How do I prove that the angular momentum is a Hermitian operator? Confirm that the operator $$\hat I_z= \left(\frac hi\right)\frac{d}{dφ},$$ where $\varphi$ is an angle, is Hermitian.
 A: You probably meant 
$$\hat{I}_z=\left(\frac{h}{i}\right)\frac{\partial}{\partial \varphi}=-ih\frac{\partial}{\partial\varphi}.$$
Let $\mathcal{H}$ be the Hilbert space of state functions.  Then, 
$$\langle \hat{I}_z f|g\rangle =\int_{\mathbb{R}^3}\overline{I_zf}(r,\theta,\varphi)\ g(r,\theta,\varphi)\ r^2\sin\theta\ dr\ d\theta\ d\varphi.$$
That is,
$$\langle \hat{I}_z f|g\rangle =\int_{\mathbb{R}^3}ih\ {\frac{\partial \bar{f}}{\partial \varphi}}(r,\theta,\varphi)\ g(r,\theta,\varphi)\ r^2\sin\theta\ dr\ d\theta\ d\varphi.$$
Using integration by parts,
\begin{align}\langle \hat{I}_z f|g\rangle &=\int_{0}^\infty\int_0^\pi\int_0^{2\pi}ih\ \frac{\partial \bar{f}}{\partial \varphi}(r,\theta,\varphi)\ g(r,\theta,\varphi)\ r^2\sin\theta\
d\varphi\ d\theta\ dr
\\
&=\int_{0}^\infty\int_0^\pi ih \left(\Big.\bar{f}(r,\theta,\varphi)\ g(r,\theta,\varphi)\Big|_{0}^{2\pi} - \int_0^{2\pi}\bar{f}(r,\theta,\varphi)\ \frac{\partial  g}{\partial \varphi}(r,\theta,\varphi)\right)\ r^2\sin\theta\
d\varphi\ d\theta\ dr.\end{align}
Since the points $(r,\theta,0)$ and $(r,\theta,2\pi)$ are the same, 
\begin{align}\langle \hat{I}_z f|g\rangle &=\int_{0}^\infty\int_0^\pi ih \left(- \int_0^{2\pi}\bar{f}(r,\theta,\varphi)\ \frac{\partial  g}{\partial \varphi}(r,\theta,\varphi)\right)\ r^2\sin\theta\
d\varphi\ d\theta\ dr
\\&=\int_{0}^\infty\int_0^\pi \int_0^{2\pi}\bar{f}(r,\theta,\varphi)\ \left(\frac{h}{i}\right)\frac{\partial  g}{\partial \varphi}(r,\theta,\varphi)\ r^2\sin\theta\
d\varphi\ d\theta\ dr
\\&=\langle f,\hat{I}_zg\rangle.\end{align}
So, $\hat{I}_z$ is self-adjoint (i.e., Hermitian).
A: I agree with the comment, $L= x \times p$, where x and p are Hermitian.  You also need to know that $[x_i,p_j]=0$ for $i \neq j$. In the case of $L_z$ we have $L_z = x_x p_y - x_y p_x$, the adjoint is $p_y x_x - p_x x_y =L_z$ because $[x_i,p_j]=0$.  It really is that easy.
