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

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

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

Thanks!

• Try putting "heat semigroup" in google and you'll get lots relevant references. – André Henriques Aug 12 '12 at 13:51
• I think no-one has mentioned yet that $e^{it\sqrt{-\Delta}}$ has an explicit kernel $c_n (-it) (|x|^2 - t^2 - 0i)^{-(n+1)/2}$, where $n$ is the dimension and $c_n$ is an appropriate constant. The kernel is singular around $|x|=t$. In particular, in dimension $n = 1$, the kernel is simply $$\operatorname{p.v.} \biggl(\frac{1}{\pi} \, \frac{-it}{x^2 - t^2}\biggr) + \frac{1}{2} (\delta_t(x) + \delta_{-t}(x)).$$ – Mateusz Kwaśnicki Feb 5 '20 at 19:52
• Also, the operator $\sqrt{-\Delta}$ is sometimes called the (quasi-)relativistic Hamiltonian for a massless particle, and has been studied heavily by, for example, Lieb and co-authors. – Mateusz Kwaśnicki Feb 5 '20 at 19:56

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. The dispersion relation is $$\omega = |\xi|$$, 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.
• You're right, $e^{it\Delta}$ isn't the heat semigroup. Editing my answer. – Nik Weaver Aug 13 '12 at 3:14
• It is not actually true that this operator has a finite speed of propagation. Finite speed = compactly supported kernel = entire analytic symbol (by Paley-Wiener). The symbol $e^{it|\xi|}$ is not entire analytic – Piero D'Ancona Feb 5 '20 at 15:51
• Note that the wave equation has entire symbol, indeed it is $\cos(t|\xi|)$ plus $\sin(t|\xi|)|\xi|^{-1}$ – Piero D'Ancona Feb 5 '20 at 15:52
You can say a lot about the function $$u(t,x)=e^{it\sqrt{-\Delta}}f$$ by noticing, as remarked by others, that $$u$$ is also solution of the Cauchy problem $$u_{tt}-\Delta u=0,\quad u(0,x)=f,\quad u_t(0,x)=i\sqrt{-\Delta}f.$$ Thus most estimates for solutions of the wave equation carry over to $$u$$. However there are some subtle differences, mostly due to the non--locality of the operator $$\sqrt{-\Delta}$$.