8
$\begingroup$

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.

$\endgroup$
  • 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. $\endgroup$ – pentavalentcarbon Mar 19 '18 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. $\endgroup$ – pentavalentcarbon Mar 19 '18 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. $\endgroup$ – pentavalentcarbon Mar 19 '18 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 $\endgroup$ – Fl.pf. Mar 19 '18 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. $\endgroup$ – pentavalentcarbon Mar 19 '18 at 22:51
2
$\begingroup$

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.

Note:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.