quantum mechanics - 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?

   
   

Answer

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.