In a calculation of for example a Li-Atom in a DZ basis we have 4 MO.

For the alpha string (containing two electrons), that means we have for a FCI calculation the determinants: |12>, |13>, |14>, |23>, |24>, |34> (electrons in MO 1 and 2, 1 and 3 and so on).

If I wanted to do a truncated CI calculation (CIS for example) only singly excited wavefunctions would contribute.

My (maybe extremely dumb) question now is: Are the determinants |23> and |24> singly excited or doubly excited? Is it the excitation $$a^\dagger_3a_1|12> = |32> = -|23>$$ or is it $$a^\dagger_2a_1 a^\dagger_3a_2|12> = |23>$$ Or is it both? And the coupling (1,-1) so to say determines whether this in used as a single or doubly excited determinant?

EDIT: From the "Molecular Electronic Structure Theory" book by Helgaker, Jorgensen and Olsen, from the chapter 11.8.5 (Direct CI algorithms for RAS Calculations) I read "In CISD theory, there are three such classes (string classes): class 0, with no electrons in virtual space; class 1, with one electron in virtual space; and class 2, with two electrons in virtual space".

This for me reads as if |23> would count as a class 1 string and not a class 2.

  • 2
    $\begingroup$ Both. If you think about it from the context of CC theory, the first determinant will come from $\hat{T}_{1}$, and the second will come from $\hat{T}_{2}$ and $\hat{T}_{1}^{2}$. The prefactors handle overcounting. $\endgroup$ Mar 19, 2018 at 15:30
  • $\begingroup$ As for the book statement, without any more context, I would view this as a partitioning that avoids the overcounting by simplifying the number of strings before generating the determinants. $\endgroup$ Mar 19, 2018 at 15:32
  • $\begingroup$ Also, to be more exact, I know $\hat{T}_{2}$ is not the same as $\hat{T}_{1}^{2}$, but I don't think it's important here. $\endgroup$ Mar 19, 2018 at 15:34
  • $\begingroup$ @pentavalentcarbon yes you're right, my error was in thinking it's both |23>, but one is |32> and the other |23>. Writing the question actually made myself realise my error 5 minutes later :D $\endgroup$
    – user37142
    Mar 19, 2018 at 22:21
  • $\begingroup$ But isn't that a separate issue? I have to admit that because I don't do CI/CC I can't immediately see what you're talking about. I also don't see there being a problem with with question, it's a good one. $\endgroup$ Mar 19, 2018 at 22:51

1 Answer 1


In configuration interaction (and in coupled cluster) we typically think of excited determinant being caused by single particle excitations of the form $\hat{t}^a_i = \hat{a}^\dagger_a\hat{a}_i$ where $i$ labels a virtual orbital and $a$ labels an occupied orbital, causing an excitation of the Hartree Fock reference determinant $\Psi_0$.

The resulting CI expansion would go: $$ \Phi_\text{CI} = \Psi_0 + \sum_{i\in occ\\ a \in vir} \hat{t}^a_i \Psi_0 + \sum_{i,j\in occ\\ a,b \in vir} \hat{t}^{ab}_{ij} \Psi_0 + \sum_{i,j,k\in occ\\ a,b,c \in vir} \hat{t}^{abc}_{ijk} \Psi_0 + \: ... $$

Hence, in the calculation on Li mentioned above, we would not consider the excitation $\hat{t}^2_1 = \hat{a}^\dagger_2\hat{a}_1$ in the first place as it would annihilate the Hartree Fock determinant $|12>$ containing a particle in orbital 2.

By this logic the determinant $|23>$ is a singly excited determinant caused by $\hat{t}^3_1= \hat{a}^\dagger_3\hat{a}_1$, or class 1 as described by Helgaker.


As determinants $|23>$ and $|32>$ only differ by a multiplicative constant (-1) only one needs to be considered and the resulting coefficient/cluster amplitude will take care of the relative sign.


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