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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 $0$?

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2 Answers 2

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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 $\langle a|L_x|a\rangle=0$.

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  • $\begingroup$ Such a neat mathematical description. If the expectation values of Lx and Ly are zero if |a> is an eigen state of Lz, does it mean that Lx|a> and Ly|a> = 0? $\endgroup$
    – Patrick
    Commented Nov 9, 2017 at 19:50
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    $\begingroup$ No it does not mean so. Expectation values vanish but $L_x|a\rangle \neq 0$, $L_y|a\rangle \neq 0$ in general. $\endgroup$ Commented Nov 9, 2017 at 20:07
  • $\begingroup$ So is there any way to find what Lx|a> is? $\endgroup$
    – Patrick
    Commented Nov 9, 2017 at 20:17
  • $\begingroup$ No there is not, too few information: every eigenvector of $L_z$ satisfies your requirement and produces different values of $L_x|a\rangle$. $\endgroup$ Commented Nov 9, 2017 at 20:43
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    $\begingroup$ The text says common eigenvector, you missed this requirement in your question. $\endgroup$ Commented Nov 9, 2017 at 20:55
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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...

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