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Wikipedia, in the article on Orbital Hybridisation, states the following:

One misconception concerning orbital hybridisation is that it incorrectly predicts the ultraviolet photoelectron spectra of many molecules. While this is true if Koopmans' theorem is applied to localised hybrids, quantum mechanics requires that the (in this case ionised) wavefunction obey the symmetry of the molecule which implies resonance in valence bond theory.

(Source; emphasis mine)

I am wondering about the bolded part. I interpret it as meaning that the wavefunction is an eigenfunction to both the Hamiltonian and to the symmetry operators of the molecule's point group. To me, the bolded part then seems equivalent to stating that the molecule emerges from the ionization process in one of the eigenstates of the ionized system, and I can understand how the symmetry requirement would hold true in that case.

However, my intuition wants to think about the ionization process in such a way that the departing electron leaves a "hole" somewhere in the wavefunction, which then evolves over time as a superposition state; i.e., the system does not emerge in an eigenstate, and the "hole" left in the wavefunction will "travel" through the molecule in some way. I am under the impression that the time-dependent wavefunction at any point in time could in principle break the symmetry imposed by the molecular point group.

So this leads me to two questions:

  1. Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry? And

  2. is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

Update: I am of course aware of the argument that we can make unitary transformations of the overall wavefunction as we please, as it leaves the total electron density distribution and energy unchanged; this is also how hybrid orbitals are formed when starting from an MO picture. However, because the orbitals involved in hybridization belong to different energy eigenvalues, this creates a time-dependent superposition state with no defined energy. In extension, hybridization silently implies that the MO energies themselves are not observables that would describe the initial states in the ionization process. And, in addition, the hybrid orbitals do not necessarily form irreducible representations of the underlying molecular point group anymore, and do thus not obey the bolded symmetry requirement in the quote above.

So here are two questions which are quite closely related to the two I had originally formulated above:

  1. What is the justification for the assumption that the MO energies are not observables? And

  2. how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

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There are really two parts to your question - one is to interpret the wikipedia entry, the other is the more general question about the transition of methane from the parent molecule to the ionized $\ce{CH4+}$.

For the first question, I believe the author has muddled the explanation a bit (though it's of course just as likely that I'm the one who is muddled). It sounds to me like they are explaining the reconciliation of molecular orbital (MO) theory and valence bond (VB) theory for $\ce{CH4}$ rather than for the cation or the ionization process itself. Here is what I think they are trying to say:

Electronic structure of $\ce{CH4}$

  1. MO theory If we carry out an analysis of the delocalized molecular orbitals of methane, we find that the four highest energy occupied orbitals are spread between two energy levels, with one orbital at the lower energy level and the other three at the higher level. The orbitals meet the quantum mechanical requirements of being orthonormal and matching the symmetry operations of the molecule. The energy levels are eigenvalues of the wave function energy operator, so we'll call them eigenstates. This representation is supported by the experimental data from PES, which shows two ionization energies, with a relative intensity of 3:1.

  2. VB theory In contrast, the valence bond view of the molecule has four orbitals of equal energy, each one corresponding to a specific C-H bond. These orbitals can be constructed as linear combinations of the MO theory orbitals and are orthonormal to each other. As such, they form an alternate basis set for the same space as described by the MO theory orbitals (that is, any linear combination of the MO theory orbitals can be described by a linear combination of the VB theory orbitals). However, they do not match the symmetry of the molecule, and their energy levels are not eigenvalues for the energy operator. This representation seems to be in disagreement with the experimental evidence, since it seems to suggest that only one ionization energy should be observed. However, it is not incorrect, as I will explain.

  3. Reconciliation The reconciliation between the VB description and the experimental result comes from realizing that "a linear combination of MO theory orbitals" is just a different way of describing a superposition of states. All quantum systems exist as a superposition of states. In the mathematical edifice that is quantum mechanics, we require that when we make a measurement of an observable of a superposition of states, the result must still be an eigenvalue for one of the states in the superposition. The probability of a specific eigenvalue appearing is a function of the contribution of the corresponding state to the superposition.

    The simplest example of this is measuring the spin of an electron. Unless a constraint is imposed, we expect to measure ($+\frac12$) 50 per cent of the time and ($-\frac12$) 50 per cent of the time. In neither case do we conclude that the electron was fixed in that spin state prior to the measurement. It was in a superposition of both spins until we made the measurement.

    Getting back to the methane molecule, we say that all 8 electrons in the valence orbitals can be viewed as existing in superpositions of states, with the only requirement being that within each individual molecule, the electrons are in states that are orthonormal to each other. The VB and MO theory representations are just two of an infinite number of possibilities. When we do the PES experiment, the results must always be eigenvalues, and the intensities will always have a 3:1 ratio, because the average contribution of the eigenstates to the population of orbitals across the population of methane molecules must remain 3:1.

Ionization of $\ce{CH4}$ With the above in mind, we can move on to the harder general question about time evolution of states during and after ionization. Since the measurement in the PES experiment returns an eigenvalue, it would seem that it must force the molecule into an eigenstate, just as measuring the spin of a free electron forces it into a single spin state.

We know that initially the molecule retains the tetrahedral geometry of $\ce{CH4}$, but within a short time it relaxes to a $C_{2v}$ symmetry. This change in symmetry must be accompanied by a change in the wave function and the energies (eigenvalues) of the states. I would assume that this time-evolution of the wave function also allows for scrambling of the eigenstates back to a superposition of states, but I don't know that process well enough to give a clear description. Perhaps someone else can jump in with that.

Conclusion Referring to your specific questions -

1.Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry?

As I described, I would say you are correct. Superpositions do not need to meet the symmetry requirement. The only requirement is that they are linear combinations of the symmetry-consistent basis set and that as a group they form an equivalent orthonormal basis set.

And is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

I think so, but I'm not certain.

What is the justification for the assumption that the MO energies are not observables?

I don't understand this question. Energy is definitely an observable, and the PES experiment measures it. An observable is anything that can be measured.

UPDATE: Rereading your updated question, I think the confusion is between "not having a fixed value" and "not being an observable". Just because the electrons do not have defined energies does not mean that their energies are not observables. In chemistry, we use language that refers to fixed values of observables like energy, but a fundamental concept of QM is that observables do not have fixed values until they are measured. The measurement process itself collapses the wave function to a specific eigenstate. I tried to clarify that with the example of the electron spin. Position is another common example. Ideally, we would preface lots of statements with "if you were to measure x. . " but that would be really cumbersome.

And how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

As I tried to explain, I believe the Wikipedia article was mixing up the parent molecule and the ionized state. The immediate effect of the ionization (ie measurement) is to force the molecule into an energy eigenstate. However, that is an eigenstate of the immediate product, which still resembles the methane parent in symmetry and orbitals. There is a time evolution to the ground state of the cation, which has a different symmetry ($C_{2v}$) and different eigenstates. Since we have performed no measurements on this molecule, it is not fixed in any of those eigenstates. Since it has evolved away from the initial $T_d$ symmetry, it is also no longer an eigenstate of the parent.

To summarize, the only point in the process in which the electrons are in a fixed eigenstate is immediately after ionization (before relaxation), and this eigenstate is described by the set of symmetry-allowed MO theory orbitals that correspond to the $T_d$ symmetry of $\ce{CH4}$, not the $C_{2v}$ symmetry of $\ce{CH4+}$.

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  • $\begingroup$ Thank you for your answer! I have updated my original question with two more questions which basically tackle the same problem from slightly different angles. If you could perhaps comment on those perspectives, it would be much appreciated. $\endgroup$ – Antimon Dec 16 '19 at 20:29
  • $\begingroup$ There is also something in your answer that I struggle with: You claim that the PES experiment must end in an eigenstate of the ionized system. Why? Then at the end, you judge my perspective on a time-dependent product state as possibly correct, but that directly implies that the ionization does not immediately end up in an eigenstate, so I'm wondering if your answer contradicts itself. In either case, my updated question 4 becomes relevant: If the process must end in an eigenstate, isn't it inconsistent to claim that it can start from hybrid orbitals, i.e. non-eigenstates? $\endgroup$ – Antimon Dec 16 '19 at 20:29
  • $\begingroup$ I'll update to try to address new questions. To answer your comment, I did not claim that the PES ends in an eigenstate of the ionized system. It is instead an eigenstate of the parent CH4 molecule. This is a requirement of any quantum mechanical measurement. Not only must the result be an eigenvalue, but the measurement itself forces the system into the eigenstate corresponding to the measurement output. Simple example is electron spin - once measures as +1/2, the electron remains in +1/2 spin state until another event occurs which scrambles it back to a superposition. $\endgroup$ – Andrew Dec 16 '19 at 23:27
  • $\begingroup$ Thanks for your updates. I'll have to go other things in my head a bit, I feel that the issue is still not settled for me somehow. Would you be open to perhaps moving the discussion to a chat sometime later? $\endgroup$ – Antimon Dec 17 '19 at 16:41
  • $\begingroup$ Sure. Feel free to open a chat whenever you want. $\endgroup$ – Andrew Dec 17 '19 at 18:32

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