A double delta potential is given by [itex]V(x) = c_+ \delta (x + \frac{L}{2}) + c_- \delta (x - \frac{L}{2})[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Use the discontinuity relation to find the boundary conditions in [itex]x = \pm \frac{L}{2} [/itex].

The general solutions are:

[itex]

\psi(x) =

\begin{cases}

Ae^{ikx} + Be^{-ikx} & x < -\frac{L}{2} \\

Ce^{ikx} + De^{-ikx} & -\frac{L}{2} < x < \frac{L}{2} \\

Ae^{-kx} & x > \frac{L}{2}\end{cases}

[/itex]

So using the discontinuity relation, we get in [itex]\frac{L}{2}[/itex]:

[itex]ikAe^{ik\frac{L}{2}} - ik Be^{-ik\frac{L}{2}} - ikCe^{ik\frac{L}{2}} + ikDe^{-ik\frac{L}{2}} = \Delta \frac{d\psi}{dx} = \frac{2mc_+}{\hbar^2}\psi(\frac{L}{2})[/itex].

And that would be the first boundary condition...is any of this correct?

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# Double delta potential - boundary conditions

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