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Delta potential barrier in a box

  1. May 27, 2008 #1
    i unfortunately couldn't find a solution to this problem although it seems like a classical textbook problem...

    how can i solve the (time independent) schroedingerequation for the following potential

    [tex]V(x) = \infty[/tex] for x<=-1
    [tex]V(x) = a\delta(x)[/tex] for -1<x<1
    [tex]V(x) = \infty[/tex] for x>=1

    so at x=0 there is a barrier.

    i thought i can set the solutions to the left and right of x=0 for a free particle but i don't know how to handle x=0.

    Last edited: May 27, 2008
  2. jcsd
  3. May 27, 2008 #2
    I haven't done this, but I think I can see how it can be done. Basically you need to define the wave function [tex]\psi(x)[/tex] in parts, for intervals -1<x<0 and 0<x<1 separately (they will be trigonometric functions), so that they vanish at the end points x=-1 and x=1, and that you get certain kind of sharp angle at the x=0. You should somehow solve how big jump there has to be in the derivative [tex]\psi'(x)[/tex] at the origin (this could be the most understanding demanding part. You can ask more questions if you don't know how to calculate with equations like [tex]D_x\theta(x)=\delta(x)[/tex] and so on). After this, substitute in some attempt of trigonometric functions with some constants, and try to figure out what set of equations you get for the constants.
  4. May 27, 2008 #3


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    You need to apply BCs to determine the coefficients of the plane wave states. But there's one trick here: while the wave function itself is continuous at x=0, its derivative is not. The discontinuity in [itex]\partial \psi/\partial x [/itex] at x=0 (due to the delta function potential there) can be found by integrating the SE over a tiny window about x=0 and taking the limit that the width of this window goes to zero.
  5. May 29, 2008 #4
    i think the idea here is this: you have and incident wave, and reflected wave and a transmitted wave. split the problem into two parts. in region (1) (ie before the potential) the wave could be of the form q(x) = Aexp(ikx) + bexp(-ikx) (+ikx for he incident, -ikx for reflected). and in region (2) (after the potential) q(x) = Cexp(ikx) (transmitted wave).
    one boundary conditon obviously is continuity of q(x), but the bc of the first derivative is found as said by Golkul43201.
  6. May 29, 2008 #5
    It is not necessary to start with plane waves. You would then have to solve suitable superpositions to get the boundary conditions at the x=-1 and x=1 correctly. But we already know that the wave function is going to be trigonometric functions that vanish at these end points. So the attempt

    \psi(x) = A\sin(B(x+1)),\quad -1<x<0

    \psi(x) = C\sin(D(x-1)),\quad 0<x<1

    gives a good start. The task that remains to be carried out, is that one must solve what constants A,B,C,D satisfy the correct boundary condition at the origin x=0.
  7. May 29, 2008 #6
    oh yeah of course, they aren't normalizable. i completely spaced on the fact that this is in a box. obviously we need bound states... sorry to the op.
  8. May 29, 2008 #7


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    Oops. I did too. Nevertheless, the procedure I outlined ought to work.
  9. May 30, 2008 #8
    thanks for your answers.

    i'll outline the solution method i found for later viewers:

    this potential can be handled exactly the same way as the infinite square well with the exception of having 2 conditions at x=0. one to have a continuous wave function so [tex]\Psi_{left}(0)=\Psi_{right}(0)[/tex] and the other one to manage the delta potential which demands for a discontinuous first derivation of [tex]\Psi(x)[/tex] at x=0. this discontinuity condition is obtained by integrating over the schroedingerequation.

    thanks again and greetings.
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