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The motivation of comparison of this two kind of operators is that,$$\partial_{tt}-\Delta=(\partial_{t}-i\sqrt{-\Delta})(\partial_{t}+i\sqrt{-\Delta})$$ From the above that a wave operator can be seen as the product of two (general type) schr$\ddot{o}$dinger operators.Indeed,compared with other kinds of differential operators(elliptic,parabolic),these two operators seem to share more in common.Such as they don't have the usual global regularity property,and the use of harmonic analysis has made a great success in these two areas.

So I'm very curious to know the intersting links or fundemental differences between this two operator in order to understand them better.

*Edit:*In the above,I'm refering to a general dispersive equation:$u_{t}-i\phi({D})u=0$.When $\phi=-|\xi|^{2}$,it's the free schr$\ddot{o}$dinger equation.

edited Aug 3 at 5:04

asked Aug 3 at 3:35

Shanlin Huang
926●2●9

	That is not a Schrodinger equation... – Chris Gerig Aug 3 at 4:14
	well, I think you can look it as a fractional schrodinger operator. – Shanlin Huang Aug 3 at 4:56

Jon · Answer 1 · 2012-08-03 09:47:15Z UTC

2

There is no direct link in the way you write it down. In order to decompose the wave equation you will need a Clifford algebra of matrices $\gamma$, such that $\gamma_i\gamma_j + \gamma_j\gamma_i = 2\eta_{ij}$, and define the operator $D=i\gamma_k\partial^k$. Then you will have $D^2=\partial_{tt}-\Delta$. But note that this operator will not apply on ordinary scalar functions but on spinors. Finally, when you consider the "massive" operator $D-I$, you are able to recover in some limit the Schroedinger equation.

answered Aug 3 at 9:47

Jon
777●1●2●6

	Well,I found a more direct link (in Terry Tao's book)between this two operators,which says that let $u\in C(R\times R^{d})$,and defines $v\in C(R\times R^{d+1})$ by $v(t,x_{1},\dots,x_{d},x_{d+1})=e^{-i(t+x_{d+1})}u(\frac{t-x_{d+1}}{2},x_{1},\dots,x_{d})$.Then v solves the d+1 dimensional wave equation if and only if u solves the d dimensioanal Schrodinger equation. – Shanlin Huang Aug 5 at 8:47
	I should look at Tao's book, please cite. Anyhow, you cannot factorize wave equation as you did. The only way, devised by Dirac in the '20s, to get the square root of a Laplacian or a wave operator is using Clifford algebra of matrices. This can also be seen by the fact that wave equation is invariant under Lorentz group SO(1,3) and Schroedinger equation is not. Dirac approach preserves this symmetry of the wave equation as it must. – Jon Aug 5 at 10:21
	Nonlinear dispersive equations: local and global analysis.Chap 2.In my case ,the square root of minus laplacian is established by spectra decomposition theorem(or through pseudo-differential operators).besides,it's an nonlocal operator. – Shanlin Huang Aug 6 at 12:50
	First three chapters of Tao's book are at math.ucla.edu/~tao/preprints/chapter.pdf (thanks to author). Of course, I agree with Tao's conclusion but you are completely off the track. Read this chapter 2 more carefully. – Jon Aug 6 at 16:52

Wave equation v.s.Schrödinger equation

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