quantum mechanics - The explicit solution of the time-dependent Schrödinger equation for a free particle that starts as a delta function

A previous thread discusses the solution of the time-dependent Schrödinger equation for a massive particle in one dimension that starts off in the state $\Psi(x,0) = \delta(x)$. This can easily be solved in the momentum representation, giving the solution as $$ \Psi(x,t) = \frac{1}{2 \pi} e^{i\frac{m}{2\hbar t}x^2} \int_{-\infty}^{\infty} e^{-i\frac{\hbar t}{2m}(k-\frac{m}{\hbar t}x)^2} \operatorname{d}k, $$ and this Fourier transform to the position representation is a Fresnel integral that can be explicitly integrated to give the explicit solution $$ \Psi(x,t)=\begin{cases}\delta(x) & t=0,\\ \sqrt{\frac{m}{2\pi\hbar |t|}} e^{-i\,\mathrm{sgn}(t)\pi/4} \exp\left[i\frac{mx^2}{2\hbar t}\right] &t\neq 0. \end{cases} \tag{$*$} $$

Now, this procedure can sometimes seem a bit back-handed, and it leaves a lingering doubt of whether the solution in $(*)$ actually satisfies the differential equation $$ i\hbar\frac{\partial}{\partial t}\Psi(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t) \tag S $$ in some suitable sense.

While it might seem to be just a case of filling in the gaps, as a function of time, $\Psi(x,t)$ is highly irregular at $t=0$, since it contains a singularity in the factor of $e^{-i\,\mathrm{sgn}(t)\pi/4}/\sqrt{|t|}=1/\sqrt{i\:\!t}$, and the exponential $$ \exp\left(i\frac{mx^2}{2\hbar t}\right) = \cos\left(\frac{mx^2}{2\hbar}\frac{1}{t}\right) +i\sin\left(\frac{mx^2}{2\hbar}\frac{1}{t}\right) $$ oscillates infinitely fast at $t\to 0^\pm$, so the behaviour of the solution at the $t=0$ line is highly irregular (and indeed it has an essential singularity). This is to be expected to a degree: the initial condition $\Psi(x,0) = \delta(x)$ is a distribution, and the Schrödinger equation calls, at the very least, the second derivative of that delta function, so to the extent that the Schrödinger equation holds, it will be only in some kind of distributional sense; this will probably be a bit challenging but otherwise it should be possible.

So, just to fill in the gaps: in what sense, and without referring to the momentum representation, is $(*)$ a solution of $(\mathrm{S})$?

Answer

Sketched proof: If we define a regularized distribution

$$ \begin{align}\Psi_{\epsilon}[f;t] ~&:=~\iint_{\mathbb{R ^2}} \!\frac{\mathrm{d}x~\mathrm{d}k}{2\pi}~f(x) \exp\left\{ ikx-\frac{\hbar k^2}{2m}(\epsilon +it)\right\} \cr~&=~\sqrt{\frac{m}{2\pi\hbar(\epsilon +it)}}\int_{\mathbb{R}} \!\mathrm{d}x~f(x) \exp\left\{ -\frac{m x^2 }{2\hbar(\epsilon +it)}\right\} ,\quad t~\in~\mathbb{R},\quad \epsilon~\in~\mathbb{R}_+, \tag{A}\end{align}$$

for a spatial test function $f$, one may show firstly via Lebesgue's dominated convergence theorem that $\Psi_{\epsilon}[f;t]$ becomes the Dirac delta distribution

$$ \lim_{\epsilon\to 0^+}\lim_{t\to 0}\Psi_{\epsilon}[f;t] ~=~ \delta[f]~:=~f(0) \quad\text{for}\quad t~\to~ 0, \tag{B}$$

and secondly that the regularized $\Psi_{\epsilon}[f;t]$ satisfies TDSE for $t\in\mathbb{R}$ and $\epsilon\in\mathbb{R}_+$. Here the spatial derivative $$\Psi_{\epsilon,xx}[f;t]~:=~\Psi_{\epsilon}[f_{xx};t]\tag{C}$$ is defined in the usual distribution sense.