## What do we know about the semigroup $e^{it\sqrt{-\Delta}}$

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I'm very interested in the properties of the semigroup $e^{it\sqrt{-\Delta}}$, it may has some fundamental differences(such as the kernel) with the well-known schrodinger semigroup $e^{it\Delta}$.

Any properties (or references or books) that related this semigroup are appreciated.

Thanks!

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Try putting "heat semigroup" in google and you'll get lots relevant references. – André Henriques Aug 12 at 13:51
Are you going to accept the answer? – timur Oct 20 at 13:21

## 1 Answer

The wave operator decomposes as $$\partial_t^2-\Delta = (\partial_t-i\sqrt{-\Delta})(\partial_t+i\sqrt{-\Delta}),$$ so you can think of $e^{it\sqrt{-\Delta}}$ as solving a "half of" the wave equation. In particular, it has a finite propagation speed. This can also be seen from the dispersion relation $\omega = |\xi|$, where $\omega$ and $\xi$ are the Fourier variables for $t$ and $x$, respectively. On the other hand, the Schrödinger propagator $e^{it\Delta}$ has the dispersions relation $\omega=|\xi|^2$, which makes it genuinely dispersive, i.e., the propagation speed depends on the frequency.

Note that $e^{it\Delta}$ is not the heat semigroup, which the other answers and comments seem to suggest.

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The downvoter care to comment? – timur Aug 13 at 0:23
You're right, $e^{it\Delta}$ isn't the heat semigroup. Editing my answer. – Nik Weaver Aug 13 at 3:14
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