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$L_x$ and $L_y$ don't commute since $[L_x,L_y] = i \hbar L_z$. From the uncertainty principle, we know that $\Delta L_x \cdot \Delta L_y \geq \frac{1}{2} \cdot |\langle i \hbar L_z \rangle|$. There would be no simultaneous eigenstates of $L_x$ and $L_y$ if $\langle L_z \rangle$ was zero for all eigenstates of $L_x$ and $L_y$. Is it possible to show that ?

Otherwise I would be happy with an other explanation of why or why not there would be simultaneous eigenstates of $L_x$ and $L_y$.

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According to the closure relationship we have for any two components of the angular momentum: $$[\hat{J}_i,\hat{J}_j]=i\hbar\epsilon_{ijk}L_k\tag{1}$$ where $\epsilon_{ijk}$ is Levi-Civita symbol and the sum is understood over the repeated indexes. A non-zero commutator implies that there is no common basis of eigenvectors.

That said, common eigenstates can still exist: in this case the state vector such that all components of angular momentum are zero. Please notice this happens only because the commutator is an operator: if it were a number (I mean a multiple of the identity operator), not a single common eigenstate would be possible. This is the case of position and momentum. Consider the 1D case: $$[\hat{x},\hat{p}]=i\hbar\mathbb{1}\tag{2}$$ Suppose there exists a common eigenstate $|\alpha\rangle$ such that: $$\hat{x}|\alpha\rangle=x|\alpha\rangle \\ \hat{p}|\alpha\rangle=p|\alpha\rangle$$ Then $$[\hat{x},\hat{p}]|\alpha\rangle=xp|\alpha\rangle-px|\alpha\rangle=0$$ that is contradiction with $(2)$.

On the other hand if the commutator were an operator as in $(1)$ it could possibly be zero for that particular state, as it happens for angular momentum.

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    $\begingroup$ The first part of your answer is good but I don't think it's wise in the second part to use $\hat x$ and $\hat p$ as an example given it is plagued by subtleties with normalization and unboundedness. $\endgroup$ Jan 3 at 17:45
  • $\begingroup$ Do you mean because their eigeinbases are non-normalizable states? $\endgroup$
    – Feynman_00
    Jan 3 at 17:48
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    $\begingroup$ In part yes. Also because in finite dimensional spaces as there is the contradiction of $\hbox{Tr}[\hat x,\hat p]=\hbox{Tr}(\hat x \hat p)-\hbox{Tr}(\hat p\hat x)=i\hbar \hbox{Tr}(\mathbb{1})$, whereas in infinite dimensional spaces of course this identity must be handled carefully because of the unbounded nature of the operators. $\endgroup$ Jan 3 at 18:38
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It is true that $\langle \alpha\vert L_z\vert\alpha\rangle=0$ if $\vert\alpha\rangle$ is an eigenstate of $L_x$ OR an eigenstate of $L_y$. Indeed, if it’s an eigenstate of $L_x$ then $\Delta L_x=0$ so the uncertainty relation yields $$ 0\times \Delta L_y=0\ge \frac{1}{2}\vert\langle \alpha\vert L_z\vert\alpha\rangle\vert \tag{1} $$ which implies the right hand side (which is necessarily real and non-negative) must be $0$. The same argument can be trivially modified to show the same result if $\vert\alpha\rangle$ is an eigenstate of $L_x$ instead.

I don’t know how this proves there are no simultaneous eigenstates of $L_x$ and $L_y$ and nothing in (1) prevents $\Delta L_y=0$.

The converse is not true: there are states with $\langle L_z\rangle=0$ which are not eigenstates of $L_x$ or $L_y$.

There is one state which is a simultaneous eigenstate of $L_x$ AND $L_y$: it is the $L=0$ state.

Now if you have a basis where $L_x$ is diagonal, then you can construct ladder operators $\tilde L_\pm$ in terms of $L_y$ and $L_z$, and in particular express $L_y$ as a combo of $\tilde L_\pm$, which do not act diagonally on the eigenstates of $L_y$. The exception is (again) the $L=0$ state, which is killed by $\tilde L_\pm$.

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One tool for finding your way towards a proof is to construct some representation matrices for a few different $L_{x,y,z}$ and see how you feel about them. Here’s a representation for spin two, which gives you a sense of how it goes. The size of the matrix is $2\ell+1$. The $z$-axis matrix has the eigenvalues on the diagonal already,

$$ L_z = \left(\begin{array}{ccccc} \ell \\ &\ell-1 \\ && \ddots \\ &&& -\ell+1 \\ &&&& - \ell \end{array}\right), $$

and the raising and lowering matrices are one-off-the-diagonal,

$$ L_+ = \left(\begin{array}{ccccc} 0& \sqrt{2\ell} & \\ &0& \sqrt{\cdots} & \\ &&\ddots \\ && \cdots & 0& \sqrt{2\ell} \\ &&&&0 \end{array}\right) \qquad L_- = (L_+)^T \renewcommand{\ket}[1]{\left|{#1}\right>} $$

so that $L_\pm\ket{\ell,m} = \sqrt{\ell(\ell+1) - m(m\pm1)} \ket{\ell, m\pm1}$. From these raising and lowering matrices $L_\pm$ you construct

\begin{align} 2 L_x &= L_+ + L_- \\ 2i L_y &= L_+ - L_- \end{align}

and then just find the eigenvectors. For spin-half (which isn’t allowed for orbital angular momentum $L$, but the algebra is similar), the $x$-axis eigenvectors are ${1}\choose{\pm1}$, with eigenvalues $\pm1$, and the $y$-axis eigenvectors are $1\choose{\pm i}$, with eigenvalues $\pm i$. Those basis vectors are orthogonal to each other. However, it is the case that the eigenstates of $\sigma_x$ have expectation value $\left<\sigma_y\right> = \left<\sigma_z\right> = 0$.

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