ABSTRACT
Nonleptic decays occur from the combination of X with the term ( p ¯ n ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq212.tif"/> in the current J. This term ( p ¯ n ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq213.tif"/> is of isotopic spin 1, so if combined with an X assumed to be pure isotopic spin 1/2 we can form only isotopic spin 1/2 and 3/2, and suggest the third rule governing strangeness changing decays:
In nonleptic decays the change of isotopic spin can be only Δ T = 1/2 or Δ T = 3/2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq214.tif"/>
This does not appear very restrictive, yet it does have consequences that we can check.
First, we can predict charge ratios for the three pion decays of the kaons. The three pions in the decay () K + → π + + π + + π − https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq215.tif"/>
seem from their momentum distribution to be in a state of zero angular momentum, and therefore the wave function of the pions is completely symmetrical. It can be shown that the only symmetrical isotopic spin states available to three particles, each of isotopic spin 1, are T = 1 and T = 3. If rule 3 is correct, the state T = 3 cannot be generated from the original T = 1/2 of the kaon, for a change of at least 5/2 would be involved. Thus the final state must be T = 1, and from the rules for combining states it is easy to show that the rate of the decay K+ → π+ + π° + π° should be 1/2 of the rate of (13–1)(except for a rate increase of about 9 per cent for each π° resulting from the small mass difference of π+ and π°). Experimentally the ratio is 0.30 ± 0.06, which is consistent with the predicted 0.25(1.2) = 0.30.
By exactly the same argument, the three pion decays of the K 2 ∘ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq216.tif"/> , () K 2 ∘ → π + + π − + π ° https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq217.tif"/> () K 2 ∘ → π ° + π ° + π ° https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq218.tif"/>
61should occur in ratio 2/3 [or corrected for π° mass difference, ratio 2(1.1)/3(1.3) = 0.56] if the final state is T = 1. Measurements on the K 2 ∘ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq219.tif"/> are just beginning to be made, and are so far in agreement with this prediction.
There is also a consequence for the two-pion disintegration of the kaon. The data are K 1 ∘ → π + + π − 78 ± 6 per cent → π ° + π ° 22 ± 6 per cent K + → π + + π 0.002 times K 1 ∘ decay rate https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq220.tif"/>
A remarkable feature is that the K+ decay is so much less rapid than the K 1 ∘ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq221.tif"/> decay.
For two pions in a symmetrical state, the total isotopic spin is T = 0 or T = 2. Only the T = 2 state is available for the case K+ of one π+ and one π°. Now this could be reached in general from the kaon T = 1/2 either by ΔT = 3/2, or by ΔT = 5/2 (we add isotopic spin as vectors). According to hypothesis 3, however, only the ΔT = 3/2 operates. That means that the rate of K+ decay gives us the relative amplitude of T = 2 in the decay of K°. Actually it gives us only the square, but we know that the amplitude is 0.052 times some complex phase for the two pions of K 1 ∘ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq222.tif"/> to be in state T = 2. This amplitude is so small that the K 1 ∘ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq223.tif"/> should decay almost purely into T = 0. If it did so, the relative proportions of π+ + π− to π° + π° should be 2:1 or 67 per cent should be charged. If the amplitude for T = 2 has its phase for maximum or for minimum interference, the percentage predictions are 72 and 62 per cent, respectively. Theoretically then, the proportion of π+ + π− in the K 1 ∘ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq224.tif"/> must lie somewhere between 62 and 72 per cent if hypothesis 3 is valid. We must wait for more precise data to see if this is true; the present results are just consistent with it. As far as we know, then, the X current can be restricted to one whose strangeness is −1 and isotopic spin is 1/2. In terms of the known particles then, it would have the form X = α ( p ¯ Λ ) + β [ – ( p ¯ ∑ ° ) + 2 ( n ¯ ∑ − ) ] + γ [ − ( K ¯ + π ° ) + 2 ( K ° π − ) ] + δ [ − ( ∑ ¯ ° Ξ − ) + 2 ( ∑ ¯ + Ξ ° ) ] + ε ( Λ ¯ Ξ − ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq225.tif"/>
with the coefficients α, β, γ, δ, and ε to be determined. This is as far as we have been able to proceed. The difficulties in proceeding further will now be pointed out.
The Question of a Universal Coupling Coefficient. In view of the apparent equality of the coefficients in J of such different terms as ( ν ¯ e ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq226.tif"/> , ( ν ¯ μ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq227.tif"/> , and ( p ¯ n ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq228.tif"/> , it is natural to suggest a kind of universality and propose that the coupling 62coefficients of all particles are equal (universal), so that all the coefficients α to ε are equal and equal to unity. [Or at least if factors 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq229.tif"/> are differently distributed, some of them may be 1 and others 1 / 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq230.tif"/> to provide some special symmetry. For example, if α = 1 / 2 = − β https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq231.tif"/> the first two terms become ( p ¯ Z ) + ( n ¯ ∑ − ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq232.tif"/> , a combination especially simple from the viewpoint of global symmetry.]
As a further example in the second strong coupling scheme suggested above, a particularly simple hypothesis would be that (1) the nonstrangeness changing Fermi current is coupled to the same combination of particles as is the π+, namely, a component of isotopic spin, and also (2) the strangeness changing Fermi current is coupled to the same combination of particles, as is the K+. This would mean α = −β, δ = ε, and possibly γ = 0. But there is direct evidence that this is not the case. If γ = 1, disregarding the other terms, the decay K+ → π° + e+ + ν could be a direct process, and its rate calculated. It comes out 170 times too fast! There may be some modification from other diagrams but it surely cannot be so drastically reduced. We deduce that either γ is 0 or of the order of 0.08 (i.e., 1 170 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq233.tif"/> , for the rate goes as γ2). If γ were 0 the process would have to be an indirect one, which we cannot calculate (although at first guess it is not easy to see how it could be so slow even if γ were 0 and the other constants were of order unity—yet no firm conclusion can be drawn this way about the other constants). Again, if α = 1, we can calculate a rate for the process Λ → p + e + ν ¯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq234.tif"/> . This process and the one with μ replacing e have not been seen, yet we predict it should appear in 16 per cent of the Λ disintegrations. Experimentally it occurs at least less than one-tenth as often as this. It is unlikely that this is an effect of interference from other diagrams, so α is probably less than 0.3. In addition no lepton decay of the Σ− has been seen again less than 10 per cent of that expected with β = 1 / 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq235.tif"/> , so β / 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq236.tif"/> must also be less than 0.3. It therefore does not look as if the X is coupled to leptons with the full coefficient expected from the universal coupling; in fact, a coefficient of order 0.1 seems more likely. (It is not possible to disprove this from the rapidity of the K + → μ + + ν decay, again because of uncertainties in all such quantitative calculations.)
We can summarize these points by the observation that, although theoretically unexpected, the data may indicate that
Leptic decays with change of strangeness are relatively much slower than those without change of strangeness (although the K + → μ + + ν is a possible violation).
But if the coefficients in X are of the order of 0.1 for lepton coupling, we should expect them to be exactly the same for the ( p ¯ n ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429492501/e6a0c676-1f8d-4684-97a0-6bf3353f7ce0/content/eq237.tif"/> coupling. This is uncomfortable because the nonleptic decays seem too fast for this. They seem to require coefficients of order unity, but we cannot be sure, for we cannot really calculate these processes because of the virtual states of strongly interacting particles that are involved.
63In addition there is a further approximate symmetry rule suggested by the data for which we have no theoretical explanation:
Nonleptic strangeness changing decays with ΔT = 3/2 are relatively much slower than those with ΔT = 1/2, i.e., weaker.
