I have seen an increase in the use of the coupled-cluster doubles (CCD) method recently. CCD uses the exponential ansatz of the coupled cluster equations, but only includes amplitudes to double excitations. I know that it is possible to derive MP2 from the coupled cluster equations, and essentially show that MP2 is capturing some portion of the double excitations.

I am curious, however, how MP2 and CCD compare in this respect? That is, should they be expected to be essentially comparable? Does MP2 implicitly include any excitations that CCD does not (perhaps some single excitations)? Or should CCD be strictly better than MP2 in terms of the amount of correlation included, presumably due to the presence of some approximate quadruple excitations (from the product of $t_{ij}$ terms)? Also, is how does the computational cost of CCD compare to MP2?

  • 1
    $\begingroup$ Related reference: sites.smu.edu/dedman/catco/publications/pdf/320.MP_review.pdf $\endgroup$
    – Tyberius
    Nov 8, 2021 at 19:17
  • $\begingroup$ "Does MP2 implicitly include any excitations that CCD does not (perhaps some single excitations)?" - It's the other way around: CCD includes excitations that MP2 doesn't, because of multiple applications of the doubles operator. See this reference, section III.C (starting on page 297). $\endgroup$
    – Antimon
    Nov 9, 2021 at 0:01
  • $\begingroup$ @Antimon Right. I understand that CCD has terms that MP2 doesn't, but I didn't know if there were terms that appear in MP2 that don't appear in CCD. Based on the review Tyberius linked, it seems to be that MP2 energies can actually be extracted from CCD calculations, so I guess CCD is strictly a superset of MP2 in terms of excitations. $\endgroup$
    – jheindel
    Nov 9, 2021 at 5:27
  • $\begingroup$ Yep, that's right. Lots of quantum chemical program packages will actually do that in practice: Run a CC calculation and you'll get MP energies reported as well. $\endgroup$
    – Antimon
    Nov 9, 2021 at 16:46
  • 3
    $\begingroup$ @Antimon The MP2 results are actually not extracted from a converged CC calculation. Rather, the CC calculation uses MP2 amplitudes as the initial guess, which is the true reason why the MP2 energy can be obtained for free from a CCSD/CCD/CCSD(T) calculation. $\endgroup$
    – wzkchem5
    Nov 9, 2021 at 18:36

1 Answer 1


Part of your question was:

Does MP2 implicitly include any excitations that CCD does not (perhaps some single excitations)?

From the expansion of the exponential form of the cluster operator, we achieve powers of the each excitation operator (see this reference, pp.297). So it's in fact the other way around: CCD includes excitations that MP2 does not.

Excitations alone don't make the final energies, though. Both MP and CC calculate a series of amplitudes to weigh their excitation terms to calculate an energy, and they go about it in different ways. Quantum chemistry programs do typically report both CC and MP energies when conducting CC calculations, however as user @wzkchem5 pointed out in response to my previous comments, those amplitudes are in fact actual MP results that are then used as starting guesses for the CC amplitudes; they are not calculated using the final CC amplitudes. (I guess it would be interesting to see what would happen if one were to use CC amplitudes to calculate MP2 energies, although that idea seem somewhat like "throwing pearls before pigs".)

As to a direct comparison of MP2 and CCD results, I found this reference:

Förner et al.: Coupled cluster studies. III. Comparison of the numerical behaviour of coupled cluster doubles with configuration interaction and perturbation theory. Basis set and geometry optimizations

From the conclusions:

The results suggest that in most cases (besides pathological ones, e.g. related to large nuclear seperations), even when MP2 gives a larger correlation energy than CCD, the CCD results are of similar quality as those of MP4(DQ).
The equilibrium properties of the investigated molecules computed with CCD are as reliable as those obtained with CID (which is variational but not size consistent) and very similar to those computed with MP4(DQ) (...)

For the small test molecules used in this paper, CCD seems to perform somewhat comparably to MP4(DQ), which I suppose shouldn't come as too big of a surprise seeing as they include some of the same excitations. For that matter, it would be interesting to see a systematic comparison with even higher MP orders such as MP6(DQH) etc.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .