(the orthogonality condition: (− (ĩ | d+ (2 |) (| Ĩ )+d | 2)) = −d+d=0). Then the transition amplitude is M (1→2) = [− (Ĩ | T | 1 ) + (2 | T | 2)]d. The results of these experiments are very puzzling and still very poorly understood. Let us discuss them briefly. The formula for M (1-2) with A0 was used to interpret the results. σ1 was taken as the well-known pion-nucleon total cross section, σ was a free parameter used to fit the coherent production cross section. The energy resolutions were too poor to have pure coherent processes but the incoherent processes were subtracted reasonably reliably because (as we stated many times) they are rather unimportant at small momentum transfers (where coherent processes are important). Realistic nuclear densities were used: c = 1.1241/3 fm, a=0.545 fm. To compute the integrated coherent production cross section for a given bin of the invariant masses of the produced systems we use the formula where q2 is a cut-off at the first maximum of the distribution. σcoh (A, M1, M2) depends very critically on A and this fact enables one to extract σ2 (in complete analogy with photoproduction of vectormesons). The amazing result was that σ2 (both for 3 production and for 5 production came out to be very small). The tables below give some of the numbers obtained from the CERN experiment [S4]. The same equation is valid for nucleon targets. Hence we can eliminate d and get M (1-2): = f21 [(2 Î) (2 | T‹N) | 2 ) − (1 | T‹~) | Ĭ) [‹2 | T‹4) | 2) — (1 | T‹4) | Ï)], which gives the production amplitude on the nucleus in terms of the production amplitude on a nucleon and the elastic scattering amplitudes of the objects 1 and 2 from the target nucleus Let us use some standard parametrization of the nucleon amplitudes, e.g., We get, neglecting the q2 dependence of the elementary amplitudes (which vary much slower with q2 than (2 | T‹4) | 2) or (I | T‹4) | Ĩ )): All these formulae are very useful (because of their simplicity) in discussing some effects which do not depend dramatically on A=0. Let us go back, however, to the case A=0 and discuss some recent experiments in which. →3′′ and π→ɔ̃π processes were measured on various nuclear targets. (A very rich literature on this subject, both from experimental and theoretical points of view, can be found in the Proceedings of the XII Cracow School of Theoretical Physics, June 1972 [S4].) These results are very puzzling because the 3 and 5′′ systems do not seem to form well defined particles. For example, at 9 GeV/c the time required by a system of 1.2 GeV mass to cross one-half of the thickness of the Pb nucleus (6.5 fm) is r≈2.9X10-24 s or r≈230 MeV (relativistic dilation included). The observed distribution is wider than 230 MeV (perhaps even as much as 500 MeV). Besides, σ2 should be smaller at 15 GeV than at 9 GeV (due to time dilation), which is not the case. In fact, it is hard to accept that these systems are resonances-they look more like a group of 3 or 5! But then their cross section should be 30 N and 50,N, respectively! 5.1. Discussion of the Anomalously Small Absorption of 3 and 5π Systems in Nuclear Matter First, let us stress that there is no satisfactory explanation of this phenomenon available. One can, nevertheless, make a few points which may advance a bit our understanding of the process. Let us start with an analysis of the process of diffractive production in the high energy limit in the language of multiple scattering. (We choose the high energy limit because it is simple and we believe that the finite longitudinal momentum has little to do with absorption properties of the outgoing systems.) First, the "one step" mechanism of production: 2 Particle I converts into 2 on a Let us introduce the profiles of individual nucleons smeared over the volume of the target nucleus: where we took only the forward amplitudes fu1 (0) and f22 (0). We proceed exactly the same way as we did in obtaining the high energy limit of the y-p amplitude M (1-2) = (c-1) factors ik / db exp (i▲·b) Σ (1− (2 | г | 2)) . . . (1 − (2 | г | 2)) (2 | r‹ | 1) c1 X (1 − (1 | T | 1)) ... (1 − (1 | T | 1)). which is the same formula (5.2) which was obtained before. As we have said before, the "weakness" of the production process makes the "one step" description plausible. Let us point out, however, that KL→K, regeneration on one nucleon is weak; nevertheless, this process cannot be described correctly as a one-step process. Indeed, from the "one step" formula (5.2) we get d2b exp (i▲ b) {exp [-1⁄2oκ'T (b)]— exp [-1⁄2ok,'T(b)]}, (5.3) which is wrong! We know the correct answer because we know the composition of | KL) = 1/√2 (| Ko) + | Řo)) and | K.)=1/√2 (| Ko) — | Ko)) which is nonperturbative, hence fundamentally different from the y— V structure of the photon. In fact, accepting the fact that Ko and Ão scatter only elastically from the nucleus we get the formula which is correct: d2b exp (i▲ b) {exp [−1⁄2oÃ°'T (b)]— exp [−1⁄2oк°'T (b)]}. (5.4) Since fKLKL=fK,K,= 1⁄2 ( ƒK°K°+ƒ°K), we can see from the numerical values given below that the profiles in (5.3) and (5.4) are different. To complete the point we are making, let us compute fKLKL=fK,K, and compare it with SKL K,=2(fKK-fKK). We use the data of Foeth et al. [26]: So, the production process on one nucleon is indeed weak! But as we now see clearly (comparing (5.3) and (5.4)), that is not enough to apply the "one step" formula; one has to know the internal structure of the objects which undergo diffractive scattering. Remark: Let us note that we may write the amplitude (5.4) so that it has the KK, regeneration amplitude on a single nucleon as a factor similarly to that appearing in the "one step" description, (5.3). d2b exp (i▲ b) {exp [−1⁄2oÃ°T (b)]— exp [−1⁄2oK°T(b)]}. (5.5) But, although this formula does seem to have the production taking place on one nucleon, in fact nothing like that takes place: there is only a "smooth scattering" of the two components (Ko and Ẩo). No one nucleon along the path plays any role distinguished from the others. (From (5.5) one can see again that the "one step" formula (5.3) is incapable of describing KL→K, regeneration on nuclei). Let us go back to the general case of incoming particle 1 and outgoing particle 2. From the above discussion we can see that (within the framework of the Good and Walker description of the diffractive production processes) we have to have weak coupling between | I) and | 2) states in order to have a "one step" description, since, when the coupling is weak, we can replace Perhaps at this stage one should point out that one could easily explain the small absorption cross section by abandoning the "one step" description. The price to pay for it would be the loss of the interpretation of the states | I) and | 2) as “almost" a one physical pion state and “almost" a three physical pion state, respectively. For instance, the following scheme (which imitates K-K, regeneration) would explain the observed effects. (Note that we are working in the high energy limit, hence our arguments are based on the assumption that the absorption properties of the objects produced are the same in this limit as at the experimentally available energies.) Let us accept that (Note that the values of the 's and the coefficients of the transformation (5.6) are independent quantities.) Then the production amplitude (as in the KL→K, regeneration) is The formula (5.7) contains the correct attenuation of the outgoing object (compare the discussion of experimental results given previously). Then the elastic scattering amplitude of | 1) (= |)) is totally determined M(1→1) = 1⁄2[(Ï | T‹4) | 1)+(2 | T‹4) | 2)] = ik 2π [ db exp (i▲·b)1⁄2{1— exp [−1⁄2o;T (b)]+1− exp [−1⁄2o;T (b)]} hence the correct -nucleus elastic scattering amplitude. Then, one has to worry about the fact that M (2→2) = M (1→1), as implied by (5.6). One can argue that M (2-2) is not measurable since | 2) decays into 3 and hence is not available as an initial beam, thus making this concern irrevelant. This example shows that one can easily explain the "anomalously" low absorption of diffractively produced objects if one assumes that the initial (one pion) and the final (something decaying into 3) states are just the different configurations of the same components. (In the weak coupling case" it is physically more accurate to describe the initial and final states as composed, approximately, of one and three particles, respectively.) π Note the following amusing point: Assuming | ) to be in the form of a superposition of two scattering eigenstates we determine uniquely the exponentials (the total cross sections). Once we have that, the production cross section is completely determined and comes out right! |