This question comes from some thoughts I had after reading this question.

First of all, is an orbital an observable? I know the answer to this question is no because there is no "single-orbital operator" or whatever you'd like to call this. I brig this up because exactly what this means should be a part of a thorough answer. Plus an orbital is a state and we don't observe the eigenvectors but the eigenvalues.

Second, is the energy of an orbital an observable? It's more tempting to think that the answer to this is yes, but if it's true that there is no "single-orbital operator" then there must be no corresponding eigenvalue which represents the energy of this orbital. As I understand it, if one runs a HF calculation and sums the energies calculated for the individual orbitals, this is the total electronic energy. In the answer linked above, however, it is said that there is an infinite number of unitary transformations for a given set of orbitals to another set. Is there any correspondence between the energies of these transformed orbitals and the original set of orbitals? By that I mean, the orbitals themselves change, but would the optimized energies of each orbital change and it is the only the wavefunction and total energy which stay constant?

Finally, how do the answers to the above questions connect with Koopmans' theorem and photoelectron spectroscopy? That is to say, Koopmans' theorem says that the energy of the HOMO as calculated from Hartree-Fock corresponds to the first ionization energy of the system. Additionally, in photoelectron spectroscopy, it sure seems like people are observing the energies of individual orbitals, but I don't really know anything about this technique in detail so that may just be a misunderstanding.

Additionally, is Koopmans' theorem still true once we perform a unitary transformation on the Hartree-Fock wavefunction?

The reason I lump all these questions together is because I think in order to sufficiently answer the question of whether or not an orbital energy is observable, all of these points should be addressed, so I just asked them all at once. I hope it's not overkill.

  • $\begingroup$ This won't answer rather than comment only the the first question "is there a mean to observe an orbital". If you equate "orbital" with "local high electron density", than I remember single diffraction crystallographic characterization of a Pt complex, nicely crystallized. After structure solution and structure refinement, the difference Fourier map still revealed significant electron density close to the late transition metal cation that was tentatively attributed to d-orbitals not involved in bonding with the ligands around. $\endgroup$
    – Buttonwood
    Commented Feb 25, 2017 at 0:20
  • $\begingroup$ Well I mean observable in the quantum mechanical sense. i.e. the eigenvalues of a Hermitian operator. $\endgroup$
    – jheindel
    Commented Feb 25, 2017 at 8:37
  • 1
    $\begingroup$ No, they are not observables, though there are many observables that can be approximately calculated from them. $\endgroup$
    – Greg
    Commented Mar 1, 2017 at 0:40
  • 3
    $\begingroup$ Just something I didn't notice at when I initially viewed this question, but the sum of the orbital energies from an HF calculation is not equal to the total energy calculated from HF because if you simply sum the orbital energies, you wind up double counting the exchange and coulomb interactions. $\endgroup$
    – Tyberius
    Commented Nov 28, 2017 at 19:20
  • 1
    $\begingroup$ Since the last part of this question got answered (chemistry.stackexchange.com/questions/69503/…), you might want to edit that in (in whatever way you wish, of course). $\endgroup$ Commented Dec 4, 2017 at 21:10

1 Answer 1


Excluding the trivial case of a one electron system, where the wavefunction for the system is an orbital (1 electron wavefunction) and so could be measured via the electron density, orbitals are not observables. There is no known Hermitian Operator that returns orbitals (or their energies) from its expectation value. So is that the end of the discussion? Not necessarily.

In a 2006 paper, Shabazian and Zahedi$^1$ argue that what constitutes an observable depends on the conventions we establish for describing a system. They give as an example the Quantum Theory of Atoms in Molecules (QTAIM) developed by Richard Bader$^2$. Strictly speaking, the normal formulation of Quantum Mechanics doesn't allow us to describe subsystems within a molecule. We form molecules from atoms, but once the molecule is formed, there are no longer atoms; rather there is just a single wavefunction which describes the whole system. This is horribly inconvenient from a classical chemists perspective, as much of our understanding of chemistry comes from subdividing molecules into important functional groups that give the major properties of the molecule. To resolve this, Bader proposed a definition of an atom in a molecule by defining boundary conditions for a atom based on flux of the electron density. This gives an unambiguous, experimentally measurable way of determining observables like atomic charge, which without this set of external conditions is not considered an observable at all. While this has as of yet not been done, Shabazian and Zahedi argue that orbitals could be made observables by establishing a similar set of external conditions.

So while orbitals are not currently observables, what is and isn't an observable can change as new theoretical models develop.

  1. Shahbazian, S. & Zahedi, M. Found Chem (2006) 8: 37. https://doi.org/10.1007/s10698-005-8247-4
  2. Bader, R. Chem. Rev., 1991, 91 (5), pp 893–928 DOI: 10.1021/cr00005a013
  • $\begingroup$ I disagree with your last paragraph. Whether or not something is an observable is an experimental question. Also, I see what you're saying about how QTAIM allow experimental electron densities to be mapped to atomic charges, but QTAIM is not unique so one cannot really say you are observing the "true" atomic charges. Also, there is no unique way to go from an election density to a set of orbitals. This, I think, is why orbitals cannot be observable. The hard part of this question I think is explaining what is really measured in photoelectron spectroscopy. $\endgroup$
    – jheindel
    Commented Dec 12, 2017 at 2:09
  • $\begingroup$ @jheindel I'm not an expert in QTAIM, but I'm curious in what sense it isn't unique. In regards to the orbitals, I want to clarify that I'm not (and the paper isn't) saying QTAIM leads to observable orbitals, but that a similar formulation could lead to an experimentally measurable quantity corresponding to something like an orbital. In terms of photoelectron spectroscopy, what we observe is the energy of moving from a ground state to an excited state or vice versa. It can be thought of as the energy to move between different PES. $\endgroup$
    – Tyberius
    Commented Dec 12, 2017 at 5:29
  • $\begingroup$ Right but the question is more whether or not these transitions correspond to orbital energies in a meaningful way. I don't think it can be said that they do. I just mean not unique in that there are many ways to partition a system into atoms and write down molecular properties in terms of these atoms. QTAIM just chooses the mathematically simplest one which leads to nice equations. $\endgroup$
    – jheindel
    Commented Dec 12, 2017 at 6:19
  • $\begingroup$ @jheindel I would say you are right. If the answer to the first and second question are no (i.e. if orbitals and their energies aren't observables) then the photoelectron spectrum can't depend meaningfully on the orbital energies or else that would constitute the orbital energies being observables. My answer really boils down to the first paragraph, but I was hoping to give somewhat of a different perspective just to avoid giving an answer that basically just said "no". $\endgroup$
    – Tyberius
    Commented Dec 12, 2017 at 17:19
  • $\begingroup$ @jheindel I think in terms of uniqueness, you could make the same argument about velocity or energy, which can vary based on what you choose as a reference frame. This is somewhat philosophical, but I think what we consider observables is more of a construction rather than something inherent about the universe. $\endgroup$
    – Tyberius
    Commented Dec 12, 2017 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.