Delta function potential

Consider a potential $V(x) = a\delta(x)$ . For $x \ne 0$ , we simply get the same behavior as for a piecewise constant potential. The interesting behavior is at the delta function. Usually we require $\psi'$ to be continuous, but since $V(0)$ is infinite, $\psi'$ can’t be continuous, otherwise we would have infinite energy.

Integrate the Schrödinger equation around the discontinuity to consider its behavior:

\begin{align*} -\frac{\hbar^2}{2m} \int_{-\epsilon}^{\epsilon} \frac{\d^2\psi}{\d x^2} \d x + \int_{-\epsilon}^\epsilon V(x) \psi(x) \d x= E \int_{-\epsilon}^\epsilon \psi(x) \d x \\ \left. \frac{\d\psi}{\d x}\right|_{\epsilon} - \left. \frac{\d\psi}{\d x}\right|_{-\epsilon} = \frac{2m}{\hbar^2} \lim\limits_{\epsilon\to0} \int_{-\epsilon}^{\epsilon} V(x)\psi(x)\d x = \frac{2m}{\hbar^2} a\psi(0). \end{align*}

We see that the slope on the left and right of the discontinuity is a function of $a$ , the “strength” of the delta function, and $\psi(0)$ the value of the wavefunction at the discontinuity.

We can combine this requirement with the constraint of a piecewise constant potential to analytically find wavefunctions for piecewise constant $V$ with delta discontinuities.