# Wave function with a delta potential

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?

## 2 Answers

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: $$\psi'_R(0)-\psi'_L(0)=+k\psi(0)$$

• Shouldn't that be $\psi'_R\left(0\right) - \psi'_L\left(0\right) = + k \psi\left(0\right)$? – Eric Angle Jan 3 '14 at 17:40
• @EricAngle: Good catch. Fixed. – Kyle Kanos Jan 3 '14 at 17:42
• @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). – Kyle Kanos Jan 3 '14 at 17:45
• 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. :) – Eric Angle Jan 3 '14 at 17:47

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 $\epsilon \rightarrow 0$.

• Sorry, I always forget to look carefully in the English Wikipedia. In Italian there are many fewer pages. – Charlie Jan 3 '14 at 22:46