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.


1 Answer 1


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 $$(AB-BA)\textbf{v} = ab\textbf{v}-ba\textbf{v} = 0$$

  • $\begingroup$ Hi Prallax, can two non commuting operators have all the eigenstates common? $\endgroup$
    – Kashmiri
    Commented Feb 1, 2022 at 10:50
  • 1
    $\begingroup$ @Kashmiri no, if two (hermitian) operators share all their eigenvectors, then they are both diagonal in some basis. Therefore they commute $\endgroup$
    – Prallax
    Commented Feb 2, 2022 at 19:59

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