19 KL→π0νν¯

The rare kaon decay $KL→π0νν¯$ may become very important in the study of CP violation. In the context of the Kobayashi–Maskawa model, it gives direct access to the parameter $η$ of the CKM matrix, with little associated theoretical uncertainty—unknown hadronic matrix elements or other poorly known parameters. On the other hand, this decay is extremely challenging for experimentalists, due to the presence of two neutrinos and no charged particle in the final state. The process $KL→π0νν¯$ has not yet been observed; the 90%-confidence-level experimental bound is (E799 Collaboration 1994)

This is six orders of magnitude above the SM prediction (Buchalla 1997; Buras 1997)

and therefore much progress on the experimental side is still needed before we are able to vindicate theory. Inverting the argument, there is ample room for the discovery of new physics by observation of a branching ratio larger than the SM prediction.

In the standard model $KL→π0νν¯$ violates CP. The argument leading to this conclusion still holds in many extensions of the SM (Grossman and Nir 1997). The first assumption is conservation of the individual lepton numbers. Then, the neutrino and the antineutrino in $KL→π0νν¯$ have the same flavour, i.e., they are the antiparticle of each other. This is a necessary condition for the final state to be an eigenstate of CP.

If CP is conserved, $KL$ does not decay into two pions and has CP-parity –1. Then, the CP-parity of $π0νν¯$ should be –1 too. In the rest frame of $KL$

where L is the relative angular momentum of $π0$ and the $νν¯$ pair. As both $KL$ and $π0$ are spinless, L is equal to J, the total angular momentum of the $νν¯$ pair. We know—see § 4.2 and 4.4—that $π0$ has CP-parity –1. Therefore,

One now assumes that the neutrino is left-handed and the antineutrino is right-handed. Then, in the rest frame of the $νν¯$ pair, the projection of angular momentum on the direction of flight of those massless particles is 1. Moreover, the dominant operator creating the $νν¯$ pair out of the vacuum is $ν¯γμγLν.$⁴⁷ Let us separate this operator into its time and space components: $ν¯γ0γLν$ and $ν¯γ→γLν$ (Kayser 1997). The time component creates a $νν¯$ pair with $J=0,$ and therefore does not contribute to $KL→π0νν¯,$ as one easily sees in the rest frame of the $νν¯$ pair. The space component has $CP=+1$ (see eqn 3.80) and creates a $νν¯$ pair with $J=1.$ The product CP $(νν¯)(−1)J$ is –1. We thus conclude that

As the CP-parities of $KL$ and of $π0νν¯$ are different, CP is violated in $KL→π0νν¯.$ Notice that eqn (19.5) depends on $ν$ and $ν¯$ being each other’s antiparticle, and on the operator which creates them being $ν¯γμγLν;$ in general, the three-particle state $π0νν¯$ would not have a well-defined CP-parity.

In the rest of this chapter we study the SM prediction of BR $(KL→π0νν¯).$ Some readers may want to skip this.

The transition $KL→π0νν¯$ corresponds to either $K0→π0νν¯$ or $K0¯→π0νν¯.$ In terms of quarks, the first decay is $s¯→d¯νν¯,$ and the second decay is $s→dνν¯.$ The standard-model effective Hamiltonian for these transitions has been computed by Inami and Lim (1981)—see Appendix B. It is

where $λα≡Vαs∗Vαd,xα≡mα2/mW2,$ and

With $mc=1.25±0.25GeV$ and $mt=175.5±5.5GeV,$ one has

We shall also need the Hamiltonian for the tree-level decay $K+→π0e+ν$—in terms of quarks, $s¯→u¯e+ν$—, which is

The crucial parameter to be computed is

Equation (19.6) tells us that

We have used CP symmetry together with eqn (19.5) to evaluate the ratio of matrix elements.

We know the value of $qK/pK$ from eqn (17.7). Thus,

The parameter $λπνν¯$ is independent of the spurious phases $ξ$⁠, as it should be.

From eqn (7.25) we know that $λπνν¯≠±1$ implies CP violation. This may happen because of indirect CP violation $(∣q/p∣≠1)$⁠, direct CP violation $(∣A¯/A∣≠1)$⁠, or interference CP violation $(sinargλπνν¯≠0)$⁠. In the case at hand there is no direct CP violation; strong final-state-interaction phases are absent, because there is no state scattering strongly to $π0νν¯$⁠; absorptive parts of Feynman diagrams could in principle be present, but only when the intermediate quark is the up quark, and the GIM suppression makes it that diagrams with intermediate up quarks hardly contribute to the decay amplitude at all (see Appendix B). There is indirect CP violation $(δ≈3.3×10−3$ does not vanish), but it is very small. The main reason for $λπνν¯≠±1$ is interference between mixing and decay: the phases of q/p and of $A¯/A$ do not match, as $λuλt∗$ and $λuλc∗$ are not real. This is different from CP violation in the two-pion decays of the neutral kaons; there, the phases of the mixing and decay amplitudes are practically equal and CP violation arises almost exclusively from $δ$⁠. CP violation in $KL→π0νν¯$ may be much larger precisely because it is mainly interference CP violation; the measured value of $δ$ would lead by itself alone to a branching ratio much smaller than the prediction in eqn (19.2). Thus, if that prediction is experimentally vindicated the superweak theory of Wolfenstein (1964) will be disproved.⁴⁸

At this point it is convenient to introduce

Then, from eqn (19.12),

Using the Wolfenstein parametrization and the fact that $X(xt)≫X(xc)$⁠, one has

The relevant decay amplitude is

We remind that $∣pK∣2=(1+δ)/2$⁠. Therefore,

We cannot compute the matrix element, but we may equate it, using isospin symmetry, to the matrix element for the tree-level decay $K+→π0e+ν$⁠. From eqn (19.9),

Thus,

On the other hand,

Remembering $g2=4πα/sw2$⁠, one has

The factor 3 is because there are three neutrino species. With $∣Vud∣≈1,∣Vus∣=λ$⁠, and $∣Vcb∣=Aλ2,$ one may write, because of the second eqn (19.15),

The rare decay $K+→π+νν¯$ also originates in the effective Hamiltonian in eqn (19.6). Once again, one must compare it to the dominant decay in order to get rid of the unknown matrix element:

where we have taken into account that, as $π0∼(u¯u−d¯d)/2$ while $π+∼d¯u$⁠, the ratio of matrix elements is equal to $2$ when isospin symmetry is exact.

Before proceeding to explicit numerical predictions, one must take into account various corrections.

Firstly, $X(xt)$ receives a QCD correction

where the function $X1$ has been computed by Buchalla and Buras (1993a,b). In practice, eqn (19.24) amounts to making $X(xt)→0.985X(xt)$⁠.

Secondly, there are QCD corrections to $X(xc)$ too. These are much more uncertain, because for $μ∼mc$ the strong interactions are largely non-perturbative. Besides, the effective Hamiltonian for $KL→π0νν¯$ depends on the neutrino flavour, because the box diagram has an internal charged-lepton propagator. In Appendix B we have assumed the charged leptons to be massless, but this is not a good approximation for the $τ$ lepton. In practice, the ensuing dependence on $mτ$ proves to be very small in the case of $X(xt)$⁠, but is important in $X(xc)$⁠. Instead of $X(xc)$ it is better to use $(2/3)XNLe+(1/3)XNLτ$ with the function $XNLl$ computed by Buchalla and Buras (1994) for any mass $ml$ of the charged lepton l.⁴⁹ In practice,

which is somewhat smaller than $X(xc)$ in eqn (19.8).

Thirdly, the isospin symmetry used to equate matrix elements is not exact. It is convenient to separate the isospin-breaking corrections (Marciano and Parsa 1996) into three factors. The first factor originates in the different phase space for different decays; the second factor comes from isospin violation in the $K→π$ form factors; the third factor stems from electromagnetic radiative corrections. The latter factor is equal to 0.979 for both rare kaon decays considered; the phase-space correction is 1.0522 for the decay of $KL$ and 0.9614 for the decay of $K+$⁠; the form-factor correction is 0.9166 for the former and 0.9574 for the latter. Thus, $Γ(KL→π0νν¯)$ is reduced by $1.0522×0.9166×0.979=0.944$ relative to the original computation, while $Γ(K+→π+νν¯)$ is reduced by $0.9614×0.9574×0.979=0.901$⁠.

After introducing these corrections we may proceed to the numerical computations. We use $BR(K+→π0e+ν)=0.0482,τ(KL)=5.17×10−8s,τ(K+)=1.2386×10−8s,α=1/127.9$⁠, and $sw2=0.2315$⁠. We get

in which we have used $∣Vcb∣=0.04,η=0.35$⁠, and $X(xt)=0.985×1.615$⁠. We thus reproduce the prediction in eqn (19.2). Notice that BR $(KL→π0νν¯)$ depends strongly on $∣Vcb∣$⁠, and is proportional to $η2$⁠.

Now consider the case of the superweak model. In that model the ratios of the decay widths of $KL$ and of $KS$ to CP eigenstates with CP-parity +1 are all equal—see § 7.3.4. Thus,

Comparing eqns (19.21) and (19.27), we see that

We take $∣ϵ∣2≈10−5/2$ from the two-pion decays and, from eqn (19.15),

We obtain

This means that experimental vindication of eqn (19.2) would disprove the superweak theory.

For the charged-kaon decay $K+→π+νν¯$ one has, from eqn (19.23), and using the Wolfenstein parametrization,

where we have used the same values as above, together with $ρ=0$ and $X(xc)=9.5×10−4$⁠. A careful analysis yields (Buchalla 1997)

The computation of $BR(K+→π+νν¯)$ has much larger theoretical uncertainties than that of $BR(KL→π0νν¯)$⁠. This is because the value of $X(xc)$ is relevant in $K+→π+νν¯$ while it is mostly immaterial in $KL→π0νν¯$⁠. This is a consequence of the CP-violating character of the latter transition; as $KL→π0νν¯$ violates CP, the top and charm quarks must contribute to it with opposite signs, and the relevant quantity is $X(xt)−X(xc)≈X(xt)$⁠.