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  1. For solving hydrogen atom energy level by $SO(4)$ symmetry, where does the symmetry come from?

  2. How can one see it directly from the Hamiltonian?

Qmechanic
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ahala
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    Related: https://physics.stackexchange.com/q/116244/2451 and http://physics.stackexchange.com/questions/tagged/runge-lenz-vector . A derivation of $SO(4)$ symmetry is e.g. given in G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 9. The pdf file is available here. – Qmechanic Dec 09 '13 at 20:34
  • After studying the answers and comments here, I think (1) this is not really a quantum mechanics question. Maybe I should study why Kepler problem has SO(4) symmetry. (2) People have studied the classical Kepler problem should be able to see the so(4) symmetry from the Hamiltonian. – ahala Dec 10 '13 at 02:18
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    Of course the $SO(4)$ symmetry is preserved in the classical limit. Note however that computationally, the proof of $SO(4)$ symmetry is a order of magnitude harder in the quantum mechanical problem than in the classical problem. – Qmechanic Dec 10 '13 at 14:55

3 Answers3

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The Hamiltonian for the hydrogen atom $$ H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r} $$ describes an electron in a central $1/r$ potential. This has the same form as the Kepler problem, and the symmetries are similar. There is an obvious $SO(3)$ generated by the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In other words, the components of $\mathbf{L}$ satisfy $$ [L_i,L_j] = i \hbar \epsilon_{ijk}L_k . $$ A more subtle symmetry is given by the Laplace-Runge-Lenz vector $$ \mathbf{A} = \frac{1}{2m} ( \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - k \frac{\mathbf{r}}{r}. $$ The commutation relations involving $\mathbf{L}$ and $\mathbf{A}$ are $$ [L_i,A_j] = i\hbar \epsilon_{ijk} A_k \\ [A_i,A_j] = -i\hbar\epsilon_{ijk} \frac{2H}{m} L_k . $$ Up to the normalization of $\mathbf{L}$ this is the commutation relations of $SO(4)$. (Here I assume that we are considering a bound state whose energy $E$ is negative. If $E>0$ the above relation generate a non-compact $SO(3,1)$ symmetry.)

Furthermore, both $\mathbf{L}$ and $\mathbf{A}$ commute with the Hamiltonian, $$ [H,L_i] = 0, \qquad [H,A_i] = 0 $$ showing that they indeed generate symmetries of the hydrogen atom.

Olof
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    See also this blog post by Lubos Motl. – Olof Dec 09 '13 at 20:41
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    Minor remark : The dynamical symmetry is $SO(4)$ for bound states ($H <0$), and $SO(3,1)$ for excited states ($H>0$). There is a very interesting discussion in "Robert Gilmore, Lie Groups, Physics and Geometry, Cambridge", Chapter $14$, Hydrogenic atoms. – Trimok Dec 09 '13 at 20:53
  • @Trimok: Looks like a very interesting reference, thanks. I implicitly assumed that $H<0$, otherwise the $[A,A]$ commutator changes sign and the symmetry is non-compact. – Olof Dec 09 '13 at 20:59
  • thanks. My question was that how can one see the physical picture behind these. what's the corresponding SO(4) rotation on H? why is there a 4d sphere involved? – ahala Dec 09 '13 at 21:10
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    @ahala: $L_i$ and $A_i$ commute with $H$, so the $SO(4)$ rotations don't act on $H$. Wikipedia has a discussion about making the $SO(4)$ symmetry manifest in the Kepler problem by mapping it to a free particle moving on a three-sphere, but I don't really think it makes the physics behind it clearer. – Olof Dec 09 '13 at 21:18
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    @ahala: For some physical intuition one can note that the energy of an eigenstate of the hydrogen atom only depends on the principal quantum number $n$, but not on the angular momentum $l$ or the magnetic quantum number $m$. Such a degeneracy is often related to additional symmetries. Of course, this doesn't tell you what the relevant symmetry should be. – Olof Dec 09 '13 at 21:22
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    @ahala : In the book of Gilmore (see previous comment), there is a discussion. One notices, that, in the momentum space, which is in a plane, momenta obey a circle equation. The next observation is that a circle in $R^3$ is promoted to a circle in $S^3$ (included in $R^4$), by a projective transformation, which is a stereographic projection, which is invertible and preserve angles (conformal), so that circles in $R^3$ are in one-to-one correspondence with circles in $S^3$. Now, obviously, $SO(4)$ is a symmetry of $S^3$, so it is also a hidden (dynamical) symmetry of the hydrogenoid atom. – Trimok Dec 10 '13 at 11:44
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It's because there is another vector quantity $A_i$ conserved in addition to the angular momentum $L_i$. Furthermore, the commutation relations of $A_i$'s and $L_i$'s are those of $SO(4)$. See for instance this reference : http://hep.uchicago.edu/~rosner/p342/projs/weinberg.pdf

Adam
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I wanted to complement the answers above. For (1) $so(4) = so(3) \times so(3)$, one $so(3)$ is from the geometric 3D symmetry of the Hamiltonian, and the other $so(3)$ is from the potential term of $\frac{k}{r}$.

For (2). the second $so(3)$ symmetry is a dynamic symmetry and only holds when potential term is inversely proportional to $r$. One has to do the calculation to find it.

ahala
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