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?

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    $\begingroup$ Aren’t $H$ and $L^2$ essentially the same operator? (up to some scalar constants) $\endgroup$ – Incnis Mrsi May 5 '15 at 20:35
  • $\begingroup$ You probably want to read up in Messiah QM vI, Ch XII,  § 15, p 456. $\endgroup$ – Cosmas Zachos Dec 27 '18 at 16:06
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Dec 27 '18 at 17:20
  • $\begingroup$ 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. $\endgroup$ – Cosmas Zachos Dec 27 '18 at 17:37

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