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Qm finite well

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Say I have a potential that is infinite from - infinity to 0, -V_0 from 0 to a, and 0 from a to infinity.

    Then I have three unknowns, two for the plane wave solution in the finite well and one for the exponential outside the well.

    But I also have four equations. Two wavefunction continuity boundary conditions and a wavefunction derivative continuity boundary conditions. I also have the normalization equation!

    How is this possible? How can the solutions depend on which 3 of the four equations I select to use?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2007 #2

    Dick

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    You already used the normalizability constraint in eliminating the exp(+ax) solution outside the well. Otherwise you would have four equations in four unknowns.
     
  4. Oct 30, 2007 #3
    But I never used the constraint that square integral over all space of the piecewise wavefunction must be 1.
     
  5. Oct 30, 2007 #4

    Dick

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    You used that it should be finite and then arbitrarily threw away a free constant.
     
  6. Oct 30, 2007 #5
    But if I use the 3 boundary conditions to determine the 3 remaining constants then I could get a wavefunction with no adjustable parameters left and with a square integral over all of space not equal to 1, right?
     
  7. Oct 30, 2007 #6

    Dick

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    No. If you've done this problem, and I know you have, you should know that the remaining three constants are only determined up to a multiplicative constant. If you want to continue bickering about this just do it with four equations and four unknowns. The same thing will happen.
     
  8. Oct 31, 2007 #7
    I do not want to bicker. I just want to understand the problem.

    I found that it is not even possible to implement the continuity of the derivatives at a which makes some sense considering the shape of the functions. So I was wrong, there are only a total of three constraints left and all is well.

    I am not sure why it makes sense that the derivatives would not be continuous since there are no delta functions at boundaries, though?
     
    Last edited: Oct 31, 2007
  9. Oct 31, 2007 #8

    Dick

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    The derivative at zero is not continuous because there is an infinite potential at x=0. That's worse than a delta function.
     
  10. Oct 31, 2007 #9
    Of course. I am talking about the derivatives at a.
     
  11. Oct 31, 2007 #10

    Dick

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    The derivatives at a are continuous.
     
  12. Oct 31, 2007 #11
    So, maybe I did something wrong. I still do not understand what you were saying about the "the remaining three constants are only determined up to a multiplicative constant".

    We have three equations and three unknowns, we should be able to uniqually determine each of the unknowns if there is a solution.
     
  13. Oct 31, 2007 #12

    Avodyne

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    The point is that your four equations will determine the three coefficients AND the set of allowed energy eigenvalues.
     
  14. Oct 31, 2007 #13
    Yes. I agree with that.
     
  15. Oct 31, 2007 #14

    Avodyne

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    So, that's the solution. If you don't use normalization, then you don't determine the values of all three coefficients, but only their ratios; the 3rd equation then determines the allowed eigenvalues.
     
  16. Oct 31, 2007 #15
    Thanks. I see.
     
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