# Question regarding radial raising/lowering operator for isotropic harmonic oscillator

I understand the symmetry structure of the 3D isotropic harmonic oscillator $$H = \frac{\mathbf{P}^2}{2\mu} + \frac{1}{2}m\omega^2\mathbf{X}^2$$ 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 $$H = \frac{1}{2\mu}\left ( P_r^2 + \frac{\mathbf{L}^2}{R^2}\right ) + \frac{1}{2}m\omega^2R^2$$, 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 $$H_\ell = \frac{1}{2\mu}\left ( P_r^2 + \frac{\ell(\ell + 1)\hslash^2}{R^2}\right ) + \frac{1}{2}m\omega^2R^2$$. Then the notes introduce the radial lowering operator $$A_\ell = \frac{1}{\sqrt{2\mu \hslash \omega}} \left ( iP_r - \frac{(\ell + 1)\hslash}{R} + \mu\omega R\right)$$. 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.

• Does $A_l$ commute with the rotation group? I believe that's one of the conditions for Schur's lemma – octonion Oct 15 '19 at 8:13
• I would think that since both $P_r$ and $R$ are rotation invariant, $A_\ell$ indeed commutes with $\mathbf{L}$. – Yegreg Oct 15 '19 at 17:03
• Well this raises another problem. If $A_l$ is rotationally invariant, how can it possibly raise the total angular momentum of a state? Simply act with $L^2$ on $A_l\,|l,m\rangle$ and commute – octonion Oct 15 '19 at 17:36
• That is a good point. So if we start with a common eigenstate $|E, \ell \rangle$ of $H$ and $\mathbf{L}^2$, then $A_\ell |E, \ell \rangle$ is still an eigenstate of $\mathbf{L}^2$ with the same eigenvalue. It's also an eigenstate of $H_{\ell + 1}$, but that is not the right Hamiltonian in this case, so it seems that $A_\ell |E, \ell \rangle$ is not an energy eigenstate after all? – Yegreg Oct 15 '19 at 20:19

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 $$H = \frac{\mathbf{P}^2}{2\mu} + \frac{1}{2}m\omega^2\mathbf{X}^2$$ 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 $$H = \frac{1}{2\mu}\left ( P_r^2 + \frac{\mathbf{L}^2}{R^2}\right ) + \frac{1}{2}m\omega^2R^2$$ 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 $$\ell + 1$$).