Question regarding radial raising/lowering operator for isotropic harmonic oscillator

Question

I understand the symmetry structure of the 3D isotropic harmonic oscillator $H = frac{mathbf{P}^2}{2mu} + frac{1}{2}momega^2mathbf{X}^2$ as follows. The energy levels are $E_N = hslash omega (N + 3/2)$, each energy eigenspace is the highest weight $(N, 0)$ irreducible $SU(3)$ representation (given by homogeneous degree $N$ polynomials on three variables), of dimension $frac{(N+1)(N+2)}{2}$. As $SO(3)$ representations these split as $(N, 0)_{SU(3)} cong (N)_{SO(3)}oplus (N-2)_{SO(3)} oplus ldots $, where the last term is either $(0)$ or $(1)$ depending on the parity of $N$, and $(ell)_{SO(3)}$ denotes the usual spin $ell$ representation of $SO(3)$.

The lecture notes at http://www.damtp.cam.ac.uk/user/dbs26/PQM/chap6.pdf point out that one may rewrite the Hamiltonian as $H = frac{1}{2mu}left ( P_r^2 + frac{mathbf{L}^2}{R^2}right ) + frac{1}{2}momega^2R^2$, where $R^2 = mathbf{X}cdot mathbf{X}$, $P_r = frac12 left(frac{mathbf{X}}{R}cdot mathbf{P} + mathbf{P}cdot frac{mathbf{X}}{R}right)$ is the radial momentum, and $mathbf{L}^2$ is total angular momentum. When restricted to an eigenspace of $mathbf{L}^2$ of total angular momentum number $ell$ (i.e. $mathbf{L}^2$ has eigenvalue $ell(ell + 1)$), the Hamiltonian can be simplified to $H_ell = frac{1}{2mu}left ( P_r^2 + frac{ell(ell + 1)hslash^2}{R^2}right ) + frac{1}{2}momega^2R^2$. Then the notes introduce the radial lowering operator $A_ell = frac{1}{sqrt{2mu hslash omega}} left ( iP_r - frac{(ell + 1)hslash}{R} + muomega Rright)$. Then some computation shows that $A_ell$ lowers the energy by $hslash omega$, while raising the total angular momentum number by $1$.

It seems that we have the following structure:

Here the $N^{th}$ row corresponds to the $N^{th}$ energy eigenspace, and the $ell^{th}$ column to the $ell^{th}$ total angular momentum eigenspace. The bold face $mathbf{0}$ denotes the zero vector space, indicating that at each energy level $A_ell$ annihilates the top total angular momentum eigenstates ($ell = N$).

Now to my confusion: supposedly the rest of the $A_ell$ should not be zero, even though they are seemingly $SO(3)$-equivariant maps $A_ell: (ell)_{SO(3)}to (ell - 1)_{SO(3)}$ between irreducible $SO(3)$ representations of different dimensions (which ought to be zero by Schur's lemma). I would be very grateful if someone could point out where I went wrong.

Yegreg · Answer

After some further offline reflection and discussion, the following understanding was developed. The confusion in the question stems from the fact that the operator $A_ell$ is better understood as acting only on the radial part of the wave function. Some of the details are explained below.

Since the Hamiltonian $H = frac{mathbf{P}^2}{2mu} + frac{1}{2}momega^2mathbf{X}^2$ is rotationally symmetric, we expect to be able to find a basis $|E, ell, m rangle$ of mutual eigenstates of $H$, $mathbf{L}^2$ and $L_z$, and in position space these eigenstates have wave functions $psi(mathbf{x})$ that are separable into radial and angular parts $psi(r, theta, varphi) = rho_E^ell (r) cdot Y_ell^m (theta, varphi)$. Here $Y_ell^m$ denotes the usual spherical harmonics.

From rewriting the Hamiltonian in the form $H = frac{1}{2mu}left ( P_r^2 + frac{mathbf{L}^2}{R^2}right ) + frac{1}{2}momega^2R^2$ as in the question, and the corresponding $H_ell$ as above, we can consider $H_ell$ as a one-dimensional Hamiltonian operator (to be thought of as acting on the radial part of the wave function). Then if $psi(r, theta, varphi) = rho_E^ell (r) cdot Y_ell^m (theta, varphi)$ is the state $|E, ell, m rangle$ then $rho_E^ell (r)$ should be an eigenstate of $H_ell$ with eigenvalue $E$.

In this setting the operator $A_ell$ maps $rho_E^ell (r)$ to (possibly a scalar multiple of) $rho_{E - hslash omega}^{ell + 1} (r)$, which corresponds to the radial part of states $|E - hslash omega, ell + 1, m rangle$. Thus the operator $A_ell$ (and its adjoint $A_ell^dagger$) are useful for moving between the radial parts of eigenstates with varying energy and total angular momentum.

To make it clearer where the confusion in the questions stems from, we can instead consider $A_ell$ as acting on the whole wave function. As pointed out by octonion, since the operator $A_ell$ is rotationally invariant, it leaves the angular part of the wave function unchanged, so acting on an eigenstate $|E, ell, m rangle = rho_E^ell (r) cdot Y_ell^m (theta, varphi)$ results in a state $rho_{E - hslash omega}^{ell + 1}(r) cdot Y_ell^m (theta, varphi)$, which is not an energy eigenstate anymore (due to the mismatch between $ell$ and $ell + 1$).

Question regarding radial raising/lowering operator for isotropic harmonic oscillator

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