Sunday 1 November 2020

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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...