= Solution
Apply the <radial momentum operator> to a smooth radial test function. More generally $[f(r),p_r]=i\hbar f'(r)$, giving the first two labelled identities. Expanding the square gives the third:
$$
\text{(i)}\ [r,p_r]=i\hbar,\qquad
\text{(ii)}\ [r^{-1},p_r]=-i\hbar r^{-2},\qquad
\text{(iii)}\ p_r^2=-\hbar^2\left(\partial_r^2+\frac2r\partial_r\right).
$$
The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A <central potential> commutes with $L^2$ and $L_z$, so choose simultaneous angular <eigenfunctions> $Y_{lm}$, with $L^2=l(l+1)\hbar^2$ and $l=0,1,\ldots$. The radial <Hamiltonian> consequently has the stated centrifugal term. The radial inner product uses measure $r^2dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial <momentum> is not thereby asserted to be a self-adjoint observable on a half-line.
Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i p_r/\hbar$. Since $D^\dagger=-D$ formally and $[D,W]=W'=q/r^2$,
$$
A_l^\dagger A_l=\frac{a^2}{2}\left[\frac{p_r^2}{\hbar^2}+\frac{l(l+1)}{r^2}-\frac2{ar}+\frac1{q^2a^2}\right].
$$
The <hydrogen radial Hamiltonian factorization> follows by matching the Coulomb coefficient, which fixes the <Bohr radius>
$$
\boxed{a=\frac{4\pi\epsilon_0\hbar^2}{me^2}},\qquad
H_l=\frac{\hbar^2}{ma^2}\left(A_l^\dagger A_l-\frac1{2(l+1)^2}\right).
$$
For the second pair of labelled relations, the opposite product changes the sign of $W'$, hence
$$
\text{(i)}\quad[A_l,A_l^\dagger]=a^2W'=\frac{a^2(l+1)}{r^2}
=\frac{ma^2}{\hbar^2}(H_{l+1}-H_l).
$$
Using $[A,A^\dagger A]=[A,A^\dagger]A$ in the <factorization> then gives
$$
\text{(ii)}\quad[A_l,H_l]=(H_{l+1}-H_l)A_l,
\qquad H_{l+1}A_l=A_lH_l.
$$
This <Coulomb angular-momentum intertwining> means applying $A_l$ to an <energy eigenstate> either gives zero or an eigenstate of $H_{l+1}$ with the same <energy>. For a normalized negative-energy state,
$$
\|A_l\psi\|^2=\frac{ma^2E}{\hbar^2}+\frac1{2(l+1)^2}\ge0.
$$
At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The <bound-state ladder termination by norm positivity> occurs at some $L$ with $A_L\psi=0$, giving
$$
\boxed{E_n=-\frac{\hbar^2}{2ma^2n^2}=-\frac{me^4}{2(4\pi\epsilon_0)^2\hbar^2n^2},\quad n=L+1=1,2,\ldots}.
$$
The terminating radial function is proportional to $r^L e^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy <scattering states> are not subject to this terminating negative-energy argument. Here $m$ is the <mass> in the supplied <Hamiltonian>; including nuclear motion replaces it by the <reduced mass>.
Back to article page
