1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

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

    And yes, T = |F/A|2.
  8. Sep 24, 2011 #7
    Thanks a lot!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook