Past exam of the physics A course of the University of Cambridge / 2016 / Paper 1 / Section B / B8

Expanding with the angular momentum commutators givesFor ${\hat{L}}_z|\psi ⟩=\alpha \hslash |\psi ⟩$, commuting ${\hat{L}}_z$ past a ladder operator givesThus the raised or lowered vector, if nonzero, is an eigenvector with the shifted eigenvalue; an angular momentum ladder endpoint can instead be annihilated. Also $[{\hat{L}}^2,{\hat{L}}_{\pm }]=0$, so ${\hat{L}}^2{\hat{L}}_{\pm }|\psi ⟩=\Lambda {\hslash }^2{\hat{L}}_{\pm }|\psi ⟩$ for every nonzero ladder result.

For orbital angular momentum, ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$ gives an azimuthal factor $e^{i\alpha \varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ requires integer $\alpha$. Positivity of ladder norms givesfor normalized states. A fixed finite $\Lambda$ therefore has upper and lower endpoints. Writing the upper endpoint as $m=\ell$, its vanishing raising norm gives $\Lambda =\ell (\ell +1)$; the lower endpoint is $m=-\ell$. Consequentlywith $\ell =0,1,2,\dots$. This integer conclusion is for orbital angular momentum; spin representations can also have half-integer labels.

For the specified Gaussian-radial state, use a Cartesian p-orbital decomposition: write $x+y+z=r(\sin \theta \cos \varphi +\sin \theta \sin \varphi +\cos \theta )$. Its angular factor isOnly $\ell =1$ occurs. The squared angular coefficients are $2,2,2$ before common normalization, so each allowed $m$ has probability one third. Therefore the simultaneous outcome isThe common normalized radial factor cancels from this ratio. The standard symmetric $m$ sequence used above resolves the duplicated central positive entry in the printed list.