# Differentials, pseudo-differentials, and locality

1. Mar 28, 2010

### jostpuur

Claim 1:

If $\psi(x,0)$ has a compact support, and
$$i\partial_t\psi(x,t) = -\partial_x^2 \psi(x,t),$$
then $\psi(x,t)$ does not have a compact support for any $t>0$.

Claim 2:

If $\psi_1$ and $\psi_2$ are the same in some environment of a point $x_0$, then
$$\partial_x^2 \psi_1(x_0) = \partial_x^2 \psi_2(x_0)$$

Claim 3:

If $\psi(x,0)$ has a compact support, and
$$i\partial_t\psi(x,t) = \sqrt{1 - \partial_x^2} \psi(x,t),$$
then $\psi(x,t)$ does not have a compact support for any $t>0$.

Question:

If $\psi_1$ and $\psi_2$ are the same in some environment of $x_0$, will it follow that
$$\sqrt{1 - \partial_x^2}\psi_1(x_0) = \sqrt{1 - \partial_x^2}\psi_2(x_0)?$$

Thoughts:

According to my understanding, at least with some assumptions, the claims 1,2,3 are all true. If I had not mentioned the claims 1 and 2, some people might have answered to my question, that the pseudo-differential operator $\sqrt{1 - \partial_x^2}$ does not possess the locality property I'm asking in the question, because we know that it is non-local in the sense of the claim 3. However, we know that the ordinary differential operator $\partial_x^2$ has the non-locality property in the sense of claim 1 too, and still it has the locality property in the sense of claim 2. So without better knowledge, it could be that the pseudo-differential operator has this locality property too. But what's the truth?

2. Apr 14, 2010

### jostpuur

I think that the answer to my question is "no", and the instantaneous spreading caused by $$\sqrt{1 - \partial_x^2}$$ is significantly faster than the instantaneous spreading caused by $$\partial_x^2$$.

I just realized that I don't yet know how to prove that the non-relativistic Schrödinger equation causes instantaneous spreading. Here's the problem:

If we want to satisfy the heat equation

$$\partial_t u(t,x) = \partial_x^2 u(t,x)$$

with some initial condition $u(0,x)\geq 0$ which has compact support, we can accomplish it with a formula

$$u(t,x) = \frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty} \exp\Big(-\frac{y^2}{4t}\Big) u(0,x-y)dy.$$

It can be seen from this formula that $u(t,x)>0$ for all $t>0$. Hence the compact support is lost instantaneously.

But the take a closer look at the non-relativistic Schrödinger equation. We want

$$i\partial_t \psi(t,x) = -\partial_x^2\psi(t,x)$$

with some initial condition $\psi(0,x)$ with compact support. The solution can be written as

$$\psi(t,x) = \frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty} \exp\Big(\frac{iy^2}{4t}\Big) \psi(0,x-y)dy.$$

But how do you prove that $\psi(t,x)$ does not have a compact support for $t>0$? It's not so obvious now because there could be some canceling in the integral.

I'm interested in rigor comments only! (Or comments that at least attempt some rigor. I'm not sure if I'm perfectly rigorous either...).

I know some physicists like to say that "well look at the propagator. It is non-zero for arbitrarily large intervals, and hence the instantaneous spreading." You might as well try to insist that Fourier transforms (or inverse transforms) will never give functions with compact support, because the plane waves extend to infinities. Of course canceling can occur!

An interesting remark about the heat equation is that if we fix some $x'$ outside the original compact support, then the Taylor series

$$u(0,x') + t \partial_t u(0,x') + \frac{1}{2!} t^2 \partial_t^2 u(0,x') + \cdots = 0$$

is identically zero, even though $u(t,x')>0$ for arbitrarily small $t>0$. This means that the spreading is extremely slow, at least for small $t$.

The same slowness does not occur for relativistic Schrödinger equation. If $\psi(0,x)$ has a compact support, then $$\hat{\psi}(0,p)$$ is analytic. But

$$i\partial_t\hat{\psi}(0,p) = \sqrt{1 + p^2}\hat{\psi}(0,p)$$

is not analytic because of the branch cuts at $p=\pm i$. Hence $\partial_t\psi(0,x)$ does not have a compact support, and the instantaneous spreading is faster than with the heat equation.