Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 2 / 18F (Quantum Mechanics)

The orbital angular momentum operators areOn smooth wavefunctions, $[x_i,p_j]=i\hslash {\delta }_{ij}$, while position operators commute with each other and momentum operators commute with each other. Applying the product rule for commutators givesUse it on the four terms in $[L_1,L_2]$: only $[x_2{p}_3,x_3{p}_1]=-i\hslash x_2{p}_1$ and $[x_3{p}_2,x_1{p}_3]=i\hslash x_1{p}_2$ survive, with the latter entering positively from the two minus signs. ThereforeThe cyclic versions similarly give $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. Thus, for $L_{\pm }=L_1\pm i{L}_2$,Now define $L^2=L_1^2+L_2^2+L_3^2$. For each $i$,because the Levi-Civita symbol is antisymmetric in $j,k$ and the operator pair is symmetric. Linearity then gives $\boxed{[L^2,L_{\pm }]=0}$. The PDF superscript here is the squared total angular momentum, not the component $L_2$.

Every orbital angular momentum operator annihilates a radial function: substituting ${\partial }_{i}f(r)=f^{\prime }(r)x_i/r$ makes the two terms cancel. It also annihilates $r^2$. Thus the radial factor can be carried through the angular operators. Since $x_3^2$ is independent of $x_1,x_2$, $L_3\psi =0$ for every value of $a$.

Direct differentiation givesConsequently $L^2{x}_3^2={\hslash }^2(6x_3^2-2r^2)$ and $L^2{r}^2=0$. For a nonzero admissible radial factor, comparison with $\lambda (x_3^2+a{r}^2)f(r)$ forces $\lambda =6{\hslash }^2$ and $6a=-2$. Therefore the simultaneous eigenfunction occurs atIt is the axially symmetric quadratic orbital eigenfunction, with quantum numbers $l=2$, $m=0$ and angular dependence proportional to the spherical harmonic $Y_{20}$. The subtraction of $r^2/3$ removes its rotationally invariant scalar component.