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Let $L_iL_j-L_jL_i = i\hbar\varepsilon_{ijk}L_k$ where $i,j,k\in\{1,2,3\}$

Let $u$ be any eigenstate of $L_3$. How might one show that $\langle L_1^2 \rangle = \int u^*L_1^2u = \int u^*L_2^2u = \langle L_2^2\rangle$ ?

I can show that $\langle u|L_3L_1L_2-L_1L_2L_3|u\rangle=\int u^*(L_3L_1L_2-L_1L_2L_3)u =0$. And I know that $L_3 $ can be written as ${1\over C}(L_1L_2-L_2L_1)$. Hence I have $\int u^*L_1L_2^2L_1u=\int u^*L_2L_1^2L_2u$. But I don't seem to be getting the required form... Help will be appreciated.

Added: The $L_i$'s are operators that don't necessarily commute.

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    $L_i$ are operators on what? $u$ is a function defined over what space?2011-09-03
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    @Willie W: The $L_i$'s operate on the space of functions. In fact, I must say this reminds me of the angular momentum operator and eigenstates/eigenfunctions (can't distinguish those 2)...2011-09-03
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    @Willie W: You are quite right. I have edited the question to give it verbatim. Hopefully it makes more sense now? :)2011-09-03
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    Are the $L_i$ assumed to be self-adjoint?2011-09-03
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    @Willie W: Yup. Sorry :S2011-09-03

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