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fast_eddie

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**I am looking at a 1-d quantum system with a delta-potential barrier in the centre (at x = 0) and an infinitely 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:

[tex]V = κ\delta(x)[/tex] 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:

[tex] -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi[/tex]

or

[tex] \frac{d^2\psi}{dx^2} = -k^2\psi[/tex]

where [tex] k = \frac{\sqrt{2mE}}{\hbar}[/tex]

And so the wavefunction in both regions has form:

[tex] \psi_1 = Ae^{ikx} + Be^{-ikx}[/tex]

[tex]\psi_2 = Ce^{ikx} + De^{-ikx}[/tex]

The boundary condition at -a gives:

[tex] Ae^{-ika} + Be^{ika} = 0[/tex]

Or equivalently:

[tex] Acos(ka) = Bsin(ka)[/tex]

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:

[tex] ik (C-D-A+B)= \frac{2m\kappa}{\hbar^2}(A+B)[/tex]

which by rearranging we can also express as:

[tex] C-D = A(1-2i\beta) - B(1+2i\beta)[/tex]

where [tex] \beta= \frac{m\kappa}{k\hbar^2}[/tex]

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.