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I have a particle and a potential $$V(x)=\frac{\hbar^2}{2m}k\delta(x),$$

where $\delta (x)$ is the Delta function, and I am interested in the solutions of the stationary Schroedinger equation.

If $\psi_1$ is the solution for $x<0$ and $\psi_2$ for $x>0$, I must have $\psi_1'(0) \neq \psi_2'(0)$, because of the delta function.

Now I read that the condition is $$\psi_2'(0) -\psi_1'(0) = -k\psi_2(0).$$

My question is: why? How do I get to this conclusion?

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2 Answers 2

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With a delta function potential, the particle is free on either side of the barrier: $$ \psi(x)=\begin{cases}\psi_L(x)=A_re^{ikx}+A_le^{-ikx} \\ \psi_R(x)=B_re^{ikx}+B_le^{-ikx}\end{cases} $$ where $A_i,\,B_i$ are constants such that $A_r+A_l=B_r+B_l$ (i.e., $\psi(x)$ satisfies the continuous function condition).

But at the barrier we have the issue that $V(0)=\infty$. So to resolve this issue, we use Schroedinger's equation and integrate it over some small region $\left[-\epsilon,\,\epsilon\right]$ and then let $\epsilon\to0$: $$ -\frac{\hbar^2}{2m}\int_{-\epsilon}^\epsilon\psi''\,dx+\int_{-\epsilon}^\epsilon V\psi\,dx=E\int_{-\epsilon}^\epsilon\psi\,dx $$ The first term is clearly $d\psi/dx$ evaluated at two points. The last term goes to zero in the limit $\epsilon\to0$ (recall that $E$ is constant and finite, so that as $\epsilon\to0$, the width goes to 0 and so does the whole value).

For the potential term, the delta function has the great property that $$ \int\delta(x-a)f(x)\,dx=f(a) $$ Thus, that middle term becomes $\left.\psi(x)\right|_{-\epsilon}^\epsilon$. We then combine these two to get $$ -\frac{\hbar^2}{2m}\left[\psi'\left(+\epsilon\right)-\psi'(-\epsilon)\right]+\left.\lambda\psi(x)\right|_{-\epsilon}^{\epsilon}=0 $$ As $\epsilon\to0$, we can get the relation you are confused over: $$ \psi'_R(0)-\psi'_L(0)=+k\psi(0) $$

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  • $\begingroup$ Shouldn't that be $\psi'_R\left(0\right) - \psi'_L\left(0\right) = + k \psi\left(0\right)$? $\endgroup$
    – Eric Angle
    Commented Jan 3, 2014 at 17:40
  • $\begingroup$ @EricAngle: Good catch. Fixed. $\endgroup$
    – Kyle Kanos
    Commented Jan 3, 2014 at 17:42
  • $\begingroup$ @EricAngle: Gah! Caught again! I blame it on the lack of sleep (had to comfort my son who was dreaming of being chased by dinosaurs last night). $\endgroup$
    – Kyle Kanos
    Commented Jan 3, 2014 at 17:45
  • $\begingroup$ Oops, I deleted my second comment (about it being $+k \psi\left(0\right)$ on the right hand side) because I saw you change it shortly thereafter. I understand wrt sleepless nights. :) $\endgroup$
    – Eric Angle
    Commented Jan 3, 2014 at 17:47
  • $\begingroup$ I wonder what would happen if instead of integrating the SE one would would like conventionally just do $<\psi|H|\psi>$? $\endgroup$ Commented Nov 28, 2022 at 10:14
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See this, beginning at "A second relation can be found by studying the derivative of the wavefunction." For your problem, $\lambda = \hbar^2 k / 2 m$.

The idea is to integrate the Schrödinger equation over the interval $\left(-\epsilon, \epsilon\right)$ and let $\epsilon \rightarrow 0$.

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  • $\begingroup$ Sorry, I always forget to look carefully in the English Wikipedia. In Italian there are many fewer pages. $\endgroup$
    – Charlie
    Commented Jan 3, 2014 at 22:46

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