Wigner pairing and marginals

The map $(x,y)\mapsto ((x+y)\mathrm{/}2,x-y)$ is an invertible linear map with absolute determinant $1$. The product of $a$, $u$, and $\bar{u}$ in the Weyl pairing is Schwartz in the transformed variables, so it is absolutely integrable. Fubini [F5] and change of variables [F8] give $⟨{\mathrm{Op}}_{ϵ}^W(a)u,u⟩=(2\pi ϵ{)}^{-d}\int e^{ir\cdot \xi \mathrm{/}ϵ}a(q,\xi )u(q-r\mathrm{/}2)\overline{u(q+r\mathrm{/}2)}dqdrd\xi \mathrm{.}$ Applying [F8] to the reflection $y=-r$ turns the last two factors and the exponential into the defining integrand of $W_{ϵ}[u]$; hence the pairing identity follows.

The function $g(x,y):=u(x+y\mathrm{/}2)\overline{u(x-y\mathrm{/}2)}$ is Schwartz on ${\mathbb{R}}^{2d}$, because $(x,y)\mapsto (x+y\mathrm{/}2,x-y\mathrm{/}2)$ is invertible linear and products of Schwartz functions are Schwartz. Its full negative-sign $2\pi$ Fourier transform $\widehat{g}(\eta ,\zeta )$ is Schwartz by [F3]; applying inverse Fourier transformation in $\eta$ by [F4] gives the partial transform $\int e^{-2\pi iy\cdot \zeta }g(x,y)dy$, Schwartz jointly in $(x,\zeta )$. The desired partial transform with phase $e^{-iy\cdot {\zeta }^{\mathrm{\prime }}}$ is this function at $\zeta ={\zeta }^{\mathrm{\prime }}\mathrm{/}(2\pi )$; frequency dilation preserves Schwartz space. Replacing ${\zeta }^{\mathrm{\prime }}$ by $\xi \mathrm{/}ϵ$ and multiplying by $(2\pi ϵ{)}^{-d}$ therefore gives $W_{ϵ}[u]\in \mathcal{S}({\mathbb{R}}^{2d})\subset L^1$. Conjugating the defining integral and applying [F8] to the reflection $y\mapsto -y$ returns the same integral, so $W_{ϵ}[u]$ is real.

Fix $x$ and set $f_x(y):=u(x+y\mathrm{/}2)\overline{u(x-y\mathrm{/}2)}$. The integrand ${\widehat{f}}_x$ is Schwartz and hence in $L^1$. Apply [F8] to the positive scaling $\eta \mapsto \xi =ϵ\eta$, whose Jacobian is ${ϵ}^d$, to obtain the substitution $\eta =\xi \mathrm{/}ϵ$ and $d\xi ={ϵ}^{d}d\eta$. Write ${\widehat{f}}_x(\eta )=\int e^{-iy\cdot \eta }{f}_x(y)dy$, the rescaled negative-sign $2\pi$ transform. Then $\int W_{ϵ}[u](x,\xi )d\xi =(2\pi {)}^{-d}\int \widehat{f_x}(\eta )d\eta =f_x(0)=\mathrm{\mid }u(x){\mathrm{\mid }}^2,$ where the middle equality follows from [F4] at zero after the change from the library frequency $\eta \mathrm{/}(2\pi )$ to $\eta$, also covered by [F8]. All integrals are absolutely convergent because $f_x,\widehat{f_x}\in \mathcal{S}$.

For the momentum marginal, $u\in \mathcal{S}\subseteq L^1$, so the double integral in $x,y$ is absolutely integrable by Fubini [F5]. The affine change $(x,y)\mapsto (q,r)=(x+y\mathrm{/}2,x-y\mathrm{/}2)$ has absolute determinant $1$ and $q-r=y$; [F8] therefore gives $\int W_{ϵ}[u](x,\xi )dx=(2\pi ϵ{)}^{-d}\int e^{-i(q-r)\cdot \xi \mathrm{/}ϵ}u(q)\overline{u(r)}dqdr=\mathrm{\mid }({\mathcal{F}}_{ϵ}u)(\xi ){\mathrm{\mid }}^2\mathrm{.}$ The last equality is the product of the scaled Fourier integral with its complex conjugate.

To see that nonnegativity fails, set $v_{ϵ}(x)=\sqrt{2\mathrm{/}ϵ}(\pi ϵ{)}^{-1\mathrm{/}4}x{e}^{-x^2\mathrm{/}(2ϵ)}$. Substituting $x=\sqrt{ϵ}s$ and applying [F6] at Gaussian parameter $1\mathrm{/}4$ gives $\int e^{-s^2}ds=\sqrt{\pi }$; one integration by parts [F7] then gives $\int s^2{e}^{-s^2}ds=\sqrt{\pi }\mathrm{/}2$, hence $\int x^2{e}^{-x^2\mathrm{/}ϵ}dx=\sqrt{\pi }{ϵ}^{3\mathrm{/}2}\mathrm{/}2$ and $\mathrm{\parallel }{v}_{ϵ}{\mathrm{\parallel }}_2=1$. At the origin, $W_{ϵ}[v_{ϵ}](0,0)=-(2\pi ϵ{)}^{-1}\frac{1}{2ϵ\sqrt{\pi ϵ}}\int y^2{e}^{-y^2\mathrm{/}(4ϵ)}dy=-\frac{1}{\pi ϵ}<0.$ Here the Gaussian second moment is obtained from [F6] at $\xi =0$ by one integration by parts [F7]. Thus $W_{ϵ}[v_{ϵ}]$ is a normalized Schwartz Wigner distribution with a negative value.