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Given is the potential :$V(x) = \frac{-\hbar^2}{m}D\delta(x+a) - \frac{-\hbar^2}{m} D\delta(x-a)$ with $a >0$ and $ D > 0$. A stream of particles from the positive $x$-axis are falling towards the potential and I want to calculate the probability of transmission. I would need help with the boundary conditions. My Ansatz is: $\psi_1(x) = Ae^{ikx}+Be^{-ikx}$ for the incoming wave, $\psi_2(x) =Ce^{ikx}+De^{-ikx}$ for the wave between the delta distributions and $\psi_3(x) = Ee^{ikx}+Fe^{-ikx}$ for the wave function $\psi > a$ with $F = 0$. These are my conditions until now: $\psi_1(-a) = \psi_2(-a)$ $\psi_2(a) = \psi_3(a)$ What are the other conditions? I was thinking of calculating the transfer matrix first and afterwards the transmission.

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Ok your ansatz about the form of the $\psi$. Now, the other conditions you need come from the discontinuity of the derivative through the delta potential. If you take the time-independent Schrödinger equation and integrate it around a delta (see, for example, here for the steps), then it can be shown that in the case of a potential like $V(x)=\gamma\delta(x-x_0)$ the derivative must possess a discontinuity given by $$ \left.\frac{d\psi}{dx}\right|_{x\to x_0^+} - \left.\frac{d\psi}{dx}\right|_{x\to x_0^-} = \frac{2m\gamma}{\hbar^2}\psi(x_0) $$ in your case $|\gamma|=D\hbar^2/m$ and hence for the two $\delta$s you have $$ \left.\frac{d\psi}{dx}\right|_{x\to -a^+} - \left.\frac{d\psi}{dx}\right|_{x\to -a^-} = -2D\psi(-a) $$ and $$ \left.\frac{d\psi}{dx}\right|_{x\to +a^+} - \left.\frac{d\psi}{dx}\right|_{x\to +a^-} = +2D\psi(+a) $$

Since you have 6 constants $A,B,\dots,F$ to determine, these 2 conditions plus the 2 you wrote plus the $F=0$ will leave us with just one free parameter that should be absorbed in the computation of the reflection and transmission probabilities.

Edit: in the last line I originally wrongly wrote $\psi(-a)$ instead of $\psi(+a)$, now it is corrected.

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  • $\begingroup$ Why do you have $+2D(-a)$ when $/gamma$ is $\frac{-D\hbar^2}{m}$? When I try to solve this, I get $A = -Be^{2ika}$ can you confirm this? I get $B=D$ and $A=C$ by Equating the coefficients and I'm not quite shure if this is allowed. I'm also not shure which $\psi$ you mean after the equal sign when you say $ \frac{d\psi}{dx}_{x\rightarrow a^+}-\frac{d\psi}{dx}_{x\rightarrow a^-} = -2D\psi(-a)$ In fact I thought I can choose which Ansatz I like for that $\psi$. $\endgroup$ – gamma Feb 27 '18 at 14:45
  • $\begingroup$ I made a typo sorry: in the last line it is a $+$ and not a $-$ sign so that it is $+2D\psi(+a)$. Probably the results you found are due to this. I will correct the answer. $\endgroup$ – ndrearu Feb 27 '18 at 14:48
  • $\begingroup$ But still I don't understand why it's $+2D\psi(a)$ instead of $-2D\psi(a)$? $\endgroup$ – gamma Feb 27 '18 at 15:14
  • $\begingroup$ In the potential you wrote there is a minus between the deltas, so in turn the minus in front of $\delta(x-a)$ gives a $+$ in the discontinuity. $\endgroup$ – ndrearu Feb 27 '18 at 15:19

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