Diagonalization of total angular momentum over creation operators for an isotropic harmonic oscillator?

You have an isotropic three dimensional quantum harmonic oscillator so the Hamiltonian is $$H=\frac{p^2}{2}+\frac{r^2}2$$ If you do the creation-annihilation operator-algebra trick and define creation operator $$a^\dagger_j=\frac{x_j-ip_j}{\sqrt{2\hbar}}$$ so that the Hamiltonian is $$H=\hbar \left (a_1^\dagger a_1+a_2^\dagger a_2+a_3^\dagger a_3+\frac 3 2 \right)$$ so that the Hamiltonian is diagonalized by states characterized by the excitation number in each direction.

BUT if we want to compute angular momentum, we could define $$3$$-rotating creation operators $$a_\uparrow^\dagger=\frac{a^\dagger_1+ia^\dagger_2}{\sqrt 2} \ \ \ \ \ \ \ \ \ a_\downarrow^\dagger=\frac{a^\dagger_1-ia^\dagger_2}{\sqrt 2}$$ and, as $$a_1^\dagger a_1+a_2^\dagger a_2=a_\uparrow^\dagger a_\uparrow+a_\downarrow^\dagger a_\downarrow$$, the Hamiltonian is still nicely diagonized; moreover we also have the 3-angular momentum operator diagonal, $$L_3=\hbar (a_\uparrow^\dagger a_\uparrow-a_\downarrow^\dagger a_\downarrow)$$, so $$L_3 \left | n_\uparrow n_\downarrow n_z \right \rangle=\hbar (n_\uparrow-n_\downarrow) \left | n_\uparrow n_\downarrow n_z \right \rangle$$.

How might you similarly diagonalize $$L^2$$ with respect to creation-annihilation? The spherical harmonics are hellish, but Schwinger's creation operator approach is so nice and intuitive. How could we compute $$L^2$$ probabilities without ever integrating again?

Edit More precisely: Are there commuting operators $$a,b,c$$ such that $$[a,a^\dagger]=\hbar$$ (etc for $$b$$ and $$c$$) and the set of states defined by $$\left | n_a n_b n_c \right \rangle=(a^\dagger)^{n_a}(b^\dagger)^{n_b}(c^\dagger)^{n_c}\left | \psi_0 \right \rangle$$ form an orthogonal basis diagonalizing $$H,L^2$$,and $$L_3$$? Here $$\left | \psi_0 \right \rangle$$ is the ground state. How can you determine $$a,b,c$$ algebraically?

• Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants) – Incnis Mrsi May 5 '15 at 20:35
• You probably want to read up in Messiah QM vI, Ch XII,  § 15, p 456. – Cosmas Zachos Dec 27 '18 at 16:06
• – Cosmas Zachos Dec 27 '18 at 17:20
• You are probably aware that the basis $|n_\uparrow, n_\downarrow,n_3\rangle$ does not diagonalize $L^2$. You should emphasize that, to explain why you are wondering about the existence of yet another basis, a,b,c. It does not exist. – Cosmas Zachos Dec 27 '18 at 17:37