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Open-Shell systems (radicals and so on) can be modeled by single-determinant quantum chemical methods using unrestricted (u) or restricted open-shell (ro) methods (I am aware that single-determinant approaches are not always reasonable for open-shell systems, but let's ignore that for a second and suppose we have to use a single-determinant method).

Unrestricted calculations may suffer from spin-contamination. Restricted open-shell calculations remedy this, they are unaffected by spin-contamination. However, restricted open-shell methods are much more computationally expansive. Also, Koopmans' theorem cannot be rigorously applied in the case of ro calculations.

My question is:

Apart from the before mentioned issues, are there any disadvantages of restricted-open-shell computations in comparison to unrestricted calculations? Are there any situations where an unrestricted calculation would be more appropriate than a restricted-open-shell calculation?

[Methods of interest are - for example - MP2, DFT and HF.]

There is already a somewhat related discussion: U- or RO-method for Singlet-Triplet Gap?. However, my question is much broader, it includes for instance: Are there cases when it is absolutely not ok to use ro methods, and you have to use unrestricted methods?

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    $\begingroup$ One rumored disadvantage of RO is less reliable and possibly slower convergence. This may also be a result of the implementation at hand. $\endgroup$
    – TAR86
    Jun 19, 2017 at 11:36

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To be more precise, it's not so much that restricted open-shell calculations are unaffected by spin-contamination as that you are forcing them to give back the multiplicity you want.

Broadly speaking, closed-shell methods, restricted open-shell methods and constrained unrestricted open-shell methods all give correct spin multiplicities, but that's because all of them assume from the beginning which is the correct multiplicity and force it into the calculation - there is always a restriction in place. For closed-shell methods, it is that the orbitals occupied by both spin populations are the same; for restricted open-shell methods, the same applies to all but the excess spin electrons from the larger spin population. But more recent developments point to restricting multiplicity after the fact - by adjusting orbital population to minimise spin contamination. There are a couple of methods to do that, but they start from an unrestricted calculation (much faster, as you mention), and then force the population in a way that minimises spin contamination - either by forcing orthogonality constraints between the $\alpha$ and $\beta$ determinants, or by parametrising spin contamination in a way that allows it to be fitted to the expected total multiplicity by constraining occupation. This retains the computational advantage and simpler mathematical implementation of unrestricted methods but makes sure that the multiplicity of the model system reflects the multiplicity of the modelled system with generally very small additional computational effort.

I'd suggest that this is going to be an active area of research that will see increased use in the next few years.

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You don't even have to consider open-shell systems to see differences between RHF and UHF. A very popular example is the dissociation of $\ce{H2}$. RHF is simply not capable of describing the dissociated fragments and will not converge with respect to the internuclear distance $R$. UHF will be spin contaminated, but you get the right energies (apart from correlation effects, which are negligible for 2 separated hydrogen atoms.)

You should be able to find this easily in standard quantum chemistry text books.

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  • $\begingroup$ Thanks for reminding me of this simple example system. Is there a way to generalize the learning from this model to larger systems? Would you - based on your answer to my broad question - argue, that unrestricted methods are always superior to ro methods? $\endgroup$ Jun 19, 2017 at 14:40
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    $\begingroup$ To cite from your question: "Unrestricted calculations may suffer from spin-contamination." So no, you cannot argue that one or the other will always be superior. In UHF you sacrifice a well defined spin quantum number to get the degree of freedom of splitting your spatial orbitals into separate spin orbitals. If such a result is not good enough, you will need to go to more sophisticated methods. $\endgroup$
    – Feodoran
    Jun 19, 2017 at 15:30
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The real question is:

What do I gain from spin contamination?

Spin contamination is usually presented as a bad thing. However, the extra (sometimes large) spin polarization that can occur when relaxing the spatial constraint on the wavefunction is what gives UHF a little bit of electron correlation beyond RHF and ROHF. This is why you should be wary of people who blindly state "there is no correlation in Hartree-Fock". They probably mean "when compared to wavefunction-based methods", but without context, it isn't true!

I will steal from Frank Jensen's book, section 4.4:

By including electron correlation in the wave function, the UHF method introduces more biradical character into the wave function than RHF. The spin contamination part is also purely biradical in nature, i.e. a UHF treatment in general will overestimate the biradical character of the wave function. Most singlet states are well described by a closed shell wave function near the equilibrium geometry and, in those cases, it is not possible to generate a UHF solution that has a lower energy than the RHF. There are systems, however, for which this does not hold. An example is the ozone molecule, where two types of resonance structures can be drawn.

The biradical resonance structure for ozone requires two singly occupied MOs, and it is clear that an RHF type wave function, which requires all orbitals to be doubly occupied, cannot describe this. A UHF type wave function, however, allows the $\alpha$ and $\beta$ orbitals to be spatially different, and can to a certain extent incorporate both resonance structures. Systems with biradical character will in general have a (singlet) UHF wave function different from an RHF.

If you look for the singlet with RHF, there will be almost no biradical character in the resulting wavefunction, unless there is a biradical-looking orbital delocalized over the whole molecule. I don't know if there is, just guessing. With the biradical you can also see how the triplet state will enter the picture; I am leaving the details to Jensen's chapter. There is an analogy between UHF including too much biradical character with MP2 overestimating the dispersion interaction; it is the first level of theory where the effect is being considered qualitatively, and should not be expected to be quantitatively accurate. Though, UHF is probably worse at describing biradical character than MP2 is at quantifying the dispersion energy.

I haven't thought much about if this is considered to be dynamic or static correlation if we perform that artificial division. For example, this singlet biradical can be described by two determinants composing a single configuration state function (so TCSCF would work, and some flavors of ROHF), but you can capture the same effect after the fact with EOM-CCSD. It might be considered a static effect, since the presence of higher spin states in the resulting wavefunction results in some configurational mixing.

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    $\begingroup$ HF describes exchange correlation exactly, so the statement "there is no correlation in Hartree-Fock" should always be considered wrong. $\endgroup$ Aug 8, 2018 at 14:53
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One particular application where it is a bad idea to use the assumptions of RO, is the simulation of NMR or EPR spectra.

If you include the coupling of nuclear and electronic spin (which you have to) you have to allow different energies and population for the inner s-orbitals.

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  • $\begingroup$ So what you really want to say is that spin polarization of the core orbitals is needed, not just the valence orbitals, especially for something like the Fermi contact interaction. One would not want to perform ROHF indirect spin-spin coupling constant calculations. $\endgroup$ Jun 19, 2017 at 21:34

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