8 The Neutral-Kaon System

The neutral kaons $K0$ and $K0¯$ are two of the eight members of the octet of light spin-0 mesons with negative parity, which also includes the charged kaons $K±$⁠, the pions $π±$ and $π0$⁠, and the $η$ (see Fig. 8.1). The kaons are strange particles, the strangeness of $K0$ and of $K+$ being +1, while that of $K0¯$ and of $K−$ is –1. In the quark model, $K0∼s¯d,K0¯∼sd¯,K+∼s¯u$⁠, and $K−∼su¯$⁠.

The neutral kaons constitute the only system in which CP violation has been observed up to now. Indeed, their mixing makes them an excellent laboratory to look for very weak effects, like CP violation and CPT violation (Kostelecký 1998). The measurement of the tiny mass difference between the long-lived and the short-lived neutral kaons is one of the most precise measurements in particle physics.

We discuss in this chapter the specifics of neutral-kaon mixing and decays, relying on notation and formulae from Chapters 6 and 7. We assume CPT invariance.

The neutral-kaon system has two features which distinguish it from other neutral-meson systems.

First feature: the lifetimes of the two eigenstates of mixing are very different. As a consequence, it is usual to distinguish the eigenstates of mixing by their lifetimes instead of distinguishing them by their masses: $KS$ is the short-lived neutral kaon and $KL$ is the long-lived neutral kaon. The corresponding masses and decay widths are given by $μS=mS−(i/2)ΓS$ and $μL=mL−(i/2)ΓL$⁠, respectively. We define

Also, $Γ≡(ΓS+ΓL)/2$ as in eqn (6.12). Experimentally,.

while

We see that $ΓS≈579ΓL$⁠. Therefore, $ΔΓ≈−ΓS$ and $Γ≈ΓS/2$⁠. Moreover, $ΓS≈2Δm$⁠. From the present point of view, the latter approximate equality is just a coincidence. Remembering eqns (6.15), we see that in the neutral-kaon system $x≈−y≈u≈1$⁠.

By definition, $ΔΓ<0$⁠. The sign of $Δm$ was experimentally determined by means of regeneration experiments; it was found that $Δm>0$⁠, as anticipated in eqn (8.3). The average mass of the neutral kaons is

Notice that the mass difference $Δm$ is fourteen orders of magnitude smaller than the average mass. This is a consequence of the fact that $Δm$ arises at second order in the weak Hamiltonian, which has a typical strength $10−7$ that of the strong Hamiltonian, responsible for $mK$⁠.

As $KL$ is heavier than $KS$⁠, we should identify $PH$ of Chapter 6 with $KL$⁠, while $PL$ is $KS.$ Accordingly, we write

or

with

Second feature: the kinematically allowed phase space for the two-pion decay channels $π+π−$ and $π0π0$ is much larger than the one for any other decay channel. If it were not for CP symmetry, the decays to two pions would be dominant for both $KS$ and $KL$⁠. As a matter of fact, the two-pion decays are dominant for $KS$⁠, but not for $KL$⁠. Experimentally,

As both the kaons and the pions are spinless, the two-pion state resulting from the decay of a neutral kaon is in an s wave. That state then has C = P = CP = +1. If CP was conserved $KS$ and $KL$ would be eigenstates of CP, one of them with eigenvalue +1 and the other one with eigenvalue –1 (remember eqn 6.76). The eigenstate with eigenvalue +1 would decay to two pions, the one with eigenvalue –1 would not. Thus, the short-lived neutral kaon $KS$⁠, which decays predominantly to two pions, would be the CP-even superposition of $∣K0〉$ and $∣K0¯〉$⁠. On the other hand, $KL$⁠, being prevented by CP symmetry from decaying to the kinematically favoured two-pion states, would automatically have a lifetime much longer than that of $KS$⁠.

Small CP violation in the kaon system slightly disturbs this state of affairs. The original discovery of CP violation (Christenson et al. 1964) consisted in the observation of two-pion decays of $KL$⁠. Intrinsically and a priori, $KL$ is equivalent to $KS$ and there is no CP violation in the fact that $KL$ by itself alone decays to two pions. CP violation rather lies in the fact that both $KS$ and $KL$⁠, which are mixtures of two CP-conjugate states, decay to the same CP eigenstate (Sachs 1987).

Thus, the dominance of the two-pion decay modes is closely related to the large difference between the lifetimes of the two eigenstates of mixing. The eigenstate $KS$⁠, which is allowed by CP symmetry to decay to the kinematically favoured two-pion states, automatically has a much smaller lifetime than $KL$⁠, which only decays to two pions because of CP violation.

As CP violation in neutral-kaon mixing is very small, $K0$ and $K0¯$ are approximately half $KS$ and half $KL$⁠. This means that the evolution of a neutral-kaon beam is characterized by two-pion decays, from the $KS$ component, at a short distance from the production vertex, followed at much larger distances by the decays of $KL$⁠, given by

Equations (6.53 and (6.54) read, in the case of neutral kaons,

From the experimental values in eqns (8.2) and (8.3), together with eqn (8.11), one obtains $∣δ∣≤6×10−2$⁠. This bound indicates that CP violation in neutral-kaon mixing is very small. The strength of this bound is a direct consequence of $∣y∣≈1$⁠.

A better bound can be obtained if we use eqn (8.10), together with the experimental results in eqns (8.8) for the two-pion decay modes, which dominate the sum. We obtain

With such a small $∣δ∣$⁠, and $Δm$ and $ΔΓ$ being of the same order of magnitude, it is reasonable to approximate eqns (6.64) and (6.65) by

From eqn (6.63), and as

we derive

The complex numbers $M12$ and $Γ12$ have a phase difference close to $π$⁠, because of eqn (6.62) and $(Δm)(ΔΓ)<0$⁠.

The leptonic asymmetries are the clearest signs of CP violation in the neutral-kaon system. They are defined by

where l may be either the electron e or the muon $μ$⁠. If CP is conserved then $KL$⁠, being a neutral particle with unique mass and decay width, must be an eigenstate of CP. If CP is conserved a CP eigenstate must decay with equal probability to two states which are CP-conjugates of each other. Therefore, $δl≠0$ is an unmistakable signature of CP violation (see § 1.4.4).

Experiment indicates that $δe$ and $δμ$ are almost equal:

The equality of $δe$ and $δμ$ follows from the universality of the weak interaction.

One should remember the ΔS = ΔQ rule for the decays of strange particles. This rule implies that $K0$ decays to $π−l+νl$ but not to $π+l−ν¯l$⁠, while $K0¯$ decays to $π+l−ν¯l$ but not to $π−1l+νl$⁠:

Thus, the semileptonic decays tag the flavour of the neutral kaon. Moreover, there is only one hadron—the charged pion—in the semileptonic final states $π±l∓νl$⁠, and therefore no final-state strong interactions may scatter those final states into or from other final states; as a consequence, CPT invariance leads to

Then, the leptonic asymmetry measures the difference between the probability of finding a $K0$ and the probability of finding a $K0¯$ in $KL$⁠. Therefore, when the ΔS = ΔQ rule is strictly valid,

Notice the closeness between the value of $δ$ in eqn (8.17) and the unitarity bound in eqn (8.12). From the derivation of the unitarity bound in Chapter 6 we learn what this means: the relevant phases

must be very close to each other.

We may also define leptonic asymmetries for the semileptonic $KS$ decays,

Their measurement may be possible at a $ϕ$ factory (Buchanan et al. 1992). CPT invariance together with the ΔS = ΔQ rule predict $δl′=δl$⁠.

Violation of the ΔS = ΔQ rule is parametrized by

As long as final-state interactions in the semileptonic decays may be neglected, the right-hand side of eqn (5.27) vanishes. Thus, $(CPT)T(CPT)−1=T†=T$ and therefore $〈f∣T∣i〉=〈fCPT∣T∣iCPT〉∗,$ where $∣iCPT〉≡CPT∣i〉$ and $∣fCPT〉≡CPT∣f〉$⁠. Thus, CPT invariance implies $xl=x¯l$⁠. The parameter $xl$ is expected to be $∼10−7$⁠, basically because $〈π−l+νl∣T∣K0〉$ and $〈π+l−ν¯l∣T∣K0¯〉$ are first-order in $ℋW$⁠, while $〈π−l+νl∣T∣K0¯〉$ and $〈π+l−ν¯l∣T∣K0〉$ are second-order. Experiment indicates that $xl$ is of order $10−2$ or smaller.

The general expression for $δl$⁠, assuming CPT invariance but allowing for violation of the ΔS = ΔQ rule, is

Thus, $δl$ must originate in mixing CP violation, $∣R12∣≠∣R21∣$⁠, even when the $ΔS=ΔQ$ rule is violated.

We define, for an arbitrary decay channel f, the parameters

The factor r is introduced in order to obtain rephasing-invariance. It is largely arbitrary, it must satisfy only two conditions. First, it must depend on the phases of the kets $∣KS〉$ and $∣KL〉$ in such a way as to offset the phase-convention dependence of the ratio $〈f∣T∣KL〉/〈f∣T∣KS〉$⁠. Only then is the phase $ϕf$ physical. Second, in the CPT invariant case, and with the phase convention of eqns (8.5), one must have r = 1. Thus, $〈K0∣KL〉=〈K0∣KS〉$ and $〈K0¯∣KL〉=−〈K0¯∣KS〉$ must imply r = 1. For instance, Kayser (1996) has suggested

and Lavoura (1991) has suggested

The exact definition of r is immaterial as long as we assume CPT invariance and the phase convention in eqns (8.5). It becomes important only when we want to study the CPT-violating case. We shall assume r = 1, but we write down r explicitly in the definitions of parameters, whenever necessary.

With r = 1,

is directly measurable.

If we define, as in eqn (7.5),

where $Af≡〈f∣T∣K0〉$ and $A¯f≡〈f∣T∣K0¯〉$⁠, then

The parameters $η$ are measured by observing the time dependence of the decays of tagged neutral kaons. For instance, if at production time a neutral-kaon beam had strangeness +1, we would use

If at production time the strangeness of the neutral-kaon beam was –1, we would use

The time t is measured in the rest frame of the neutral kaons. Notice the relative minus sign between the interference terms in eqns (8.29) and (8.30). The interference pattern in those equations, which is displayed in Fig. 8.2, is one of the best experimental demostrations of quantum mechanics in the realm of particle physics.

Summing eqns (8.29) and (8.30) over all decay modes f, by means of the unitarity eqns (6.48)–(6.50), one obtains

The difference between the two decay curves in eqns (8.31) depends only on the CP-violating parameter $δ$⁠.

For the two main decay channels, $π+π−$ and $π0π0$⁠, the parameters $η$ are named $η+−$ and $η00$⁠, respectively. Their measured values are

Particularly important are the ratio

and the phase difference

If the $ΔS=ΔQ$ rule is valid, then $ηπ−l+νl=+1$ and $ηπ+l−ν¯l=−1$⁠.

This section and the next one consider specific methods for the experimental study of the neutral-kaon system, and may be skipped without loss of continuity.

Suppose that we have a beam of neutral kaons and, after letting it evolve for a proper time much longer than $τS$ but much shorter that $τL$⁠, we have it incident on a thin slab of material, called a regenerator. When it is incident on the regenerator, the beam is almost exclusively $KL,$ because the $KS$ component has decayed away. Inside the regenerator kaons are scattered by strong interaction with the nuclei in the regenerator. The strong interaction distinguishes between $K0$ and $K0¯$⁠, which have different forward-scattering amplitudes off the regenerator. As a consequence, from the opposite side of the regenerator emerges a superposition of $K0$ and $K0¯$ different from the incident one. Emerging from the regenerator we will not have a $KL$ beam, rather a beam of neutral kaons with a regenerated $KS$ component.

We may describe the process in the following way. We name $〈KLR∣T∣KLR〉$ the amplitude for a $KL$ incident on the regenerator to emerge as $KL$⁠. The amplitude for the process in which a $KL$ incident on the regenerator emerges as $KS$ is $〈KSR∣T∣KLR〉$⁠. Thus, the neutral-kaon state which emerges from the regenerator is

We now observe the decays of $Kr$ to the channel f as a function of the proper time t, measured with the initial time being the instant at which the kaon beam emerges from the regenerator. We get

where

Notice that $υf$ is a rephasing-invariant quantity, and therefore its phase $θf$ is measurable.

Regeneration is a way of measuring the parameters $ηf$ whenever we are able to make a reliable theoretical computation of $〈KLR∣T∣KLR〉/〈KSR∣T∣KLR〉$⁠, so that we are able to extract $ηf$ from $vf$⁠.

We consider in this section the decays of $K0K0¯$ pairs in an antisymmetric correlated state. This is important because such correlated states will be copiously produced in the upcoming $ϕ$ factories, in particular at DAΦNE. The resonance $ϕ$ has spin 1, and upon its decay the resulting $K0K0¯$ pair is in a p wave. This correlated state is C- and P-odd and is written, in the rest frame of the decaying $ϕ$⁠,

Notice the absence of $∣KL(k→)〉⊗∣KL(−k→)〉$ and $∣KS(k→)〉⊗∣KS(−k→)〉$ components. This is because the p wave is antisymmetric, and two identical bosons in an antisymmetric state would violate Bose symmetry. This fact holds not only for the initial instant at which $ϕ$ decays into a kaon pair, but also for any later time, even after the neutral kaons have oscillated back and forth into each other. The antisymmetry of the wave function is preserved by the linearity of the oscillation. This holds even when CPT is violated in the mixing.

Let the kaon with momentum $k→$ decay to the state f at time $t1$ and the kaon with momentum $−k→$ decay at time $t2$ to the state g. The density of probability for this decay is

If f and g are eigenstates of CP with the same CP-parity, this decay is forbidden by CP symmetry. Indeed, in that case the eigenstates of mass would coincide with the eigenstates of CP; $KS$ would have $CP=+1$ and $KL$ would have $CP=−1$⁠. Equation (8.38) tells us that, if in one side of the detector we have $KS$⁠, in the opposite side of the detector we must have $KL$⁠. Thus, the occurrence of two final CP eigenstates with the same CP-parity in both sides of the detector is forbidden by CP symmetry.

Suppose that experimentally we do not observe $t2$⁠. Then, it is adequate to integrate eqn (8.39) from $t2=0$ to $t2=∞$ and obtain a distribution dependent only on $t1$⁠:

where $ϕ~sw≡arctan(Δm/Γ)≈43.39°$⁠. If we do not observe $t1$ either, we obtain the probability for the decays to f and g to occur at any time:

Summing this expression over all decay channels f and g we obtain 1 as we should, after using the unitarity relations.

We may also sum eqn (8.40) over all decay channels g and obtain the time distribution of the decays of the meson with momentum $k→$ to the channel f: