involve: 1. Treatment of a decaying particle. 2. Superposition of states with different masses.

Transcription

1 Physics 195a Course Notes The K 0 : An Interesting Example of a Two-State System F. Porter 1 Introuction An example of a two-state system is consiere. involve: 1. Treatment of a ecaying particle. 2. Superposition of states with ifferent masses. Interesting complexities 2 The K 0 Meson an its Anti-particle The K 0 meson is a pseuoscalar state consisting (for present purposes) of a quark an an s antiquark: K 0 = s. (1) Its antiparticle is the K 0 : K 0 = s. (2) If we efine a strangeness operator, S, (which counts strange quarks), these states are eigenstates, with: S K 0 = K 0, (3) S K 0 = K 0. (4) We may write S as the two-by-two matrix ( ) in the K 0, K 0 basis, but a convenient basis-inepenent form is: S = K 0 K 0 K 0 K 0. (5) These are not eigenstates of C, the charge conjugation operator (which changes particles to antiparticles). It is convenient to pick the antiparticle phases such that: C K 0 = K 0, (6) C K 0 = K 0. (7) 1

2 If we multiply the C operator by the parity operator P,wehave: We thus have, in the K 0, K 0 basis: CP K 0 = K 0, (8) CP K 0 = K 0. (9) ( ) 0 1 CP =, (10) 1 0 which we may also express in the basis-inepenent form: CP = K 0 K 0 + K 0 K 0. (11) The eigenstates of CP are (the choice of nomenclature will shortly be motivate): KS 0 = 1 ( K 0 + K 0 ), with CP =+1, (12) 2 K 0 L = 1 2 ( K 0 K 0 ), with CP = 1. (13) We remark that the K 0 (or K 0 ) is the lowest mass particle containing the strange quark. Thus, the only permitte ecays must be via the weak interaction. To a goo approximation (but not exactly!), CP is conserve in the weak interaction (an even more so in the strong an electromagnetic interactions); we shall assume this here. A neutral K meson (K 0 or K 0 ) is observe to ecay sometimes to two pions an sometimes to three pions. For example, consier the observe process K 0 π 0 π 0. Since all of the particles in this ecay are spinless, the ecay must procee with zero orbital angular momentum ( S-wave ecay). Thus, the parity of the π 0 π 0 system in the final state must be positive. But we sai that the K 0 is a pseuoscalar particle, i.e., has negative parity. Thus, this is a parity-violating ecay. The weak interaction is known to violate parity (i.e., parity is not conserve in the weak interaction), so this is all right. The π 0 is its own anti-particle, hence the π 0 π 0 final state is an eigenstate of C with eigenvalue +1. Thus, the π 0 π 0 final state is also an eigenstate of CP with eigenvalue +1. Uner our approximation that CP is conserve in the weak interaction, we therefore conclue that the observation of a K 0 π 0 π 0 ecay projects 2

3 out the KS 0 component of the K0 meson (likewise for the K 0 ). The 2π ecay moe is favore by phase space over ecays to greater numbers of pions. However, the KL 0 2π ecay is forbien by CP conservation. Hence, the KL 0 3π ecay is important for K0 L. Because the phase space is consierably suppresse, the KL 0 ecay rate is much slower than the KS 0 rate. The observe lifetimes of the KS 0 an K0 L are, respectively: τ S = s (90 ps), (14) τ L = s (50 ns). (15) 3 Time Evolution of a Kaon State Suppose that at time t = 0we have the state ψ(0) = K 0 S. (16) How oes this state evolve in time? We shoul have, at time t, ψ(t) =e iht KS. 0 (17) For a free particle, the energy is ω S = p 2 + m 2 S,wherem S is the mass of the KS 0. But if we just use this for H, we won t have a particle which ecays in time. We know that, if we start with a particle at t = 0the probability to fin it unecaye at a later time t if it has a lifetime τ S =1/Γ S is: P (t) =e Γ St. (18) Thus, the amplitue shoul have an exp( Γ S t/2) time epenence, in aition to the phase variation: ψ(t) =e iω St Γ S t/2 KS. 0 (19) Letting ω L = p 2 + m 2 L,wherem L is the mass of the KL 0,anΓ L =1/τ L, we similarly have for an intial KL 0 state (ψ(0) = K0 L ): ψ(t) =e iω Lt Γ L t/2 K 0 L. (20) In the KS 0, K0 L basis, the Hamiltonian operator is: ( ) ωs iγ H = S /2 0. (21) 0 ω L iγ L /2 3

4 This requires some further iscussion. For example, how i I know that H is iagonal in this basis (an not, perhaps, in the K 0, K 0 basis)? The answer is that we are assuming that CP is conserve. Hence, [H, CP] = 0. The Hamiltonian cannot mix states of iffering CP quantum numbers, so there are no off-iagonal terms in H in the KS, 0 KL 0 basis. The secon point is that we have allowe the possibility that the masses of the two CP eigenstates are not the same (having alreay note that the lifetimes are ifferent). This might be a bit worrisome the C operation oes not change mass. 1 However, the KS 0 an K0 L are not antiparticles of one another, so there is no constraint that their masses must be equal. So, we allow the possibility that they may be ifferent. We will aress shortly the measurement of the mass ifference. Now suppose that at time t = 0we have a pure K 0 state: ψ(0) = K 0. (22) Experimentally, this is a reasonable proposition, since we may prouce such states via the strong interaction. For example, if we collie two particles with no initial strangeness (perhaps a proton an an anit-proton), we make strange particles in associate prouction, i.e., in the prouction of s s pairs. Thus, we might have the reaction pp nλk 0 (see Fig. 1). The presence of the Λ, which contains the s quark, tells us that the kaon prouce is a K 0, since it contains the s quark. So, we can realistically imagine proucing a K 0 at t =0. Butthetimeevolution to later times is governe by the Hamiltonian, which is not iagonal in the K 0, K 0 basis. Thus, we might expect that at some later time we may observe a K 0. What is the probability, P K 0(t) thatak 0 meson is observe at time t, given a pure K 0 state at t = 0? The answer, noting that ψ(0) = K 0 =( K 0 S K0 L ) / 2, is: P K 0(t) = K 0 ψ(t) 2 (23) = 1 2 K0 KS e 0 iω St Γ S t/2 K 0 KL e 0 iω Lt Γ L t/2 2 = 1 { } e ΓSt + e ΓLt 2e Γ S +Γ L 2 t cos [(ω S ω L )t]. (24) 4 1 Actually, this is only an assumption here. But it is a funamental theorem in relativistic quantum mechanics that particle an anti-particle have the same mass (as well as the same total lifetime). 4

5 p p { u u { u u } u Λ s s } K 0 } u n Figure 1: A possible reaction to prouce an K 0 meson. The lines inicate flow of quark flavors from left to right. No interactions are shown. Note that the prouction of the antibaryon tells us that it is a K 0,notaK 0. By measuring the frequency of the oscillation in the last term, we may measure the mass ifference between the KS 0 an the K0 L. When the momentum is small, ω S ω L m S m L. Because this ifference is very small, it is experimentally intractable to attempt this with irect kinematic measurements. Measurements of the oscillation frequency yiel a mass ifference of m S m L = s 1 (25) = s fm/s ev-fm = 3 µev, (26) a ifference comparable to the energy of a microwave photon. Since the mass of the kaon is approximately 500 MeV, this is a fractional ifference of orer one part in 10 14! We remark that this example shows that sometimes, even in non-relativistic quantum mechanics, the rest mass term in the energy must be inclue. This is because we may have a superpostion of states with ifferent masses, an the time evolution of the components is corresponingly ifferent, such that there is a time-epenent interference. 5

6 P K 0 (t) t/ τ s Figure 2: Upper curve: the K 0 K 0 oscillation probability as a function of time (in units of τ S ). Lower curve: the oscillation probability if m S = m L. 4 Exercises 1. Fin the neutral kaon Hamiltonian in the K 0, K 0 basis. Is the symmetry of your result consistent with the notion that the masses of particles an antiparticles are the same? Same question for the ecay rates? 2. Repeat the erivation of Eqn. 24, but work in the ensity matrix formalism. We i not consier the possibility of ecay when we evelpe this formalism, so be careful you may fin that you nee to moify some of our iscussion. 3. In this note, we iscusse the neutral kaon (K) meson, in particular the phenomenon of K 0 K 0 mixing. Let us think about this system a bit further. The K 0 an K 0 mesons interact in matter, ominantly via the strong interaction. Approximately, the cross section for an interaction with a euteron is: σ(k 0 ) = 36 millibarns (27) σ( K 0 ) = 59 millibarns, (28) 6

7 at a kaon momentum of, say, 1.5 GeV. Note that a barn is a unit of area equal to cm 2. (a) Consier a beam of kaons (momentum 1.5 GeV) incient on a target of liqui euterium. Let λ be the K 0 interaction length, i.e., the average istance that a K 0 will travel in the euterium before it interacts accoring to the above cross section. Similarly, let λ be the K 0 interaction length. To a goo enough approximation for our purposes, you may treat the euterium as a collection of euterons (why?). The ensity of liqui euterium is approximately ρ =0.17 g/cm 3.Whatareλan λ, in centimeters? (b) Suppose we have prepare a beam of KL 0 mesons, e.g., byfirst creating a K 0 beam an waiting long enough for the KS 0 component to ecay away. If we let this KL 0 beam traverse a euterium target, the K 0 an K 0 components will interact ifferently, an we may en up with some KS 0 mesons exiting the target. Let us make an estimate for the size of this effect. Since the kaon is relativistic, we nee to be a little careful compare with our iscussion in the note: In the KL 0 rest frame, the amplitue epens on time t accoring to: exp( im L t Γ L t /2), (29) where Γ L =1/τ L is the K 0 L ecay rate. In the laboratory frame, where the kaon is moving with spee v, anγ =1/ 1 v 2, t t/γ, where t is the time as measure in the laboratory frame. In the lab frame, we have t/γ = x/γv, an we may write the amplitue as for the K 0 L as: exp( im L x/γv Γ L x/2γv), (30) Let us consier a euterium target, of thickness w, alongthebeam irection. At a istance x into the target, an interaction may occur, resulting in a final state: 1 2 (f K 0 f K 0 ), (31) 7

8 where, for example, the amplitue f for the K 0 component traversing istance x is just: f = e x/2λ 1 x 2λ. (32) Put all this together an fin an expression for the probability to observe a K 0 S to emerge from the euterium, for a K 0 L incient. Assume that w λ. You may wish to use m m L m S, Γ S,L 1/τ S,L,an Γ Γ L Γ S Γ S (c) Suppose w = 10cmanγv = 3. What is the probability to observe a KS 0 emerging from the target? What is the probability to observe a KL 0?Youmayuse: Γ S = s 1, (33) m = s 1. (34) You have been investigating a phenomenon often calle regeneration by passing through material, a KS 0 component to the beam has been regenerate. A similar consieration has been propose to help explain the solar neutrino problem. 8