Feb17-13, 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?
mfb
#7
Feb17-13, 11:20 AM
Mentor
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.
Bill_K
#8
Feb17-13, 11:36 AM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
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 KS and KL must be by definition the eigenstates, even though they won't be exact eigenstates of CP.
mfb
#9
Feb17-13, 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.


Register to reply

Related Discussions
Why doesn't Diffeomorphism Invariance lead to Scale Invariance? Special & General Relativity 40
Can a theory have local Lorentz invariance but not diffeo invariance? Beyond the Standard Model 9
Oscillations: Neutral Kaon vs neutrino High Energy, Nuclear, Particle Physics 5
Neutral Kaon oscillations High Energy, Nuclear, Particle Physics 0
Lorentz Invariance and Non-Galilean Invariance of Maxwell's Equations Classical Physics 4