Given the Schrodinger equation:

$\displaystyle -\frac{\hbar ^2}{2m} \frac{d^2 \psi}{dx^2} + V(x) \psi (x) = E \psi (x)$

and $\displaystyle V(x) = -a \delta (0)$

find the energy eigenvalue, E, for a bound state in terms of a > 0.

To make an appropriate energy eigenfunction we require that $\displaystyle \psi (x)$ is continuous on the real line, $\displaystyle \lim_{x \to \pm \infty} \psi (x) = 0$ and $\displaystyle \lim_{x \to \pm \infty} \frac{d \psi }{dx} = 0$. Also, since the potential contains a delta function, the first derivative of the wavefunction will not be continuous.

-Dan