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Incidentally, only the spin of the deuteron as a whole is essential. The qualitative effect is independent of the spin of the incident particle (the M(A, s) operator does not act on spin quantum numbers.) All the other spin effects are presumably not important. (iii) Calculations such as the one above, as well as more sophisticated calculations, have always

produced cross sections in excellent agreement with experiment. (We are not considering here backward scattering, where the above model does not apply (see also [48])). There is only one exception: the experiment performed at CERN by Bradamante et al. [20]. In this experiment the discrepancy with theory occurs at a fairly large momentum transfer (42–2 GeV). What is the cause? Perhaps some relativistic effects? There is no good answer, so far. Without going into any explanation of this discrepancy, let us emphasize the following point:

It is important to realize that when we use the same internal wave function in the initial and final states, we exclude, by doing this, any possible relativistic deformations of the recoiling target (we are still discussing only elastic processes). For large momentum transfers (A?/M~1) this is probably not a good approximation. Take the deuteron example. In the standard Glauber model, it is enough to have p(s) = 8 dzo* (s, z) $ (3, 2) to compute the cross section. Suppose there is some deformation in the final state:

00* (s, 2)+00'*(A, S, 2)

*

(one can assume that the deformation is defined by the momentum transfer A). Then we should replace

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The interesting fact is that in exactly the same form one can write the Delbrück amplitude

MDelbrück

(A)~ | ab exp (ia. b)I”(A, 8) (1– exp [ixo-(6–48) +ix+(b+98) }}

where we are the Coulomb phase shifts of the electron-positron pair and I'(A, s) is constructed from

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the "relativistic wave functions” in an analogous way to that shown above in the case of the deuteron. Here the possibility of a well-defined procedure of introducing relativistic deformations occurs-modeled on QED! These and other related problems have been discussed in a series of papers by Cheng and Wu [21, 22] (see also [30], [84]).

4. Diffractive Dissociation and Diffractive Excitation

Diffractive processes-a brief characterization.

(i) they do not vanish in the limit E(ii) the target plays a passive role (except in double diffraction, but in any case: no quantum

numbers are exchanged).

Examples: in QED; elastic electron (positron) scattering from a Coulomb field, Delbrück scattering,

Compton scattering, etc. in hadron physics; all kinds of elastic hadron-hadron scattering do

(experiments seem to indicate that the =f ()

differential cross sections depend weakly DA?

on energy)

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The nucleon and nuclear targets supplement each other because the nuclear medium amplifies the scattering of the produced objects.

The model of diffractive processes described below is based on: M. L. Good and W. D. Walker (1960) [23]. The article which discusses some very early papers on the subject is: E. L. Feinberg and I. Pomerančuk (1956) [24]. For more recent discussions of many experimental and theoretical aspects of diffractive processes in hadron physics see refs. [S4] and the article by A. Bialas in [25]. We shall describe diffractive production processes in very close analogy to diffractive dissociation phenomena which are well known in the case of systems where degeneracy exists.

us start with an example taken from optics. Consider the absorption of polarized light by an anisotropic absorber. The incident wave is polarized in the direction n (perpendicular to the z direction).

n= (Nz, ny)

V=n,y,+ny, where V, is the wave polarized in the x direction and y, is the wave polarized in the y direction.

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Suppose the target is a Nicol prism oriented in such a way that it stops all light polarized in the y direction. Hence, the only component which goes through is n, Vz. But it can be decomposed into n and nXe, components. Hence due to the process of absorption, a new object is created: the wave which is polarized in the direction n Xez.

Let us compute the elastic and inelastic scattering amplitudes. Since the transmitted wave is y=n,V, the wave which goes into scattering and production is

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The "undisturbed" wave is yn everywhere, hence the scattered wave is

(0,

2<0
Yn-y=
Vn-4

2>0.
Actually it is more important for our purposes to introduce partial absorption (in general different
for the two components (x, y)).

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This formula shows that we always produce inelastic scattering, except in two cases:

(i) when the incident wave is polarized either along the x or y axis (hence either n;=0 or ny=0) (ii) when the absorption coefficients are equal (the absorber is isotropic).

We shall extend this description to diffractive production processes of hadronic systems. We consider the incident hadron to be a superposition of some states which get eaten up at different rates during the passage through the target; the new combination emerging from the collision then contains, in general, a new particle (or a collection of new particles).

First we introduce the physical states of the system, I fi) (which are analogous to the states yn and ynxe, of the photon). We want to compute (il T | li). We expand Ni) into a set of states | 1;) whose scattering and absorption in the target we assume known:

14:)= dij| ;) Ili), 1 n,) form

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The states 1:) are assumed to be eigenstates of T in the following sense:

1

T | 1;)= (1 – n;) | 1;)+ Erik Mx)

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We have obtained the result completely analogous to the one obtained for the optical diffractive production: for itj the production amplitude is proportional to the difference in absorptions of the i, j components. This is a very general property. All specific models of diffractive production processes I know of exhibit this property. Otherwise the formula is so general that it has virtually no predictive power.

The difficulty in applying it to any realistic process is the determination of absorption parameters ni because the states 1) are not observed in scattering experiments. (The process of KK,' regeneration given below is an example where we know ni's however!) The situation changes when we accept that diffractive production processes are weak compared to elastic scattering. This may mean that the transformation from Xi) to 1 1.) differs little from unity:

3 One must, however, keep in mind that in the case when the coefficients dij are zero or of the same order of magnitude one does predict some characteristics of production processes from the knowledge of elastic scattering. Compare the end of this section.

Cij=dij teij
+ tij small, hence e terms

(4.3) di;=dij - Eij) can be neglected.

Remark: The minus sign guarantees the property

dic= Cij dji= a (dij+eij) (832— 6;1)

=

=

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Hence the absorption parameters ni are determined by elastic scattering of real particles. The inelastic amplitude (il T | ;)= (1-ni) ei;— (1-ni) tij

(4.4) is proportional to the difference between the absorption of the produced particle and the absorption of the incident particle.

One still faces the problem of specifying the absorption parameters ni and the coefficients tij. The coefficient ni of the incident particle is, as a rule, easy because this is a well-known particle which can form a beam and its scattering (elastic) properties are known reasonably well. The trouble is with the outgoing objects; e.g., when 37 are produced in the 131 reaction: are then its n's given by the absorption of 31 in the target? In fact, one usually determines them experimentally (see the end of this section). So far as the tij are concerned, they are small-hence some perturbation theory can be used to compute them. We shall give some examples further in the text.

How does one implement this program? There are strongly interacting particles which realize precisely the above outlined scheme and we even know dij's and n,'s: the neutral K mesons. Because of the relation

charge baryon no. strangeness isospin
Q 12NB.

128 + Tz the partners of K+, Ko particles are K-, Ko antiparticles. Hence there are two different neutral K mesons which can be produced in the collision of strongly interacting particles: Ko and Ko (they are different because they have opposite strangeness, unlike pions where to are identical to 7°!) which have the same masses, and thus can be considered as a two component degenerate system. When left in empty space, however, both Ko and K, decay weakly with two different lifetimes as if they were made up of two different particles, which is indeed the case. These two particles are the following superpositions of | Ko) and ð) states

+

|K,)=

1
V2(1+1 812)

[(1+0) | K°)– (1–0) | KO)](1 KO)- | Ř°)) = | K7°)

1

1 |K1°)=

[(1+8) | K°)+(1-5) | K°)]= (| Ko)+ | K°)) = | K2°), V21+ 8 )

V2

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