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- Homework Statement
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- Relevant Equations
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To show ##Y_{1,1}(\theta,\phi)## is an eigenfunction of ##\hat{L}^2## we operate on ##Y_{1,1}(\theta,\phi)## with ##\hat{L}^2##

\begin{equation}

\hat{L}^2Y_{1,1}(\theta,\phi)=\hat{L}^2\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)

\end{equation}

\begin{equation}

=-\hbar^2\Big[\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\Big(sin\theta\frac{\partial}{\partial\theta}\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)\Big)+\frac{1}{sin^2\theta}\frac{\partial^2}{\partial\phi^2}\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)\Big]

\end{equation}

\begin{equation}

=\hbar^2\sqrt{{\frac{3}{8\pi}}}\Big[ \frac{1}{sin\theta}\frac{\partial}{\partial\theta}\Big(sin\theta cos\theta\Big)e^{i\phi}+\frac{1}{sin\theta}\frac{\partial^2}{\partial\phi^2}e^{i\phi}\Big]

\end{equation}

\begin{equation}

=\hbar^2\sqrt{{\frac{3}{8\pi}}}\Big[\frac{1}{sin\theta}(cos^2\theta-sin^2\theta)-\frac{1}{sin\theta}\Big]e^{i\phi}

\end{equation}

\begin{equation}

=\hbar^2\sqrt{{\frac{3}{8\pi}}}\Big[\frac{cos^2\theta-sin^2\theta-1}{sin\theta}\Big]e^{i\phi}

\end{equation}

\begin{equation}

\hat{L}^2Y_{1,1}(\theta,\phi)=2\hbar^2\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)=2\hbar^2Y_{1,1}(\theta,\phi)

\end{equation}

so ##Y_{1,1}(\theta,\phi)## is an eigenfunction of ##\hat{L}^2## with a corresponding eigenvalue of ##2\hbar^2##. Next we work out how ##\hat{L_z}## operates on ##Y_{1,1}(\theta,\phi)##

\begin{equation}

\hat{L_z}Y_{1,1}(\theta,\phi)=-i\hbar\frac{\partial}{\partial\phi}(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi})

\end{equation}

\begin{equation}

=i\hbar\sqrt{{\frac{3}{8\pi}}}sin\theta\frac{\partial}{\partial\phi}e^{i\phi}=\hbar\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)=\hbar Y_{1,1}(\theta,\phi)

\end{equation}

and we find that ##Y_{1,1}(\theta,\phi)## is an eigenfunction of ##\hat{L_z}## with a corresponding eigenvalue of ##\hbar##.