Szabo/Ostlund list the CI matrix elements between singlet symmetry-adapted configurations (SAC) in Table 4.1 of their book:

$$ \langle ^1\Psi^r_ a \lvert \mathcal{H} - E_ 0 \rvert ^1\Psi^r_ a \rangle, \langle \Psi_ 0 \lvert \mathcal{H} \rvert ^1\Psi^{rr}_ {aa} \rangle, \langle \Psi_ 0 \lvert \mathcal{H} \rvert ^1\Psi^{rs}_ {aa} \rangle, \langle \Psi_ 0 \lvert \mathcal{H} \rvert ^1\Psi^{rr}_ {ab} \rangle, \langle \Psi_ 0 \lvert \mathcal{H} \rvert ^A\Psi^{rs}_ {ab} \rangle, \langle \Psi_ 0 \lvert \mathcal{H} \rvert ^B\Psi^{rs}_ {ab} \rangle, \langle ^1\Psi^{rr}_ {aa} \lvert \mathcal{H} - E_ 0 \rvert ^1\Psi^{rr}_ {aa} \rangle, \langle ^1\Psi^{rs}_ {aa} \lvert \mathcal{H} - E_ 0 \rvert ^1\Psi^{rs}_ {aa} \rangle, \langle ^1\Psi^{rr}_ {ab} \lvert \mathcal{H} - E_ 0 \rvert ^1\Psi^{rr}_ {ab} \rangle, \langle ^A\Psi^{rs}_ {ab} \lvert \mathcal{H} - E_ 0 \rvert ^A\Psi^{rs}_ {ab} \rangle, \langle ^B\Psi^{rs}_ {ab} \lvert \mathcal{H} - E_ 0 \rvert ^B\Psi^{rs}_ {ab} \rangle, \langle ^A\Psi^{rs}_ {ab} \lvert \mathcal{H} \rvert ^B\Psi^{rs}_ {ab} \rangle. $$

The expressions are given in terms of restricted, canonical MOs. While I have been able to arrive at the same results for all listed elements I tried so far, I am apparently unable to correctly derive elements that are not listed. I do not believe that they are all $0$.

I am testing my toy SCF/MP2/CIS/CID implementation against ORCA and am able to reproduce RHF, UHF, RMP2, and RCIS/TDHF results for different systems with good precision. However, CID in general eludes me thus far. $\ce{H_2}$ in a minimal basis set (single-$\zeta$) works correctly, as does $\ce{He}$ in double-$\zeta$. However, my results for $\ce{He}$ in triple-$\zeta$ are rather far off.

I am looking for the correctly derived off-diagonal elements of the CISD matrix. Lacking this, concrete pointers on which other, freely available QC suite will print the matrix are also welcome. Unfortunately, I am reduced to the status of a hobbyist without access to literature.

Edit: In the comments, it was asked how I was confident in the integrals. The AO integral code is ancient and has been verified for RHF and UHF against ORCA and Turbomole. While performing the CID calculation, the RHF and RMP2 energies are being calculated at the same time, and match the ORCA results. I just performed a calculation for $\ce{BeH2}$ with matching results - given the number of AO/MO involved, I feel confident in my AO-MO-transformation.

A Szabo, NS Ostlund Modern Quantum Chemistry, Dover Publications, first edition, 1996.

  • 2
    $\begingroup$ So your CID in a DZ basis for He matches ORCA in the nano-Hartree digit, but TZ is far off? How are you 100% sure the integrals are the same? ORCA doesn't print an FCIDUMP file (or any ASCII integrals) right? Are you able to make sure your integrals match the integrals from PySCF, OpenMOLCAS, or some other open-source software that prints integrals in a readable format? $\endgroup$ Aug 11, 2020 at 0:11
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    $\begingroup$ I appreciate you offering the generous bounty! We've unfortunately had some bad experience with bounties (not nearly as many people showed up to try to earn them, in my experiences so far). To get the very most out of your points spent, you might wish to get into as much detail as possible: for example, maybe your integrals in FCIDUMP format so that people can test with their own programs, or at least give the actual basis sets (not just DZ / TZ) and energies (my comment asked about how far off the energies were with DZ, if they disagree at the nH digit, something is wrong & could get worse) $\endgroup$ Aug 18, 2020 at 0:18
  • $\begingroup$ If you tweet this question @StackMatter, several people here will re-tweet it for you to help the question get more attention. $\endgroup$ Aug 18, 2020 at 0:18
  • $\begingroup$ I know that Rubin Pauncz has a book on this, and if I remember correctly, I read in his book on "spin eigenfunctions" that FA Matsen was the one to come up with this idea of spin free quantum chemistry, I think there was also a citation to Isha Shavitt's work. @NikeDattani Do you happen to know the exact reference that contains these closed form expressions for CSFs mentioned in Szabo? $\endgroup$
    – user784
    Jan 23, 2022 at 16:20

1 Answer 1


An easier route to using symmetry-adapted CI with hand-derived matrix elements is to implement CI with determinant strings. That is, you construct bitstrings of which orbitals are occupied in the determinant, and you don't care about adapting your basis for $\hat{S}^2$. This is the way most codes work, since the resulting algorithm is easy to make very fast, whereas spin-adaptation may require you to limit your expansion length.

The string based CI works in the space of spin-orbitals. For instance, the Hartree-Fock state would be $|{\rm HF}\rangle = (1, 1, \dots, 1, 1, 0, 0, \dots, 0, 0)^{\rm T}$, and the first excited determinant would be $(1, 1, \dots, 1, 0, 1, 0, \dots, 0, 0)^{\rm T}$. What you need to do is just build all the determinants that have the wanted $\hat{S}_z$, i.e. the correct number of $\alpha$ and $\beta$ electrons.

Now, building the CI Hamiltonian is very straightforward: once you have the bitstrings $|i\rangle$, you can obtain the matrix elements $\langle i | H | j \rangle$ with e.g. the Slater-Condon rules. The matrix element is zero unless $|i\rangle$ and $|j\rangle$ differ by at most a double excitation; you can find out the difference in the states' occupations with a bit-wise XOR operation. Counting the bits in the result of the XOR gives 0 if the determinants match, 2 if they differ by a single excitation, 4 in the case of a double excitation, and you don't care about the rest since the matrix element is zero.

You can build the sparse Hamiltonian in memory, if you don't go to huge active spaces just by looping over the pairs of bit strings, and use library implementations of sparse matrix diagonalizers.

The best thing about using this method, in addition to being relatively straightforward to implement, is that you can do arbitrary level CI with it: CID, CISD, CISDT, CISDTQ, ..., all the way to FCI and CASSCF.


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