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Quantum rectangular barrier

  • Thread starter judonight
  • Start date
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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:

Answers and Replies

32
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1. Homework Statement



2. Homework 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?
 
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233
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The transmission coefficient represents the probability flux of the transmitted wave relative to that of the incident wave.
I hope this help you
 
32
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The transmission coefficient represents the probability flux of the transmitted wave relative to that of the incident wave.
I hope this help you

Ok, so the probability of incident wave over transmitted?

How do I get there from general solutions though?

(A*A)/(E*E) ?
 
32
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The transmission coefficient represents the probability flux of the transmitted wave relative to that of the incident wave.
I hope this help you

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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