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I would expect them to be very similar, with differences on

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 quantitative level"half of" the wave equation. Both solve dispersive equationsIn particular, only it has a finite propagation speed. This can also be seen from the dispersion relations relation $\omega = |\xi|$, where $\omega$ and $\xi$ are differentthe 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.

2 added 111 characters in body

I would expect them to be very similar, with differences on a quantitative level. Both solve dispersive equations, only the dispersion relations are different.

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

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I would expect them to be very similar, with differences on a quantitative level. Both solve dispersive equations, only the dispersion relations are different.