Bound states of Dirac Delta function in infinite well

Question

If there is a potential of $-\alpha\delta(x)$ for $-a<x<a$ and $\infty$ elsewhere, and the energy of the system is less than 0, then I'm trying to find the wave function. From the Schrodinger Equation, I've worked that wave function is

$\begin{align} &-\frac{\hbar}{2m}\frac{d^2}{dx^2}\psi = E\psi \\ &k^2 = \frac{-2mE}{\hbar^2} \\ &\frac{d^2}{dx^2}\psi - k^2\psi = 0 \\ \end{align} $

Which I solved to two parts

$A e^{kx}+ B e^{-kx}$ for $-a<x<0$

$C e^{kx} + D e^{-kx}$ for $a>x>0$.

From here, I tried the boundary conditions at $x=0,a,-a$:

$\begin{align} &x=0: A+B=C+D \\ &x=-a: A e^{-ka} + B e^{ka} = 0 \\ &x=a: C e^{ka} + B e^{-ka} = 0 \\ \end{align}$

But I'm not sure where to go from here. I can show a relationship between the boundary conditions at x=-a and x=a, but I don't think that would give me anything valuable.

Answer 1

You should also match the derivatives at $x=0$ so that they took into account the $\delta$-function. If you take smooth well $V_\epsilon(x)$ and consider small region near zero $(-\epsilon,\epsilon)$ where the "meat" of the well is concentrated you may then integrate the Schrodinger equation at that region, $$-\frac{\hbar^2}{2m}\int\limits_{-\epsilon}^{+\epsilon}\psi'' dx+\int\limits_{-\epsilon}^{+\epsilon}V_\epsilon\psi dx=E\int\limits_{-\epsilon}^{+\epsilon}\psi dx$$. Consider now the limit of $\epsilon\rightarrow 0$. The rhs will disappear and the first integral will give the difference between the derivatives from the right and the left. So if $V_\epsilon(x)\rightarrow -\alpha\delta(x)$ you get the needed matching condition, $$-\frac{\hbar^2}{2m}\Big(\psi'|_{0+}-\psi'|_{0-}\Big)=\alpha\psi(0)$$