Can we consider the laplacian as a shift operator. If so, are there citations to this effect?
The short answer is no. Shift operator must formally generate the Taylor expansion of the function on which it is applied at the point of application. The Laplacian operator does not achieve this.
One way I like to think of the Laplacian operator is that it it meaures the difference of the value of a function f(x) at a point x from the average value of f in the neighborhood of x. This can be seen by approximating the second derivative by a second-order finitie difference.
Here is a little diversion from the subject under discussion. Following is a link to an article in which spin is directly introduced into the Laplacian operator and Direc energy levels for the Hydrogen atom are obtained.
Article Quarkonium and hydrogen spectra with spin-dependent relativi...
Dear Robert, thank you for bringing up the interesting perspective. I have however a slight problem with your argument. The heat equation (your first equation) is not an identity, but as an equality applies only to the solution of the heat equation, subject to the appropriate initial and boundary conditions. It follows that your second equation is not applicable to a general function, but only to the solutions of the heat equation. In contrast to this, the shift operator is to induce shift of argument when applied to arbitrary functions. Similarly, the Hamiltonian operator induces time shift in the solutions of the corresponding time-dependent Schrödinger equation, subject to the appropriate initial and boundary conditions, and not in an arbitrary function of time.
A shift operator is exp(a d/dx) in one spatial direction this gives a shift of a or exp(t d/dt) in time or exp(a nabla) where a is a vector in R^n . In all cases these are well defined on functions in regions where they are C^infinity i.e. infinitely differentiable. Shift operators for functions that have discontinuities need more care and proper analysis
@ Jim N Mcelwaine - Agreed. However, by considering distributions, as opposed to mere functions, the problems that you are referring to are automatically taken care of (this is achieved by the limitations imposed on the space of test functions underlying distributions).
Write so its fine for x>0 or x< 0 but goes wrong at the essential singularirty
Does "essential singularity" have any meaning for real functions or is it only meaningful for complex ones?
Dear Carringtone, for a brief answer i say yes. Actually for Laplacian operator defined in a "continuum" sense, the previous answers in this thread are perfectly right, However in the graph theory applied in frequency analysis of signal processing , a Laplacian operator acts on functions as a shift operator because here the discrete notion of analysis taken into account , for a complete review see this paper in signal processing: arXiv:1307.0468v2 [cs.SI] 18 Nov 2013.Finally the question for a physicist implies that Laplacian operator absolutely could not be considered as a shift operator, Nevertheless in a general view including graph theory and its application, Laplacian operator appears as a shift operator acting on graphs nodes and edges.Mathematics is a broad field !
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