# Delta-potential scattering problem

1. Jan 22, 2013

### fast_eddie

I am looking at a 1-d quantum system with a delta-potential barrier in the centre (at x = 0) and an infintely high wall on one side of this barrier (at x = -a), while the system is open on the other side.

So the potential V is equal to:
$$V = κ\delta(x)$$ at x = 0, κ being some constant and δ being the Dirac delta-function
V = ∞ at x = -a, this is where the wall is
and V = 0 everywhere else

Splitting the system into region 1 (the bounded part to the left of the delta-barrier, -a < x < 0) and region 2 (the open part to the right of the delta-barrier, x>0), we get V=0 in both regions and the Schrodinger equation will be of the same form:
$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi$$

or
$$\frac{d^2\psi}{dx^2} = -k^2\psi$$
where $$k = \frac{\sqrt{2mE}}{\hbar}$$

And so the wavefunction in both regions has form:
$$\psi_1 = Ae^{ikx} + Be^{-ikx}$$
$$\psi_2 = Ce^{ikx} + De^{-ikx}$$

The boundary condition at -a gives:
$$Ae^{-ika} + Be^{ika} = 0$$
Or equivalently:
$$Acos(ka) = Bsin(ka)$$

And using the standard trick of integrating the Schrodinger equation on a small integral around the origin [-ε,ε] and taking the limit ε→0, we get another condition:
$$ik (C-D-A+B)= \frac{2m\kappa}{\hbar^2}(A+B)$$

which by rearranging we can also express as:
$$C-D = A(1-2i\beta) - B(1+2i\beta)$$
where $$\beta= \frac{m\kappa}{k\hbar^2}$$

And at this stage I am a bit stuck about what to do next with these boundary conditions, in order to solve for the constants A,B,C,D and get the S-matrix components. I have already solved this same problem in the case where there is a wall on either side of the delta-barrier, and in the case where the system is open on either side, but I'm not sure how to proceed with this semi-open case. Any help, hints or advice would be greatly appreciated, thanks.

2. Jan 22, 2013

### CFede

What did you do to determine the constants in the case where its open on either side? I'm guessing you can follow a similar procedure here for the open side. That plus normalization should be enough.

3. Jan 22, 2013

### fast_eddie

In that case I simply took the constant D to be 0, since it represents the amplitudes of waves coming in from the right side, and in that example I made the arbitrary assumption to consider waves being propagated from the left only. This made everything a lot easier, but here I don't think I can do the same thing, since I should expect some reflected waves coming back in the opposite direction due to the wall. Am I right in thinking this?

4. Jan 22, 2013

### CFede

That's true in the region between the infinite wall and the delta potential, however, for x>0 you can assume that there are no waves propagating from the right since there's no potential to the right of the delta function were they could be reflected.