Energy eigenvalue of hydrogen-like atoms using Laplace-Runge-Lenz vector

Question

I have a basic question about a few calculations involving the quantum mechanical Laplace-Runge-Lenz vector.

In classical mechanics there is the Laplace-Runge-Lenz vector, which for a hydrogen-like atom can be written as $$ \mathbf{A} = \frac{4\pi\epsilon_0}{Ze^2\mu}\left(\mathbf{p}\times\mathbf{L}\right) - \mathbf{\hat{r}}, $$

where $\mu$ is the reduced mass and $Z$ the atomic number. I read that in quantum mechanics, we can similarly define an LRL vector, but since now $[\mathbf{L},\mathbf{p}]\neq 0$, we have to let $\mathbf{p}\times\mathbf{L}\to\frac{1}{2}\left(\mathbf{p}\times\mathbf{L}-\mathbf{L}\times\mathbf{p}\right)$. Wikipedia gives the result (up to constants due to the unit system in use) $$ \mathbf{A} = \left(\frac{4\pi\epsilon_0}{Ze^2\mu}\right)\left(\mathbf{L}\times\mathbf{p}-i\hbar\mathbf{p}\right) - \mathbf{\hat{r}}. \quad \text(*) $$

Then it defines the raising/lowering (ladder) operators associated with $\mathbf{A}$ as $$ A_\pm = A_x\pm iA_y, $$

and gives the following properties:

• (equ. 1) $$ [A_\pm,L_z]=\mp \hbar A_\pm. $$
• (equ. 2) $$ [A_\pm,L^2] = \mp2\hbar^2A_\pm \mp 2\hbar A_\pm L_z\pm 2\hbar A_z L_\pm. $$
• (Pauli equ. III) $$ 1-\mathbf{A}\cdot\mathbf{A} = -\frac{2E}{\mu Z^2 e^4}\left(L^2+\hbar^2\right). $$
• (Pauli equ. IV) $$ \left(\mathbf{A}\times\mathbf{A}\right)_j = -\left(\frac{2i\hbar E}{\mu Z^2e^4}\right)L_j. $$

The wiki page linked to then derives the energy eigenvalue $$ E = -\frac{\mu Z^2 e^4}{(4\pi\epsilon_0)^2 2\hbar^2n^2} $$ using the ladder operators defined above. I am stuck trying to prove the above equations, nor am I able to see why in (*) the operator $\mathbf{p}\times\mathbf{L}$ has been changed to $i\hbar\mathbf{p}$.

I cannot find resources that detail the calculation, since most of them simply say "the calculations are involved, but you'll get so and so once you finish the calculation". I also cannot read Pauli's original 1926 paper, which is in German. Any help or hints will be appreciated.

#Edit: As naturallyInconsistent's comment pointed out, there are some typos in the original question which are now fixed. Also, I realised there are some typos in the commutators given by Wikipedia, and in Pauli's equs. III & IV above, I omitted a factor of $(4\pi\epsilon_0)^2$.

Answer 1

Your first task is to absorb all superfluous constants into your nondimensionalized variables, and do the same for the nice review by Valent which is required reading, if you cannot follow WP or Pauli. You may reintroduce your unfriendly units here, to the nondimensionalized answers, by fecklessly repeating the simple calculations below lugging pointless units, after you appreciate the point... I then understand you cannot derive elaborate commutators, but this should be part of your training: this should not be a tutorial of commutation techniques!

Start with the Hermitian version of $\mathbf{p}\times\mathbf{L}$, namely $\frac{1}{2}\left(\mathbf{p}\times\mathbf{L}-\mathbf{L}\times\mathbf{p}\right)=\mathbf{p}\times\mathbf{L}-i\hbar\mathbf {p} $. Get it through use of $$ \epsilon^{ijk} L^j p^k= \epsilon^{ijk}p^k L^j +\epsilon^{ijk}[L^j,p^k]\\ =-\epsilon^{ijk}p^j L^k +2i \hbar p^i ~~~ \leadsto \\ \frac{1}{2}\left(\mathbf{p}\times\mathbf{L}-\mathbf{L}\times\mathbf{p}\right)=\mathbf{p}\times\mathbf{L}-i\hbar\mathbf {p} , $$ the final "quantum correction" expression (*) being deeply counterintuitive and unfriendly, but if you must know what it means... (You might rewrite it as $= (\mathbf {p\cdot p})\mathbf {r}-(\mathbf {p\cdot r})\mathbf {p}$.)

Proceed to slog through the commutators (1) and (2), using the above rules, $[L^i,A^j]= i\hbar \epsilon^{ijk} A_k $, so $\mathbf A$ rotates like a vector.

The entire construction relies on Valent's (6a,b,c) (7a,b), and Bohm's book referenced in WP will provide partial insight & shortcuts, as will refs 8 and 9 of Valent. But, just give $\mathbf {A\cdot A}$ a try...