Molecular Hamiltonian - Ammonia

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Konte
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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.
 
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Konte said:
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?
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.
 
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DrClaude said:
What operator is that?
I mean to say : the double-well potential of ammonia.
You answered my question, I think.
Thanks.

Konte