quantum mechanics - 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?

Answer

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