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I have found the eigenvalues of the following systems: $H=-\frac{1}{2}\Delta+V_1$ and $H=-\frac{1}{2}\Delta+V_2$, using NDEigensystem by Wolfram Mathematica.

In the first case the potential is Coulomb one ($V_1=-\frac{1}{\sqrt{\rho^2+z^2}}$ (is written in a cylindrical coordinate system)). The second potential has a more complex structure $V_2=-\frac{1}{\sqrt{\rho^2+z^2}}-\frac{1}{2\sqrt{\rho^2+z^2}}e^{\frac{1}{2\sqrt{\rho^2+z^2}}}$.

The codes is written in Wolfram Mathematica. In the both codes I renamed $\rho≡r$

1.

ClearAll["Global`*"]
rmax = 20;
zmax = 20;

V1[r_, z_] := -1/Sqrt[r^2 + z^2]
{valsc, funsc} = 
  NDEigensystem[{(-1/2*
        Laplacian[\[Psi][r, z], {r, \[Theta], z}, "Cylindrical"] + 
       V1[r, z]*\[Psi][r, z]) + \[Psi][r, z]*0.5}, \[Psi][r, z], {r, 0,
     rmax}, {z, -zmax, zmax}, 20, 
   Method -> {"SpatialDiscretization" -> {"FiniteElement", \
{"MeshOptions" -> {"MaxCellMeasure" -> 0.05}}}, 
     "Eigensystem" -> {"Arnoldi", "MaxIterations" -> 10000}}];
Sort[valsc] - 0.5

20 first eigenvalues obtained from the code 1:

{-0.503427, -0.125434, -0.125001, -0.0591382, -0.0575661, \
-0.0567415, -0.0366165, -0.0348045, -0.0331543, -0.0305338, \
-0.0194174, -0.011569, -0.00383252, -0.000633831, 0.0115625, \
0.0130426, 0.0228641, 0.0287679, 0.0380814, 0.0405388}

As follows from the general theory, the levels are degenerate:

-0.503427 — 1s state
-0.125434, -0.125001 — 2s and 2p states
-0.0591382, -0.0575661, -0.0567415 — 3s, 3p and 3d states
...

2.

ClearAll["Global`*"]
rmax = 20;
zmax = 20;

V2[r_, z_] := -1/Sqrt[r^2 + z^2] - 
  1/2/Sqrt[r^2 + z^2]*Exp[-Sqrt[r^2 + z^2]]
{valsc1, funsc1} = 
  NDEigensystem[{(-1/2*
        Laplacian[\[Psi][r, z], {r, \[Theta], z}, "Cylindrical"] + 
       V2[r, z]*\[Psi][r, z]) + \[Psi][r, z]*0.5}, \[Psi][r, z], {r, 
    0, rmax}, {z, -zmax, zmax}, 20, 
   Method -> {"SpatialDiscretization" -> {"FiniteElement", \
{"MeshOptions" -> {"MaxCellMeasure" -> 0.05}}}, 
     "Eigensystem" -> {"Arnoldi", "MaxIterations" -> 10000}}];
Sort[valsc1] - 0.5

20 first eigenvalues obtained from the code 2:

{-0.816105, -0.152359, -0.134395, -0.0649383, -0.0597383, \
-0.0570375, -0.0381872, -0.0375616, -0.0340771, -0.0326892, \
-0.0220107, -0.0128179, -0.00388086, -0.00246523, 0.0106794, \
0.011968, 0.0221767, 0.0277981, 0.0328003, 0.0358925}

As can be seen in this case, the eigenvalues are different — the levels are not degenerate:

-0.816105 — 1s state
-0.152359 — first excited state
-0.134395 — second excited state
...

Question:
Why are the levels not degenerate in the second case? Like the first potential ($V_1$), the second potential ($V_2$) has spherical symmetry, hence the system is spherically symmetric. Shouldn't any spherically symmetric system give levels that are degenerate in the orbital quantum number?

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    $\begingroup$ Only the degeneracy in the quantum number $m$ follows from spherical symmetry. The degeneracy in $l$ and $m$ is a consequence of an additional symmetry of the Coulomb potential (the classical analogy is that the Runge-Lenz vector is conserved). (And I guess that the degeneracy counts for $m$ are only very approximate is some numerical error of your code, but I haven't checked.) $\endgroup$ Jul 9, 2023 at 22:34
  • $\begingroup$ @Sebastian Riese, thanks a lot for the explanation! The code 1 doesn't degenerate in the $m$ quantum number, and I think it's because I don't enter the angle into the wave function. Could yo correct me if I'm wrong. But if I enter the angle, the code stops working. I asked this question here mathematica.stackexchange.com/questions/287427/…. If you have no trouble and are familiar with Wolfram Mathematica, could you please explain why the code stops working in this case? $\endgroup$
    – Mam Mam
    Jul 9, 2023 at 22:47
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    $\begingroup$ Your problem is spherically symmetric, and since you cannot seem to get it to work, why do you make the computer work so hard? You should just start with spherical symmetry and state the problem with spherical harmonics, so that you just solve only for the radial solution in terms of $n$ and $\ell$, guaranteeing $m$ degeneracy. $\endgroup$ Jul 10, 2023 at 1:24
  • $\begingroup$ @naturallyInconsistent, thanks! I have asked a question on a similar problem in spherical coordinates mathematica.stackexchange.com/questions/287446/…. Сould you please comment on it $\endgroup$
    – Mam Mam
    Jul 10, 2023 at 9:06

1 Answer 1

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The Coulomb problem has an extra symmetry related to the Laplace-Runge-Lenz vector. The full symmetry group of the Hamiltonian is $SO(4)$ [which is locally isomorphic to $SO(3)\otimes SO(3)$], not just $SO(3)$. After some work, one can show that in fact the components of the Laplace-Runge-Lenz vector are generators of transformations that augment $SO(3)$ to $SO(4)$ and lead to the extra degeneracy.

A good discussion can be found in many places, but the textbook by Gordon Baym is very clear on this topic.

This explains why the Coulomb problem had additional degeneracies NOT present in other central potentials, for which there is no Laplace-Runge-Lenz symmetry.

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  • $\begingroup$ Thanks for the explanation! $\endgroup$
    – Mam Mam
    Jul 10, 2023 at 5:19

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