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Quantum rectangular barrier

  1. Feb 9, 2007 #1
    Ok, so I've been dealing with this problem for a while and can't figure it out. (I tried to clean it up, but I don't know LaTeX, hopefully it is more clean in post #2; problem stated in #1, and my work in #2)

    --Consider a (plane-wave) particle tunneling through a rectangular barrier potential w/ height v, and width a (the particle has E<v)....

    a) write general solutions of stationary state functions in each 'region' of the potential.

    --Done, no problem.

    b)Find four relations among the five arbitrary constants in part a).

    --Again, no problem

    c) Use relations in part b) to equate transmission coefficient.


    --Part c) is the problem, I have no idea what to do...
     
    Last edited: Feb 9, 2007
  2. jcsd
  3. Feb 9, 2007 #2
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    a)

    Region I: x<0 : psi(x)= A exp[i k x] + B exp [-i k x]

    Region II: 0<x<a: psi(x)= C exp [K x]+ D exp [-K x]

    Region III x>a : psi(x)= E exp [i k x]


    [tex]\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }i \hbar \pd{\Psi}{t}{} =- \frac{\hbar^2}{2 m} \ \pd{\Psi}{x}{2} + V \Psi[/tex]

    k^2= 2mE/hbar^2
    K62= 2m(E-v)/hbar^2

    c) took derivative of psi I = derivative psi II, at x=o.

    Derivative psi II = derivative psi III, at x=a

    psi(o)I=psi(0)II

    psi(a)II=psi(a)III

    3. The attempt at a solution

    I tried to linearly add:

    psi(0)I + d/dx psi(o)I = psi(o)II + d/dx psi(o)II

    and same for when x=a for psi II and III.

    I attempted to eliminate C and D constants, and solve for E. I let A =1 (thinking other waves than the incident have some multiple of amplitude.)

    I thought that abs(E)^2= transmittion coeficient, is this not true? And B would be the reflection coefficient?

    Where am I going wrong?
     
    Last edited: Feb 9, 2007
  4. Feb 9, 2007 #3
    The transmission coefficient represents the probability flux of the transmitted wave relative to that of the incident wave.
    I hope this help you
     
  5. Feb 10, 2007 #4

    Ok, so the probability of incident wave over transmitted?

    How do I get there from general solutions though?

    (A*A)/(E*E) ?
     
  6. Feb 10, 2007 #5

    I finally figured it out.

    Eliminate coefficients of general solution for inside the barrier by applying boundry conditions at x=0, x=a.

    Solve for coefficient of transmission, take abs, square it, bam, it's done.
     
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