Ground state energy

  • Thread starter greisen
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  • #1
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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
 

Answers and Replies

  • #2
Gokul43201
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1. Write down the eigenfunctions of the unperturbed hamiltonian, [itex]H _0 = L^2/2I[/itex] (recall from the hydrogen atom), and it's eigenvalues (already written above).

2. From the eigenfunctions, [itex] \phi _n ^{(0)} [/itex] and the eigenvalues, [itex]E_n^{(0)} [/itex], what expression gives you the first order energy shift due to the perturbing hamiltonian, [itex]H _1 [/itex]?
 
  • #3
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so the change will be given as
E_1 = <psi_0|V|psi_0>
where V is the small term?
 
  • #4
dextercioby
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That's right, that's the first order correction to the ground state energy.
 
  • #5
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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?
 
  • #6
dextercioby
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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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