# I Molecular Hamiltonian - Ammonia

1. Dec 6, 2016

### Konte

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. Dec 6, 2016

### 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.

3. Dec 6, 2016

### Konte

I mean to say : the double-well potential of ammonia.
You answered my question, I think.
Thanks.

Konte