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

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?

• Related: physics.stackexchange.com/q/116244/2451 and 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. Dec 9 '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. Dec 10 '13 at 2:18
• 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. Dec 10 '13 at 14:55

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
Dec 9 '13 at 20:41
• 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. Dec 9 '13 at 20:53
• @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 9 '13 at 21:18
• @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 9 '13 at 21:22
• @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. Dec 10 '13 at 11:44

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

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.