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I've been working on Bourgain's paper 'Moment inequalities for trigonometric polynomials with spectrum in curved hypersurfaces' for my Master's thesis and everything was going great until I reached the last section where he begins with applications on the Laplacian; and introduces $e^{it\Delta} $, the periodic Schrödinger group ...

I realize this might not be the best place to ask this; and I tried searching Bourgain's earlier papers for a definition to no avail, so would someone please tell me what $e^{it\Delta} $ is?

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    $\begingroup$ I would imagine it's operator notation. $e^x = 1 + x + x^2/2 + \cdots $ so $e^{it\nabla} = 1 + it \nabla - t^2/2 \nabla^2 + \cdots $ $\endgroup$
    – Ryan Budney
    Commented Mar 22, 2013 at 17:57
  • $\begingroup$ Whoops, that should have been $\Delta$ rather than $\nabla$. $\endgroup$
    – Ryan Budney
    Commented Mar 22, 2013 at 17:58
  • $\begingroup$ Thanks a lot! I figured I should write it that way, but I guess the name caught me a bit off guard... $\endgroup$
    – Sara
    Commented Mar 22, 2013 at 18:09
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    $\begingroup$ Probabilistically the exponent of a Laplacian (=the Markov geberator of the Brownian motion) is just the Markov transition operator. You may think of this as of a transition matrix. Of course, for that you need to take $\Delta$ the negative definite (I.e. possibly change the sign) $\endgroup$
    – Leonid Petrov
    Commented Mar 22, 2013 at 18:29
  • $\begingroup$ This is an instance of Stone's theorem en.wikipedia.org/wiki/… $\endgroup$
    – Alain Valette
    Commented Mar 23, 2013 at 9:07

2 Answers 2

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$e^{it\Delta}$ is the Fourier multiplier $e^{-4it\pi^2\vert D\vert^2}$, i.e. the operator defined by $$ (e^{it\Delta} u)(x)=\int_{\mathbb R ^d} e^{2i\pi x \xi}e^{-4it\pi^2\vert \xi\vert^2}\hat u(\xi) d\xi. $$ It is also the convolution with $E(t)$, say for $t>0$, $$ E(t)(x)=e^{-i(d-2)\pi/4}(4\pi t)^{-d/2}e^{i\vert x\vert^2 /(4t)},\quad (e^{it\Delta} u)(x)=(E(t)\ast u)(x). $$ The Fourier multiplier definition is the simplest.

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  • $\begingroup$ Thanks Bazin! The Fourier multiplier definition is EXACTLY what I was looking for. $\endgroup$
    – Sara
    Commented Mar 22, 2013 at 23:27
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Heuristically it is what Ryan Budney suggests, but of course it does not make sense as a power series, only for so called analytic vectors (which are dense though in this case).

Usually you define this via the spectral theory of selfadjoint operators, see for example

Weidmann: Linear operators in Hilbert space, Section 7.6.

The construction to show that the spectral theorem in this case reduces to the operator being a Fourier multiplier is described in Chapter 10.

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  • $\begingroup$ Well, the Fourier multiplier is in this case just an other name for the spectral theorem. $\endgroup$
    – András Bátkai
    Commented Mar 23, 2013 at 8:19

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