 Solution

For a normalized state and a self-adjoint operator, the expectation value and dispersion areUse the physics convention that the inner product is conjugate-linear in its first argument, and assume the displayed products and second moments are defined. Self-adjointness gives $\overline{⟨\psi |A\psi ⟩}=⟨A\psi |\psi ⟩=⟨\psi |A\psi ⟩$, so the expectation is real. MoreoverIt vanishes exactly when $\psi$ is an eigenstate, with eigenvalue its expectation.

The product rule gives $XPf=-i\hslash x{f}^{\prime }$ and $PXf=-i\hslash (f+x{f}^{\prime })$, hence $\boxed{[X,P]f=i\hslash f}$ on their common differentiable domain. For general operators, put $z=⟨A\psi |B\psi ⟩$. ThenThe Cauchy-Schwarz inequality yieldsApply this to centered position and momentum operators, whose commutator is unchanged. Their squared norms are the variances, givingEquality requires both proportionality in Cauchy-Schwarz and a purely imaginary inner product. Thus for real means $x_0,p_0$,with real $\gamma$. Solving this first-order equation givesNormalizability on the real line forces $\gamma >0$. Writing ${\sigma }^2=\hslash /(2\gamma )$, all normalized equality states are thereforeThey have $\Delta X=\sigma$ and $\Delta P=\hslash /(2\sigma )$. A nonconstant quadratic phase would increase the product, so the full family here has a real positive Gaussian width and only a linear spatial phase.