I have been using Gaussian16 por metal complexes SCF single-point calculations involving triplets to investigate phosphorescence. I have actually run into trouble (very long computational time, even with 36 processors and quite a lot of memory) when using the regular SCF convergence criterion as default in the code.

As a result from that, I thought of using the QC keyword, which made the calculations much faster and successfully converged. However, I am not sure if there is evidence on the effect of such an approximation or the lack of accuracy that might be obtained.

I have also thought of using XQC keyword, which would make me able to run regular SCF procedure up to a number of cycles and then switch to QC.

How safe am I when using QC? Am I losing a lot of accuracy, in general, when compared to the regular SCF procedure?


2 Answers 2


The SCF=QC keyword in Gaussian actually pertains to the choice of algorithm on how find the solution in the SCF procedure while SCF=Tight is an option related to SCF cycle convergence. Gaussian SCF criterion depends on density matrix changes, consequently in energy.

So using QC algorithm should not cost accuracy if the convergence criterion is the same.

If you refer to Gaussian website, you will see SCF=TightLinEq. They described this option as having 'tight' convergence throughout QC algorithm which I think makes this a safe choice.

I am also sharing the link to Gaussian SCF page for future reference.

  • $\begingroup$ I heard that scf=qc is more prone to converging to unstable wavefunctions than calculations without scf=qc. If this is indeed true, then scf=qc does increase the likelihood that the converged result is wrong. However I'm yet to see any formal publication that support this view $\endgroup$
    – wzkchem5
    Nov 17, 2021 at 13:22
  • $\begingroup$ I have not heard of this. But shouldn't the stability of wavefunctions be related to convergence criteria? From the write up of Gaussian, it seems to me, they have different convergence criteria in SCF=QC. This could explain the unstable wavefunctions. $\endgroup$
    – danjodc
    Nov 17, 2021 at 13:29
  • $\begingroup$ Maybe. Nevertheless the convergence criteria (or even combined with the initial guess) do not uniquely determine the probability that the converged wavefunction is unstable $\endgroup$
    – wzkchem5
    Nov 17, 2021 at 13:39

As @danjodc already pointed out, SCF=QC is not an approximation but rather a way of solving the SCF equations. The SCF equations arise from the extremality condition for the orbitals that minimize energy (see e.g. our open access review https://www.mdpi.com/1420-3049/25/5/1218). Instead of solving the SCF equations by iterative re-diagonalization, you can just optimize the orbitals directly. This is essentially what SCF=QC does in Gaussian.

The diagonalization based approach tends to be very fast for "easy" systems, because of Pulay's DIIS procedure that is often able to find the ground state solution quickly. However, in cases where DIIS fails to lead to reliable convergence, energy minimization may be a better alternative.

I don't remember by heart what kind of algorithm Gaussian uses in SCF=QC, but the most famous quadratic convergence algorithm i.e. Newton-Raphson algorithm is also infamous for converging to saddle-point solutions. Moreover, direct energy minimization may only lead you to the closest minimum, not the global minimum.

If you are experiencing convergence problems, I recommend the XQC algorithm which is sort of the best of both worlds: it uses DIIS for starters, and only when DIIS does not converge it switches over to quadratic convergence. If you're dealing with metals, you might also try toying around with SCF=Fermi which might help in locating the best orbitals. Moreover, running the calculation in a smaller basis first and reading it in as guess for the "production calculation" may also help in finding the converged solution.

On a final note, metal complexes may prefer higher spin states. It might be good to check how the energy behaves with respect to the spin state: does it go down when you do e.g. singlet -> triplet -> quintet -> septet -> nonet -> etc?

  • $\begingroup$ From the description on the Gaussian website, it looks like it does use Newton-Raphson (mostly) when specifying QC. $\endgroup$
    – Tyberius
    Nov 18, 2021 at 0:22
  • $\begingroup$ +10 for your first answer since October, welcome back and thank you again for your contributions here! They're always appreciated! $\endgroup$ Nov 18, 2021 at 0:22

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