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Rectangular Potential Barrier Where E=V0

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    V(x) = 0 for x < 0; V0 for 0 < x < a; 0 for x > a

    Particle of mass m


    Asks for general solution and transmission coefficient.

    2. Relevant equations

    Time-Independent Schrodinger Equation

    3. The attempt at a solution

    I've found region I (the left region where x <0) to be Aeikx+Be-ikx.

    Middle region to be C+Dx (not sure where this comes from but I read that this is the solution to the middle region when E=V0 so any help with this is appreciated).

    Right region where x > a to be Feikx+Ge-ikx.

    My textbook has the right region in an example as just Feikx. It goes through the explanation of limits mattering as x-> infinity or -infinity. I understand this for when E<V0, but the case seems completely different when E=V0.

    Any help is appreciated.

  2. jcsd
  3. Sep 24, 2011 #2


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    You do the same thing you did to find Aeikx+Be-ikx for region I, except this time the potential is V(x)=E.
    Do you know what eikx and e-ikx physically represent? There's nothing different between the E<V0 case and E=V0 case.
  4. Sep 24, 2011 #3
    For region I, the potential is zero. So I just get the Aeikx+Be-ikx answer. My question for this portion was about the limits. I know that if this was a finite square well, and if the potential was >0 for x < 0, then Be-ikx would have to be removed, because that would mean that the wave function would go to infinity.

    What's confusing me is how in my book, it has psiIII=Ceikx for when x > a (to the right of the barrier). I'm getting Ceikx+De-ikx. Why do they remove the "D" term? The potential is zero in that region.

    No I don't know what eikx and e-ikx physically represent.

    As for the cases being different: http://en.wikipedia.org/wiki/Rectangular_potential_barrier#E_.3D_V0

    I don't know where they get the C+Dx from.
  5. Sep 24, 2011 #4


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    If you apply the momentum operator to one of these functions, you get[tex]\hat{p}e^{ikx} = \frac{\hbar}{i}\frac{d}{dx}e^{ikx} = \hbar k\ e^{ikx}[/tex]so you can see it is an eigenfunction of the momentum operator with eigenvalue [itex]p=\hbar k[/itex]. In other words, it represents a particle with a definite momentum p. The one with the positive sign corresponds to a particle moving to the right, and the other, to a particle moving to the left.

    In these types of a problems, you have a particle incident on the barrier from the left. There will be a reflected wave and a transmitted wave. Can you figure out the rest from here?

    You say you find in region I that [itex]\psi_\mathrm{I} = Ae^{ikx} + Be^{-ikx}[/itex]. How did you get that?
    Last edited: Sep 24, 2011
  6. Sep 24, 2011 #5
    I see. I think I understand why region III only has the Ceikx. It is because the only direction a particle will be going is to the right. The other regions can have a particle being reflected.

    I find the wave functions for each region by using the Schrodinger Equation.


    For region I, V=0, so it becomes


    where k2=2mE/[itex]\hbar^2[/itex]

    which produces Aeikx+Be-ikx since it is a second order diff. eq.

    When V=V0, that gives


    where [itex]\alpha^2[/itex]=2m(V0-E)/[itex]\hbar^2[/itex].

    If V0=E, [itex]\alpha[/itex] becomes zero. So,


    Which would give C+Dx? Since it would be a first order polynomial? That makes sense to me.


    Is using F2/A2 the way to find the transmission coefficient? Ratio of transmitted particles to incident particles? (jtrans/jinc)
  7. Sep 24, 2011 #6


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    Yes, exactly. That's how you get Cx+D.

    And yes, T = |F/A|2.
  8. Sep 24, 2011 #7
    Thanks a lot!
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