1
$\begingroup$

I've come across numerous articles asserting that orbital optimized methods, where minimization of the energy is achieved through a series of unitary rotations (achieved by exponentiated single-electron excitations from the ground state reference) of the reference determinant, are exceptionally good at removing spin contamination. I've seen the numbers they report, and I believe them! I get the gist that orbital optimization allows us to overcome the error introduced by the UHF orbitals, but nowhere can I find the actual explanation of why these rotations effectively cure spin contamination.

Both references and explanations warmly welcomed.

$\endgroup$
3
  • $\begingroup$ Could you please list some papers where you saw the numbers? $\endgroup$
    – Frank
    Commented Apr 16, 2020 at 5:30
  • $\begingroup$ For example, Neese, Schwabe, Kossmann, et al. J. Chem. Theory Comput. 2009, 5, 3060–3073. $\endgroup$
    – jezzo
    Commented Apr 17, 2020 at 17:06
  • $\begingroup$ Or Bozkaya, J Chem Phys. 135, 224103 (2011) $\endgroup$
    – jezzo
    Commented Apr 17, 2020 at 17:07

2 Answers 2

2
$\begingroup$

Upon further thought, I've summarized it as follows. If anyone comes along with a better explanation, they'll get the cake!

Spin contamination is a fictitious effect arising from not achieving the full CI solution to the Schrödinger equation. For, had we found the exact solution, then the solution we found would be an exact eigenfunction of the ${\hat{S}}{}^2$ operator, as $[\hat H, {\hat{S}}{}^2]=0$.

In a general sense, orbital optimization can be viewed as simply allowing for more correlation to be recovered on top of whatever other improvements you've applied to your initial Hartree-Fock reference. So, by recovering more correlation, you've brought your solution closer to FCI. In doing so, you've brought your solution closer to a solution that contains no spin contamination.

However, there seems to be a stronger statement that could be made. This type of correlation seems especially potent in removing spin contamination.

$\endgroup$
1
$\begingroup$

I agree with the statement that FCI has no spin contamination, however, it's not necessarily true that if one reference is closer to FCI than other, it has less spin contamination (which is why UHF can produce better energies but lose spin symmetry). One way to think about it is by using the Thouless Theorem to relate any two non-orthogonal references

$$| \tilde{\Phi} \rangle = e^{T_1} |\Phi\rangle$$

where $\langle \tilde{\Phi} | \Phi \rangle \neq 0$ and the single-excitation operator is $T_1 = \sum_{ai} t_{a}^{i} a_a^+ a_i$ where $a$ denotes unoccupied MOs and $i$ denotes occupied MOs. As long as the references are not of different symmetries or spin multiplicities, they are equivalent up to an orbital rotation. This is how orbital optimization works - by changing $T_1$, I can move over the entire manifold of different references. For instance, an optimization condition of $T_1 = 0$ produces something called Brueckner orbitals. This also means that I can rotate from RHF to UHF or to something in between like ROHF. If you read some of the ROHF papers by Scuseria et al. for instance, you can use this framework to define references with spin constraints, such as preserving $\langle S^2 \rangle$ by applying a penalty for deviations from $S = 0$ or $S = 1$. That will easily produce orbitals that are both a good description and free (for the most part) of spin contaminatination.

However, that's all to do with the reference. The paper you've cited is about applying orbital-optimized MP2 which essentially rotates an existing reference and then applies MP2. The reason why is probably works so well is that the convergence of MP2 is very sensitive to the choice of reference. So, by rotating first, you obtain a more suitable reference and then perturbation theory gets you much closer to FCI than usual, hence, decreasing spin contamination for the reason you noted. It could be cool to compare orbital-optimized MP2 to CCSD. The term $e^{T_1}$ is built into the cluster operator $T = T_1 + T_2$ which is why CCSD (and all CC methods) do not exhibit this sensitivity to the choice of reference, thus you would expect that they would have the same or lower spin contamination right off the bat.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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