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Use the canonical commutation relations $[{\hat{x}}_i,{\hat{x}}_j]=[{\hat{p}}_i,{\hat{p}}_j]=0$ and $[{\hat{x}}_i,{\hat{p}}_j]=i\hslash {\delta }_{ij}I$. For compactness write products without hats in the calculation, on a common domain such as smooth compactly supported wavefunctions. Expanding the commutator givesThe middle two commutators vanish because all the relevant indices differ. The first is $-i\hslash x_2{p}_1$, since $[p_3,x_3]=-i\hslash I$; the last is $i\hslash x_1{p}_2$, since $[x_3,p_3]=i\hslash I$. ThusThis is one of the orbital angular momentum commutation relations.

The total angular momentum operator isUsing the cyclic orbital angular momentum commutation relations,while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,{\mathbf{L}}^2]=0}$. In Cartesian coordinates,Because $L_z(x+iy)=\hslash (x+iy)$ and it annihilates $z$ and every radial function,Replacing $i$ by $-i$ gives eigenvalue $-m\hslash$.

The six-dimensional space splits into the five trace-free quadratic spherical harmonics with $\ell =2$ and one radial state with $\ell =0$. A simultaneous eigenbasis for $L_z$ and ${\mathbf{L}}^2$ iswhere the eigenvalue $\ell (\ell +1){\hslash }^2$ of ${\mathbf{L}}^2$ was used. Every displayed state is a complex linear combination of the original six energy eigenstates, so it is still an energy eigenstate at the same level.

The orbital angular momentum commutation relations giveConsequentlyThe same commutation relations imply $[L^2,L_x]=[L^2,L_y]=0$, and henceThese are the defining identities for the angular momentum ladder operators.

Ifthen, whenever $L_{\pm }\varphi \ne 0$,Thus $L_{\pm }\varphi$ has quantum numbers $(\ell ,m\pm 1)$.

Use $[AB,C]=A[B,C]+[A,C]B$ and the canonical commutation relation. With $L_i={\epsilon }_{iab}{x}_a{p}_b$,Using these two transformation laws on $L_j={\epsilon }_{jab}{x}_a{p}_b$ givesThese are the orbital angular momentum commutation relations.

NowThe factor $[L_j,L_i]$ is antisymmetric in $j$ and its remaining product is symmetric after the two terms are combined, so the contraction vanishes:

The same commutators show that $p^2=p_j{p}_j$ and $r^2=x_j{x}_j$ are rotational scalars: their commutators with every $L_i$ vanish by contraction of the antisymmetric ${\epsilon }_{ijk}$ with a symmetric product. Therefore $U(r)$ also commutes with every $L_i$, and forwe have $[H,L_i]=0$, henceIn particular $H,L^2,L_3$ are pairwise commuting Hermitian operators. By simultaneous diagonalization, they admit a common eigenbasis, subject to the usual spectral-domain qualifications for unbounded operators. This is the rotational invariance of a central-potential Hamiltonian.