## W 4 J0

one incoming photon

where k is the momentum of the incoming photon. Hence

dt (f\Hl(t)\i)=- -== — / dt e>(Tn+Tp + c-ul)t 00 V2im m J_x frd3R

d3rd3R

where we have expressed the answer in terms of the dimensionless quantity 7 =

To complete the calculation, we must find the wave function of the deuteron. One of the important features of photodisintegration near threshold (the threshold energy is the energy at which the process just becomes physical, which in this case is u = e) is that it is insensitive to the details of the deuteron wave function, and hence a reliable prediction is possible without knowing much about the short range structure of the nuclear force.

To see why this is so, we estimate the wave function using the Hulthen model for the nuclear potential. This is a crude model which nevertheless is very useful for such estimates. The model assumes that the nuclear force at long range (i.e., for large internucleon separation r) is dominated by the exchange of a single pion (a good approximation) and that spin dependence of the interaction can be neglected (which is not too bad an approximation for interactions which depend only on the charge but overlooks many features of the deuteron, such as the D state).* Under these assumptions the potential is a Yukawa potential with a range of the pion mass (denoted by fi). The Hulthen model approximates this potential as follows:

This approximation captures the correct behavior of the potential at both long and short range and permits us to solve the Schrodinger equation for S states exactly. The equation for the relative coordinate is

---^<t>(r)--^-<t>(r) + V(r)J>(r) = e4>(r) . (3.117) marL rar

Substituting a wave function of the form

o-0T

into this equation gives a solution, provided

The momentum space wave function is the Fourier transform [worked out in Eq. (4.52)], and hence m

*The one-pion exchange force will be derived from field theory in Sec. 9.9.

Note that the second term is very small. For example, if p = 0, the two terms are in the ratio of

Therefore a very good estimate is obtained by using the asymptotic wave function only (in which case the answer does not depend on the use of the Hulthen model). The normalization constant for the asymptotic wave function is N = ^/b/'lir, and the square of the asymptotic wave function, evaluated at p2 = m(u> — e), is then

^asy(p)- Kp2 + me Substituting this into (3.115) gives finally da (e2\ 1 (7-l)3/2 . 2 f\

We emphasize that this result only includes the contributions from the proton charge (the electric dipole interaction) and that only the contributions from the asymptotic deuteron wave function were retained. It might appear that corrections from the interior part of the deuteron wave function would be uncertain and hard to estimate, but Bethe and Longmire [BL 50] showed that these additional contributions can be expressed in terms of the effective range for the scattering of two nucleons in the 3 5i channel, a quantity which is readily measured. There are also additional contributions from the magnetic interactions of the nucleons which contribute an angular independent background term which dominates at energies within 0.1-0.2 MeV of the threshold but contribute only a few percent to the cross section at higher energies.

Because of the simplicity of this process and its insensitivity to the details of the nuclear force, deuteron photodisintegration has been of considerable interest for many years. Recent precise measurements of the angular distribution at low energies (see, for example, [De 85]) show the large sin2 6 dependence expected and are in good agreement with theory.

## Post a comment