Common eigenstate of incompatible observables In many resources I have seen that incompatible observables cannot have a common eigenbasis set, but may share one or few eigen states. I followed the thread Can incompatible observables share an eigenvector?, where through a matrix example it is proven to be true.
I want to know whether this can actually be true for physical incompatible observables like position-momentum in a direction, or angular momentum in two perpendicular directions and so on. If it happens to be true then in that particular state, measurement of both will be precise and the uncertainty principles:
$$\Delta x\Delta p_x\ge\frac{\hbar}{2} \hspace{1cm}\text{or}\hspace{1cm}\Delta L\Delta \theta\ge\frac{\hbar}{2}$$
seem to be violated.
There may be non-commuting matrices that can share an eigenvector. But whether those matrices can represent physical observables in some basis is a point of concern. Looking for opinions.
 A: The Heisenberg inequality for any two observables $\hat A$ and $\hat B$ is :
$$ \Delta a \Delta b \geq \frac{1}{2}|\langle [\hat A ,\hat B]\rangle|$$
If $\hat A$ and $\hat B$ have a common eigenvector, then in this state we have $\Delta a= \Delta b = 0$, and therefore $|\langle [\hat A ,\hat B]\rangle| = 0$. We can actually explicitly check that the commutator has a zero eigenvalue : let $\hat  A|\psi \rangle = a|\psi\rangle$ and $\hat B |\psi\rangle = b|\psi\rangle$. Then :
$$[\hat A,\hat B]|\psi\rangle = (\hat  A\hat B -\hat B\hat A)|\psi\rangle = (ab-ba)|\psi\rangle = 0$$
In the case of $\hat x$ and $\hat p$ (or more generally any two conjugate variables), we have $[\hat x,\hat p]  =i\hbar$, which prevents the situation above from occurring.
Edit : Compatible and incompatible observables
Compatible observables are observables which commute, so that there is a basis of common eigenvectors.
Incompatible observables do not commute. They may or may not have common eigenvectors. If their commutator is non-degenerate, which is the case of most usual commutations relations (eg $\hat x$ and $\hat p$, spin operators, etc.), then they have no common eigenvectors.
It is, however, easy to construct matrices which do not commute and have a common eigenvector. This is done in this answer to the question linked in OP above : take
\begin{align}
A &= \begin{pmatrix} 1 & 0 &0 \\
0& 1 & 0 \\
0 & 0 & -1 \end{pmatrix}\\
B &= \begin{pmatrix}  1 & 0 & 0 \\
0&0&1 \\
0&1&0 \end{pmatrix}\\
\psi &= \begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}
\end{align}
Then, $A\psi = B\psi = \psi$, but :
$$[A,B] = \begin{pmatrix} 0&0&0 \\
0 & 0 & 2 \\
0& -2 & 0\end{pmatrix} \neq 0$$
