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Variational method in a finite square well

  1. Oct 10, 2009 #1
    I am trying to prove that there is always one bound state for a finite square well using variational method, and I am stuck. I've tried using e^(-bx^2) as my trial wave function, but I am left with E(b)=(hbar^2)b/2m - V, where V is the depth of the well. In this equation, taking the derivative in terms of b and setting it to zero doesn't lead to a critical value of b, am I doing something wrong?

    Thanks,
    Jacob
     
  2. jcsd
  3. Oct 10, 2009 #2
    I think you evaluated your potential wrong. If your well goes from 0 to a, then you need to evaluate
    [tex]
    -V \int_0^a \psi^*(x) \psi(x) dx
    [/tex]
    which should give you error functions.
     
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