Penetrating the Rectangular Potential Barrier, E < V0

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SUMMARY

The discussion focuses on a code implementation in Octave for solving the problem of penetrating a rectangular potential barrier where the energy E is less than the potential V0. The user is experiencing discrepancies in concavity between their computed solution and the known correct result. The code utilizes constants such as hbar and mass m, and applies wavenumber formulas to derive transmission coefficients. The user seeks assistance in identifying errors in their calculations to correct the concavity issue before simplifying to the known answer.

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  • Understanding of quantum mechanics, specifically potential barriers.
  • Familiarity with Octave programming and syntax.
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  • Basic grasp of numerical methods for solving differential equations.
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  • Learn about numerical integration techniques for differential equations.
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Students and researchers in quantum mechanics, physicists working on potential barrier problems, and programmers using Octave for numerical simulations in physics.

Chip
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The following code compares my result to the known correct result. Can anyone find where my error is? If you run the octave code you will see that my answer has positive concavity whereas the correct one has negative concavity...I've checked this so many times and can't find my error. I wanted to see if I could correct it at this stage before trying to simplify to the known answer...thanks for any help.

****

%I wrote this to check my work thus far on solving the GIANT algebra problem of penetrating the rectangular barrier, where E < V0
clear;close;clc;
% setting some constants

hbar = 1.0545718*10^(-34);
m = 9.10938356 * 10^(-31);
a = 10^-20; %can be purely arbitrary

V0 = 100; %can be purely arbitrary
inc = 1;
E = inc:inc:(V0-inc);

%wavenumber formulas
k1 = (sqrt(2*m*E))/hbar;
k2 = (sqrt(2*m*(V0-E)))/hbar;

%this section contains the solution
TSoln = (1 + (((sinh(k2*a)).^2)./((4*E/V0).*(1-E/V0)))).^(-1)

%this section contains my work thus far

iVal = i*k2/k1;
BLABLA1 = ( 1 - ((1-iVal).*(e.^(-2*k2*a) )./(1+iVal) ) ) + iVal.* (1 + ((1-iVal).*(e.^(-2*k2*a) )./(1+iVal)));
BLABLA2 = ( 1 - ((1+iVal).*(e.^(2*k2*a) )./(1-iVal) ) ) - iVal.* (1 + ((1+iVal).*(e.^(2*k2*a) )./(1-iVal)));

C = (e.^(-i*k1*a)).* ((e.^(-k2*a)./ ( BLABLA1 ) ) + (e.^(k2*a)./ ( BLABLA2 )));
TChip = 4*(C.*conj(C))

plot (E, [TSoln;TChip]);
 
Physics news on Phys.org
Why not numerically integrate the differential equation. If the sign of the second derivative is right, the concavity of the solution will also be right.
 

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