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Quantum Tunnelling Problem

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A beam of particles of energy E is incident from the left on a rectangular step of infinite width and height U0, situated at the origin, with E > U0, as shown in the diagram.

    http://img39.imageshack.us/img39/5918/problemda.jpg [Broken]

    (a) Write down the Schrodinger equation for a particle in Region 1, and give a general expression for its wave function in terms of the wave number k1 in Region 1. Give the expression for k1 also. Repeat the same process for Region 2 and write an expression for k2.


    (b) By applying the boundary conditions at the origin, show that the reflection coefficient R is given by

    [itex]R=\frac{(k_1-k_2)^2}{(k_1+k_2)^2}[/itex]

    2. Relevant equations

    Relevant boundary conditions:

    [itex]A+B=C+D[/itex]

    [itex]ikA-ikB=-\alpha C + \alpha D[/itex]

    3. The attempt at a solution

    (a) In region 1 I believe the potential is 0 so the Schrodinger equation in this region is:

    [itex]-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi[/itex]

    or

    [itex]\frac{d^2 \psi}{dx^2} =k^2 \psi[/itex]

    [itex]k_1= \frac{\sqrt{-2mE}}{\hbar}[/itex]

    A general solution to this would be:

    [itex]\psi (x) = Ae^{ik_1x} + Be^{-ik_1x}[/itex]

    Now for Region 2 the potential is U(x)=U0 so the equation is

    [itex]-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2}-U_0 \psi = E \psi[/itex]

    [itex]\implies \frac{d^2 \psi}{dx^2} = -k^2 \psi[/itex]

    [itex]k_2= \frac{\sqrt{2m(E+U_0)}}{\hbar}[/itex]

    And wave function is [itex]\psi(x) = Ce^{-\alpha x} + De^{\alpha x}[/itex]

    where [itex]\alpha= \frac{\sqrt{2m(U-E)}}{\hbar}[/itex]

    I hope everything is correct so far. I would appreciate it if anyone could correct me if I'm wrong.

    (b) So I'm not sure if my approach is correct, but since in my notes the reflection coefficient is defined as

    [itex]R= \frac{|B|^2}{|A|^2}[/itex]​

    I have substituted k1 and k2 into (k1-k2)2/(k1-k1)2 to see if we can simplify it to yield the same expression:

    [itex]\frac{\left(\frac{\sqrt{-2mE}}{\hbar} - \frac{\sqrt{2m(E+U_0)}}{\hbar} \right)^2}{\left(\frac{\sqrt{-2mE}}{\hbar} + \frac{\sqrt{2m(E+U_0)}}{\hbar} \right)^2}[/itex]

    [itex]= \frac{-4mE+2mU_0}{2mU_0}[/itex]

    So how can I show that this is equal to the reflection coefficient R? :confused:
    Any guidance is appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
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