= Solution
For the allowed $n=2k$ states, the normalized standing wave is a <quantum superposition> of two opposite <angular momentum eigenstates>:
$$
\Phi_{2k}(\phi)=\frac{\sin(k\phi)}{\sqrt\pi}
=\frac{u_k(\phi)-u_{-k}(\phi)}{i\sqrt2}.
$$
The <Born rule> therefore predicts
$$
\boxed{l_z=+k\hbar\ \text{or}\ -k\hbar,\qquad P(+k\hbar)=P(-k\hbar)=\tfrac12.}
$$
The <expectation value> is zero, but a single measurement never gives zero for these $k\ge1$ states. Their uncertainty is $\Delta l_z=k\hbar$. The two components have the same rotor energy $\hbar^2k^2/(2I)$, so a definite energy does not imply definite angular momentum.
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