Ground State Energy of Diatomic Molecule

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SUMMARY

The discussion focuses on calculating the ground state energy of a diatomic molecule using the Hamiltonian H = l²/2I + F*d*cos(theta), where d represents the dipole moment. The ground state energy is expressed as E_0 = ħ*l*(l+1)/2I. The user seeks clarification on the correctness of this energy expression and guidance on applying first-order perturbation theory to determine the energy shift due to the perturbation term. The first-order correction to the ground state energy is given by E_1 = , where V is the perturbing Hamiltonian.

PREREQUISITES
  • Understanding of quantum mechanics, specifically Hamiltonians and eigenvalues.
  • Familiarity with perturbation theory in quantum mechanics.
  • Knowledge of angular momentum in quantum systems, particularly the role of l (angular momentum quantum number).
  • Basic concepts of dipole moments and their significance in molecular interactions.
NEXT STEPS
  • Study the derivation of eigenfunctions and eigenvalues for the unperturbed Hamiltonian H_0 = L²/2I.
  • Learn about first-order perturbation theory and its applications in quantum mechanics.
  • Explore the implications of dipole moments in molecular quantum mechanics.
  • Investigate scenarios where excited states may exhibit lower energy than ground states in specific quantum systems.
USEFUL FOR

Students and researchers in quantum mechanics, particularly those focusing on molecular physics and perturbation theory, will benefit from this discussion.

greisen
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I am looking at a diatomic molecule where the Hamiltonian is given as

H = l²/2I + F*d*cos theta

where d is the dipole moment. The term F*d*cos theta is small. I write the energy of ground state as

E_0 = \hbar*l*(l+1)/ 2I

Than I have to determine how much the ground-state energy changes as a result of interaction with the field. I have two questions:

1. Is the ground state energy correct - it should not be <psi_0|H|psi_0)?

2. How to proceed using first-order perturbation theory


Thanks in advance
 
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1. Write down the eigenfunctions of the unperturbed hamiltonian, H _0 = L^2/2I (recall from the hydrogen atom), and it's eigenvalues (already written above).

2. From the eigenfunctions, \phi _n ^{(0)} and the eigenvalues, E_n^{(0)}, what expression gives you the first order energy shift due to the perturbing hamiltonian, H _1?
 
so the change will be given as
E_1 = <psi_0|V|psi_0>
where V is the small term?
 
That's right, that's the first order correction to the ground state energy.
 
Thanks could there be a situation where an electronic excited state could be "ground state" for a molecule - having lower energy than a none excited state?
 
The ground state is the state with minimum energy. If the energy is quantized (which happens only in bound states), then states with higher energy are called "excited states". So your question is meaningless...
 

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