There is no measurement uncertainty iff the eigenstate is in the kernel of the operator

i.e. ${\Delta }_{\psi }A=0\iff A\psi =0$, where

• A is an Operator
• $\psi$ is a Quantum state.
• ${\Delta }_{\psi }A$ is the Measurement uncertainty

Proof

$\Leftarrow$ is trivial. To prove $\Rightarrow$, we first define $B:=A-{⟨A⟩}_{\psi }$, where ${⟨\cdot ⟩}_{\psi }$ is the Expectation operator, so that

• ${⟨B⟩}_{\psi }=0$
• ${\Delta }_{\psi }B={\Delta }_{\psi }A$

$0={\Delta }_{\psi }B={⟨B^2⟩}_{\psi }=⟨\psi ,B^2\psi ⟩=⟨B\psi ,B\psi ⟩=\Vert B\psi \Vert$. Hence $A\psi =(B+{⟨A⟩}_{\psi })\psi =B\psi +\lambda \psi =\lambda \psi$, where we defined $\lambda :={⟨A⟩}_{\psi }$. Finally, $0={\Delta }_{\psi }A=⟨\psi ,A\psi ⟩=⟨\psi ,\lambda \psi ⟩=\lambda \Vert \psi \Vert =\lambda$, so $A\psi =\lambda \psi =0\psi =0$.

For this prove we used the fact that Quantum operators are self-adjoint, and that $\psi$ is normalised.