 Solution

Let $U(t)=e^{-iHt/\hslash }$ and take an observable with no explicit time dependence. In the Schrödinger picture, ${\psi }_S(t)=U(t){\psi }_0$ and $A_S=A$. In the Heisenberg picture, ${\psi }_H={\psi }_0$ and $A_H(t)=U(t{)}^{†}AU(t)$. For the interaction picture choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and defineThen $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$ and ${\dot{A}}_I=(i/\hslash )[H_0,A_I]$. Unitarity gives the identical expectation in all three descriptions:Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.

Differentiate $U^{†}AU$ to obtain the Heisenberg equation of motionFor an explicitly time-dependent observable there is additionally $U^{†}({\partial }_{t}A)U$. Thus $H_H=H$ is constant. In a normalized energy eigenstate, $⟨[H,A_H]⟩=E⟨A_H⟩-E⟨A_H⟩=0$, so the expectation of its Heisenberg derivative vanishes whenever the products are defined.

For the harmonic oscillator, the canonical commutator givesThe last identity follows from the operator product rule without commuting $X$ and $P$. Although $XP$ is not itself self-adjoint, the same calculation can be applied to the self-adjoint dilation observable $(XP+PX)/2$ and gives the same derivative.

Stationarity of the three expectations gives $⟨P⟩=⟨X⟩=0$ and $⟨P^2⟩=m^2{\omega }^2⟨X^2⟩$. Substitution into $E=⟨P^2⟩/(2m)+m{\omega }^2⟨X^2⟩/2$ yields the quantum virial theoremThe vanishing means make the variances equal to the second moments, soThis satisfies the position-momentum Heisenberg uncertainty relation, with equality only for the ground state.