 Solution

In the position representation, the momentum operators are ${\hat{p}}_x=-i\hslash {\partial }_x$ and ${\hat{p}}_y=-i\hslash {\partial }_y$. Thereforesince ${\partial }_{\varphi }=-y{\partial }_x+x{\partial }_y$ at fixed radius. Solving the eigenvalue equation on the circle and imposing $2\pi$ periodicity givesTheir normalization uses the angular measure $d\varphi$ on $[0,2\pi )$.

For the half-ring sine-squared angular state, write $C=\sqrt{8/(3\pi )}$. The stated wavefunction is normalized because $C^2{\int }_0^{\pi }{\sin }^4\varphi d\varphi =1$. It and its first derivative match continuously to zero at the boundaries, so the integrations by parts have no boundary contributions. The angular momentum expectation and second moment areHence $\mathrm{Var}(L_z)=4{\hslash }^2/3$. The jumps in the second derivative do not invalidate this result: the periodic wavefunction has a square-integrable second derivative and a continuous first derivative.

The probability of $L_z=\hslash$ is the squared projection onto $u_1$:The odd sine term gives the imaginary overlap; a real wavefunction need not have zero probability of nonzero angular momentum. Its positive and negative angular-momentum probabilities are equal.^[ref]