# 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 $0$?

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 $\langle a|L_x|a\rangle=0$.
• No it does not mean so. Expectation values vanish but $L_x|a\rangle \neq 0$, $L_y|a\rangle \neq 0$ in general. Commented Nov 9, 2017 at 20:07
• No there is not, too few information: every eigenvector of $L_z$ satisfies your requirement and produces different values of $L_x|a\rangle$. Commented Nov 9, 2017 at 20:43
By the definition of expectation value: $$\langle\hat{A}\rangle=\langle a|\hat{A}|a\rangle$$ where $\hat{A}$ is the observable you are interested in (that is, $L_x, L_y$ for this case). Using also the expressions for $L_x, L_y$: $$L_x=\frac{L_++L_-}{2}, \quad L_y=\frac{L_+-L_-}{2i}$$ 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...