Double delta potential - boundary conditions

[SoggyBottoms]

A double delta potential is given by $V(x) = c_+ \delta (x + \frac{L}{2}) + c_- \delta (x - \frac{L}{2})$.
Use the discontinuity relation to find the boundary conditions in $x = \pm \frac{L}{2}$.

The general solutions are:

$\psi(x) = \begin{cases} Ae^{ikx} + Be^{-ikx} & x < -\frac{L}{2} \\ Ce^{ikx} + De^{-ikx} & -\frac{L}{2} < x < \frac{L}{2} \\ Ae^{-kx} & x > \frac{L}{2}\end{cases}$

So using the discontinuity relation, we get in $\frac{L}{2}$:

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

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

 

[eljose79]

Yes, your approach is correct. Using the discontinuity relation, we can equate the difference in the derivative of the wave function at the boundary point with the potential function at that point. This gives us the boundary condition for the wave function at that point. In this case, we have two boundary points at x = ±L/2, so we will have two boundary conditions.

At x = -L/2:

ikAe^{ik(-L/2)} - ik Be^{-ik(-L/2)} = \Delta \frac{d\psi}{dx} = \frac{2mc_-}{\hbar^2}\psi(-\frac{L}{2})

At x = L/2:

ikCe^{ik(L/2)} - ikDe^{-ik(L/2)} = \Delta \frac{d\psi}{dx} = \frac{2mc_+}{\hbar^2}\psi(\frac{L}{2})

These boundary conditions will help us determine the values of the coefficients A, B, C, and D in the general solutions. So, in short, your approach and equations are correct. Keep up the good work!