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Finite double potential barrier transmission coefficient

Physics Asked by fintallrik on December 21, 2020

TL;DR: I want to calculate the transmission coefficient of a particle travelling into a finite double potential barrier system and I think I’ve got stuck by the fact that I have 9 unknown variables (amplitudes) and only 8 equations. How do I manage to solve this?

Problem

I have a particle (an electron) with energy $E$ travelling in from the left into an area with two potential barriers. The potential is defined by

$$ V(x) = V_1cdot[Theta(x)- Theta(x-a_1)] + V_2cdot[Theta(x-(a_1 + L) – Theta(x-(a_1 + a_2 + L]$$
Where $Theta(x)$ is the Heaviside step function, $a_1$ is where the first barrier stops (i.e its length), $L$ is the width of the separation between the two barriers and $a_2$ is the width of the second barrier.

The known quantities are:

  • $V_1, V_2, E$
  • $a_1, a_2, L$
  • The particle mass $m$ is not given by I assume it can be said to be the rest mass of the electron.

The goal is to calculate the transmission coefficient $T$.

My work

I solved the equations for the different sections and got the following solutions to the time-independent Schrödinger equation

$$ Psi_1 = Ae^{ikappa x} + Be^{-ikappa x}
Psi_2 = Ce^{ilambda x} + De^{ilambda x}
Psi_3 = Fe^{ikappa x} + Ge^{-ikappa x}
Psi_4 = He^{mu x} + Ie^{-mu x}
Psi_5 = Je^{ikappa x}$$

Where $kappa = frac{sqrt{2mE}}{bar{h}}, lambda = frac{sqrt{2m(E-V_1)}}{bar{h}}, mu = frac{sqrt{2m(V_2-E}}{bar{h}}$ and ${A,..,J}$ are the amplitudes of the different waves. I have excluded the second solution to $Psi_5$ as I assume there is no wave travelling in from the right.

If I apply boundary conditions to $Psi_i$ and $Psi_i’$ at the points $x = {0, a_1, a_1 + L, a1 + a2 + L}$ I get 8 separate equations, and the goal is to calculate $T = frac{|F|^2}{|A|^2}$. As I have 9 unknown variables and 8 separate equations I do not see how I will be able to solve this. Any help is appreciated and if possible I don’t want the answer outright, just some guidance. 🙂

One Answer

Well, since scattering states are not normalizable, the wavefunction has an arbitrary overall normalization factor. The reflection coefficient $R=frac{|B|^2}{|A|^2}$ and transmission coefficient $T=frac{|J|^2}{|A|^2}$ depend only on the relative amplitudes. In other words, we can e.g. put the amplitude $A=1$ of the incoming right-mover w.l.o.g.

Answered by Qmechanic on December 21, 2020

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