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In my atomic lecture notes we prove (at least for electric dipole transitions), that only one electron can 'jump' at once.

The proof is as follows. The matrix element (assuming distinguishable electrons for now with understand that the real state would involve a Slater determinant) for a multi electron transition is e.g. \begin{align} & \propto \langle \psi_1(1s) \psi_1(2p)|\vec{r}_1 + \vec{r}_2|\psi_1(3p)\psi_2(3d)\rangle \\ & =\langle \psi_1(1s)|\vec{r}_1|\psi_1(3p)\rangle \times\langle\psi_2(3p)|\psi_2(3d)\rangle \\ & \qquad + \langle \psi_2(2p)|\vec{r}_2|\psi_2(3d)\rangle \times\langle\psi_1(1s)|\psi_1(3p)\rangle \\ & =0+0=0 \end{align}

My question is whether this property of 'only single electron jumps allowed' is unique to just electric dipole transitions, or just for treating atoms under central field approximation etc, or is this a general property of transitions. If multiple electron transitions are possible, in what cases are they possible?

Emilio Pisanty
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Alex Gower
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    Just as a note (which doesn't aswer the math in your question): In any multielectron system, if one electron changes its orbital, the potential felt by the other electrons changes. Hence they also need to adapt the shape of their orbitals. This is why one doesn't talk about single electrons changing their state, but the whole atom does so. – A. P. Feb 11 '21 at 09:17

3 Answers3

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This is a general property of transitions $-$ in the Hartree-Fock formalism.

If the eigenstates of both the initial and the final states of the transitions are (i) given by single Slater determinants which are (ii) made out of single-particle orbitals taken from the same basis set, then the selection rule you have derived is exact.

As it happens, this Hartree-Fock approach generally works extremely well. To be clear, there is no guarantee that atomic eigenstates will be given by single Slater determinants. This is definitely not implied by the fermionic particle-exchange symmetry, and you can have a state like $$ |\Psi⟩ = \frac{|\phi⟩|\psi⟩-|\psi⟩|\phi⟩}{2} + \frac{|\chi⟩|\varphi⟩-|\varphi⟩|\chi⟩}{2}, %\tag{$*$} $$ with perfect antisymmetry, which cannot be reduced to a single determinant. But, as it turns out, for most atoms, the ground state and the 'reasonable' excited states don't really need this, and they can largely be reduced to single Slater determinants.

But, that said, this is not universal, either. If you go looking for them, then it's not hard to find states that buck this trend. One good place to do this is the NIST ASD levels database, looking in particular at heavy atoms and ions. So, for example, if you look at the level scheme for boron, you'll find plenty of states with nontrivial correlations.

  • This starts already with the ground state, which is assigned a principal configuration of $\rm 2s^22p$ $\rm {}^2P^o_{1/2}$, but this is only about 95% of the population, with the next leading state, at 4%, being the $\rm 2p^3$ configuration.
  • An even more notable example is the $\rm 2s2p^2$ $^2\rm D_{}$ excited states, where that leading configuration only accounts for just under 80% of the population, with the next leading state, $\rm 2s^23d$, only accounting for 7% of the remainder (i.e. with the implication that there's a sizeable zoo of additional Slater determinants involved in that eigenstate).

For full clarity, this is what it means, in real technical down-to-the-calculation-ground terms, when we say things like "if one electron changes its orbital, the potential felt by the other electrons changes; hence they also need to adapt the shape of their orbitals" (as mentioned by A.P. in the comments). The existing answer's assertion that this is handled by "higher order matrix elements in the Fermi Golden Rule" isn't really correct:

  • It is possible to have transitions between states that differ in more than one orbital by having a sequential multi-photon process where, e.g. a $\rm 1s2p$ configuration is excited first to a $\rm 1s 3d$ and from there to the final $\rm 3p 3d$ state (the correct description of which is via second- or higher-order perturbation theory, at which point the simpler structures of Fermi's Golden Rule stop having equivalents and you just use the full brunt of the theory).

  • However, this is not what was meant in the comments $-$ the "rearrangement" coming from the changes in the orbitals of other electrons in response to changes in the transition corresponds to the maths I described above, not to those multi-photon transitions.

To wrap things up: the assertion in your lectures,

only one electron can 'jump' at once

corresponds to the Hartree-Fock picture, and it is correct in that setting (but it can break where HF breaks). It is also possible to have multi-photon processes where several electrons jump, by having them jump one-at-a-time over several chained single-photon processes, but that's not addressed either by your textual quote or by the math that underlies it.

Emilio Pisanty
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  • I don't really appreciate your categorical statements about other answers being "correct" or not - they are unfriendly and unkind, which is a reason for flagging. While I respect your technical expertise, and acknowledge that your theoretical/mathematical background is not bad for an experimentalist, you don't seem to be qualified for making such judgements. – Roger V. Apr 20 '21 at 14:35
  • @Emilio Pisanty. So is your point that it's less that the multi electron transitions appear due to higher orders in Fermis golden rule but more than if we consider higher orders in the energy eigenstates they don't have well defined individual electron orbitals? – Alex Gower Apr 20 '21 at 14:48
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    @Vadim Technical criticism of content is what this site is for. If you want a more detailed explanation of why I think the quoted text is wrong, you should explain in more detail what you mean by it -- at present, the text is too vague to be strictly wrong. – Emilio Pisanty Apr 20 '21 at 14:48
  • The question is mostly semantic and not technical in nature. Its unclear what is meant with "multiple electron transition". One can interpret it either as referring to higher order terms in time-dependent perturbation theory or it can be interpreted as referring to a Configuration interaction expansion of electronic states and transitions between these states or a combination of both. It seems that the first answer uses the first approach while this answer is based on the second. – Hans Wurst Apr 20 '21 at 14:50
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    ... and, that said, if you want to go into personal attacks on my credentials ("your theoretical/mathematical background is not bad for an experimentalist"), you should make sure that you actually attack the right credentials. I have no idea what gave you the impression that I'm an experimentalist, but I can assure you that I'm not. – Emilio Pisanty Apr 20 '21 at 14:51
  • @Hans Higher order terms in a perturbation theory expansion (corresponding to multi-photon processes) are indeed a thing. But the text of the OP is very clearly not discussing those - the quote at stake deals with the fact that single-photon transitions are enacted by single-particle operators, which is what I address here. – Emilio Pisanty Apr 20 '21 at 14:53
  • @EmilioPisanty your answer contains a claim that mine is incorrect, without any supporting technical evidence. – Roger V. Apr 20 '21 at 15:03
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    @Vadim It wasn't really necessary, but see the expanded explanation of why your answer addresses the wrong part of the problem. If you want further responses, though, I'm still waiting for an apology for the ad-hominem insult you started with. (And, for full clarity: no, statements that other answers are incorrect are not "unfriendly" or "unkind", and are not a reason for flagging.) – Emilio Pisanty Apr 20 '21 at 15:33
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    @AlexGower The statement that "only one electron can jump at once" is correct within HF (and breaks when HF breaks). Multi-photon processes can chain multiple jumps, but that's not the same thing. – Emilio Pisanty Apr 20 '21 at 15:48
  • @EmilioPisanty Indeed, it was unnecessary, which is why I took an offense. I see no reason to apologize before you do so yourself and edit your answer in more professional terms. – Roger V. Apr 20 '21 at 16:28
  • @EmilioPisanty so would you also say 'only one electron can jump at once' is only perfectly true within the central field approximation (as, if I'm not mistaken, that's equivalent to saying correct within Hartree fock) as was one of the guesses in my original question? – Alex Gower Apr 20 '21 at 16:38
  • @AlexGower I suspect you're confusing the central field approximation with the mean field approximation. The latter is related to HF but they're not identical. This formalism is completely independent of whether the field is central or not. – Emilio Pisanty Apr 20 '21 at 19:08
  • What is the crucial difference that I am confusing between them? I have read there exists several types of central field approximation - either using a single $U(r)$ for all the electron or using an individual $U_i(r)$ for each electron (accounting for the fact they 'see' different fields). Of course neither will account for exchange correlations (exccept for Fermi), but I thought that was also the one thing that Hartree Fock doesn't account for, so I'm trying to see the difference? – Alex Gower Apr 20 '21 at 19:11
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Dipole approximation
Interaction between electrons and the electromagnetic field is a one-particle operator. The dipole approximation has to do with approximating the exact form of the electron-field coupling:

  • first replacing the potential term by a dipole-like interaction (which is actually exact when the field is a plane wave) $$ V(\mathbf{r},t)\longrightarrow -\mathbf{d}\cdot\mathbf{E}(\mathbf{r},t) $$
  • and then expanding the electric field to the first order in position: $$ \mathbf{E}(\mathbf{r},t)=\mathbf{E}_0e^{i\mathbf{k}\mathbf{r}} + c.c.\approx \mathbf{E}_0(1 + i\mathbf{k}\mathbf{r}) $$ (keep in mind that $\mathbf{E}_0$ is a complex vector with a non-zero phase).

In other words, even if we did not perform this approximation, we would still be dealing with a one-particle operator, which is what proven in the notes, cited in the OP. (I say one-particle, from the point of view of the election system, ignoring the photon.) What would be more complicated is the selection rules, which under the dipole approximation heavily rely on the symmetries of $\mathbf{r}$.

Other electrons
It was correctly pointed out in the comments that electrons in the atom are interacting via the Coulomb interaction, and a transition of one of them inevitably perturbs the others. This does not contradict what I said previously, but calculating such a process requires higher order matrix element in the Fermi golden rule, when calculating the transition cross-section. Note that even excitons in the solid state, although essentially excitations of an interacting many-particle system, are well modeled using hydrogen-like model. A well-known example of a multi-electron de-excitation process is Auger recombination.

Two-electron transitions
By going to even higher orders of the perturbation theory, one can study transitions where more than one electron change their orbitals to higher energy one (on the scale of photon energy), but these have low probability in comparison to one-electron transitions, and hard to observe in practice.

Update
To answer to criticisms in respect to my answer voiced by @EmilioPisanti.

  • Hartree-Fock is a mean-field approach, which reduces a system interacting via Coulomb forces to one-particle description. Although self-consistent HF may account for changes in electron density following an electron transition, higher-order processes, such as excitation of two electrons following an absorption of a photon cannot be described in this way, for the reasons outlined above. At best, HF can serve as a zero-th order approximation for calculating higher order matrix elements. The critic seems to admit as much:
To wrap things up: the assertion in your lectures,
only one electron can 'jump' at once

corresponds to the Hartree-Fock picture, and it is correct in that setting (but it can break where HF breaks)

  • HF approximation indeed works extremely well for atoms, and as well as for some phenomena in solid state. For example, it ios behind the hydrogen-like description of excitons that I referred to in my answer (for the full justificatuion of such a description see Knox's Theory of excitons). On the other hand, Auger processes cannot be described within HF.
  • A purely semantic point is my use of the term Hartree-Fock, which @EP seems to narrowly interpret as a formula using the first-order matrix element, while I would go as far as to apply the term to the formulas used for calculating the scattering cross-sections in the QFT. While one may disagree on the exact meaning of specific terms, the meaning should be clear from my above answer, where I more tha once referred to higher-order calculations. The critic admits that much:
the correct description of which is via second- or higher-order 
perturbation theory, at which point the simpler structures of Fermi's 
Golden Rule stop having equivalents and you just use the full brunt of the theory

(Btw, second and third order time-dependent perturbation theory is not that complex - it properly should be studied in basic QM textbooks.)

To summarize: there may be obviously different interpretations of the question OP. Mine was tilted towards the headline: Can multiple electrons transition simultaneously?, which I interpreted as more than one electrons transitioning to energy levels, separated from their initial state by the energy of the order of the photon energy. Such a process is not possible within the HF framework, which can describe merely rearrangement of electrons via modification of the self-consistent one-electrons states.

The @EP's answer seems to address different points. However, I consider their answer complimentary to mine, as they do give valuable examples of what is possible in atoms (although this is by no means the case for solid state.)

Roger V.
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Phrases like "electrons jump between orbitals" or "electron doing a transition" are misleading in my opinion. The physical objects that matter are electronic states. An electronic state of a system with multiple interacting electrons, always describes all electrons at once.

The next thing that is often misused are orbitals. Orbitals are not physically relevant in systems with multiple interacting electrons. They should be viewed as mathematical tools to describe the physical electronic states. Orbitals are not unique and the orbital basis can be changed , while the electronic states remain the same. The problem is that solving the time independent Schrödinger equation for electronic states is very difficult. And for most systems approximations have to be made.

One method to model electronic states is the Hartree-Fock method. The HF method makes use of orbitals to model electronic states of interacting systems. Within this approximate method we can look at orbitals that contribute to an electronic state, which is represented by a Slater determinant, a properly symmetrized linear combination of products of orbitals.

The electronic states obtained by the HF method can now be used to model spectroscopy. The equations of spectroscopy are again in most cases to complicated to be solved without approximations and the standard approximation for spectroscopy is time dependent perturbation theory. Time dependent perturbation theory(=TDPT) is in principle independent of HF. The connection between time dependent perturbation theory is made when we want to model transitions between electronic states, which requires a model for the electronic states. We use the HF method to obtain the states in question.

First order TDPT requires the calculation of matrix elements of the dipole operator between electronic states. This leads, when using HF as model for the electronic states, to expressions as presented in the question. We can easily deduce by expanding the determinant and the fact that the dipole operator is a one particle operator, that matrix elements vanish when two determinants differ in more than one pair of orbitals.

The phrase "only one electron can 'jump' at once" means that the electronic states may only be different in one orbital that is part of the Slater determinant, otherwise the matrix element is zero. It is a very specific statement about the transition dipole moment operator matrix elements between two electronic states that are modeled with the HF approximation.

The central field approximation and the dipole approximation are not relevant to this.

Lastly "If multiple electron transitions are possible, in what cases are they possible?" First we have to define what we mean with multiple electron transitions. I assume that this is supposed to mean transitions where more than two orbitals differ in the determinants that model the electronic states. The short answer is No. Hartree-Fock is a single determinant method and as mentioned before, all matrix elements vanish in this case when more than on pair of orbitals is different. But if we consider more advanced methods like configuration interaction, that uses linear combinations of determinants to model electronic states, the answer is yes. Examples are shown in the answer from Emilio Pisanty.

P.S.

I made this discussion to show that the question requires are clear definition of what one means with "electron doing a transition" and that a lot of context should be given to clarify the meaning.

Hans Wurst
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  • Thank you. Why do you say central field approximation is not relevant to this though when I thought Hartree Fock essentially just uses the central field approximation? – Alex Gower Apr 20 '21 at 17:33
  • @AlexGower The mean field approximation is not the same as the central field approximation. And Hartree Fock uses the mean field approximation not necessarily the central field approximation. It could be that they are the same for atoms though. I would need to check that. My background is chemistry where the HF method is used for molecules. – Hans Wurst Apr 20 '21 at 17:35
  • I guess it also depends on what central field approximation is used. Just a single radial function assumed identical for all electrons, or, as sometimes used, a separate function for each electron – Alex Gower Apr 20 '21 at 17:47
  • This previous Q&A of mine is relevant regarding your second paragraph. – Emilio Pisanty Apr 20 '21 at 19:12