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A particle of mass $m$ is confined to move in a one-dimensional and Dirac delta-function attractive potential $$V(x)=-\frac{\hbar}{m}\alpha\delta(x)\text{ $,\alpha>0$ }.$$

Show that the function $\psi(x)$ is discontinuous at $x=0$, and determine its jump at the point.

Showing its discontinuous at zero is easy. It follow from the following: $\frac{-\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}\frac{d^2\psi}{dx^2}+\frac{-\hbar^2\alpha}{m}\int_{-\epsilon}^{\epsilon}\delta(x)\psi(x)=0\implies \lim_{\epsilon\to 0} \frac{d\psi}{dx}\bigg|_-+\frac{d\psi}{dx}\bigg|_+=2\alpha\psi(0)$

As for the jump, that's where I'm stuck. I assumed it was at $y=\alpha$. But I believe I'm wrong.

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    $\begingroup$ Suggestion to the question (v1): Replace [discontinuity of the wave function] with [discontinuity of derivative of the wave function]. $\endgroup$ – Qmechanic Jul 1 '13 at 9:30
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The wave function itself can never be discontinuous. It is the derivative what it is discontinuous at $x=0$, and that discontinuity can be calculated integrating the Schrödinger equation between $(+\epsilon,-\epsilon) $ and taking the limit $\epsilon \to 0$. All terms but the proportional to the delta vanish, giving you

$$\left. \frac{d\psi(x)}{dx}\right|_{\epsilon=0^+} -\left. \frac{d\psi(x)}{dx} \right|_{\epsilon=0^-}=-2\hbar \alpha\psi(0) $$

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