Transmission over a linear barrier (QM)

[cc94]

Homework Statement

Suppose we have a potential such that $$
V =
\left\{
\!
\begin{aligned}
0 & \text{ if } x<0\\
\mathcal{E}x & \text{ if } x>0, x<L\\
\mathcal{E}L & \text{ if } x>L
\end{aligned}
\right.
$$

for some electric field $\mathcal{E}$. I'm trying to find the transmission coefficient $T$ for a plane wave incoming from the left.

Homework Equations

Solving Schrodinger's equations, I believe we have:
$$
\psi =
\left\{
\!
\begin{aligned}
Ae^{ikx} + Be^{-ikx} & ; x<0\\
C\text{Ai}(\zeta) + D\text{Bi}(\zeta) & ; x>0, x<L\\
Fe^{i\kappa x} & ; x>L
\end{aligned}
\right.
$$

Where $\zeta$ is a change of variable involving $x$, $E$, and $\mathcal{E}$, and we only keep a forward traveling wave for the region $x>L$. Since one of the coefficients is arbitrary, we can choose $F$ = 1

Then we have boundary conditions:
$$\psi_I(0) = \psi_{II}(0) \\
<=> A + B = C\text{Ai}(\zeta_0) + D\text{Bi}(\zeta_0)$$
$$\psi_I'(0) = \psi_{II}'(0) \\
<=> ik(A - B) = C\text{Ai}'(\zeta_0)(\zeta'_0) + D\text{Bi}'(\zeta_0)(\zeta'_0)$$
$$\psi_{II}(L) = \psi_{III}(L) \\
<=> C\text{Ai}(\zeta_L) + D\text{Bi}(\zeta_L) = e^{i\kappa L}$$
$$\psi_{II}'(L) = \psi_{III}'(L) \\
<=> C\text{Ai}'(\zeta_L)(\zeta'_L) + D\text{Bi}'(\zeta_L)(\zeta'_L) = i\kappa e^{i\kappa L}$$

The Attempt at a Solution

Then we get the matrix (using a little bit of shorthand):

$$\begin{bmatrix}
1 & 1 & -\text{Ai}_0 & -\text{Bi}_0 & 0 \\
ik & -ik & -\text{Ai}'_0\zeta '_0 & -\text{Bi}'_0\zeta '_0 & 0 \\
0 & 0 & \text{Ai}_L & \text{Bi}_L & e^{i\kappa L} \\
0 & 0 & \text{Ai}'_L\zeta '_L & \text{Bi}'_L\zeta '_L & i\kappa e^{i\kappa L}
\end{bmatrix}$$

I used MATLAB to solve this symbolically, and then we know that $T = \lvert\frac{F}{C}\rvert ^2 = \frac{1}{|C| ^2}$. So I plugged in various energies above the highest V and plotted T. But the transmission turns out to always be larger than 1, which makes no sense. Is my math wrong, or do I need to find a typo in my code somewhere? Has anyone solved this problem?

 

[Simon Bridge]

This is something you can do by hand... so do it that way and compare.
Also see how other people do it.
https://en.wikipedia.org/wiki/Rectangular_potential_barrier

... I don't know why you changed variable. As you switch about it is easy to mislay terms.

 

[cc94]

I finally found the answer here: http://aapt.scitation.org/doi/10.1119/1.11582[/URL]. I don't know how he got it, because even using a diff eq solver I couldn't get that result. But I'll assume it's correct and I'm just missing something.