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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.

Take a normalized vector in the required common operator domains. Expanding the squared norm and using self-adjointness givesThus its final term is $\hslash t⟨C⟩$. Set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=⟨C⟩$. If $b>0$, minimizing the nonnegative quadratic at $t_{\ast }=-\hslash c/(2b)$ givesEquality holds precisely when the minimum norm vanishes, that is $(A-i{t}_{\ast }B)\psi =0$. If $b=0$, however, then $B\psi =0$ and $c=0$, so equality is automatic even when $A\psi \ne 0$. The printed equality criterion needs this zero-vector exception. For example $A=I$, $B=0$, $C=0$ saturates the inequality but admits no such real $t$. The exact general criterion is linear dependence of $A\psi ,B\psi$ together with vanishing real part of their inner product, as in the uncentered second-moment commutator inequality.

Apply the same calculation to $A-⟨A⟩I$, $B-⟨B⟩I$, whose commutator is unchanged. This gives the Robertson uncertainty principleIf $\psi$ is an eigenvector of $A$, its dispersion is zero; finite dispersion of $B$ then forces $\boxed{⟨C⟩=0}$.

In the position representation, $X\psi =x\psi$ and $P\psi =-i\hslash {\psi }^{\prime }$. The product rule gives $[P,X]\psi =-i\hslash \psi$, hence $\boxed{[P,X]=-i\hslash I}$. Since $V(X)$ commutes with $X$,For an energy eigenstate, $⟨[H,X]⟩=E⟨X⟩-E⟨X⟩=0$, so $\boxed{⟨P⟩=0}$. For the oscillator $[H,P]=i\hslash m{\omega }^{2}X$, and the same argument gives $\boxed{⟨X⟩=0}$. These expectation identities require the stated finite moments and commutator domains; formal position eigenvectors are not normalizable exceptions.

Apply the Robertson uncertainty principle to each angular-momentum pair:If $\psi$ is an eigenvector of $L_3$, then $\Delta L_3=0$. The inequalities for $(L_2,L_3)$ and $(L_3,L_1)$ force $\boxed{⟨L_1⟩=⟨L_2⟩=0}$, assuming the relevant moments exist.

Set $u=A\psi$, $v=B\psi$ and $z=⟨u,v⟩$. The self-adjointness of $A,B$ impliesThe Cauchy-Schwarz inequality followed by $|z{|}^2\ge (\mathrm{Im}z{)}^2$ givesBoth inequalities must be equalities. The exact conditions are $A\psi$ and $B\psi$ are linearly dependent, and $\mathrm{Re}⟨A\psi ,B\psi ⟩=0$. Equivalently, equality holds if either vector is zero; when both are nonzero it holds precisely whenThe zero-vector cases matter: they saturate the inequality but cannot all be expressed by solving for $B\psi$ as a multiple of $A\psi$. The real-part condition can also be written ${\mathbb{E}}_{\psi }(AB+BA)=0$, using the anticommutator. This is the uncentered second-moment commutator inequality.