- #1

jostpuur

- 2,112

- 18

If [itex]\psi(x,0)[/itex] has a compact support, and

[tex]

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

[/tex]

then [itex]\psi(x,t)[/itex] does not have a compact support for any [itex]t>0[/itex].

Claim 2:

If [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are the same in some environment of a point [itex]x_0[/itex], then

[tex]

\partial_x^2 \psi_1(x_0) = \partial_x^2 \psi_2(x_0)

[/tex]

Claim 3:

If [itex]\psi(x,0)[/itex] has a compact support, and

[tex]

i\partial_t\psi(x,t) = \sqrt{1 - \partial_x^2} \psi(x,t),

[/tex]

then [itex]\psi(x,t)[/itex] does not have a compact support for any [itex]t>0[/itex].

Question:

If [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are the same in some environment of [itex]x_0[/itex], will it follow that

[tex]

\sqrt{1 - \partial_x^2}\psi_1(x_0) = \sqrt{1 - \partial_x^2}\psi_2(x_0)?

[/tex]

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 [itex]\sqrt{1 - \partial_x^2}[/itex] 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 [itex]\partial_x^2[/itex] 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?