# Orbital Energy states

1. Nov 16, 2015

### UchihaClan13

Guys
I have a doubt
When we calculate the trial function
We do it for the wave function of the orbitals
Right in order to get the total orbital energy
(Which included the energy of the electron) and that of the orbital
Well my question
Is does the orbital possess
Some energy even if the electron is not there
And would it have any essence??
Help is much appreciated! !:)

2. Nov 17, 2015

### UchihaClan13

And guys
Another doubt
How to calculate the energy of an orbital without the electron

3. Nov 17, 2015

### Staff: Mentor

Empty orbital doesn't have any energy - it is an electron occupying an orbital that has the energy.

4. Nov 17, 2015

### Staff: Mentor

I think that this is easier to understand in the case of a single electron. What you are solving for are the energy eigenstates of the Hamiltonian for the nucleus + electron system. The energy you get is the energy of the electron when it is in that state. The orbitals themselves do not have any energy or "essence".

5. Nov 17, 2015

### UchihaClan13

thanks both of you
i overlooked that the energy was a eigenvalue of the wave function
and a eigenstate of the hamiltonian
Thanks again!!

6. Nov 17, 2015

### Staff: Mentor

Wave functions don't have eigenvalues. Operators have eigenvalues and eigenstates.

7. Nov 17, 2015

### UchihaClan13

sorry

8. Nov 18, 2015

### DrDu

Yes and no. An orbital is a solution of the Schroedinger equation with a corresponding eigenvalue of energy. So you can calculate an energy irrespective of whether the orbital is occupied or not. Sometimes this is useful, e.g. to calculate possible excitation energies, although you can argue that this involves the orbital being occupied at some point of time.

9. Nov 18, 2015

### Staff: Mentor

This is bordering philosophy

10. Nov 18, 2015

### DrDu

Admittedly, but e.g. to apply the aufbau principle, you first have to calculate the energies before you can fill the orbitals.

11. Nov 18, 2015

### Staff: Mentor

Yes, but do we calculate energy of "an empty orbital" or of "an electron occupying the orbital"? Schroedinger equation describes an electron interacting with other charges, doesn't it? At least at far as I remember the way Hamiltonians are constructed, they contain these interactions, so it seems to me the calculations are based on the assumption electron is there.

12. Nov 19, 2015

### DrDu

I fear it is not so easy, as orbitals need not be either completely filled or empty as in QM we may also observe superpositions, or, to say it differently, an electron need not be in an eigenstate of the Hamiltonian.
For example, if you bring suddenly a proton close to a hydrogen atom, then the electron can be described as being in a 50-50 superposition of the bonding and the antibonding MO. It will start to oscillate between the two atoms with a frequency given exactly as $\nu=\Delta E/h$ where $\Delta E$ is the energy difference between the bonding and anti-bonding orbital.
It can be shown that it would lead to logical contraditions if we were to assume that the electron is "in reality" in a definite energy eigenstate, so we can't ascribe the energy to the electron. Nevertheless, we can observe the energetic splitting with high accuracy. So I would rather ascribe the energy to the orbital than to the electron.