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Scattering-Transmission-Reflection in box potential? Help. Please. Please.

  1. May 17, 2004 #1
    I'm trying to analyze the reflection of a plane wave of energy E on the box potential where

    v(x)= 0 for x<-a
    -Vnaught for -a<x<0
    infinity for 0<x

    and I'm trying to solve the schroedinger eqn. through:

    Phi(x) = e^ikx + r(k)*e^-ikx in region 1
    A(k)sin(k'x) in region 2

    Continuity of phi and phi'/phi at x=-a.

    I'm trying to solve for r(k) and A(k), so I've gone and derived all the nastiness, and I get something like r=2ikr'- r'' and A''/A= (-hbar^2/2m)-Vnaught-E


    I'm supposed to be able to tell what it means physically that the abs. value of r(k) = 1? I can'tget r(k) to equal one, and I can't find A max. How would I do this?? :cry:
     
  2. jcsd
  3. May 18, 2004 #2

    turin

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    I don't know the answer, but I can offer suggestions:

    - Define your regions. We can guess/figure out what you mean by "region I" etc., but we don't really have any way of knowing.

    - Why is there a continuity of phi'/phi. I've never heard of that. I suppose that says the same thing as continuity of phi', since continuity of phi has already been established?

    - What are your B.C.'s at the other boundary (x = 0)?

    - Is Phi(x) due to an incident particle at a particular momentum, or is this just a stationary state? I don't understand why you have r as a function of k unless Phi(x) is an integral over k.

    - Should A(k) really be A(k')?
     
  4. May 18, 2004 #3

    Dr Transport

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    if the potential at x=0 is infinite, the boundary condition requires that the wave function is zero, in this case for all x > 0. the condition [tex] \frac{\psi '(x)}{\psi(x)} [/tex] is called the logrithmic continuity of the wave function.

    The original question was a tiny bit sloppy (sorry clumsy9irl) but I recognized it fairly quickly. Be careful in your notation, it is the little things that will jump up and bite you later.

    keep working in this problem, the answer will come out. It is analogous to a 3-d shperical poential well.
     
  5. May 18, 2004 #4

    turin

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    What condition? Did you mean something more like:

    (1) [tex] \frac{\psi_1 '(x_0)}{\psi_1(x_0)} = \frac{\psi_2 '(x_0)}{\psi_2(x_0)} [/tex]

    What I don't get is, if we impose:

    (2) [tex] \psi_1(x_0) = \psi_2(x_0) [/tex]

    then what does (1) say that (3) does not:

    (3) [tex] \psi_1 '(x_0) = \psi_2 '(x_0) [/tex]
     
  6. May 19, 2004 #5

    Dr Transport

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    it is just a way to express the continuity of the wavefunction........
     
  7. May 19, 2004 #6

    (1) is nothing more than equation (3) divided by equation (2). Any two of these equations are enough to pin down the Boundary Conditions.
    Cheers,
    Norm
     
  8. May 19, 2004 #7

    turin

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    Thanks, Dr Transport and Norman. I just wanted to make sure I wasn't missing something.
     
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