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Standing waves in a funny Potential distribution

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data

    The description of the potential distribution is given in the attached image.
    The particle arrives from the left with E>V0.

    write the solutions to the S.E in regions x<o and x between o and a


    2. Relevant equations
    I believe psi(x)= e^ikx+Re^-ikx in x<0
    and psi(x)=Ae^iqx+Be^-iqx for x b/w o and a.


    3. The attempt at a solution
    My question is, since there is complete reflection occuring at x=a, can A=B in region x b/w 0 and a? If so, there will be destructive interference in the region, giving R=1, which is what we are asked to prove in the question. Is this approach of equating coefficients of wave traveling in +-x directions in this region applicable?
     

    Attached Files:

  2. jcsd
  3. Dec 19, 2008 #2

    turin

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    Homework Helper

    The potential:
    V=0 for x<0
    V=V0 for 0<x<a
    V->infinity for x>a
    You have one boundary condition at x=a that relates A and B. You have two boundary conditions at x=0 that relate A, B and R. Once you have determined k and q (which I'm assuming you know how to do), then I believe you simply apply these boundary conditions.
     
  4. Dec 19, 2008 #3
    thanks turin, I understand the problem well. My question is regarding the relation of the coefficients, A and B; with the respective intensities.

    Since R=1 at x=a, I would immediately assume that B=A
    instead of having to use Boundary conditions to find coefficients.
    Would this be correct logic?
     
  5. Dec 19, 2008 #4

    turin

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    No. You're contradicting yourself. You specified R for x<0, and now you want to talk about R at x=a, which doesn't even make sense, unless this is somehow a different R than the coefficient of the exponential that you originally gave. It's been a while since I solved one of these problems, so I can't remember if it should turn out that A=B is, in fact true; however, your logic to arrive at this conclusion is flawed.
     
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