1 Introduction 2 the K0 Meson and Its Anti-Particle

Physics 195a Course Notes The K0: An Interesting Example of a “Two-State” System 021029 F. Porter

1 Introduction

An example of a two-state system is considered. Interesting complexities involve: 1. Treatment of a decaying particle. 2. Superposition of states with diﬀerent masses.

2TheK0 Meson and its Anti-particle

The K0 meson is a pseudoscalar state consisting (for present purposes) of a d quark and ans ¯ antiquark: |K0 = |ds¯. (1) 0 Its antiparticle is the K : 0 |K = |ds¯ . (2) If we deﬁne a “strangeness” operator, S, (which counts strange quarks), these states are eigenstates, with: S|K0 = |K0, (3) 0 0 S|K = −|K . (4) S 10 |K0 |K0 We may write as the two-by-two matrix 0 −1 in the , basis, but a convenient “basis-independent” form is: 0 0 S = |K0K0|−|K K |. (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: 0 C|K0 = −|K , (6) 0 C|K = −|K0. (7)

1 If we multiply the C operator by the parity operator P ,wehave: 0 CP|K0 = |K , (8) 0 CP|K = |K0. (9)

0 We thus have, in the |K0, |K basis: 01 CP = , (10) 10 which we may also express in the basis-independent form: 0 0 CP = |K0K | + |K K0|. (11)

The eigenstates of CP are (the choice of nomenclature will shortly be moti- vated): 0 1 0 0 |KS = √ |K + |K , with CP =+1, (12) 2 0 1 0 0 |KL = √ |K −|K , with CP = −1. (13) 2 0 We remark that the K0 (or K ) is the lowest mass particle containing the strange quark. Thus, the only permitted decays must be via the weak interaction. To a good approximation (but not exactly!), CP is conserved in the weak interaction (and even more so in the strong and electromagnetic interactions); we shall assume this here. 0 A neutral K meson (K0 or K ) is observed to decay sometimes to two pions and sometimes to three pions. For example, consider the observed process K0 → π0π0. Since all of the particles in this decay are spinless, the decay must proceed with zero orbital angular momentum (“S-wave” decay). Thus, the parity of the π0π0 system in the final state must be positive. But we said that the K0 is a pseudoscalar particle, i.e., has negative parity. Thus, this is a parity-violating decay. The weak interaction is known to violate parity (i.e., parity is not conserved in the weak interaction), so this is all right. The π0 is its own anti-particle, hence the |π0π0 final state is an eigenstate of C with eigenvalue +1. Thus, the |π0π0 final state is also an eigenstate of CP with eigenvalue +1. Under our approximation that CP is conserved in the weak interaction, we therefore conclude that the observation of a K0 → π0π0 decay projects

2 0 0 0 out the KS component of the K meson (likewise for the K ). The 2π decay mode is favored by phase space over decays to greater numbers of pions. 0 However, the KL → 2π decay is forbidden by CP conservation. Hence, the 0 0 KL → 3π decay is important for KL. Because the phase space is considerably 0 0 suppressed, the KL decay rate is much slower than the KS rate. The observed 0 0 lifetimes of the KS and KL are, respectively:

−11 τS =9× 10 s(90ps), (14) −8 τL =5× 10 s(50ns). (15)

3 Time Evolution of a Kaon State

Suppose that at time t = 0we have the state

0 ψ(0) = |KS. (16) How does this state evolve in time? We should have, at time t,

−iHt 0 ψ(t)=e |KS. (17) 2 2 For a free particle, the energy is ωS = p + mS,wheremS is the mass of 0 the KS. But if we just use this for H, we won’t have a particle which decays in time. We know that, if we start with a particle at t = 0the probability to ﬁnd it undecayed at a later time t if it has a lifetime τS =1/ΓS is:

P (t)=e−ΓS t. (18)

Thus, the amplitude should have an exp(−ΓSt/2) time dependence, in addi- tion to the phase variation:

−iωSt−ΓS t/2 0 ψ(t)=e |KS. (19) 2 2 0 Letting ωL = p + mL,wheremL is the mass of the KL,andΓL =1/τL, 0 0 we similarly have for an intial KL state (ψ(0) = |KL):

−iωLt−ΓLt/2 0 ψ(t)=e |KL. (20)

0 0 In the |KS, |KL basis, the Hamiltonian operator is: ωS − iΓS/20 H = . (21) 0 ωL − iΓL/2

3 This requires some further discussion. For example, how did I know that 0 H is diagonal in this basis (and not, perhaps, in the |K0, |K basis)? The answer is that we are assuming that CP is conserved. Hence, [H, CP]=0. The Hamiltonian cannot mix states of differing CP quantum numbers, so 0 0 there are no off-diagonal terms in H in the |KS, |KL basis. The second point is that we have allowed the possibility that the masses of the two CP eigenstates are not the same (having already noted that the lifetimes are different). This might be a bit worrisome – the C operation does not 1 0 0 change mass. However, the |KS and |KL 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 different. We will address shortly the measurement of the mass difference. 0 Now suppose that at time t = 0we have a pure K state:

0 ψ(0) = |K . (22)

Experimentally, this is a reasonable proposition, since we may produce such states via the strong interaction. For example, if we collide two particles with no initial strangeness (perhaps a proton and an anit-proton), we make strange particles in “associated production”, i.e., in the production of ss¯ pairs. Thus, 0 we might have the reactionpp ¯ → nΛK (see Fig. 1). The presence of the Λ, 0 which contains thes ¯ quark, tells us that the kaon produced is a K , since it contains the s quark. 0 So, we can realistically imagine producing a K at t =0. Butthetime- evolution to later times is governed by the Hamiltonian, which is not diagonal 0 in the |K0, |K basis. Thus, we might expect that at some later time we 0 0 may observe a K . What is the probability, PK0 (t)thataK meson is 0 observed at time t, given a pure√K state at t = 0? The answer, noting that 0 0 0 ψ(0) = |K =(|KS−|KL) / 2, is:

0 2 PK0 (t)=|K |ψ(t)| (23)

1 0 0 −iωSt−ΓS t/2 0 0 −iωLt−ΓLt/2 2 = |K |KSe −K |KLe | 2 ΓS+ΓL 1 −ΓSt −ΓLt − t = e + e − 2e 2 cos [(ωS − ωL)t] . (24) 4 1Actually, this is only an assumption here. But it is a fundamental theorem in relativistic quantum mechanics that particle and anti-particle have the same mass (as well as the same total lifetime).

4 _ d _ _ u Λ _ _ } _ d_ s p u { _ u _ _ 0 s K d }

u p { u d d u } n d

0 Figure 1: A possible reaction to produce an K meson. The lines indicate ﬂow of quark ﬂavors from left to right. No interactions are shown. Note that 0 the production of the antibaryon tells us that it is a K ,notaK0.

By measuring the frequency of the oscillation in the last term, we may 0 0 measure the mass difference between the KS and the KL. When the momentum is small, ωS − ωL ≈ mS − mL. Because this difference is very small, it is experimentally intractable to attempt this with direct kinematic measurements. Measurements of the oscillation frequency yield a mass difference of

10 −1 |mS − mL| =0.5 × 10 s (25) 0.5 × 1010 s−1 = 200 × 106 eV-fm 3 × 1023 fm/s =3µeV, (26) a difference comparable to the energy of a microwave photon. Since the mass of the kaon is approximately 500 MeV, this is a fractional difference of order one part in 1014! We remark that this example shows that sometimes, even in non-relativistic quantum mechanics, the rest mass term in the energy must be included. This is because we may have a superpostion of states with different masses, and the time evolution of the components is correspondingly different, such that there is a time-dependent interference.

5 P (t) K 0

0.30

0.25

0.20

0.15

0.10

0.05

0.00 0 5 10 15 20 25 30 35 t/τ s

0 Figure 2: Upper curve: the K → K0 oscillation probability as a function of time (in units of τS). Lower curve: the oscillation probability if mS = mL.

4Exercises 0 1. Find the neutral kaon Hamiltonian in the |K0, |K basis. Is the symmetry of your result consistent with the notion that the masses of particles and antiparticles are the same? Same question for the decay rates?

2. Repeat the derivation of Eqn. 24, but work in the density matrix formalism. We did not consider the possibility of decay when we develped this formalism, so be careful – you may ﬁnd that you need to modify some of our discussion.

3. In this note, we discussed the neutral kaon (K) meson, in particular the phenomenon of K0 − K¯ 0 mixing. Let us think about this system a bit further. The K0 and K¯ 0 mesons interact in matter, dominantly via the strong interaction. Approximately, the cross section for an interaction with a deuteron is:

σ(K0d) = 36 millibarns (27) σ(K¯ 0d) = 59 millibarns, (28)

6 at a kaon momentum of, say, 1.5 GeV. Note that a “barn” is a unit of area equal to 10−24 cm2.

(a) Consider a beam of kaons (momentum 1.5 GeV) incident on a target of liquid deuterium. Let λ be the K0 “interaction length”, i.e., the average distance that a K0 will travel in the deuterium before it interacts according to the above cross section. Similarly, let λ¯ be the K¯ 0 interaction length. To a good enough approximation for our purposes, you may treat the deuterium as a collection of deuterons (why?). The density of liquid deuterium is approximately ρ =0.17 g/cm3.Whatareλ and λ¯, in centimeters? 0 (b) Suppose we have prepared a beam of KL mesons, e.g.,byfirst 0 0 creating a K beam and waiting long enough for the KS compo- 0 nent to decay away. If we let this KL beam traverse a deuterium target, the K0 and K¯ 0 components will interact differently, and 0 we may end up with some KS mesons exiting the target. Let us make an estimate for the size of this effect. Since the kaon is relativistic, we need to be a little careful com- 0 pared with our discussion in the note: In the KL rest frame, the amplitude depends on time t∗ according to:

∗ ∗ exp(−imLt − ΓLt /2), (29) 0 where ΓL =1/τL is the KL decay rate. In the laboratory√ frame, where the kaon is moving with speed v,andγ =1/ 1 − v2, t∗ → t/γ,wheret is the time as measured in the laboratory frame. In the lab frame, we have t/γ = x/γv, and we may write the 0 amplitude as for the KL as:

exp(−imLx/γv − ΓLx/2γv), (30) Let us consider a deuterium target, of thickness w,alongthebeam direction. At a distance x into the target, an interaction may occur, resulting in a ﬁnal state: 1 √ (f|K0−f¯|K¯ 0), (31) 2

7 where, for example, the amplitude f for the K0 component tra- versing distance dx is just: dx f = e−dx/2λ ≈ 1 − . (32) 2λ Put all this together and ﬁnd an expression for the probability to 0 0 observe a KS to emerge from the deuterium, for a KL incident. Assume that w λ.Youmaywishtouse∆m ≡ mL − mS , ΓS,L ≡ 1/τS,L,and∆Γ≡ ΓL − ΓS ≈−ΓS (c) Suppose w =10cmandγv = 3. What is the probability to 0 observe a KS emerging from the target? What is the probability 0 to observe a KL?Youmayuse:

10 −1 ΓS =1.1 × 10 s , (33) ∆m =0.5 × 1010 s−1. (34)

You have been investigating a phenomenon often called “regeneration” 0 – by passing through material, a KS component to the beam has been “regenerated”. A similar consideration has been proposed to help ex- plain the “solar neutrino problem”.