- #1

Konte

- 90

- 1

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.