# I Atomic orbitals: change during excitation?

1. Apr 20, 2017

### pierce15

Sorry if this question has been asked before or is common knowledge. It seems to me that when one or more electrons in an atom is excited to a higher energy state, then the effective potential experienced by other electrons should be different from the potential in the ground state. Hence the energy of each state should shift slightly. Is that wrong, or has such an effect been observed?

2. Apr 20, 2017

### hilbert2

It's not really appropriate to talk about states of individual electrons in an atom, the correct description is to talk about the atom as a whole. However, if you use some approximate orbital model of an atom, then the effective nuclear charge seen by an electron depends on what orbitals the other electrons occupy.

3. Apr 20, 2017

### pierce15

Sorry, I'm a bit rusty. Could you say whether these statements are true/false so that I can ask the question correctly:

1. The total wavefunction of the atom (ignoring nuclear spin) is equal to the antisymmetrization of the wavefunction of each electron
1.5. The atom in an excited state can be completely described by which orbital each electron is in, right?
2. The "atomic orbitals" are eigenstates for each electron of the total hamiltonian due to the nucleus + other electrons
3. The ground state of the atom is an eigenstate of the hamiltonian of the entire system
4. If statement 2 is true: the energy of the ground state is equal to the sum of the energies of each state occupied by an electron

4. Apr 20, 2017

### hilbert2

If you have something like a lithium atom, its state is not described by saying "electron 1 has wavefunction $\psi_1 \left(\mathbf{r_1}\right)$, electron 2 has wavefunction $\psi_2 \left(\mathbf{r_2}\right)$ and electron 3 $\psi_3 \left(\mathbf{r_3}\right)$. Rather, we say that the state of the whole atom (assuming stationary nucleus) is some function $\psi (\mathbf{r_1},\mathbf{r_2},\mathbf{r_3},s_1 ,s_2 , s_3)$, where the $s_i$ are the spin variables.

In the orbital approximation (where electron correlation is ignored), however, it is assumed that a good approximation of the energy states can be obtained by making a product form wavefunction $\psi_1 \left(\mathbf{r_1}\right)\psi_2 \left(\mathbf{r_2}\right)\psi_3 \left(\mathbf{r_3}\right)$ and applying an antisymmetrization operator on it.

5. Apr 20, 2017

### pierce15

Ok, I see it now. The Hamiltonian for the system defines a set of energy eigenstates for the atom.

Does an ionized atom then have different energy levels from the original atom since it has a different hamiltonian?

6. Apr 20, 2017

### hilbert2

It's a different system after ionizatìon, so yes, it has different energy levels and spectral lines.

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