Usually the first step in describing the electronic wave function of a molecule or atom is to describe it with a single Slater determinant. I get that this is an independent particle approximation, and that in the context of Hartree-Fock, each electron moves in the average field of the other electrons. That's fine.

But when we correlate the electrons, this is done by generating excitations out of the Slater determinant, and then taking weighted combinations of them. For example, configuration interaction mixes the ground state Slater determinant with singly, doubly, triply, etc excited determinants. A fancier way of doing this is coupled cluster, where cluster operators generate excitations out of the Slater determinant.

Hermann Kummel says that we can think about electron correlation this way:

"The first thing one may imagine happening is that two particles mutually interact, thereby lifting themselves out of the Fermi sea, so that after the interaction both are in unoccupied orbitals"

Kümmel, Hermann. "Origins of the coupled cluster method." Theoretica chimica acta 80.2-3 (1991): 81-89.

Why can we (or should we) think about electron correlation that way?

  • $\begingroup$ In what way? In terms of linear combinations of Slater determinants or in terms of 'interacting electrons being in unoccupied orbitals'? $\endgroup$ Commented Feb 25, 2014 at 19:49
  • $\begingroup$ @LordStryker Both, really. The standard mathematical way to treat dynamic electron correlation is to "mix in" excited Slater determinants, which is equivalent to saying that when two electron interact, they "excite" each other out of their ground state occupied orbitals and into previously unoccupied virtual orbitals. In other words, an excitation from occupied to unoccupied. $\endgroup$
    – jjgoings
    Commented Feb 25, 2014 at 22:23
  • 1
    $\begingroup$ @jjgoings, for a simple physical reasoning watch this video by professor Jack Simons. You could also read almost identical material in his book "Quantum Mechanics in Chemistry" (Chapter 8) freely available here. $\endgroup$
    – Wildcat
    Commented Oct 22, 2014 at 13:03
  • $\begingroup$ I guess, it is obvious why additional determinants are used to describe non-dynamical correlation (sort of, by definition of non-dynamical correlation), so I interpret the question as it is about dynamical correlation (i.e. electron avoidance). $\endgroup$
    – Wildcat
    Commented Sep 27, 2016 at 13:23
  • $\begingroup$ Don't want to spam, but this seemed like the easiest way to contact you. I don't know if you had seen it, but there is a new site on the network specifically about computational chemistry/materials: Matter Modeling SE. One of your papers came up in an answer I wrote there and I realized you might be interested in the site. $\endgroup$
    – Tyberius
    Commented Jul 22, 2020 at 2:34

3 Answers 3


As I mentioned in comments, professor Jack Simons offers a relatively simple physical interpretation of the idea of mixing in excited detrminant for treating dynamical correlation in his book "Quantum Mechanics in Chemistry" (Chapter 8) freely available here, as well as in this video which is also authored by him.

An important mathematical finding is that a linear combination of a reference determinant and a doubly-exicted one can be expressed as linear combination of two other determinants. Namely,

\begin{multline} c_1 | \dotsc \ \phi_a \alpha \ \phi_a \beta \ \dotsc | - c_2 | \dotsc \ \phi_r \alpha \ \phi_r \beta \ \dotsc | = \\ \frac{c_1}{2} \Big( | \dotsc \ (\phi_a - c \phi_r) \alpha \ (\phi_a + c \phi_r) \beta \ \dotsc | - | \dotsc \ (\phi_a - c \phi_r) \beta \ (\phi_a + c \phi_r) \alpha \ \dotsc | \Big) \, , \end{multline} where $c = \sqrt{c_2/c_1}$.1

Here, the determinants to the left differ by a doubly occupied spatial orbital $\phi_a$ being replaced by a doubly occupied spatial orbital $\phi_r$, while the determinants to the right describe the singlet $(\phi_a - c \phi_r)^1 (\phi_a + c \phi_r)^1$ state. So, a state created by adding the doubly-excited $| \dotsc \ \phi_r \alpha \ \phi_r \beta \ \dotsc |$ determinant to the reference $| \dotsc \ \phi_a \alpha \ \phi_a \beta \ \dotsc |$ one is equivalent to a state in which one electron occupies $\phi_a - c \phi_r$ spatial orbital (being in any spin state) while another electron occupies $\phi_a + c \phi_r$ spatial orbital (also being in any spin state). And this is the way electrons "avoid" each other: by occupying these different spatial orbitals.

For example, $\pi^2 \rightarrow \pi^{*2}$ configuration mixing in alkenes or $\mathrm{2s^2} \rightarrow \mathrm{2p^2}$ configuration mixing in alkaline earth atoms produce left-right polarized and top-bottom polarized spatial orbital pairs shown below.

enter image description here

Here one electron stays closer to the left carbon atom by occupying $\pi^2 + c \pi^{*2}$ orbital, while another avoids it staying closer to the right carbon atom by occupying $\pi^2 - c \pi^{*2}$ orbital.

enter image description here

In this case one electron stays closer to the top of an atom by occupying $\mathrm{2s} + c \mathrm{2p}$ orbital, while another avoids it staying closer to the bottom of the atom by occupying $\mathrm{2s} - c \mathrm{2p}$ orbital.

1) Here is the proof for $2 \times 2$ determinants, where for brevity $\phi_i = \phi_i \alpha$ and $\phi_i^* = \phi_i \beta$. \begin{align} c_1 | \phi_a \ \phi_a^* | - c_2 | \phi_r \ \phi_r^* | &= \frac{c_1}{2} \Big( 2 | \phi_a \ \phi_a^* | - 2 c^2 | \phi_r \ \phi_r^* | \Big) \\ &= \frac{c_1}{2} \Big( 2 (\color{red}{\phi_a \phi_a^*} - \color{green}{\phi_a^* \phi_a}) - 2 (\color{blue}{c^2 \phi_r \phi_r^*} - \color{purple}{c^2 \phi_r^* \phi_r}) \Big) \\ &= \frac{c_1}{2} \Big( 2 (\color{red}{\phi_a \phi_a^*} - \color{blue}{c^2 \phi_r \phi_r^*}) - 2 (\color{green}{\phi_a^* \phi_a} - \color{purple}{c^2 \phi_r^* \phi_r}) \Big) \\ &= \frac{c_1}{2} \Big( (\color{red}{\phi_a \phi_a^*} - \color{blue}{c^2 \phi_r \phi_r^*}) - (\color{green}{\phi_a^* \phi_a} - \color{purple}{c^2 \phi_r^* \phi_r}) \\ &\phantom{=\frac{c_1}{2}}- (\color{green}{\phi_a^* \phi_a} - \color{purple}{c^2 \phi_r^* \phi_r}) + (\color{red}{\phi_a \phi_a^*} - \color{blue}{c^2 \phi_r \phi_r^*}) \Big) \\ &= \frac{c_1}{2} \Big( (\color{red}{\phi_a \phi_a^*} + c \phi_a \phi_r^* - c \phi_r \phi_a^* - \color{blue}{c^2 \phi_r \phi_r^*}) - (\color{green}{\phi_a^* \phi_a} - c \phi_a^* \phi_r + c \phi_r^* \phi_a - \color{purple}{c^2 \phi_r^* \phi_r}) \\ &\phantom{=\frac{c_1}{2}}- (\color{green}{\phi_a^* \phi_a} + c \phi_a^* \phi_r - c \phi_r^* \phi_a - \color{purple}{c^2 \phi_r^* \phi_r}) + (\color{red}{\phi_a \phi_a^*} - c \phi_a \phi_r^* + c \phi_r \phi_a^* - \color{blue}{c^2 \phi_r \phi_r^*}) \Big) \\ &= \frac{c_1}{2} \Big( (\phi_a - c \phi_r) (\phi_a^* + c \phi_r^*) - (\phi_a^* + c \phi_r^*) (\phi_a - c \phi_r) \\ &\phantom{=\frac{c_1}{2}}- (\phi_a^* - c \phi_r^*) (\phi_a + c \phi_r) + (\phi_a + c \phi_r) (\phi_a^* - c \phi_r^* - c^2 \phi_r \phi_r^*) \Big) \\ &= \frac{c_1}{2} \Big( | (\phi_a - c \phi_r) \ (\phi_a^* + c \phi_r^*) | - | (\phi_a^* - c \phi_r^*) \ (\phi_a + c \phi_r) | \Big) \, . \end{align}


It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!

The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra chemistry that we want to add is electron correlation. The extra maths we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.

(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)

  • $\begingroup$ Unfortunately I have to agree with you -- it seems to be one of those places that we have to rely on math instead of intuition. The HF Slater determinant forms a basis for an N-electron wavefunction, just like our single electron basis functions form a basis for an independent particle wavefunction. $\endgroup$
    – jjgoings
    Commented Feb 25, 2014 at 22:26
  • $\begingroup$ Regarding your point about single electron excitations: they do show up in many correlated methods (CCSD and CISD, for example)! While you are correct in saying that single excitations don't mix with the ground state directly (e.g. Brillouin's theorem), they do mix with higher excitations. According to Thouless, single excitations create any single Slater determinant out of any other single Slater determinant. Basically it optimizes your reference in response to including dynamic electron correlation. It's a big reason CCSD is pretty insensitive to reference (which makes it a robust method). $\endgroup$
    – jjgoings
    Commented Feb 25, 2014 at 22:31
  • $\begingroup$ Of course you are correct that my statement in the final paragraph only applies to the ground state, and I’ve edited to make this clear. Apologies if the level of explanation was rather too basic for you! $\endgroup$
    – Aant
    Commented Mar 2, 2014 at 18:17
  • $\begingroup$ The concept of 'mixing' these determinants (i.e. ground and some excited state) is fine with me mathematically. However, the physical interpretation of this mathematical approach is very dissatisfying to me. $\endgroup$ Commented Mar 28, 2014 at 19:47
  • $\begingroup$ @LordStryker, there is some simple physical interpretation. Please, watch this video. $\endgroup$
    – Wildcat
    Commented Oct 22, 2014 at 13:07

Actually there is a reason for this.

The exact mathematical way to expand a function of N variables $\Psi(x1, x2...)$ - in terms of a basis - is a multi-determinant expansion, given the requirement below.

The determinant appears "magically" when the function is required to be anti-symmetrical wrt the exchange of coordinates of any two electrons.

$$\Psi(1,2) = -\Psi(2,1)$$

See Szabo and Ostlund, formulas 2.70-2.73

Analogy: we can expand any well behaved, monoelectronic orbital (a single variable function) as a set of infinite basis functions, and we can expand any multi-electronic system as an infinite number of determinants.

  • $\begingroup$ Downvoters are invited to say why, and get an explanation. There is even a reference to the exact equations in a book that answer the question. $\endgroup$
    – Mah Neh
    Commented Sep 15, 2023 at 20:01
  • 1
    $\begingroup$ The only value of this answer is a reference to a book you need to own and which isn't even cited properly. This is not an answer, this is a comment. $\endgroup$ Commented Sep 15, 2023 at 22:15
  • $\begingroup$ @Martin-マーチン Its a bit of a rude comment. The explanation is literally the second paragraph. Most readers will know the book I lazily cited. I would expect that a sharp eye would value it more. I may include more details, but I think it is a good start. You dont, no problem. $\endgroup$
    – Mah Neh
    Commented Sep 16, 2023 at 0:04
  • 1
    $\begingroup$ I don't think my comment was rude, but I'll pass your complaint on to the other moderators to check. I still want to clarify though. I know the book, I have read it, parts of it I understood, but since then I've passed it on. I do not recall the equation you referred to. You could have taken my explanation as a hint to fill it in, but you have chosen a different path. What appears to you as a complete answer doesn't to me. This is how the voting works on the site, and now that my view is locked in, I cannot (and also would not) change it $\endgroup$ Commented Sep 16, 2023 at 7:41
  • $\begingroup$ I have to agree with Martin. This seems to answer the question "why does a determinant provide an appropriate electronic wavefunction?" rather than answer the original posted Q. $\endgroup$
    – Buck Thorn
    Commented Sep 19, 2023 at 9:12

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.