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I'm trying to understand the quantum mechanical solution to this potential:

[tex]V(x) = \left\{\begin{array}{cc}\infty & \mbox{ for } x < 0,\\-\lambda\delta(x-d) & \mbox { for } x > 0\end{array}\right.[/tex]

A particle of mass m is constrained to move on the half straight line [itex]\{x \in \mathbb{R}: x > 0\}[/itex] under this potential. I want to examine the condition for existence of bound states, and its dependence on [itex]\lambda[/itex] and [itex]d[/itex], where both [itex]\lambda[/itex] and [itex]d[/itex] are positive.

For x < 0, [itex]\psi(x) = 0[/itex]

For 0 < x < d, [itex]\psi(x) = A\sin(kx)[/itex] where [itex]k = \sqrt{2mE}/\hbar[/itex].

For x > d, [itex]\psi(x) = Ce^{ikx} + De^{-ikx}[/itex].

Enforcing continuity at x = d, [itex]A sin(kd) = Ce^{ikd} + De^{-ikd}[/itex].

Integrating Schrodinger's equation over a small interval around [itex]x = d[/itex]

[tex]-\frac{\hbar^2}{2m}\int_{d-\epsilon}^{d+\epsilon}\frac{d^2 \psi}{dx^2} dx - \lambda\int_{d-\epsilon}^{d+\epsilon}\delta(x-d)\psi(x) dx = E\int_{d-\epsilon}^{d+\epsilon}\psi(x) dx[/tex]

But I don't get a condition relating [itex]\lambda[/itex] and [itex]d[/itex] for the existence of a bound state. What have I missed?

Thanks in advance.

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# Delta well + infinite barrier -> bound state

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