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Question on step potentials and schrodingers equatio

  1. Jul 9, 2009 #1
    I have a problem. I am trying to selfteach myself QM but I got stumped with this.

    Basically, in the issue of 1D step potentias, I know that you have to divide the problem into regions and solve the schrodinger equation for each one. However, when you solve it, you always get two solutions due to the nature of the second order linear differential equation. (One representig a current oming from the left and the other from the right) However, I know sometimes one of the solutions vanishes. For ezample. if V>E in a<x<infinity, the solution will be an exponential wih negative exponent because the positive one vanishes (due to the physicallity of quantum tunneling. In certain instances there will be two solutions because of reflection, etc. However, how do I know when a solution vanishes? Are they rules of thumb or you simply solve for the coefficients using initial conditions?
     
  2. jcsd
  3. Jul 9, 2009 #2
    A specific question would be if a potential well with infinite potenial in one side, zero in the middle, and a finite potenial in the right side would have both current going to the right and the left?
     
  4. Jul 9, 2009 #3

    JK423

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    Gold Member

    In this case (in "bold"):
    Lets say the wave is sent from the negative x-axis. In the region x<a, you will have two solutions because you have a wave travelling to the right (you sent it) and a reflected one from the potential. However, in the region a<x<oo we cant have two solutions. We only have a wave travelling to the right, no wave is travelling to the left since there is no other point (like a step potential) to reflect your wave.
    So imagine a wave travelling at x=a, then some been reflected and other transmitted. The one thats been transmitted will be "alone" in that region (x>a).
     
  5. Jul 9, 2009 #4

    diazona

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    Homework Helper

    Also, a solution that is non-normalizable has to vanish. This usually happens when it goes to infinity at some point, for example a solution of [itex]e^{bx}[/itex] on the domain [itex]a<x[/itex]. The wavefunction would go to infinity as [itex]x\to\infty[/itex] so there's no way to normalize it.
     
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