Suppose an electron is in a state described by the wave function

$$
\psi=\frac{1}{\sqrt{4 \pi}}\left(e^{i \phi} \sin \theta+\cos \theta\right) g(r)
$$

where

$$
\int_{0}^{\infty}|g(r)|^{2} r^{2} d r=1
$$

and $\phi, \theta$ are the azimuth and polar angles respectively.

What is the expectation value of $L_{z}$ ?
As

$$
Y_{10}=\sqrt{\frac{3}{4 \pi}} \cos \theta, Y_{1, \pm 1}=\mp \sqrt{\frac{3}{8 \pi}} \sin \theta e^{\pm i \phi},
$$

the wave function can be written as

$$
\psi=\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right) g(r)
$$

Hence the possible values of $L_{z}$ are $+\hbar, 0$.

Since

$$
\begin{aligned}
\int|\psi|^{2} d r &=\frac{1}{4 \pi} \int_{0}^{\infty}|g(r)|^{2} r^{2} d r \int_{0}^{\pi} d \theta \int_{0}^{2 \pi}(1+\cos \phi \sin 2 \theta) \sin \theta d \phi \\
&=\frac{1}{2} \int_{0}^{\pi} \sin \theta d \theta=1
\end{aligned}
$$

the given wave function is normalized. The probability density is then given by $P=|\psi|^{2}$. Thus the probability of $L_{z}=+\hbar$ is $\left(\sqrt{\frac{2}{3}}\right)^{2}$ or $2 / 3$ and that of $L_{z}=0$ is $\left(\frac{1}{\sqrt{3}}\right)^{2}$ or $1 / 3$.

$$
\begin{aligned}
\int \psi^{*} L_{z} \psi r^{2} \sin \theta d \theta d \phi d r=& \int\left[\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right)\right]^{*} \\
& \times \hat{L}_{z}\left[\sqrt{\frac{1}{3}}\left(-\sqrt{2} Y_{11}+Y_{10}\right)\right] \\
& \times|g(r)|^{2} r^{2} d r \sin \theta d \theta d \phi \\
=& \frac{2}{3} \hbar \int_{0}^{\pi} d \theta \int_{0}^{2 \pi} Y_{11}^{2} d \phi=\frac{2}{3} \hbar
\end{aligned}
$$