direct proof that schrodinger's equation kernel corresponds to delta-function initial value [closed]

Question

I want to show directly, that the kernel for the n-dimensional free linear schrodinger equation, if taken to time t=0, is dirac's $\delta $ function. I can show that the integral is constant, but it seems to me that it blows to infinity for all $x \in R^n $, rather then to go to zero for all $ x > 0 $

I should add that the proof I've seen for this question in the books and papers always "went around the bushes" with the Fourier transform. While correct, it doesn't show me really how come the "infinite phase" term make way for a $\delta$ function.

Answer 1

The Schrödinger equation is $ \frac1{i}\frac{\partial}{\partial{t}}-\Delta_{x}. $ It looks similar to the heat equation, but is in fact drastically different. We define a distribution $E$ by the following $$ \mathscr D(\mathbb R^{n+1})\mapsto \int_{0}^{+\infty} e^{-i(n-2)\frac\pi 4}(4\pi t)^{-n/2} \left(\int_{\mathbb R^n}\Phi(t,x)e^{i \frac{\vert{x}\vert^2}{4t}} dx \right) dt =\langle{E},{\Phi}\rangle. $$ $E$ is a distribution in $\mathbb R^{n+1}$ of order $\le n+2$ which is a fundamental solution to the Schrödinger equation. You may note that the integration with respect to $x$ must be done before the time integration for this bracket of duality to make sense. The best way to get this is Fourier transform the equation with respect to $x$, to get that the Fourier transform of $E$ is $v/i$ with $$ \partial_{t}v+i4\pi^2\vert\xi\vert^2 v=i\delta_{0}(t). $$ We see that we can take $ v(t,\xi)= iH(t) e^{-i4\pi^2 t\vert \xi\vert^2}. $ We have indeed $ \frac1iv(0_+,\xi)=1, $ i.e. $E(0,x)=\delta_0(x)$. There is indeed a subtlety in the determination of phase factor $$e^{-i(n-2)\frac\pi 4}$$but intuitively the vanishing of $E$ outside 0 is due to the strong oscillations outside $x=0$ of $e^{i \frac{\vert{x}\vert^2}{4t}}$. It is not easy to devise a rigorous statement without using Fourier transformation with respect to the space variables.