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Solving schrodinger, reflection coefficient

  1. Mar 7, 2012 #1
    Consider the potential

    [tex]
    V(x) =
    \begin{cases}
    0, & x < -a & (I) \\
    +W, & -a < x < a & (II) \\
    0, & x > a & (III)
    \end{cases}
    [/tex]

    for a particle coming in from the left ([itex]-\infty[/itex]) with energy E (0 < E < W). Give the solution to the Schrodinger equation for I, II and III and use these to calculate the reflection coefficient.

    I have the answer to this problem in front of me, but I don't understand. First they calculate the solution to the Schrodinger equation for I, II and III:

    [itex]\psi_I(x) = Ae^{ikx} + Be^{-ikx}, \ \mbox{with} \ k = \frac{\sqrt{2mE}}{\hbar}[/itex]

    [itex]\psi_{II}(x) = Ce^{\kappa x} + De^{-\kappa x}, \ \mbox{with} \ \kappa = \frac{\sqrt{2m(E - W)}}{\hbar}[/itex]

    [itex]\psi_{III}(x) = Fe^{i k x}, \ \mbox{with} \ k = \frac{\sqrt{2mE}}{\hbar}[/itex]

    I understand [itex]\psi_I[/itex], but not [itex]\psi_{II}[/itex] and [itex]\psi_{III}[/itex]. Why is there no i in [itex]\psi_{II}[/itex]? And why is [itex]\psi_{III}[/itex] only a single term? I imagine it has something to do with the particle coming from the left?
     
    Last edited: Mar 7, 2012
  2. jcsd
  3. Mar 8, 2012 #2

    Nabeshin

    User Avatar
    Science Advisor

    Because these are classically forbidden solutions, and the probability of the particle being found in such a region exponentially decays.

    Correct, the other term would represent some particle coming in from the right (which is usually not assumed).
     
  4. Mar 8, 2012 #3
    If I were to solve the equation for II, then [itex]Ce^{i \kappa x} + De^{-i \kappa x}[/itex] is mathematically still a valid solution right? I don't understand why it is not allowed.

    But why do I and II have two terms if the second term represents a particle coming from the right? Shouldn't they have only one term too then?
     
  5. Mar 8, 2012 #4

    Nabeshin

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    Science Advisor

    Nope. You should go through and solve why these are the solutions here, but essentially you have something like (don't quote me exactly on this) [itex]\kappa \sim \sqrt{E-V}[/itex] So in the regions I and III, V=0 and this is real and fine. But in the region where V=V0>E, this is imaginary, which is why the thing you would normally write, e^(i*k*x), goes to e^(-kx) (Where now k in this expression is understood to be the real part). [I think there is a typo in what you wrote somewhere, and the sign inside your definition of kappa should be switched so that it is by definition real]


    In region I, the term moving to the left represents the reflected wave off of the barrier between regions I/II. In region II, similarly, the term moving to the left represents the reflected wave off the barrier between II/III.
     
  6. Mar 8, 2012 #5
    Thanks!
     
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