Bound state problem

  • Thread starter amjad-sh
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  • #26
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There are bound, not continuous functions.
A continuous function over a compact support has to be bound, however.
and once it is bound the definite integral of it from ##-\epsilon \to \epsilon## will approach zero if epsilon approaches zero
Sure.
 
  • #27
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There are bound, not continuous functions.
A continuous function over a compact support has to be bound, however.Sure.
I in fact posted a thread couple of days concerning a problem somehow similar to this problem but much more complicated.
The difference in that problem is that the wavefunctions are scattering states and not bounded states.
The problem asked to derive the scattering coefficients by explicitly solving the Hamiltonian.
I used the same approach I used for this problem but didn't get an answer.
I would be glad if you can take a look at it: www.physicsforums.com/threads/solving-scattering-problem-including-spin-flip.952575/#post-6035381
some of the detailed steps may be wrong but I want know if my general approach is right.
There are bound, not continuous functions.
A continuous function over a compact support has to be bound, however.
I will appreciate it if you can also send for me a pdf that explains these concepts in details.
 
  • #28
George Jones
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But why the second derivative is not bound?

What is the derivative of a step function?
 
  • #30
George Jones
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What is the derivative of a step function?

the dirac delta function.


Since
Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0.

the first derivative has a jump discontinuity at ##x=0##. This means that the second derivative (the derivative of the first derivative) involves a Dirac delta function, and, consequently, the second derivative is unbounded.
 

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