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


1 Answer 1


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?

  • $\begingroup$ 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. $\endgroup$ Commented Oct 28, 2015 at 13:36
  • $\begingroup$ 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. $\endgroup$
    – udrv
    Commented Oct 28, 2015 at 18:03
  • $\begingroup$ I just realized that I incorrectly typed the equation. I edited it. I am really confused because $\psi$ depends on $\int \psi$ $\endgroup$ Commented Oct 30, 2015 at 1:00
  • $\begingroup$ 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. $\endgroup$
    – udrv
    Commented Oct 30, 2015 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.