Kronig-Penney model only with cosh

  • Thread starter 8Apeiron8
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  • #1
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Hey!

Why i can't use as approach for the eigenfunctions in the regions of high potential energy for bound states only Acosh(Qx+m) instead of A'cosh(Qx)+B'sinh(Qx), where m is the midpoint of the wall?

If i put the origin of the coordinate system in this midpoint and use the mirror-symmetry of the chrystal, the sinh term must vanish.

Many thanks,

Felix
 
Last edited:

Answers and Replies

  • #2
DrDu
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You can do so for the state of lowest energy or k=0. For states with different wavevector k, the symmetry is broken by the exp(ikx) term.
 
  • #3
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Sry but i don't understand this. The exp(ikx) term is part of the approach for the region with low potential energy, why it breaks the symmetry in the other region?

Could the reason be, that only the probability density must be mirror symmetric but not the wavefunction?

sinh^2 is mirror symmetric around the origin!!
 
Last edited:
  • #4
DrDu
Science Advisor
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Maybe you could specify exactly the hamiltonian and the wavefunctions you are talking about?
 

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