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Suppose I am given some geometry data (say, of a water dimer) from CCSD(T).
If I were to do a single-point energy calculation to generate orbitals for the system, wouldn't CCSD give something more representative of the correlation in the system than DFT?

I see "CCSD(T) orbitals don't exist - used DFT orbitals instead" kind of statements in papers (quotations below), and wasn't sure why they don't even mention CCSD. Perhaps it's just due to speed/availability, but that would lead to another question "why is it not worth it to use CCSD orbitals with CCSD(T) geometry".

The same question can be asked for DFT orbitals being used with MP2 geometries, but MP2 perturbs about HF ... so clearly DFT is needed to capture correlation.

[edit: more context]
What I'm trying to do is "export" these orbitals to a diffusion quantum Monte Carlo (DMC) calculation which doesn't do geometry optimization by itself but augments the imported orbitals with explicit correlations (eg. something that depends directly on electron pair distances). Usually the final DMC energy is not very sensitive to whether the orbitals are from CC or DFT (I have outputs from both), but I was checking here to make sure I wasn't missing any "deeper" reason.

From ChemPhysChem 2018, 19 (15), 1886–1894:

Since MP2 orbitals are nonexistent, the charge-transfer interactions were computed using the B3LYP functional with the aug'-cc-pVTZ basis set at the MP2/aug'-cc-pVTZ geometries, so that at least some electron correlation effects could be included.

and J. Phys. Chem. A 2017, 121 (49), 9544–9556:

Due to the nonexistence of CCSD(T) orbitals, ΔE(del) was calculated with ωB97XD/aug-cc-pVTZ.

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    $\begingroup$ I'm guessing that a lot of times, this comes down to the necessary methods being very costly and/or not implemented. The methods that then consume these orbitals are considered to be not even semi-quantitatively correct by many, so DFT orbitals are probably sufficient most of the time. $\endgroup$ – TAR86 Mar 11 at 17:02
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    $\begingroup$ I thought I had a straightforward answer to this, but I believe the difficulty is associated with the fact both these methods are perturbative in nature. In principle, one should be able to generate a new set of orbitals, but I believe in practice the orbitals from perturbative QM methods are actually somewhat arbitrary as the energy is only minimized with respect to excitation amplitudes out of the reference determinant which does not guarantee the orbitals are stationary. Thus, properties in these methods are computed via linear response rather than as expectation values over the orbitals. $\endgroup$ – jheindel Mar 11 at 17:45
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    $\begingroup$ The real question is "what do you mean by orbitals?", which is partially related to "what do you want to look at them for?" Certainly all of these methods have natural orbitals which can be obtained by diagonalizing the corresponding 1-particle density matrix (which will have non-integer occupation numbers), and also certainly there are reference orbitals used to compute these methods. Those cited statements are, to some extent, meaningless without the answers to those questions. $\endgroup$ – levineds Mar 11 at 23:13
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    $\begingroup$ And to clarify, if you run a single-point, you have generated an energy, not orbitals. All orbitals are an implementational tool to obtain this energy and, in that sense, no orbitals "exist", in that they are not observable and can be replaced with an alternative method to obtain the same energy. So in order to obtain orbitals you need to clarify what you mean by that. $\endgroup$ – levineds Mar 11 at 23:23
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    $\begingroup$ I also don't know what you mean by using "X orbitals with Y geometries". All of these calculations are done with the Born-Oppenheimer approximation. You fix the nuclei, then solve for the electronic energy. How you obtained the geometry has nothing to do with what method you subsequently use to compute energies or properties. $\endgroup$ – levineds Mar 11 at 23:23
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Pulling together some of my comments into an answer:

The principal question at issue is "what do you mean by orbitals?", which is partially related to "what do you want to look at them for?" You can obtain an corresponding 1-particle density matrix from all of these methods, which can be diagonalized to yield natural orbitals (which will have non-integer occupation numbers. There are also certainly reference orbitals used to compute these methods (usually Hartree-Fock, but there are methods like orbital optimized-MP2 (OO-MP2) and Brueckner doubles (an orbital optimized CC variant)). Those cited statements are, to some extent, meaningless without the answers to those questions.

If you run a single-point, you have generated an energy, not orbitals. All orbitals are an implementational tool to obtain this energy and, in that sense, no orbitals "exist", in that they are not observable and can be replaced with an alternative method to obtain the same energy.

For your purpose, since you are just exporting to DMC, yeah wouldn't expect the results to vary too much by orbital, the DMC procedure just needs a mostly reasonable starting point. CC natural orbitals would probably produce a more compact wavefunction and therefore maybe run faster than DFT orbitals, but they are enough more expensive to compute that it's probably not worth it. Probably the orbitals you got from the CC calculation you did are just the HF reference orbitals. It's also worth pointing out that natural orbitals are probably not so different from the DFT orbitals unless there is a lot of correlation (in which case, DFT method will probably produce poor orbitals). For systems where the reference state is qualitatively wrong (e.g. diradical system that might suffer from a lot of spin-contamination in HF), OO-MP2 would improve that and those orbitals would probably be substantially better and that would be a case where the cost would probably be justifiable.

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The orbitals used for the coupled clusters variants that you mentioned are (almost always) those from the Hartree-Fock method, which is performed as a first step. Note that the latter does not account for correlation energy. The case of MP$n$ is exactly the same.

In the case of DFT the orbitals are obtained from the Kohn-Sham method. I suppose that they are preferred due to this.

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