Showing that the angular momentum operators and Laplace-Runge-Lenz operator together are generators of $SO(4)$

Question

This is a question that popped up while reading Greiner's Quantum Mechanics Symmetries. For the sake of clarity I will omit the hat ($\hat{A}$) symbol on operators.

The quantum mechanical Laplace-Runge-Lenz vector can be defined as $$ \mathbf{A} = \frac{4\pi\epsilon_0}{\mu Z e^2}\left( \mathbf{p}\times\mathbf{L}-i\hbar\mathbf{p} \right) - \hat{\mathbf{r}}. $$

Together with the angular momentum operator $\mathbf{L}$, we can write down the following commutator relations (for details, see e.g. this, or this question I recently posted):

$$ [L_j, L_k] = i\hbar\varepsilon_{jkl}L_l, $$ $$ [A_j, L_k] = i\hbar\varepsilon_{jkl}A_l, $$ $$ [A_j, A_k] = -i\left(\frac{32\pi^2\epsilon_0^2\hbar}{\mu Z^2 e^4}\right)H\varepsilon_{jkl}L_l, $$ where $H$ is the Hamiltonian operator.

We know that $\{L_1,L_2,L_3\}$ form a closed algebra and are generators of the Lie group $SO(3)$. But if we add in $\{A_1,A_2,A_3\}$ they no longer form a closed algebra, although they do come close; the problem is with the operator $H$. Now Greiner says

However, given that $H$ is independent of time and commutes with $\mathbf{L}$ and $\mathbf{A}^1$, we can restrict ourselves to a subspace of the Hilbert space that corresponds to a particular eigenvalue $E$ of $H$. Then $H$ in (14.13) can be replaced by its eigenvalue $E$.

After this substitution, we find that the six components above form a closed algebra, and generate the Lie group $SO(4)$. However, I am having a hard time understanding the quoted paragraph.

I would like to ask "why" we can carry on this procedure, namely why time-independence and commutativity allows us to pick the said subspace, and what such a subspace corresponding to some fixed $E$ would look like.$^2$

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$^1$ The symbol used in the book is $\mathbf{M}$ instead of $\mathbf{A}$.

$^2$ I believe the the answer in Is this really SO(4) algebra? doesn't explicitly answer this question. Also, I don't know what Poisson brackets have to do with this in the answer, and what "as good as a constant" precisely means.

Answer 1

This is not the math site. Just follow the strategy of wikipedia and, noting that H commutes with L and A, it can be moved out of their commutators just like a constant number. (In its diagonal representation, it may be thought of as a diagonal array of its eigenvalues, positive or negative, discrete or continuous.) So, then, define $$ D^j\equiv \frac{ A^j}{\sqrt{\frac{32\pi^2\epsilon_0^2H}{\mu Z^2 e^4}}} $$ to immediately obtain $$ [L_j, L_k] = i\hbar\varepsilon_{jkl}L_l, \\ [D_j, L_k] = i\hbar\varepsilon_{jkl}D_l, \\ [D_j, D_k] = -i \hbar \varepsilon_{jkl}L_l. $$

This should remind you of the two ideals of the Lorentz group algebra, $$ {\mathbf S}\equiv { { \mathbf L} +i { \mathbf D}\over 2\sqrt{\hbar} },\\ {\mathbf R}\equiv { { \mathbf L} -i { \mathbf D}\over 2\sqrt{\hbar} }, \leadsto \\ [S_j, S_k] = i\varepsilon_{jkl}S_l,\\ [R_j, R_k] = i\varepsilon_{jkl}R_l,\\ [S_j, R_k]=0, $$ the well-known $\mathfrak{su(2)}\oplus \mathfrak {su(2)}$ Lie algebra of SO(4).

As a consequence, Pauli (1925) explained why the spectrum of Hydrogen has this exceptionally large degeneracy: the generators S and R connect all states within the subspace of such sharing an eigenvalue E.