# Truncated CI Wavefunction

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.

• 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. Mar 19, 2018 at 15:30
• 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. Mar 19, 2018 at 15:32
• 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. Mar 19, 2018 at 15:34
• @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
– user37142
Mar 19, 2018 at 22:21
• 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. Mar 19, 2018 at 22:51

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.