The driven oscillator: exact classical response of a coherent preparation and stationary fluctuations

[Free coherent rotation] By [L2] with $\tau =t-s$ and $w=z$: $\mathrm{\Phi }:=e^{-i\tau H_{\mathrm{o}\mathrm{s}\mathrm{c}}\mathrm{/}\mathrm{\hslash }}{\mathrm{\Psi }}_z=e^{-i\omega \tau \mathrm{/}2}{\mathrm{\Psi }}_{z(\tau )}$ with $z(\tau )=z{e}^{-i\omega \tau }$; hence $⟨Q\mathrm{\Phi },\mathrm{\Phi }⟩=\sqrt{2}\mathrm{\ell }\mathrm{Re}z(\tau )$ and $⟨P\mathrm{\Phi },\mathrm{\Phi }⟩=\frac{\sqrt{2}\mathrm{\hslash }}{\mathrm{\ell }}\mathrm{Im}z(\tau )$, while $\mathrm{\parallel }(Q-⟨Q⟩I)\mathrm{\Phi }{\mathrm{\parallel }}^2={\mathrm{\ell }}^2\mathrm{/}2$ and $\mathrm{\parallel }(P-⟨P⟩I)\mathrm{\Phi }{\mathrm{\parallel }}^2={\mathrm{\hslash }}^2\mathrm{/}(2{\mathrm{\ell }}^2)$.

[Displacement shifts the means, not the spreads] Let $\mathrm{\Phi }\in Y$ be unit and let $q,p\in \mathbb{R}$; $D:=D(q,p)$ is unitary and preserves $Y$, and $D^{\ast }QD=Q+qI$, $D^{\ast }PD=P+pI$ on $Y$ by [L1], so $⟨QD\mathrm{\Phi },D\mathrm{\Phi }⟩=⟨D^{\ast }QD\mathrm{\Phi },\mathrm{\Phi }⟩=⟨Q\mathrm{\Phi },\mathrm{\Phi }⟩+q,⟨PD\mathrm{\Phi },D\mathrm{\Phi }⟩=⟨P\mathrm{\Phi },\mathrm{\Phi }⟩+p,$ and unitarity gives $\mathrm{\parallel }QD\mathrm{\Phi }-⟨QD\mathrm{\Phi },D\mathrm{\Phi }⟩D\mathrm{\Phi }\mathrm{\parallel }=\mathrm{\parallel }(Q-⟨Q\mathrm{\Phi },\mathrm{\Phi }⟩I)\mathrm{\Phi }\mathrm{\parallel }$, so the spreads of $D\mathrm{\Phi }$ equal those of $\mathrm{\Phi }$.

[Both time orientations] For all $s,t\in [a,b]$ use oriented integrals $q(t;s)=(m\omega {)}^{-1}{\int }_s^t\sin (\omega (t-r))f(r)dr$ and $p(t;s)={\int }_s^t\cos (\omega (t-r))f(r)dr$, where ${\int }_s^t=-{\int }_t^s$ for $t<s$. Differentiating gives the declared equations and zero data in either direction; the energy $p^2\mathrm{/}(2m)+m{\omega }^2{q}^2\mathrm{/}2$ of the difference of two solutions is constant and initially zero, proving uniqueness. Set $S(t,s)={\int }_s^{t}f(r)q(r;s)\mathrm{/}2dr$ and $W(t,s)=e^{iS(t,s)\mathrm{/}\mathrm{\hslash }}D(q(t;s),p(t;s))e^{-i(t-s)H_{\mathrm{o}\mathrm{s}\mathrm{c}}\mathrm{/}\mathrm{\hslash }}$. The displacement identities of [L1] apply along every real $C^1$ trajectory, and the free group preserves $Y$; their product derivative on each $y\in Y$ is justified by expanding $D(t+h)F(t+h)y-D(t)F(t)y$ as $D(t+h)(F(t+h)-F(t))y+(D(t+h)-D(t))F(t)y$, using strong continuity and unit norm of $D$ and the fixed vector $F(t)y\in Y$. Writing $E=p^2\mathrm{/}(2m)+m{\omega }^2{q}^2\mathrm{/}2$, [L1] gives $i\mathrm{\hslash }{\mathrm{\partial }}_{t}W=[H_{\mathrm{o}\mathrm{s}\mathrm{c}}+(-p^{\mathrm{\prime }}-m{\omega }^{2}q)Q+(q^{\mathrm{\prime }}-p\mathrm{/}m)P+(E+(q{p}^{\mathrm{\prime }}-p{q}^{\mathrm{\prime }})\mathrm{/}2-S^{\mathrm{\prime }})I]W$. The differential equations give the $Q,P$ coefficients $-f,0$ and $E+(q{p}^{\mathrm{\prime }}-p{q}^{\mathrm{\prime }})\mathrm{/}2=fq\mathrm{/}2=S^{\mathrm{\prime }}$, so $W$ solves the strong equation with $W(s,s)=I$ on either side of $s$. The common-domain uniqueness in [L1] (or the constant-norm difference argument) identifies $W=U$ on $Y$, then on $L^2$ by density and unitarity. This is also the oriented extension proved in Explicit unbounded oscillator propagator Proof, the paragraph beginning “Let $v(t)$ and $w(t)$ be two $Y$-valued strong solutions”. For $a=b$, all integrals vanish and $W(a,a)=I$ directly.

[Total response] By step 1.3, $U(t,s){\mathrm{\Psi }}_z=e^{iS\mathrm{/}\mathrm{\hslash }}D(q(t;s),p(t,s))\mathrm{\Phi }$ with $\mathrm{\Phi }$ as in step 1.1, and the phase $e^{iS\mathrm{/}\mathrm{\hslash }}$ is irrelevant to means and centered norms; step 1.2 applied to the unit vector $\mathrm{\Phi }$ (unit by unitarity of the free evolution) and step 1.1 give $⟨Q⟩(t)=\sqrt{2}\mathrm{\ell }\mathrm{Re}\left(z{e}^{-i\omega (t-s)}\right)+q(t;s),⟨P⟩(t)=\frac{\sqrt{2}\mathrm{\hslash }}{\mathrm{\ell }}\mathrm{Im}\left(z{e}^{-i\omega (t-s)}\right)+p(t;s),$ with spreads $\mathrm{\ell }\mathrm{/}\sqrt{2}$ and $\mathrm{\hslash }\mathrm{/}(\mathrm{\ell }\sqrt{2})$ and product $\mathrm{\hslash }\mathrm{/}2$. These are clauses (a) and (b).

[Undriven check and scope] For $f\equiv 0$ the classical zero-data solution is $q\equiv p\equiv 0$ (uniqueness of the linear initial-value problem), so clause (a) reduces to the free rotation, as in [L2]; the computation used only the declared model and its cited mathematics, and no laboratory, damping or measurement claim entered. These are clauses (c) and (d).