The hydrogen atom with Hamiltonian obviously has $SO(3)$ symmetry since it just depends on the radius.

$$ H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}$$

This is generated by angular momentum $\mathbf{L} = \mathbf{r}\times \mathbf{p} $.

In quantum mechanics class we learn there is $SO(4)$ symmetry due to the Runge-Lenz vector:

$$ \mathbf{A}= \frac{1}{2m}(\mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - k \frac{\mathbf{r}}{r}$$

Even classical symmetry like gravity has this kind of symmetry.

I remember reading one time there is even greater symmetry for hydrogen atom. Possibly $SO(4,2)$ as in this article by Hagen Kleinert.

Has anyone heard of this?

How can one see that the Hydrogen atom has $SO(4)$ symmetry?


$SO(4,2)$ is called the full dynamical group of the Kepler (or Hydrogen atom problem). The $SO(4)$ , $SO(3,2)$ and $SO(4,1)$ subgroups of $SO(4,2)$ are called partial dynamical groups.

Unlike symmetry groups which commute with the Hamiltonian, dynamical groups do not. They have the following properties:

  1. The system's phase space is a coadjoint orbit of the group, or equivalently,

  2. The system's Hilbert space is spanned by an irreducible representation of the group.

  3. In many cases, although, it is not necessary, the Hamiltonian itself is a generator of the dynamical group.

The equivalence of the points 1. and 2. above stems from the fact that in the case under study there is a correspondence between coadjoint orbits and irreducible representations.

The partial dynamical groups span only part of the Hydrogen atom spectrum through their irreducible representations:

An $SO(4)$ irreducible representation spans the state vectors corresponding to a single energy shell of a bound state (fixed (and quantized) $n$ and varying $l$, $m$) .

An $SO(3,2)$ irreducible representation spans the full continuous spectrum and an irreducible representation of $SO(4,1)$ spans the full bound spectrum.

$SO(4,2)$ is the smallest group whose irreducible representations span both the continuous and the discrete spectrum.

The use of dynamical group representations reduces the problem of finding the Hamiltonian spectrum to an algebraic problem, instead of a solution of differential equations.

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  • $\begingroup$ Do you know of a reference where the SO(4,2) dynamical symmetry group is correctly presented? Both the books by Wybourne and by Barut and Raczka are full of small errors in their treatment of this topic, so that I was unable to reconstruct the full detials of the represntation. $\endgroup$ – Arnold Neumaier Mar 15 '16 at 19:55
  • $\begingroup$ @Arnold.N : you might like Barut & Bornzin 1971 $\endgroup$ – Cosmas Zachos May 17 '16 at 19:32

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