Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 6B (Quantum Mechanics)

The orbital angular momentum linear operator is the cross product of the position operator and the momentum operator:In position space, $\hat{\mathbf{p}}=-i\hslash \nabla$, so $\hat{\mathbf{L}}=-i\hslash \mathbf{r}\times \nabla$. The Levi-Civita symbol gives ${\hat{L}}_1={\hat{x}}_2{\hat{p}}_3-{\hat{x}}_3{\hat{p}}_2$ and its cyclic analogues.

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 product rule for a commutator gives $[A,B^2]=[A,B]B+B[A,B]$. Using the permitted cyclic angular momentum commutation relations,The first two expressions cancel, even though the factors need not commute. HenceThe squared angular momentum operator commutes with each of its components by the same cyclic calculation.

Distinct eigenvalues of the self-adjoint angular momentum operator imply $⟨{\psi }_1,{\psi }_2⟩=0$. The given combination is not normalized: its squared norm is $({\cos }^2\omega t+3{\sin }^2\omega t)/4$, which is always positive. Therefore its properly normalized expectation value isIt vanishes when ${\tan }^2\omega t=1/3$. The first positive solution has $\omega t=\pi /6$, soNormalization changes the value of the expectation value away from zero, but not its zero times.