Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2B / Section B / 35 / b / ii

The azimuthal derivative is ${\partial }_{\varphi }\psi =-i\psi$, so ${\hat{J}}_z\psi =(-i\hslash )(-i)\psi =-\hslash \psi$. For the polar part of the angular Laplacian,Their sum is $-2\sin \theta e^{-i\varphi }=-2\psi$, hence ${\hat{J}}^2\psi =2{\hslash }^2\psi$. Comparing these two eigenvalues with $M\hslash$ and $J(J+1){\hslash }^2$ gives $J=1$, $M=-1$. The sign of $M$ follows from both minus signs in the ${\hat{J}}_z$ calculation.