# Can one always take a single-particle picture (i.e. DFT) irrespective of the strength of correlations?

Most textbooks on Kohn-Sham density functional theory will assert that it is exact assuming that one has the 'appropriate' exchange-correlation functional. To my mind, this is completely astounding and would imply that irrespective of the strength of correlations there is always a single particle picture. Will this always be true in the limit of very strong correlations e.g. at quantum critical points? Aside from this, are there physical situations in which the exchange-correlation functional is sufficiently ill-behaved so as for the Hohenberg-Kohn theorems to not apply?

EDIT: I have realized that my question is rather vague. The question that I meant to ask is whether the exact exchange-correlation functional can exist for a strongly correlated system. My justification for believing this might be correct is that I would expect a perturbative expansion of the self-energy $\Sigma (\mathbf k)$ to diverge rapidly in a highly correlated system.

• There are no counterexamples to a theorem. With the exact functional, a single-particle description is always possible & exact. Of course, you are just shifting the difficulty of the many-body problem into the functional. In practice, we only have approximations to the functional (and this is a fundamental limitation), which work more or less well depending on the situation. – Norbert Schuch Jan 27 '17 at 20:17

Actually, original density functional theory (original DFT) does not support a single particle picture, which just tells you there exists a universal functional (has nothing to do with concrete materials) and hence justifies a universal one-to-one correspondence between density and the wavefunction, namely

$$\text{density} \leftrightarrow \text{external potential} \leftrightarrow \text{wavefunction} .$$

so you can extract information about your system just through density rather than wavefunction (unsolvable in general for a many-particle system).

But DFT does not deliver a method to find the universal functional. I think what you are talking is Kohn-Sham DFT (of course effective in the picture of single-particle), which is the most popular version of DFT and making much progress in many fields. But there is another version called Orbital Free DFT, which is more closely related to the spirit of the original H-K theorems or original DFT.

If you want to consider the strong correlation materials in the Kohn-Sham DFT framework, currently the popular approach is LDA+U in electronic structure theory. But in the framework of Orbital-free DFT, seemingly there aren't relevant progress.

Hope it helps.

• Thank you for your answer. Of course you are absolutely right, I was indeed referring to Kohn-Sham DFT. I find it very easy to conflate the two. I know that DFT+U can describe some correlated effects such as the Mott transition (albeit by breaking spin and orbital symmetry). But I was more interested in whether, given a very strongly correlated material, there will always be a Kohn-Sham mapping i.e. whether an exact exchange-correlation functional can even exist in principle? – guy Jan 28 '17 at 10:48
• @guy Do you have any reason to believe that such a mapping does not exist? Isn't that exactly what the theorem states, that such a mapping always exists? – Norbert Schuch Jan 29 '17 at 1:42