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Scattering off of a bi-local potential

Physics Asked on September 4, 2021

I am trying to figure out the scattering wave function for the following potential:

$$V(x,x’)=-A phi(x)phi^*(x’)$$
Such that the SE can be written as
$$[frac{hbar^2partial^2_x}{2m}-E]psi = Aphi(x)int dx’phi^*(x’)psi(x’)$$
This has a solution
$$psi(x)=alpha e^{ikx}+beta e^{-ikx}+lambda[int dx’ K(x,x’;E)phi(x’)int dx”phi(x”)psi(x”)]$$
Where $K$ is the propagator as defined in Sakurai:
$$K(x,x’;E)=frac{2m}{hbarsqrt{2mE}}e^{i|x-x’|sqrt{2mE}/hbar}$$
Back to the question, I am lost with. Based on this information how can I find a $psi$ that satisfies the boundry conditions:

$$psi(xrightarrow-infty)=e^{ikx}+re^{-ikx}$$
$$psi(xrightarrowinfty)=te^{ikx}$$

Not completely sure how to solve this. Supposedly, it can be assumed that $phi$ goes to 0 as $x$ goes to $infty$, which immediately implies the boundary conditions, but that does not seem clear to me why that happens

One Answer

Look at the integral over $K$ in your solution. It has the form
$$ int dx' K(x,x';E)phi(x') ;;text{~};; int dx'e^{i|x-x'|sqrt{2mE}/hbar}phi(x') = = int_{-infty}^x dx'e^{i(x-x')sqrt{2mE}/hbar}phi(x') + int^{infty}_x dx'e^{-i(x-x')sqrt{2mE}/hbar}phi(x') $$ If you take out the factors in $e^{pm ix sqrt{2mE}/hbar}$ in each of the last 2 integrals above, what do you see? What can you tell about the remaining integral factors? What about the solution itself?

Answered by udrv on September 4, 2021

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