The potential is given by: $$ V(r) = {1\over 2} \omega^2 r^2 $$ and we are solving the radial Dirac equation (in atomic units): $$ c{d P(r)\over d r} + c {\kappa\over r} P(r) + Q(r) (V(r)-2mc^2) = E Q(r) $$ $$ -c{d Q(r)\over d r} + c {\kappa\over r} Q(r) + P(r) V(r) = E P(r) $$ What is the analytic expression for the eigenvalues $E$ in atomic units?
It is ok to provide source code (any language) to obtain it if one needs to solve some simple analytic equation. Here are the (I think correct) energies from my numerical code that I would like to compare against analytic solution (for $c = 137.03599907$ and $\omega=1$):
n l k kappa E
1 0 0 -1 1.49999501
2 0 0 -1 3.49989517
2 1 0 -2 2.49997504
2 1 1 1 2.49993510
3 0 0 -1 5.49971547
3 1 0 -2 4.49983527
3 1 1 1 4.49979534
3 2 0 -3 3.49994176
3 2 1 2 3.49987520
4 0 0 -1 7.49945592
4 1 0 -2 6.49961564
4 1 1 1 6.49957571
4 2 0 -3 5.49976206
If one only solves the radial Schroedinger equation, then the analytic formula is $$E_{nl} = \omega (2n - l - {1\over 2})$$ I am looking for the relativistic version.
I found for example the paper: Qiang Wen-Chao: Bound states of the Klein-Gordon and Dirac equations for scalar and vector harmonic oscillator potentials. Vol 11, No 8, 2002, Chin. Phys. Soc., but it only shows a formula for nonzero scalar and vector potentials in Dirac equation (above we only have the scalar potential, the vector potential is zero).
