For solving hydrogen atom energy level by $SO(4)$ symmetry, where does the symmetry come from?
How can one see it directly from the Hamiltonian?
3 Answers
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.
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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.
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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
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