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).


Have you tried to search "Dirac Oscillator"? there's Moshinsky and Szczepaniak article ...

  • $\begingroup$ Yes. The article you linked does not provide a formula for the energies. $\endgroup$ – Ondřej Čertík Aug 30 '12 at 6:11
  • $\begingroup$ My apologies --- they do provide an energy, the equation 14a, 14b. Would you be willing to edit your answer to include these? If it reproduces the correct energies, then I would like to give you the bounty. $\endgroup$ – Ondřej Čertík Aug 30 '12 at 6:18
  • $\begingroup$ I've implemented their formula, see my script: gist.github.com/3523473, unfortunately, it doesn't agree with my numbers above (I am quite sure that my numerical numbers are correct). @nate, any ideas? $\endgroup$ – Ondřej Čertík Aug 30 '12 at 6:50
  • $\begingroup$ Since nobody gave a better answer, the bounty is yours. But my question is not answered, until I find a formula that gives the correct energies against my numerical code. $\endgroup$ – Ondřej Čertík Aug 30 '12 at 20:50
  • $\begingroup$ @Ondřej Čertík, I'm sorry for not replying earlier, I'm out of town and just flew for more than 20 hours to vacation... I'd check articles that cite the Moshinsky and Szczepaniak article... $\endgroup$ – bla Aug 31 '12 at 14:08

I found the spectrum in the end of the Appendix in the article http://arxiv.org/ftp/arxiv/papers/1203/1203.2458.pdf (the case of $\epsilon=0$, in order to get external field-free solution) Unfortunately I feel a bit lost about what are exact meaning of the scalar and vector potential in the terms of the hamiltonian there. I mean, the hamiltonian will be in the form

$$H = c \cdot \alpha \cdot p + \beta \cdot M c^2 + \text{WW}$$

And if I will express the potential WW as block-matrix (distinguishing large and small components in the Standard representation), the WW would look like? $$\begin{pmatrix} S(r) & ?V(r)\\ ?V(r) & -S(r) \end{pmatrix}$$

Where $S(r) = 1/2 M \omega_s r^2, V(r) = 1/2 M \omega_v r^2$ ? I suppose the scalar potential to give bilinear form $\bar \psi S(r) \psi $ transforming as scalar (of course, the $r^2$ should have to be completed by $-c^2 t^2$ to be Lorentz invariant, but for stacionary states it doesn't matter). I am not sure about the vector potential though.

  • 1
    $\begingroup$ Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. However, if you have a new question, please post it separately. $\endgroup$ – Kyle Kanos Mar 13 '16 at 18:00
  • $\begingroup$ @Jakub, Here is a notebook where I am trying to test the formula: nbviewer.jupyter.org/gist/certik/c0d8dc5417fc579d6158, it doesn't seem to work, also it doesn't contain the constant $c$. However, the equation (21) looks promising, but it also seems to give wrong energies. Maybe I implemented it incorrectly in the notebook. The energies I posted above should be correct, I checked those using a shooting method as well as finite elements. $\endgroup$ – Ondřej Čertík Mar 14 '16 at 4:03
  • $\begingroup$ I am sorry, I have read the ArXiv article again and more precisely, they discuss the case of $\endgroup$ – Jakub Mar 15 '16 at 16:37
  • $\begingroup$ I am sorry, you are right, one cannot apply the (A9) formula from the ArXiv article for your case, they have $WW = & \left( \begin{array}{cc} V(r)+S(r) & 0 \\ 0 & V(r)-S(r) \\ \end{array} \right) $ where either V(r)=S(r) ("spin sym.") or V(r)=-S(r) ("pseudospin") and both V and S are $\omega_V r^2 + V_0$ and $\omega_S r^2 + S_0$ respectively. For those cases, analytical solutions are known. You have investigated the S(r)=0 case, where elimination of neither small nor large bispinor radial component is straightforward and exact solution $\textit{may not}$ exist. $\endgroup$ – Jakub Mar 15 '16 at 17:26
  • $\begingroup$ errata $WW = \begin{array}{cc} V(r)+S(r) & 0 \\ 0 & V(r)-S(r) \\ \end{array} $ It is very interesting system even for the S(r)=0 case (or V(r)=0) case, so I will google and try myself to find the exact solution. I would bet that for the most complicated case, where both V and S are non-zero with different frequencies the exact solution truly might not exist. $\endgroup$ – Jakub Mar 15 '16 at 17:37

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