[[JW] >p'] = [iViV, -¿V,] = -V'V = -Ze2i3(r) . (3.98) Hence the "final" result is

AEaohs = A££amb = Z«2 l^a(0)|2 In where a = e2/47r = 1/137 is the fine structure constant. Discussion

We draw the following conclusions from Eq. (3.99):

• Only for S-states is ipa{fy / 0. Hence the shift is largest for S-states. There is a much smaller shift for other states with L ^ 0 arising from small terms which we have not calculated.

• The shift is positive because Emax > \Ea - ¿?b|.

3.5 DEUTERON PHOTODISINTEGRATION 81

This, of course, is only a rough estimate. There are numerous other terms, and the result can be calculated without a cutoff once the theory is fully renormalized.

The comparison of precise calculations of the Lamb shift (and many other energy shifts) with precise measurements continues to be an active area of research and is a good way to test the validity of the quantum theory of radiation (which becomes Quantum Electrodynamics when we also treat the particles as relativistic quantum fields; see Chapter 10). Recent theoretical and experimental results for the Lamb shift in hydrogen are*:

where the numbers in parentheses are an estimate of the errors. So far none of these tests have led to any clear failures; QED is a remarkably successful theory.

As a final application of the quantum theory of radiation, consider the photodisintegration of the deuteron. The deuteron is the only two-body bound system of two nucleons (the neutron and proton) and is therefore the simplest nuclear system. Its binding energy is 2.23 MeV, a very large number when compared with atomic binding energies but quite small on the nuclear scale; it is only about 2% of the mass of a nucleon. Deuteron photodisintegration can occur when a photon (7) ■ v'ith an energy greater than 2.23 MeV strikes a deuteron (d) and breaks it into its constituent nucleons:

where p is the proton with momentum pp and n the neutron with momentum pn. The observation of this reaction in 1935 was an early confirmation of the conversion of "energy" to "mass" as predicted by relativity [CG 35].

To calculate this reaction, we use the Hamiltonian (3.5), with the proton replacing the electron as the charged particle which interacts with the electromagnetic field (the neutron has a magnetic moment which can also interact with the EM field, but this is a small contribution which can be ignored in a first calculation). Then the interaction Hamiltonian is

Theory: 1057857(14) kHz [KS 84] Experiment: 1057845(9) kHz [LP 86] 1057851(2) kHz [PS 83]

mp where mp is the proton mass and

For a recent summary see [BG 87], with Hn the Hamiltonian of the neutron-proton system. The scattering matrix S and reduced amplitude /, in lowest order perturbation theory, are

Sfi = -i2ir6(EI-Ei)jpffi

where Eq. (3.50) has been used to relate the reduced amplitude / to S, |i) is the initial state consisting of a deuteron and a photon, and |/) is the final free neutron and proton.

The nuclear system consists of two particles, the proton at rp and the neutron at rn. The center of mass (CM) and relative coordinates for these two particles are R = (rp + rn)/2 and r = rp — rn. If the initial deuteron is at rest, its wave function is

where <t>(r) is the internal wave function of the deuteron and the factor L~3/2 is the wave function for the center of mass of the deuteron (obtained from a plane wave with box normalization and zero total momentum). We will discuss the internal wave function shortly. The wave function for the final neutron-proton pair will be approximated by a plane wave

vhere P = pp + pn is the total momentum and p = (pp - pn)/2 is the relative momentum of the outgoing pair. In this notation, the incoming and outgoing states are therefore

Was this article helpful?

## Post a comment