# Scattering off of a bi-local potential

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 $$[\frac{\hbar^2\partial^2_x}{2m}-E]\psi = A\phi(x)\int dx'\phi^*(x')\psi(x')$$ 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: $$K(x,x';E)=\frac{2m}{\hbar\sqrt{2mE}}e^{i|x-x'|\sqrt{2mE}/\hbar}$$ Back to the question, I am lost with. Based on this information how can I find a $\psi$ that satisfies the boundry conditions:

$$\psi(x\rightarrow-\infty)=e^{ikx}+re^{-ikx}$$ $$\psi(x\rightarrow\infty)=te^{ikx}$$

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

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?
• Those integrals both go to zero (if I am thinking correctly), because they are localized. This means that the solutions at infinity will just be the plane waves? In trying to relate $\alpha \, \beta$ to r and t, I see that \alpha at infinity should be zero, but nonzero at -infinity. – yankeefan11 Oct 28 '15 at 13:36
• Careful there: the integrand goes to zero at infinity. The integrals themselves are generally finite and non-vanishing. They do depend on x, but when $x\rightarrow \pm\infty$ only one survives. That makes the entire solution a superposition of plane waves at $\pm \infty$. I don't know what $f"(x)$ is, but otherwise you should have everything you need to calculate the $\alpha$, $\beta$ for your particular case. – udrv Oct 28 '15 at 18:03
• I just realized that I incorrectly typed the equation. I edited it. I am really confused because $\psi$ depends on $\int \psi$ – yankeefan11 Oct 30 '15 at 1:00
• I think you actually have $\int{dx"\phi^*(x")\psi(x")}$ in the expression for $\psi(x)$. In any case, it doesn't change the reasoning because it contributes simply as a number that can be determined in terms of $K$ and $\phi$: If you multiply both sides of the eq. for $\psi(x)$ by $\phi^*(x)$ (or $\phi(x)$ if the integral really contains $\phi(x)$) and take the integral over $x$, the result is an equation for $\int{dx"\phi^*(x")\psi(x")}$ (or $\int{dx"\phi(x")\psi(x")}$). The rest goes as before. – udrv Oct 30 '15 at 4:44