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Qm finite well

  • Thread starter ehrenfest
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  • #1
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Homework Statement


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?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #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.
 
  • #3
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But I never used the constraint that square integral over all space of the piecewise wavefunction must be 1.
 
  • #4
Dick
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You used that it should be finite and then arbitrarily threw away a free constant.
 
  • #5
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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?
 
  • #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.
 
  • #7
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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:
  • #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.
 
  • #9
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Of course. I am talking about the derivatives at a.
 
  • #10
Dick
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The derivatives at a are continuous.
 
  • #11
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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.
 
  • #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.
 
  • #13
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Yes. I agree with that.
 
  • #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.
 
  • #15
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Thanks. I see.
 

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