 Solution

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.