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I Molecular Hamiltonian - Ammonia

  1. Dec 6, 2016 #1
    Hello everybody,

    The general expression of molecular Hamiltonian operator for any molecule is:
    $$\hat{H} = \hat{T}_n+\hat{T}_e+\hat{V}_{ee}+\hat{V}_{nn}+\hat{V}_{en}+\hat{f}_{spin-orbit} $$
    where:
    ##\hat{T}## correspond to kinetic energy operator
    ##\hat{V}## correspond to potential energy operator
    ##e## and ##n## subscripts correspond to electrons and nucleus which compose the molecule.
    My question is : when I try to apply this to the ammonia molecule case, it seems like incomplete for me because the famous "bi-stable" potential energy operator of ammonia is missing. How to introduce it?

    Thank you everybody.

    Konte.
     
  2. jcsd
  3. Dec 6, 2016 #2

    DrClaude

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    Staff: Mentor

    What operator is that???

    The double-well potential of ammonia comes about simply from solving the electronic problem with the Hamiltonian you gave, in the Born-Oppenheimer approximation. What you get then is a multidimensional surface, and there is a particular "cut" in that surface that corresponds to the "inversion coordinate," which is a double well.
     
  4. Dec 6, 2016 #3
    I mean to say : the double-well potential of ammonia.
    You answered my question, I think.
    Thanks.

    Konte
     
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