The Neutral Kaons
Originally, the existence of the anti neutral kaon was not known. So when they looked at the kaons they saw the set with strangeness… S(K ) 1 S(K ) 1 S(K 0 ) 1 ? Gell-Mann suggested that the K0 should have an antiparticle with strangeness S = 1. This anti kaon was found in the reaction: p p K K 0 This reaction has S = 0 The observed neutral kaon can not be a K0 since that would imply S = 2 which is assumed to be not allowed in Gell-Mann’s strangeness scheme – even for the weak interactions. (S = 1 for the weak interactions.) The Neutral Kaons
The two neutral kaons have a very unique property.
They differ only in strangeness S (and T3) and nothing else. This is a fairly unique situation. Both the K 0 and the K 0 can decay to two pions via the weak interaction. i.e. K 0 and K 0 Both have S 1 Therefore there can be a process that connects the two kaons, such as: Weak interaction S 1 K 0
We can think of these as virtual pions. The Neutral Kaons Therefore it is possible to have the neutral kaons “oscillate” between being the particle and the antiparticle. i.e. K 0 K 0 K 0 via the weak interaction. The particles are distinct at the time of production. This is because they are produced by the strong interaction in which strangeness is conserved. i.e. p K 0 If the weak interaction could be turned off, then the K 0 and K 0 p p K K 0 would remain distinct. So, after production, we cannot distinguish between the
and K 0 or K 0 S = 1 S = ?
S = 1 S = ? The Neutral Kaons
When Gell-Mann proposed the strangeness scheme, Fermi said “I won’t believe in your scheme until you have a way of distinguishing the K 0 from the K 0.”
Because, when we observe the decay of the neutral kaon, we cannot tell if it is a K 0 or a K 0, maybe it is really some other particle state, which is a linear superposition of the and .
Equivalently we can consider the neutral kaons as being linear superpositions of other states. This may not be such a strange idea if we think of the kaons as not fundamental particles, but just different states of a system. What kind of particle states would they be? CP Invariance
We have seen that parity is not conserved in the weak interactions. e.g. in the reaction the muon always comes out with “left handed” helicity. i.e. p p never s s Left handed muon Right handed muon Right handed Helicity Projection of spin on momentum direction is positive i.e. in the same direction as the momentum. Left handed Helicity Projection of spin on momentum direction is negative i.e. in the opposite direction as the momentum. CP Invariance
It is through experiments such as the decay of the pion that we know that all neutrinos have left handed helicity and all antineutrinos have right handed helicity.
i.e. p p s s Left handed neutrino Left handed muon
Indeed, if the neutrino is massless, then a right handed neutrino can never be observed. Since no observer can be moving faster than the neutrino to make it appear that the spin is in the same direction as the momentum. Then the helicity would be Lorentz invariant. CP Invariance Charge conjugation is also not conserved in the weak interactions. e.g. the charge conjugated version of the + decay is
p p Cˆ s s Left handed anti-neutrino Left handed muon But anti neutrinos are right handed so in fact the muon is actually observed to be right handed in this reaction. p s Observed Right handed anti-neutrino Right handed muon CP Invariance
Therefore Parity and Charge conjugation are separately not conserved in the weak interaction. But if we define the “mirror image” operation to be one that swaps particles with anitparticles as well as inverting the coordinate system, then this operation would be conserved in the weak interaction, as well as the strong and EM interaction. This operation would be equivalent to applying the parity operation (as we have defined it) and the charge conjugation operation. i.e. applying the operator CˆPˆ The conservation of CP would be consistent with the CPT theorem, assuming that time reversal is conserved. CP Invariance
e.g. left handed neutrino left handed muon
Applying Cˆ left handed antineutrino left handed antimuon
Not observed. Applying Pˆ right handed antineutrino
right handed antimuon Observed. CP Invariance
For all neutrinos: Only these are observed in nature. 푃 Both parity and charge 휈퐿 휈푅 conjugation are maximally violated. i.e. It is not that there is a 퐶 푃 퐶 preference for left- handed neutrinos over right-handed neutrinos – right handed neutrinos 휈퐿 휈푅 are never observed. Similarly left-handed anti- neutrinos are never observed. The Neutral Kaons For the neutral kaons: CˆPˆ K 0 Cˆ(1) K 0 K 0
CˆPˆ K 0 Cˆ(1) K 0 K 0 Thus the neutral kaons are not eigenstates of CˆPˆ i.e. CˆPˆ K 0 K 0 Perhaps we should be considering the K 0 and K 0 as linear combinations of states that are eigenstates of . Defining: K 1 [ K 0 K 0 ] The 1 ensures that 1 2 2 K1 K1 1 K 1 [ K 0 K 0 ] 2 2 and K2 K2 1 Then ˆ ˆ i.e. K and K are eigenstates CP K1 K1 1 2 of both and of the ˆ ˆ CP K2 K2 Hamiltonian of the system. i.e. CP is conserved. The Neutral Kaons
0 0 The relationship between K1 and K2, and K and K is analogous to regarding linearly polarized light as a superposition of left and right polarized light of equal amplitude and definite phase. So far we have considered and as the “real particles”,
and K1 and K2 as the “derived particles”. Do K1 and K2 actually exist as real particles? i.e. particles whose signature we can observe. Consider their decay of the neutral kaon: We know that K 0 We first consider the angular momentum. since s 0 Ji 0 J f Lrel s s lrel 0
lrel is the relative angular momentum q.n. in the final state. The Neutral Kaons K 0 For the final state: ˆ ˆ ˆ CP C(1)(1) since lrel 0 CˆPˆ Cˆ since Cˆ and Cˆ i.e. CP = +1 in the final state. Therefore since ˆ ˆ CP K1 K1 ˆ ˆ CP K2 K2 and we assume CP is conserved in this weak decay,
the two pion final state can only come from the decay of a K1. 0 0 This applies also for the decay K1
So K1 2 The Neutral Kaons For the decay: K 0 0 For the final state: ˆ ˆ 0 ˆ 0 CP C(1)(1)(1) since lrel 0 CˆPˆ 0 Cˆ 0 0 since Cˆ and Cˆ 0 0 i.e. CP = 1 in the final state. ˆ ˆ Therefore since CP K1 K1 ˆ ˆ CP K2 K2 and we assume CP is conserved in this weak decay, the three pion final state can only come from the decay of a K2.
0 0 0 This applies also for the decay K2
So K2 3 The Neutral Kaons
So, there is a way to tell if the decay is from a K1 or a K2 but we cannot distinguish between a K 0 and K 0. It is expected that the 2 decay will be faster than the 3 decay because the energy released is greater.
i.e. K 2 T1 short Some books label: K1 KS 1 2
K2 K L K2 3 T1 long 2 But we will see that these are slightly different later. When this was predicted in 1956, only one half life for the neutral kaon was known (~ 1010 s). A search was made and indeed a second half life (~ 5 x 108 s) was found. The neutral kaon has two half lives! (Because it is really a mixture of two particle states.) The Neutral Kaons K 1 [ K 0 K 0 ] Inverting the previous equations 1 2 1 0 0 0 K [ K K ] we get: K 1 [ K K ] 2 2 2 1 2
These decay to 2 These decay to 3 10 8 with T1/2 ~ 10 s with T1/2 ~ 5 10 s
K 0 1 [ K K ] Similarly: 2 1 2
Since K1 and K2 have separate half lives and decay modes we should perhaps consider them to be the “real” particles.
Furthermore the K1 and K2 have slightly different masses. The K 0 and K 0 are mixtures of these with a well defined phase relationship. 0 But K and are antiparticles of each other; the K1 and K2 are not. The Neutral Kaons
Let us suppose we produce a beam of K0 particles, say by p K 0 p 0 decays 2 100% K 50% K 1 50% K 0 50% K2 B A 0 100% K2 50% K
After sufficient flight time the K1‘s have all decayed away 0 0 and we are left with only K2. This is a mixture of K and K . Therefore, just after production, at position A, particles cannot be produced by allowing the beam to hit a proton target: K 0 p S 2 But after drifting for a while, at position B, particles can be produced by: K 0 p S 0 The Neutral Kaons
We can use an experimental arrangement like this to measure the number of K 0 at a time t after production of K 0 particles.
ln 2 Since the K1 decays with decay constant 1 T1 ,1 2 2 2 iE1t / 1t we can write (K1,t) 0e e
Number of K1 at t Number of K1 if they did not decay = Number of K1 at t = 0 0 = time independent part of K1 wave function So (switching notation) the time-dependent wave function
may be written iE1t / 1t / 2 K1,t K1 e e
time independent part of K1wave function The Neutral Kaons
2 E1 is the energy of the K1 in its rest frame, therefore E1 m1c 2 im1c t / 1t / 2 Therefore K1,t K1 e e 2 im2c t / 2t / 2 Similarly K2 ,t K2 e e K 0 1 [ K K ] Now since: 2 1 2 2 2 K 0 ,t 1 K eim1c t / e1t / 2 K eim2c t / e2t / 2 2 1 2 But we also know that K 1 [ K 0 K 0 ] 1 2 K 1 [ K 0 K 0 ] 2 2
2 2 0 1 0 0 im1c t / 1t / 2 0 0 im2c t / 2t / 2 K ,t 2 K K e e K K e e
2 2 1 0 im1c t / 1t / 2 im2c t / 2t / 2 2 K e e e e
2 2 1 0 im1c t / 1t / 2 im2c t / 2t / 2 2 K e e e e The Neutral Kaons
We can write K 0 ,t a(t) K 0 b(t) K 0 a(t) is the probability amplitude for finding a K 0 at time t. b(t) is the probability amplitude for finding a K 0 at time t. 2 2 1 im1c t / 1t / 2 im2c t / 2t / 2 a(t) 2 e e e e 2 2 1 im1c t / 1t / 2 im2c t / 2t / 2 b(t) 2 e e e e Note: at t = 0, a(0) =1 and b(0) = 0 as it should.
2 2 1 im2c t / imc t / 1t / 2 2t / 2 with m m m b(t) 2 e e e e 1 2 Therefore the probability of finding a at time t is 2 2 2 0 1 1t 2t (1 2 )t / 2 imc t / imc t / P(K ) b(t) 4 e e e (e e )
1 1t 2t (1 2 )t / 2 2 4 e e 2e cos(mc t / ) The Neutral Kaons
0 1 1t 2t (12 )t /2 2 Plotting this: P(K ) 4 e e 2e cos(mc t / )
0.6 Note: This plot contains 0.5 arbitrary values of 1, 0 2, and m to show the 1 (1 2 )t 0.4 4 2e P(K ) for m 0 form of the result.
0.3 0 1 2t P(K ) for m 0 4 e 0.2
1 1t 0.1 4 e
0 0 1 2 3 4 5 6
P(K 0 ,t 0) 0 Time The Neutral Kaons
We see that if m 0 there is an oscillation in the probability of seeing a K 0 as a function of time. The period of this oscillation is a function of the mass difference between the K1 and K2. Measurements have shown that there is a mass difference. Best Measurement is m 3.183 0.006106 eV/c2 This is very small compared to the mass of a Kaon (~500 MeV/c2). These oscillations are call hypercharge oscillations or strangeness oscillations, since it is the strangeness of the kaon that is oscillating. The Neutral Kaons
The kaons provide a testing ground for CP invariance. By using a long enough drift distance for an initial beam of K0 particles (i.e. long enough drift time), an arbitrarily pure beam of the long-lived K2 kaons can be produced. Then if any decays to 2 are observed (which can only come from a K1 decay) this is an indication of CP violation. This was observed in 1964. It is a small effect but the result indicates that the long lived kaons are not purely K2 but must have a small admixture of K1. 1 3 i.e. The K-long is KL K2 K1 with 2.2410 1 2 Since then, additional examples of CP violation have been found. CP violation provides evidence of the unequal treatment of matter and antimatter and may be the reason why matter is dominant over antimatter in the universe.