# Can incompatible observables share an eigenvector?

I was recently introduced to the concept of compatible and incompatible observables and specifically the generalized uncertainty principle, which is written in my textbook as:

$$\sigma_A^2\sigma_B^2 \geq \left(\frac{1}{2i} \langle [A, B] \rangle \right)^2$$

where $$A$$ and $$B$$ are some observables. If $$A$$ and $$B$$ do not commute then they cannot have a complete set of common eigenfunctions and this is given as an exercise to prove in my textbook. My question is can they even share a single eigenvector?

If the wavefunction points along the common eigenvector, wouldn't both standard deviations be equal to zero and the equation not be satisfied, as the left-hand side of the inequality would be equal to zero, and the right side would be a positive number (as the operators don't commute). But my textbook seems to imply that non-commuting operators can share an eigenvector.

It is possible for two non commuting operators to have an eigenvector in common.

Consider two operators $$A$$ and $$B$$, that in some basis can be written as

$$A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{pmatrix}$$

$$B = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}$$

$$AB-BA = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix} \neq 0$$

Yet, the vector $$\textbf{v} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ is a common eigenvector.

But the inequality is still satisfied, because on the RHS there is the expected value of $$[A,B]$$ over the state. $$\textbf{v}$$ is an eigenstate of $$[A,B]$$ with eigenvalue 0, therefore the expected value is 0 and the inequality holds.

This is a general result. If two operators share a common eigenvector, it will be also an eigenvector of their commutators, with eigenvalue 0.

Proof: If $$A\textbf{v} = a\textbf{v}$$ and $$B\textbf{v} = b\textbf{v}$$, then $$(AB-BA)\textbf{v} = ab\textbf{v}-ba\textbf{v} = 0$$