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Schrodinger equation for potential step

  1. Jan 15, 2016 #1
    1. The problem statement, all variables and given/known data
    There is a stream of electrons with energy E, incident from x = -∞ on a potential step such that V(x) = ##V_{0}## for x<0 and 0 for x>0.
    E>##V_{0}##>0.
    Write the T.I.S.E for x<0 and x>0 and find the general solution for both.
    2. Relevant equations


    3. The attempt at a solution
    My main problem is I don't know how to account for the potential dropping, as it does here - in other examples, the particles always seem to start in the region of 0 potential and hit an area with potential V. So if an adaptation has to be made, I'm not sure how to do it. Here's my attempt for x<0:

    ##-\frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x^2}} + V_{0}\psi = E\psi##, I think...If that's it, then the general solution is ##Ae^{-kx}+Be^{kx}##. I know more can be done to that, I think I read something about how this wavefunction would basically always be infinite so you set B = 0. Or maybe it was A...

    And for x>0, I think
    ##-\frac{\hbar^2}{2m} \frac{\partial^2{\psi}}{\partial{x^2}} = E \psi##.

    For this, the solution would be ##Ae^{ikx}+Be^{-ikx}##.

    There are so many variations on this question, I feel like it's really important to understand how to adapt the Schrodinger equation! I'd hugely appreciate some general guidelines for different cases, for example E<V and E>V, stepping to lower or higher potential, that kind of thing. Because I don't really understand the changes made for each situation.

    Of course, if these sorts of guidelines aren't really possible I'd still really appreciate help just on this specific question! :)
     
  2. jcsd
  3. Jan 15, 2016 #2

    mfb

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    E>V0, the particle is not in a forbidden region. Your first approach should have "i" in the exponent as well, unless you want to deal with complex k (possible but doesn't help). At this point, you cannot set a component to zero.

    Don't use the same labels A and B for both regions, they are different parameters.

    To reduce free parameters: What do you know about the wavefunction directly at x=0?
     
  4. Jan 16, 2016 #3
    I'm not sure, I don't know if the potential is 0 or V.
    I'm not sure, I don't know if the potential is zero or V. It's discontinuous, isn't it? So maybe the wavefunction is zero?

    The question suggests a boundary condition can be inferred for x tending to infinity. Is there something special about that region? As far as I can see the potential would stay zero, so there's nothing particular about that can be said about the wavefunction...
     
    Last edited: Jan 16, 2016
  5. Jan 16, 2016 #4

    mfb

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    It changes there, but that is not the point. There are statements about the wave function that do not depend on (finite) potential values. Properties where the limit from both sides is the same...
    The wave function does not have to be zero, but it has to satisfy two other constraints.
    It does not, and there is nothing special about that region. The conditions come from the connection at x=0 and from the additional knowledge that nothing is incoming from the positive x-direction.
     
  6. Jan 16, 2016 #5
    Oops, sorry, I forgot to post that part. But what it literally says is:
    Explain how the boundary conditions at x tending to infinity let you determine one of the constants in the general solution for the region x>0.

    So the limit as x tends to zero has to be the same for both equations?
     
  7. Jan 16, 2016 #6

    mfb

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    The limit of what?

    Okay, strange phrasing. Well, in the region x>0 you have two terms. Can you describe the physical meaning of those terms individually?
     
  8. Jan 16, 2016 #7
    Sorry, limit of the wavefunction. Another thing: the wavefunction wouldn't tend to zero as x tends to infinity or negative infinity, would it? Doesn't that mean it can't be normalised?

    As to the physical meaning, are the two parts related to different directions of travel in some way? If they are, then I don't think there should be any motion in the positive x direction, and the coefficient of the term representing travel from right to left should be zero... So the ##e^{-ikx} term should have a coefficient of zero?
     
  9. Jan 16, 2016 #8

    mfb

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    Correct. That is one condition. Can you say something about the derivative as well?
    Right, it cannot be normalized.
    Right.
    If "right" is larger x, why do you expect no contribution here? Particles are coming from negative infinity and "fall down" a step, surely something will keep on going to the right?
    There is nothing moving in the opposite direction: nothing comes from +infinity.
     
  10. Jan 16, 2016 #9
    Oh, sorry, I'm starting to get very mixed up - I've read a lot of these types of question now. Yes, nothing comes from right to left is what I meant. So it's the coefficient of the other term that should be zero.
     
  11. Jan 16, 2016 #10
    About the derivative - that should be continuous, I presume.
     
  12. Jan 16, 2016 #11

    vela

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    You had the right term (##e^{-ikx}##) but wrong/inconsistent description.

    Back to your original post, do you understand mathematically why you get ##e^{\pm kx}## in some cases and ##e^{\pm ikx}## in others?
     
  13. Jan 16, 2016 #12
    I feel like I'm missing some pretty fundamental stuff. I don't think I do understand why, no. Based on the first post by mfb, is it related to allowed and forbidden regions?
     
  14. Jan 16, 2016 #13

    vela

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    Rearranging the Schrödinger equation slightly, you get
    $$\frac{\partial^2 \psi}{\partial x^2} = \frac{2m(E-V_0)}{\hbar^2} \psi.$$ Try plugging in a solution of the form ##\psi(x) = e^{rx}##. How is ##r## related to ##E## and ##V_0##?
     
  15. Jan 16, 2016 #14
    ## r = \frac{\sqrt{2m(E-V_0)}}{\hbar}##
    And doing the same for ##e^{ikx}## gives

    ##r = \frac{\sqrt{-2m(E-V_0)}}{\hbar}##
    So ##e^{ikx}## is a solution for a different form of the Schrodinger equation, although I'm not sure which form.
     
  16. Jan 16, 2016 #15

    vela

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    What can you say about ##r = \sqrt{2m(E-V_0)}/\hbar## when ##E > V_0## and when ##E < V_0##?
     
  17. Jan 17, 2016 #16
    Complex when ##E<V_0##, and real when ##E>V_0##. I can sort of see how that might relate to allowed and forbidden regions, a particle approaching a potential step V with energy E<V probably shouldn't be transmitted. And it has a complex wavenumber. Although actually, if the wavenumber is purely imaginary (which it is), then overall that wavefunction would be real.
     
  18. Jan 18, 2016 #17

    blue_leaf77

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    That's right in a sense that the probability current in ##x>0## vanishes if ##E<V_0##.
    The wavefunction in ##x>0## is still complex even if the solution in this region is of decaying function. The complex-ness of the wavefunction comes as a consequence of the continuity condition at the boundary (the transmission coefficient is complex).
     
  19. Jan 18, 2016 #18

    vela

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    Where?
    The solutions are of the form ##e^{rk}##. In the classically forbidden region, ##E<V_0##, the roots ##r## are real. You write ##r=\pm k##, and the solutions are of the form ##e^{\pm kx}##. When ##E>V_0##, ##r## is imaginary. You write ##r = \pm ik## where ##k## again is real, and the solutions are of the form ##e^{\pm ikx}##. So you have exponentially growing and decaying solutions. In the region where ##E>V_0##, you have sinusoidal solutions.

    Recall that the wave function is a linear combination of the solutions, so even if the solutions you choose are real, the wave function can be still be complex.
     
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