# Infinite propagation speed for the Schrodinger operator

Question related to: On the propagation of singularities in PDE and Hypoellipticity and singular support.

in what sense is to interpret the sentence the schrodinger operator has infinite propagation speed. And what about the fact that if there is a singularity at $t=0$, then at $t=\epsilon$ there isn't anymore?

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## 1 Answer

Consider the Schrodinger equation with an initial condition supported in a bounded region. At any positive $t$, the wave function is nonzero arbitrarily far away. In that sense, the particle can travel (albeit with small probability) at arbitrarily high speeds.

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Ok, thank you, that explains very well the infinite speed. What about singularities? Is it true that if the initial condition is not smooth, the solution becomes smooth 'istantaneously'? – balestrav Jun 28 '12 at 19:47
It may do so. Or it may start out smooth and then get a singularity at some later time. – Robert Israel Jun 28 '12 at 21:54
Could you provide some examples of this behavior? – balestrav Jun 28 '12 at 22:11