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When reading about Schrödinger's fundamental solution in 1D,

$$u(t,x)=\frac{1}{\sqrt{4\pi it}} \int_\mathbb{R} u_0(y) e^{\frac{i(x-y)^2}{4t}}dy$$

the author says thus Schrödinger evolution is instantaneously smoothing for localised data: if $u_0$ is so localised as to be absolutely integrable but is not smooth, the previous shows that at all other times $t>0$, $u(t)$ is smooth (but not localised).

I see that when differentiating under the integral sign, if $u_0$ is compactly supported (not necessarily smooth), the integral converges and thus the interchange is legitimate.

  1. What can localised mean apart from compactly supported? Schwartz space would do, of course, but what non-smooth initial data are considered localised?

  2. I'd like a good interpretation of the following too, please: high frequencies (Fourier space) travel very fast (we know dispersion is proportional to frequency in Schrödinger equation) and radiate quickly away from the origin (of physical space?) where they are initially localised, leaving only the low frequencies, which are always smooth, to remain near the origin. I see this when breaking an initial data in a finite number of waves which "sum" to it, but it seems to me that smoothness away from the origin wouldn't be guaranteed.

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What is the source? – Did Jan 5 '13 at 19:27
"Nonlinear Dispersive Equations. Local and Global Analysis" Terence Tao (CBMS Number 106). – sheriff Jan 6 '13 at 0:40

I think this is wave-particle duality. This is the free particle with initial wave-function $u_0(x)$. If the particle was localized (i.e. "compactly supported"), the contributing frequencies should be spread out in momentum space. The time-evolved wavefunction $u(x,t)$ should have components that have gone arbitrarily far.

More rigorously, an integral trasnform is defined by $u_0(x) \mapsto \frac{1}{\sqrt{4\pi it}} \int_\mathbb{R} u_0(y) e^{\frac{i(x-y)^2}{4t}}dy$ As $t \to 0$ this is just dirac-delta. As $t\to \infty$ the wavefunction spreads over all of $\mathbb{R}$. For any positive time, the new wavefunction will be nonzero for arbitrarily large positions.

The smoothing has to do with convolving with a Gaussian as long $u_0(x)$ did not explode (probably it's in $L^2(\mathbb{R})$ or Schwartz space).

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Why should the contributing frequencies be spread out in momentum space? (mathematically). Could this not happen if $u_0$ is only Schwartz? And, why should the time-evolved components gone arbitrarily far? And why does that mean the new wavefunction will be nonzero everywhere? – sheriff Jan 7 '13 at 14:29

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