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Delta wall and infinite square well potentials ,and 2 other questions

  1. Mar 2, 2014 #1
    Consider the following potential function: V=αδ(x) for x=0 and V=∞ for x>a and x<-a , solve the shroedinger equation for the odd and even solutions.

    solving the shroedinger equation I get

    ψ(x)=Asin(kx) +Bcos(kx) for -a<x<0

    and

    ψ(x)=Asin(kx) +Bcos(kx) for 0<x<a


    is it correct that I tried to solve the Shroedinger equation independently for both halves of the potential disregarding even or odd solution approaches (ψ(x)=ψ(-x) and ψ(x)= -ψ(-x))?

    and also is it correct to assume that ψ(0)=0, since there's an infinite potential there?

    final question (not related to this problem): consider the Delta potential well. In the book I am studying from, when the author was solving the Shroedinger equation he implicitly assumed that the energy of the particle before and after the well are the same . But isn't it careless , for the lack of a more polite word, to assume that? Isn't it possible that the particle loses or gains some energy as it crosses the well?
     
  2. jcsd
  3. Mar 2, 2014 #2

    DrDu

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    Science Advisor

    No, as long as alpha is finite, psi(0) will in general be different from 0, at least for the even solution. Try to integrate the Schroedinger equation from -ε to +ε to obtain a relation between psi at x=0 and it's derivative.

    To your second question: Where should the energy come from? A potential change already gives you the change of energy and the potential before and after the well are equal.
     
  4. Mar 3, 2014 #3
    I see. Thank You .
     
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