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I need to find energies of a particle for which the reflection coefficient from the system $U(x)=\alpha(\delta(x)+\delta(x-a))$ will be maximum. I am struggling with this task for 2 days. Here is my try:

Particle is coming from the left. We know that the reflection coefficient is $R=\frac{1}{|B_1|^2}$ where $B_1$ is the amplitude of the reflected wave. The energy spectrum of the particle is continuous (maybe not?). So the only thing we need to find is $B_1$.

We have 3 areas:

  • area 1 for $-\infty<x<0$

  • area 2 for $0<x<a$

  • and area 3 for $x>0$

The solutions of the stationary Sh. Eq. for these areas are ($k=\frac{\sqrt{2\mu E}}{\hbar}, \mu$-is the mass of the particle):

  • $\psi_1=e^{ikx}+B_1e^{ikx}$, we assume that $A_1=1$

  • $\psi_2=A_2e^{ikx}+B_2e^{ikx}$

  • $\psi_3=A_3e^{ikx}$, as there is no reflected wave

We also have 4 boundary conditions:

  1. $\psi_1(0)=\psi_2(0)$
  2. $\psi_2(a)=\psi_3(a)$
  3. $-\frac{\hbar^2}{2\mu}\int_{-\epsilon}^{+\epsilon}\frac{\partial^2\psi}{\partial x^2}dx+\alpha \int_{-\epsilon}^{+\epsilon}(\delta(x)+\delta(x-a))\psi dx=E\int_{-\epsilon}^{+\epsilon}\psi dx$
  4. $-\frac{\hbar^2}{2\mu}\int_{a-\epsilon}^{a+\epsilon}\frac{\partial^2\psi}{\partial x^2}dx+\alpha \int_{a-\epsilon}^{a+\epsilon}(\delta(x)+\delta(x-a))\psi dx=E\int_{a-\epsilon}^{a+\epsilon}\psi dx$

Condition 3 and 4 gives us the behavior of the $\psi$-function near the delta-potentials.

From 1 condition we get: $1+B_1=A_2+B_2$

From 2 condition we get: $A_2e^{ika}+B_2e^{-ika}=A_3e^{ika}$

From 3 condition we get: $-\frac{ik\hbar^2}{2\mu}(1-B_1-A_2+B_2)+\alpha (1+B_1)=0$

From 4 condition we get: $-\frac{ik\hbar^2}{2\mu}(A_2e^{ika}-B_2e^{-ika}-A_3e^{ika})+\alpha (A_2e^{ika}+B_2e^{-ika})=0$

This is the place where I dont know what to do to get the answer with minimum actions. I got one complex transcendental equation for $k$ and also got some complex and huge equation for $B_1$, I am almost sure that they are not correct. Can somebody show me the best way to get to the answer?

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closed as off-topic by AccidentalFourierTransform, Jon Custer, heather, user108787, user36790 Nov 20 '16 at 18:43

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I left that link. In it is solved a very similar double delta potential step by step.

http://www.physicspages.com/2012/08/01/double-delta-function-well/

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  • $\begingroup$ Please remember that, Manuel has not yet achieved the rep of 50 required to leave comments, (AFAIK) and that he is not answering a homework question, in line with site policy, instead he is pointing the OP towards a resource . ( that the OP could have found easily, I have to say) $\endgroup$ – user108787 Nov 20 '16 at 17:12
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    $\begingroup$ Sorry, I thought I could be taken as an answer. I will take in account for the next time. $\endgroup$ – user326159 Nov 20 '16 at 17:28
  • $\begingroup$ No, I think you would need to expand on it a bit for an answer, I found the same link as you, when I came back to post it, you were there ahead of me:). If you can put as a comment instead , I would do so though. Link answers are frowned upon, sorry. $\endgroup$ – user108787 Nov 20 '16 at 17:32

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