Single-Photon Emission of a Hydrogenlike Atom

Abstract

Implementing a previously obtained, original solution of the Dirac equation for an electron in the field of a nucleus (Ze) expressed in terms of the wave function of the corresponding Schrodinger equation and its derivatives in spherical coordinates and the spin projection operator Σ3 associated with the eigenfunction, taking into account in each component of the spinor the leading term of the expansion in the small parameter (Zα), α = e 2 / ħc ≈ 1 / 137, the partial probabilities W of emission of a photon (Zα)* → (Zα) + γ have been calculated. Here two orthogonal states of the linear polarization of the photon, and also the spin states of the electron, which previously had not been taken into consideration, have been taken into account in the transverse gauge. It turns out that the probabilities W of emission of a photon per unit time for any allowed transitions are proportional to (Zα)4, as was previously accepted, and the selection rules for the quantum number m have the usual form ∆m = 0,±1. It was found that a spin flip does not take place during emission. In contrast to the customary situation with the selection rules for the quantum number l being of the form ∆l = ±1, for ∆m = ±1 there also exist integrals over dcosθ which are not equal to zero for undetermined odd values of ∆l. In this, and also in a fundamentally different dependence of the amplitude on the quantum numbers consist the differences from the traditional approach to the problem. Necessary conditions are formulated, under the fulfillment of which the selection rules for l are not changed, having values ∆l = ±1 for arbitrary ∆m, but it was not possible, however, to give a complete proof of these rules.