Expectation Values of $L_x$ and $L_y$ operators in $L_z$-eigenstates Let's say the $z$-component of angular momentum $L_z$ has an eigenstate $\vert a\rangle$. How do I go about proving that the expectation values of $L_x$ and $L_y$ in the state $\vert a\rangle$ is $0$? 
 A: Another, more sophisticated, answer uses the fact that $L_z$ generates the rotations around the $z$ axis. 
With a rotation of $\pi$ around $z$ you can reverse the sign of $L_x$ (or of the projection of $\vec{L}$ along any unit vector normal to $z$): $$e^{-i\pi L_z} L_x e^{i\pi L_z} = -L_x$$
As a consequence
 $$\langle a|e^{-i\pi L_z} L_x e^{i\pi L_z}|a \rangle = -\langle a|L_x|a\rangle\tag{1}$$
But $e^{i\pi L_z}|a \rangle =  e^{i\pi a}|a \rangle$ so that (1) cen be re-written
 $$\langle a|e^{-i\pi a} L_x e^{i\pi a}|a \rangle = -\langle a|L_x|a\rangle\:, 
$$
that is
 $$\langle a| L_x e^{-i\pi a} e^{i\pi a}|a \rangle = -\langle a|L_x|a\rangle\:, 
$$
namely
$$\langle a| L_x |a \rangle = -\langle a|L_x|a\rangle\:, 
$$
that implies $\langle a|L_x|a\rangle=0$.
A: By the definition of expectation value:
\begin{equation}
\langle\hat{A}\rangle=\langle a|\hat{A}|a\rangle
\end{equation}
where $\hat{A}$ is the observable you are interested in (that is, $L_x, L_y$ for this case).
Using also the expressions for $L_x, L_y$:
\begin{equation}
L_x=\frac{L_++L_-}{2},  \quad L_y=\frac{L_+-L_-}{2i}
\end{equation}
Given that you are in an eigenstate of $L_z$ (an arbitrary |a$\rangle$) acting with this operators on ket you will get the $|a+1\rangle$, $|a-1\rangle$ respectively, which are orthogonal states with respect to $|a\rangle$. So your result will be 0 in both cases...
