= Solution
Let $U(t)=e^{-iHt/\hbar}$ 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 $A_H(t)=U(t)^\dagger A U(t)$. For the <interaction picture> choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-iH_0t/\hbar}$, and define
$$
\psi_I=U_0^\dagger\psi_S,\qquad A_I=U_0^\dagger A U_0,\qquad V_I=U_0^\dagger VU_0.
$$
Then $i\hbar\dot\psi_I=V_I\psi_I$ and $\dot A_I=(i/\hbar)[H_0,A_I]$. Unitarity gives the identical <expectation> in all three descriptions:
$$
\boxed{\langle\psi_S,A\psi_S\rangle=\langle\psi_0,A_H\psi_0\rangle=\langle\psi_I,A_I\psi_I\rangle.}
$$
Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.
Differentiate $U^\dagger AU$ to obtain the <Heisenberg equation of motion>
$$
\boxed{\dot A_H=\frac{i}{\hbar}[H,A_H].}
$$
For an explicitly time-dependent observable there is additionally $U^\dagger(\partial_tA)U$. Thus $H_H=H$ is constant. In a normalized <energy> <eigenstate>, $\langle[H,A_H]\rangle=E\langle A_H\rangle-E\langle A_H\rangle=0$, so the <expectation> of its Heisenberg derivative vanishes whenever the products are defined.
For the <harmonic oscillator>, the canonical <commutator> gives
$$
\boxed{\dot X=P/m,\qquad\dot P=-m\omega^2X,\qquad\frac d{dt}(XP)=P^2/m-m\omega^2X^2.}
$$
The 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 $\langle P\rangle=\langle X\rangle=0$ and $\langle P^2\rangle=m^2\omega^2\langle X^2\rangle$. Substitution into $E=\langle P^2\rangle/(2m)+m\omega^2\langle X^2\rangle/2$ yields the <quantum virial theorem>
$$
\boxed{\left\langle\frac{P^2}{2m}\right\rangle=\frac E2=\frac12m\omega^2\langle X^2\rangle.}
$$
The vanishing means make the <variances> equal to the second moments, so
$$
\boxed{(\Delta P)^2=mE,\qquad(\Delta X)^2=\frac E{m\omega^2},\qquad\Delta X\Delta P=\frac E\omega=(n+\tfrac12)\hbar\geq\hbar/2.}
$$
This satisfies the position-momentum <Heisenberg uncertainty relation>, with equality only for the <ground state>.
Back to article page
