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A pure state is a unit vector in a complex Hilbert space, with vectors differing by a global phase representing the same physical state. An observable is a self-adjoint operator, with measurement probabilities determined by its spectral projections. For vectors in the relevant operator domains,A squared norm vanishes exactly when its vector vanishes. Therefore zero dispersion is equivalent to $A\psi =E_{\psi }[A]\psi$.

In the position representation, $(X\psi )(x)=x\psi (x)$ and $(P\psi )(x)=-i\hslash {\psi }^{\prime }(x)$. On a common differentiable invariant domain,so $\boxed{[X,P]=i\hslash I}$.

For two observables set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=E_{\psi }[i[A,B]]$. For real $t$,If $b>0$, nonpositivity of its discriminant yields $\boxed{4ab\ge c^2}$. Equality holds exactly when the minimum squared norm vanishes, namelyThe degenerate case must also be included: if $B\psi =0$, then $b=c=0$ and equality holds for every $A\psi$. Thus the complete criterion is that $B\psi =0$, or that the displayed real-$t$ relation holds. This includes $A\psi =0$ by choosing $t=0$ when $b>0$.

Apply the inequality to centred operators $A=P-E_{\psi }[P]I$, $B=X-E_{\psi }[X]I$. Their squared norms are the variances and their commutator is $[P,X]=-i\hslash I$, givingAn eigenvector of either operator would have its corresponding dispersion zero, contradicting this positive lower bound if the other dispersion were finite and the stated domain conditions held. In the usual $L^2(\mathbb{R})$ representation there are in fact no normalizable exact position or momentum eigenvectors; their generalized eigenvectors are not counterexamples. The equality condition above is the uncentered second-moment commutator inequality.