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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.

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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 (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.


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

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.


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.

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  • $\begingroup$ > The approximation is known as the Hartree-Fock (HF) model || HF model does NOT give energies of individual electrons, or even orbitals. HF wavefunction is antysimmetrical by electron position exchange, so there is no way to say if some particular electron occupies some orbital, only that 'orbital' is occupied or not. You are probably refering to Koopmans' theorem. It is not about energy of particular electron, but about energy required to build system with a 'hole' relatively to original one. $\endgroup$ – permeakra Jun 28 '15 at 17:48
  • $\begingroup$ @permeakra, no, now I'm pretty sure, you're confused. Speaking of which particular electron occupies a particular orbital is meaningless, but speaking about the energy of an electron at particular orbital is perfectly sensible. Because an electron that occupies some orbital does have a well defined energy, no matter which particular electron it is. $\endgroup$ – Wildcat Jun 28 '15 at 18:01
  • $\begingroup$ @permeakra, the fact that electrons are indistinguishable does not forbid us to speak about individual electrons. For instance, in the HF approximation we talk about individual electrons occupying spin-orbitals, and that is perfectly legal (approximate, but legal). What can not be done is some sort of identification that a particular electron (say, twelfth one) occupies a particular orbital (say, $\mathrm{3s}$). But I surely can say that an electron occupies this orbital and that it will have a particular energy. $\endgroup$ – Wildcat Jun 28 '15 at 18:10
  • $\begingroup$ > For instance, in the HF approximation we talk about individual electrons occupying spin-orbitals || Nope, HF approximation has nothing to say about individual electrons or orbital energy, it never appears in the formalism. There is so known 'Koopman's theorem', which does not hold true and is about ionization potential, most likely it is the energy printed by provided input. Again, it is not about individual electrons or even electron on the orbital. Read a good QC book. $\endgroup$ – permeakra Jun 29 '15 at 3:26
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    $\begingroup$ It would be great if you'd be a little bit nicer to each other. Every argument can be lead on a non-personal basis. Some readers might interpret this as a hostile argument. $\endgroup$ – Martin - マーチン Jun 29 '15 at 11:57
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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.

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  • $\begingroup$ How does indistinguishability of electrons cause inability to calculate their energies? $\endgroup$ – Wildcat Jun 28 '15 at 11:56
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    $\begingroup$ @Wildcat since electrons are indistinguishable, one cannot attribute energy to some particular electron, since any exchange of two electrons gives raise to antisymmetric wavefuction with exactly same energy, that is, from observer's POV indistinguishable from original (and QM does not have internal variables). If you prefer more strict description, QM formalism gives only total energy of the system. It is possible, using some approximations, to give energy to some orbitals, but not individual electrons. $\endgroup$ – permeakra Jun 28 '15 at 17:44
  • $\begingroup$ What you're saying sounds wrong for me. Look, even in the HF approximation, despite what you're saying, electrons are indistinguishable. That's why our trial HF wave function is antisymmetric product of spin-orbitals and not a simple product of them. Yes, I obviously couldn't say which particular electron occupies which orbital, but I know what is the energy of an electron is at each and every orbital. $\endgroup$ – Wildcat Jun 28 '15 at 17:56
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    $\begingroup$ Besides, systems of many non-interacting particles are considered in QM, and for them the many-particle wave function perfectly separates into the product of one-particle wave functions without any approximations. Yes, such systems are rather ideal, but QM in principle doesn't seem to mind that we speak about each and every particle as being in its own one-particle quantum state. There is nothing wrong with that. Of course, for interacting particles such description is approximate, but that's another story. $\endgroup$ – Wildcat Jun 28 '15 at 18:17
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    $\begingroup$ The "energy of an electron in an orbital" is equivalent to the energy of an orbital" so electron indistinguishability is irrelevant. Orbitals are distinguishable up to symmetry. $\endgroup$ – Jeff Jun 28 '15 at 18:26
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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.

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  • $\begingroup$ No, I disagree. I do not miss anything. I know & perfectly understand what you're talking about in your answer and I have never said anything different. I agree that OP used a bit strange language ("the $\mathrm{3s^2}$ electron"), but that his problem and not mine. $\endgroup$ – Wildcat Jul 2 '15 at 6:25
  • $\begingroup$ So, I repeat my reasoning her point by point, and then you tell me what I'm missing. 1) We do have orbitals in the HF approximation. 2) HF orbitals are occupied by electrons. 3) I couldn't specify which particular electron occupies which orbital, but I could tell what is the energy of an electron on any orbital is. 4) This energy usually referred to as the orbital energy is also know as the one-electron energy, which clearly identifies its meaning. 5) Case closed. $\endgroup$ – Wildcat Jul 2 '15 at 6:28
  • $\begingroup$ Specifically, for the answer in question, what I say is that I could not tell which particular electron out of total 6 occupies $\mathrm{3s_{\beta}}$ spin-orbital, but I can tell what is the energy of this electron. And since chemistry is ultimately based on the RHF picture, this energy will be the same as the energy of $\mathrm{3s_{\alpha}}$ electron, thus, I simply say "the energy of $\mathrm{3s}$ electron": there are two of them but they have exact same energy in the restricted picture. $\endgroup$ – Wildcat Jul 2 '15 at 6:32
  • $\begingroup$ The meaning of the phrase "the energy of $\mathrm{3s_{\alpha}}$ electron" is perfectly well defined: it is the energy of an electron which occupies the $\mathrm{3s_{\alpha}}$ spin-orbital. One more time: I couldn't specify which electron it is, but whichever electrons occupies the $\mathrm{3s_{\alpha}}$ spin-orbital, I surely can tell its energy. $\endgroup$ – Wildcat Jul 2 '15 at 6:35
  • $\begingroup$ And another short remark since I notice you talked about "the lowest levels being occupied": conceptually there is no virtual orbitals in the HF approximation. Initially you have as many spin-orbitals and as many HF equations as many electrons there are in your system. In the RHF case where all spatial orbitals are doubly occupied you have as many spatial orbitals and as many RHF equations rewritten in terms of them as many pairs of electrons there are in your system. $\endgroup$ – Wildcat Jul 2 '15 at 6:45

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