Is it possible to calculate energy of electron in any orbital and atom in Schrödinger wave model theory?

Is it possible to calculate the energy of electron in any orbital and atom in the Schrödinger wave model theory? If so, how? E.g. energy of the $3s^2$ electron of the $\ce{Na-}$ ion.




It is surely *possible*. Note though, that for many-electron systems orbital picture is an *approximate* description, but that is an approximation on which the whole general chemistry is based on, so that is fine from that perspective. The approximation is known as the [Hartree-Fock](https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method) (HF) model, and thus, to get the energies of electrons occupying some orbitals all you need to do is to solve the Schrödinger equation in this approximation.




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As a quick exercise one can do HF calculation in, say, Gaussian, with the following input file:



```
#P HF/aug-cc-pVTZ Pop=Full

Na-

-1 1
 Na                 0.00000000    0.00000000    0.00000000

```

to get this picture of occupied orbitals:


![enter image description here](https://i.stack.imgur.com/edzNn.png)


The HOMO with the energy of $-0.01288 \, \mathrm{Hartree} = -0.3505 \, \mathrm{eV}$ is basically the $\mathrm{3s}$ orbital you're looking for (trust me). Note, however, that this energy is approximate since the basis set (aug-cc-pVTZ) is finite. We could do better than that, but that is a different story.




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In response to permeakra (since he does not trust his colleagues) I quote an authoritative reference in the field which explicitly talks about individual electrons occupying individual spin-orbitals and orbital energies. As a reference I choose the infamous book entitled "Modern Quantum Chemistry" written by Attila Szabo and Neil S. Ostlund.


Quote #1 (Szabo & Ostlund, p. 50)



> 
> This Slater determinant has $N$ electrons occupying $N$ spin orbitals
>  $(\chi\_i, \chi\_j, \dotsc, \chi\_k)$ without specifying which electron
>  is in which orbital.
> 
> 
> 


Quote #2 (Szabo & Ostlund, p. 54)



> 
> $$ f(i) \chi(x\_i) = \varepsilon \chi(x\_i) \tag{2.52} $$ [...] 
> 
> 
> The
>  solution of the Hartree-Fock eigenvalue problem (2.52) yields a set 
>  $\{\chi\_k\}$ of orthonormal Hartree-Fock spin orbitals with orbital
>  energies $\{\varepsilon\_k\}$.
> 
> 
> 


Below I also quote what Szabo & Ostlund have to say about Koopmans' theorem, because I already know that won't trust me.



> 
> Quote #3 (Szabo & Ostlund, p. 110)
> 
> 
> The first theorem [Koopmans' theorem] constitutes an interpretation of
>  the Hartree-Fock orbital energies as ionization potentials and 
>  electron affinities.
> 
> 
> 


See, orbitals with their energies & occupancies do exist in the Hartree-Fock theory. It is the interpretation of orbital energies as ionization potentials which requires an additional theorem.


 12

No. In QM electrons are indistinguishable.


Still, it is possible to associate a somewhat characteristic value, using various spectroscopic methods, i.e virtually 'moving' some electron from one orbital to another (or removing it entirely) and calculating the corresponding energy. This energy, however, is unprecise, as other electrons 'feel' the move of the 'moved' one. Still, something is better than nothing, so people use what they can.


 0

User permeakra's observation is right, but the inference is not. User Wildcat's answer is spot-on, but missed on one minor point, which I would explain at the end.


This is just a summary of Wildcat's explanation: The idea is that, we have a system comprising of indistinguishable fermions, which is why we need to use the antisymmetrized wavefunction. The LCAO-MO approximation results in a linear combination of AOs resulting in MOs, which can be allowed to occupy, given by electron densities. Variationally, the ground state would lead to the all the lowest levels being occupied, resulting in HF ground state wavefunction. The HOMO, which is $3s$ orbital, would be at a certain energy level which is definitely well defined for a given **level of theory**.


So, to answer your question, yes, you can calculate the energy of the $3s$ orbital for $\ce{Na^{-}}$. But, what is not possible is to calculate the energy of **the** $3s^{2}$ electron, because the electrons occupying the $3s$ (or any of the other) energy levels are indistinguishable. Another thing to note is that, when you talk of using a wavefunction to denote electrons, they are highly delocalized, and can occupy more than one energy level at once. Think about the case of Boron atom. Which orbital do you think the $2p$ electron occupies? The $2p\_x$ or $2p\_y$ or $2p\_z$? They are all symmetric in nature, and hence you cannot point at one particular orbital. This turns into a multiconfigurational problem, which is beyond the scope of this discussion. Hence, the implication here is that, the electrons occupying these orbitals would be at specified energy levels, depending on which orbital they occupy.


 0