The wave function may be rewritten in spherical coordinates as

$$
\psi=K r(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta) e^{-\alpha r}
$$

its angular part being

$$
\psi(\theta, \phi)=K^{\prime}(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta),
$$

where $K^{\prime}$ is the normalization constant such that

$$
K^{\prime 2} \int_{\theta}^{\pi} d \theta \int_{0}^{2 \pi} \sin \theta(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta)^{2} d \phi=1 \text {. }
$$

Since

$$
\cos \phi=\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right), \quad \sin \phi=\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right)
$$

we have

$$
\begin{aligned}
\psi(\theta, \phi) &=K^{\prime}\left[\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right) \sin \theta+\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right) \sin \theta+\cos 2 \theta\right], \\
&=K^{\prime}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$

The normalization condition and the orthonormality of $Y_{l}^{m}$ then give

$$
K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}+\frac{1}{2} \cdot \frac{8 \pi}{3}+4 \cdot \frac{4 \pi}{3}\right]=1
$$

or

$$
K^{\prime}=\sqrt{\frac{1}{8 \pi}},
$$

and thus

$$
\begin{aligned}
\psi(\theta, \phi)=& \sqrt{\frac{1}{8 \pi}}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}\right.\\
&\left.+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$

The total angular momentum of the particle is

$$
\sqrt{\left\langle\mathbf{L}^{2}\right\rangle}=\sqrt{l(l+1)} \hbar=\sqrt{2} \hbar .
$$

as the wave function corresponds to $l=1$.The $z$-component of the angular momentum is

$$
\begin{aligned}
\left\langle\psi^{*}\left|L_{z}\right| \psi\right\rangle=& K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3} \cdot \hbar\left(Y_{1}^{1}\right)^{2}+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\left(Y_{1}^{-1}\right)^{2}\right.\\
&\left.+4 \cdot \frac{4 \pi}{3}(0)\left(Y_{1}^{0}\right)^{2}\right] \\
=& \frac{1}{8 \pi}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}(+\hbar)+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\right]=0 .
\end{aligned}
$$The probability of finding $L_{z}=+\hbar$ is

$$
\begin{aligned}
P &=\left|\left\langle L_{z}=+\hbar \mid \psi(\theta, \phi)\right\rangle\right|^{2} \\
&=\frac{1}{8 \pi} \cdot \frac{1}{2} \cdot \frac{8 \pi}{3}=\frac{1}{6} .
\end{aligned}
$$
파동 함수는 다음과 같이 구형 좌표로 다시 작성할 수 있습니다.

$$
\psi=K r(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta) e^{-\알파 r}
$$

각진 부분은

$$
\psi(\theta, \phi)=K^{\prime}(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta),
$$

여기서 $K^{\prime}$는 다음과 같은 정규화 상수입니다.

$$
K^{\프라임 2} \int_{\theta}^{\pi} d \theta \int_{0}^{2 \pi} \sin \theta(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta)^{2} d \phi=1 \text {. }
$$

Since

$$
\cos \phi=\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right), \quad \sin \phi=\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right)
$$

우리는

$$
\begin{aligned}
\psi(\theta, \phi) &=K^{\prime}\left[\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right) \sin \theta+\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right) \sin \theta+\cos 2 \theta\right], \\.
&=K^{\prime}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$

정규화 조건과 $Y_{l}^{m}$의 정규성은 다음과 같습니다.

$$
K^{\소수 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}+\frac{1}{2} \cdot \frac{8 \pi}{3}+4 \cdot \frac{4 \pi}{3}\right]=1
$$

또는

$$
K^{\prime}=\sqrt{\frac{1}{8 \pi}},
$$

따라서

$$
\begin{aligned}
\psi(\theta, \phi)=& \sqrt{\frac{1}{8 \pi}}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}\right.\\
&\left.+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$

입자의 총 각운동량은 다음과 같습니다.

$$
\sqrt{\left\langle\mathbf{L}^{2}\right\rangle}=\sqrt{l(l+1)} \hbar=\sqrt{2} \hbar .
$$

파동 함수는 $l=1$에 해당하므로 각운동량의 $z$ 성분은 다음과 같습니다.

$$
\begin{aligned}
\왼쪽\각도\psi^{*}\왼쪽|L_{z}\오른쪽| \psi\오른쪽\각도=& K^{\프라임 2}\왼쪽[\frac{1}{2} \cdot \frac{8 \pi}{3} \cdot \hbar\left(Y_{1}^{1}\right)^{2}+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\left(Y_{1}^{-1}\right)^{2}\right.\\
&\left.+4 \cdot \frac{4 \pi}{3}(0)\left(Y_{1}^{0}\right)^{2}\right] \\
=& \frac{1}{8 \pi}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}(+\hbar)+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\right]=0 .
\end{aligned}
$$L_{z}=+\hbar$를 구할 확률은 다음과 같습니다.

$$
\start{aligned}
P &=\left|\left\langle L_{z}=+\hbar \mid \psi(\theta, \phi)\right\rangle\right|^{2} \\
&=\frac{1}{8 \pi} \cdot \frac{1}{2} \cdot \frac{8 \pi}{3}=\frac{1}{6} .
\end{aligned}
$$