How can you derive the spacetime Fourier transform of the free Schrodinger evolution rigorously?

Question

I'm trying to compute the spacetime Fourier transform of the free Schrodinger evolution. Consider $f\in L^2(\mathbb{R}^d)$ and $e^{it\Delta}f=:\mathcal{F}^{-1}(e^{-it|\xi|^2}\hat{f}(\xi))$ its free Schrodinger evolution. I wish to find its spacetime FT $\mathcal{F}_{t,x}(e^{it\Delta}f)$, defined in $\mathcal{S}'(\mathbb{R}\times\mathbb{R}^d)$ by $$ \langle \mathcal{F}_{t,x}(e^{it\Delta}f),\varphi \rangle=\langle e^{it\Delta}f,\mathcal{F}_{t,x}(\varphi) \rangle $$ for all $\varphi\in \mathcal{S}(\mathbb{R}\times\mathbb{R}^d)$, where \begin{equation} \mathcal{F}_{t,x}(\varphi)(\tau,\xi):=\int_x\int_t e^{-i(t\tau+x\cdot\xi)}\varphi(t,x)dtdx \text{ (1)} \end{equation} How can one go about this rigorously? I expect to find $$ \mathcal{F}_{t,x}(e^{it\Delta}f)(\tau,\xi)=\delta(\tau+|\xi|^2)\hat{f}(\xi) $$ interpreted as $$ \langle \delta(\tau+|\xi|^2)\hat{f}(\xi),\varphi\rangle=\int_{\{\tau=-|\xi|^2\}}\frac{\hat{f}(\xi)\varphi(\tau,\xi)}{2\sqrt{\tau^2+|\xi|^2}}d\xi d\tau $$ although I'm not completely sure if this is actually the right definition (or particularly rigorous in itself).

Part of my problem is in expressing $\mathcal{F}_{t,x}(e^{it\Delta}f)$ as an integral similar to (1) (since I am starting only from the dual definition of the FT) from which one could apply some oscillatory integral techniques.

Answer 1

I think I have an answer so I will post it here lest it be helpful to anyone else:

NB: Throughout I will use the notation $\tilde{u}=\mathcal{F}_{(t,x)}u$.

Suppose first $f\in \mathcal{S}(\mathbb{R}^d)$, set $u(x,t):=e^{it\Delta}f \in L^\infty_tL^2_x$. Fix $\chi\in C_c^\infty(\mathbb{R}\times \mathbb{R}^d)$ with $\chi(0,0)=1$ and set $\chi_\epsilon(t,x):=\chi(\epsilon t,\epsilon x)$. Then \begin{align*} \langle \tilde{u},\varphi\rangle\equiv&\langle u, \tilde{\varphi}\rangle\\ =&\lim_{\epsilon\to0}\langle u,\chi_{\epsilon}\tilde{\varphi}\rangle\\ =&\lim_{\epsilon\to0}\int_t\int_x u(x,t)\chi_{\epsilon}(x,t)\left(\int_\xi\int_\tau e^{-it\tau-ix\cdot\xi}\varphi(\xi,\tau)d\xi d\tau\right)dxdt\\ =&\lim_{\epsilon\to0}\int_\tau\int_\xi \varphi(\xi,\tau) \left(\int_t\int_x e^{-it\tau-ix\cdot\xi}u(x,t)\chi_{\epsilon}(x,t)dxdt\right)d\xi d\tau\\ =&\lim_{\epsilon\to0}\langle \widetilde{u\chi_{\epsilon}},\varphi\rangle \end{align*} where we were able to use Fubini's theorem thanks to the integrability of $\varphi$, $\chi_{\epsilon}$ and $u$. Now, \begin{align*} \int_t\int_x e^{-it\tau-ix\cdot\xi}e^{it\Delta}f(x,t)\chi_{\epsilon}(x,t)dxdt=&\int_{t,x}e^{-it\tau-ix\cdot\xi}\chi_{\epsilon}(x,t)\left(\frac{1}{(2\pi)^d}\int_{\lambda\in\mathbb{R}^d}e^{ix\cdot\lambda}e^{-it|\lambda|^2}\hat{f}(\lambda)d\lambda\right)dxdt\\ =&\frac{1}{(2\pi)^d}\int_\lambda \hat{f}(\lambda)\int_{x,t}e^{-ix\cdot(\xi-\lambda)}e^{-it(\tau+|\lambda|^2)}\chi_\epsilon(x,t)dxdt d\lambda\\ =&\frac{1}{(2\pi)^d}\int_\lambda \hat{f}(\lambda)\tilde{\chi}_\epsilon(\xi-\lambda,\tau+|\lambda|^2)d\lambda \end{align*} Note that $\widetilde{\chi}_{\epsilon}(\xi,\tau)=\epsilon^{-(1+d)}\widetilde{\chi}(\epsilon^{-1}\xi,\epsilon^{-1}\tau)$, so \begin{align*} \langle \tilde{u},\varphi\rangle=&\lim_{\epsilon\to0}\int_{\tau,\xi}\varphi(\xi,\tau)\left(\frac{1}{(2\pi)^d}\int_\lambda\hat{f}(\lambda)\epsilon^{-(d+1)}\tilde{\chi}(\epsilon^{-1}(\xi-\lambda),\epsilon^{-1}(\tau+|\lambda|^2))d\lambda\right) d\xi d\tau\\ =&\lim_{\epsilon\to0}\int_\lambda\hat{f}(\lambda)\epsilon^{-(d+1)}\left(\frac{1}{(2\pi)^d}\int_{\tau,\xi}\varphi(\xi,\tau)\tilde{\chi}(\epsilon^{-1}(\xi-\lambda),\epsilon^{-1}(\tau+|\lambda|^2))d\xi d\tau\right)d\lambda\\ =&\lim_{\epsilon\to0}\int_\lambda \hat{f}(\lambda)\left(\frac{1}{(2\pi)^d}\int_{\tau',\xi'}\varphi(\epsilon \xi'+\lambda,\epsilon\tau'-|\lambda|^2)\widetilde{\chi}(\xi',\tau')d\xi'd\tau'\right)d\lambda\\ =&2\pi\int_{\lambda}\hat{f}(\lambda)\varphi(-|\lambda|^2,\lambda)d\lambda\\ \equiv&\langle 2\pi\hat{f}(\lambda)\delta(\tau+|\lambda|^2),\varphi\rangle \end{align*} where we used the dominated convergence theorem and the fact that $\int_{\tau',\xi'}\tilde{\chi}(\xi',\tau')d\xi'd\tau'=(2\pi)^{d+1}\chi(0,0)=(2\pi)^{d+1}$ to take the limit.

We can then extend this to any $f\in L^2(\mathbb{R}^d)$ be density.

Remarks:

1. This is not quite the expected solution thanks to the factor of $2\pi$. Since the $\delta$-function actually has oscillatory integral representation $$ \delta_0=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}e^{ix\cdot\theta}d\theta $$ I think the factor of $2\pi$ is actually correct. Nonetheless if someone sees a mistake in the above or can confirm the factor of $2\pi$ I would be grateful!

2. Note this is not the exact same definition for the $\delta$ function as given in the question statement, however this definition seems more natural anyway, and avoids problems involving the zeros of the denominator. (For functions with non-vanishing gradient, I believe one could show the two definitions are the same by means of the co-area formula).