Can we have a physical interpretation for a time independent Schrodinger equation of this form?

I am interested in a time independent Schrodinger equation of this form. $$F*\psi - \frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x^2}} = E\psi$$

Here the product $V\psi$ is replaced by the convolution $F*\psi$. What I want to know is that is there any such $F$ where we can assign some kind of an intuitive physical meaning to it? It may be field or quantum field or whatever exotic thing.

The product replaced by convolution enables a lot of mathematically beautiful things to happen, but to begin with I strongly need a physical interpretation.

PS : My approach is to take a mathematically beautiful/appealing objects/equations and try to make sense of them in physical applications.

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Not sure if any useful physical interpretation could come about, but feel free to explore to your heart's content. In the special case $F(x)=\Delta E \delta(x)$ the equation simply encodes an overall energy shift $E\rightarrow E-\Delta E$. Likewise $F(x)\propto\delta''(x)$ can be interpreted as a mass shift. I'd have to think a little more about less trivial examples, but since the convolution operation is equivalent to the wavefunction being acted upon by a shift-invariant operator, I suspect there's a trivial general interpretation. – DumpsterDoofus Dec 31 '13 at 4:25
Can you briefly explain what mathematical beauty does the convolution give? The convolution seems to make the TISE explicit non-local. – hwlau Dec 31 '13 at 7:44
@hwlau : This makes the operator acting on $\psi$ shift invariant, which means it can be used for a particle interacting with itself! – Rajesh Dachiraju Dec 31 '13 at 8:17
I am not sure what "shift invariant" means. But if you want a particle interact with itself, the best thing is to let $V=|\psi|^2$ – hwlau Dec 31 '13 at 8:28
@hwlau : shiftinvariance means if $\psi(x)$ is a solution then $\psi(x-\tau)$ is also a solution – Rajesh Dachiraju Dec 31 '13 at 8:29

The Fourier transformed equation wrt. $\mathbf{x}$ is

$\tilde{F}(\mathbf{k})\tilde{\psi}(\mathbf{k})+\frac{\hbar ^{2}}{2m}\mathbf{k}% ^{2}\tilde{\psi}(\mathbf{k})=E\tilde{\psi}(\mathbf{k}),$

so the effect is a $\mathbf{k}$-dependent contribution to the energy $E$.

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Sounds interesting. Will look at it later. – Rajesh Dachiraju Jan 1 '14 at 16:53
@Trimok and Urgje : The force on an electron in a magnetic field depends on its velocity, So can we link up $\tilde{F}(k)$ in some way to Magnetic potential or magnetic vector potential? – Rajesh Dachiraju Jan 4 '14 at 3:46
uniform magnetic field. – Rajesh Dachiraju Jan 4 '14 at 3:51
@Rajesh D. I do not think so. The only thing the F-term does is to modify the kinetic energy of an otherwise free particle. Note that the left hand side commutes with the momentum operator. – Urgje Jan 4 '14 at 10:35

I don't understand why the partial derivative ? If the model is one dimensional, then one can have ordinary derivative w.r.t. x. If not, one can use Laplacian.

In Fourier analysis and elsewhere, the basic idea behind convolution is "Superposition of Waves" in some sense, so to bring a convolution of of an operator and a function, one needs to bring both of them in equally footing, if not mathematically, at least physically, so one must specify "What F is " ? (I think F replaces potential ! but one must specify what is the status of F is ? Operator or what ?) If F belongs to the same space (or class) as the eigen function belongs, then it is "kind of" integrodifferential equation, one can attach several physical meaning to the above equation, one of them is "The wave is influenced by several superposition of wave like "Sources" and glued together via convolution". One can provide more concrete information on the basis of explicit situation.

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: $F$ is an operator. – Rajesh Dachiraju Dec 31 '13 at 13:38
If F is an operator then one needs to specify exactly what is it's form, on what kind of space it acts, what is it's spectral properties, and on that basis one can do "Fourier Analysis" on the space where F acts and can define "convolution", otherwise it is not possible to extract physical interpretations from it. – user36293 Dec 31 '13 at 18:04