Isotopic spin selection rules and the decay KL → 2π

Abstract

Recently, measurements have been performed on the decay of KL ~ 2n ° [1], and it has been found that in all cases the final 2n state in K L 2n contains a rather large I = 2 admixture. Effectively there are two solutions [2], namely 1) The final state of two pions is dominantly I = 0 with a 25% I = 2 admixture.2) The final state of two pions is practically a pure I = 2 state.Where possibility 1) is compatible with a good many proposed theories, notably with a theory of C-violating semi-strong interactions obeying AI= = 0 [3], or with a theory of C-violating e.m. interactions with AI = 0 or 1 [4], the possibility 2) does not seem to correspond clearly to any proposed model. Logically, in an analysis of the properties of CP-violating interactions one must face possibility 2) also; it is the purpose of this paper to investigate in some detail the consequences of 2) [5]. Before doing so, however, we feel obliged to point out that in any case Weinberg's cancellation effect [6] will tend to suppress (how much is anybody's guess) an eventual K L ---, 2~ (I= 0) transition with respect to KL ~ 2~ (I = 2). For this reason, even if the solution 2) is experimentally established with some accuracy, one cmmot take up assumption b) of the next section too confidently, but must rather see it as a possibility to be tested experimentally.In order to derive some properties concerning the/-spin behaviour of the CP-violating interactions we make the following assumptions, consistent with experiment:a) The final 2n state in KL ~ 2n is pure I= 2".The CP-conserving non-leptonic strangeness changing interactions obey the rule AI= ½"*.* We neglect effects of second or higher order in the CP-viol ating interaction. ** The conclusions of this paper are stable against a small AI = ~2 or~ admixture.Assumptions a), b) and c) imply that ~ =0 [2]. This in turn implies that the CP-violating interaction must be such that the mass-matrix governing the composition of K L and K S in terms of K o and Ko is not affected*. Thus K S = K1, KL=K2, as if there were no CP-violation. We must now distinguish the various possibilities for the CPviolating interaction. The CP-violating interaction obeys AS = 0(i.e. it belongs to the class of C-violating strong interactions). The mass-matrix is determined by second order weak transitions from a K-meson or anti-meson to a K-meson or anti-meson. From assumption c), and the fact that the K-meson belongs to an I= ½ multiplet we conclude that the requirement of no C-violating effects in the mass-matrix leads to the rule AI ~ 3 for the Cviolating interaction. Allowing K ~ 2n (I= 2) to first order in weak and C-violating interactions leads to the conclusion that at least part of the C-violating interaction must obey AI < 3. Thus we conclude for the C-violating interaction:No AI = 0,1,2;There must be a part with AI = 3 t=.The immediate consequence of these selection rules is that no C-violating effects should exist in the decay ~7 ~ 3n, as only AI = 0 or 2 C-violating interactions can contribute to such effects [3].The C-...