 Solution

Use the normalized $L_3$ eigenstates from part (ii). The quantum state found in part (iii) has the decompositionThere is no ${\chi }_0$ component. Equivalently, its inner product with ${\chi }_0$ contains ${\int }_0^{2\pi }\cos \varphi d\varphi =0$. The Born rule therefore givesFor completeness the only outcomes are $L_3=\pm \hslash$, each with probability $1/2$. Thus $⟨L_3⟩=0$ does not mean that zero is a possible result in this state: its quantum uncertainty in $L_3$ is $\hslash$, even though $L_1$ is exactly zero.