Exact harmonic-oscillator transport of a Gaussian Wigner density

The scaled oscillator eigenfunctions are polynomial multiples of $e^{-x^2\mathrm{/}(2ϵ)}$. To verify the core hypothesis of Fixed-time Weyl Egorov estimate for subquadratic potentials, take $f\in \mathcal{S}(\mathbb{R})$ and set $Lf=-{ϵ}^2{f}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2+x^{2}f\mathrm{/}2\in L^2$. Integration by parts against each Hermite eigenfunction ${\psi }_n^{ϵ}$, whose polynomial-Gaussian tails make the endpoint products vanish, gives $⟨Lf,{\psi }_n^{ϵ}⟩=⟨f,L{\psi }_n^{ϵ}⟩=ϵ(n+\frac{1}{2})⟨f,{\psi }_n^{ϵ}⟩\mathrm{.}$ Parseval for the complete basis [F1] yields ${\sum }_{n\ge 0}{ϵ}^2(n+\frac{1}{2}{)}^2\mathrm{\mid }⟨f,{\psi }_n^{ϵ}⟩{\mathrm{\mid }}^2=\mathrm{\parallel }Lf{\mathrm{\parallel }}_2^2<\mathrm{\infty }\mathrm{.}$ Thus $f\in D(H_{ϵ})$. The coefficients of $H_{ϵ}f$ in the Hermite basis are $ϵ(n+\frac{1}{2})⟨f,{\psi }_n^{ϵ}⟩$, the same coefficients as $Lf$; completeness of that basis gives $H_{ϵ}f=Lf$. The finite Hermite span is contained in $\mathcal{S}$ and is a graph core by [F1]; since $\mathcal{S}\subset D(H_{ϵ})$, $\mathcal{S}$ is also a core. This checks every operator hypothesis of [F4].

The Gaussian integral [F5] and the positive scaling in [F8] give $\mathrm{\parallel }{\psi }_{q,p}^{ϵ}{\mathrm{\parallel }}_2^2=(\pi ϵ{)}^{-1\mathrm{/}2}\int e^{-(x-q{)}^2\mathrm{/}ϵ}dx=1$. In its Wigner integral, the product of the two Gaussian factors is $(\pi ϵ{)}^{-1\mathrm{/}2}\exp \text{\hspace{0.17em}⁣}(-\frac{(x-q{)}^2}{ϵ}-\frac{y^2}{4ϵ})\exp \text{\hspace{0.17em}⁣}\left(\frac{ipy}{ϵ}\right)\mathrm{.}$ Multiplying by $e^{-iy\xi \mathrm{/}ϵ}$ and applying [F3] with $\eta =\xi -p$ gives $W_{ϵ}[{\psi }_{q,p}^{ϵ}](x,\xi )=(2\pi ϵ{)}^{-1}(\pi ϵ{)}^{-1\mathrm{/}2}{e}^{-(x-q{)}^2\mathrm{/}ϵ}2\sqrt{\pi ϵ}{e}^{-(\xi -p{)}^2\mathrm{/}ϵ},$ which is the stated nonnegative density. Its integral is $1$ by [F5] and [F8]. For its phase-space moments, apply [F8] to $x=q+\sqrt{ϵ}s$ and $\xi =p+\sqrt{ϵ}r$. The centered first moments vanish because $s{e}^{-s^2}$ and $r{e}^{-r^2}$ are odd; the cross-moment vanishes for the same reason. Integration by parts [F5] gives $\int s^2{e}^{-s^2}ds=\frac{1}{2}\int e^{-s^2}ds$, so each centered variance is $ϵ\mathrm{/}2$. This verifies the mean and covariance definitions in the Statement. For the operator uncertainties, direct differentiation gives $P{\psi }_{q,p}^{ϵ}=(p+i(x-q)){\psi }_{q,p}^{ϵ}$, while $Q{\psi }_{q,p}^{ϵ}=x{\psi }_{q,p}^{ϵ}$. Thus ${\bar{Q}}_{\psi }=q$, ${\bar{P}}_{\psi }=p$, and the two squared standard deviations are both $\int (x-q{)}^2\mathrm{\mid }\psi {\mathrm{\mid }}^{2}dx=ϵ\mathrm{/}2$ by the same Gaussian integration by parts. To prove the lower bound for every normalized $\phi \in \mathcal{S}$, first note that $Q$ and $P$ are symmetric on $\mathcal{S}$: for $P$ this is [F5] applied to decaying Schwartz factors, and for $Q$ it follows from real multiplication. Hence ${\bar{Q}}_{\phi },{\bar{P}}_{\phi }\in \mathbb{R}$. Put $A=Q-{\bar{Q}}_{\phi }I$ and $B=P-{\bar{P}}_{\phi }I$. Cauchy–Schwarz [F6], the commutator $[A,B]=[Q,P]=iϵI$ from [F2], and symmetry give ${\mathrm{\Delta }}_{\phi }Q{\mathrm{\Delta }}_{\phi }P\ge \mathrm{\mid }⟨A\phi ,B\phi ⟩\mathrm{\mid }\ge \mathrm{\mid }\mathrm{Im}⟨A\phi ,B\phi ⟩\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }⟨[A,B]\phi ,\phi ⟩\mathrm{\mid }=\frac{ϵ}{2}\mathrm{.}$ Indeed, under the first-variable-linear convention, $⟨A\phi ,B\phi ⟩=⟨BA\phi ,\phi ⟩$ and its conjugate is $⟨AB\phi ,\phi ⟩$. Thus the Gaussian attains the lower bound.

The classical Hamilton equations for $h(x,\xi )=(x^2+{\xi }^2)\mathrm{/}2$ solve to ${\mathrm{\Phi }}_t(x,\xi )=(x\cos t+\xi \sin t,\xi \cos t-x\sin t)\mathrm{.}$ This matrix is orthogonal with determinant $1$, and ${\mathrm{\Phi }}_t(q,p)=(q_t,p_t)$. For $a\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$, [F4] gives $⟨W_t,a⟩=⟨{\mathrm{Op}}_{ϵ}^W(a\circ {\mathrm{\Phi }}_t){\psi }_{q,p}^{ϵ},{\psi }_{q,p}^{ϵ}⟩={\int }_{{\mathbb{R}}^2}a({\mathrm{\Phi }}_{t}z)W_{ϵ}[{\psi }_{q,p}^{ϵ}](z)dz,$ where the last equality is [F2]. The integrand is in $L^1$ because $a$ is compactly supported and the Gaussian is integrable. Apply [F8] to the rotation $z^{\mathrm{\prime }}={\mathrm{\Phi }}_{t}z$; its determinant is $1$, so the resulting density is $W_{ϵ}[{\psi }_{q,p}^{ϵ}]({\mathrm{\Phi }}_{-t}{z}^{\mathrm{\prime }})$. Orthogonality gives $\mathrm{\mid }{\mathrm{\Phi }}_{-t}(x,\xi )-(q,p){\mathrm{\mid }}^2=\mathrm{\mid }(x,\xi )-{\mathrm{\Phi }}_t(q,p){\mathrm{\mid }}^2,$ so this density is exactly the displayed $W_t$. The same change of variables sends its mean $(q,p)$ to $(q_t,p_t)$ and transforms covariance as $R_t(ϵI_2\mathrm{/}2)R_t^T=ϵI_2\mathrm{/}2$, since $R_t$ is orthogonal. This proves the asserted exact transport and covariance for every compactly supported smooth test symbol and hence for the distributional Wigner functional defined in the Statement.