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The Mott insulator is a system that due to strong electron-electron interactions is an insulator which be a metal by formal charge counting of electrons in the unit cell.

Often, the Mott insulator is defined to be the ground state of the Hubbard model at half filling

$$H = -t \sum_{i} (c^{\dagger}_{i+1} c_{i} + c^{\dagger}_{i} c_{i+1}) + U\sum_{i} n_{i\uparrow} n_{i \downarrow}$$

Presumably, the Mott insulator is the ground state of the Hubbard model. Yet, in a paper by Robert Laughlin (Nobel prize in Condensed Matter Physics 1998), Phys. Rev. B 89, 035134 (2014), he states the following (emphasis mine):

Unfortunately, the phenomenological definition of a Mott insulator has always been somewhat difficult to state [...].

The enormous amount of theoretical work stimulated by the cuprate discovery has now built up a strong case that the Mott insulator does not exist as a distinct zero-temperature state of matter.

Laughlin says similar things in Phys. Rev. Lett. 112, 017004 (2014)

Unfortunately, the cuprates are so anomalous phenomenologically that they have thus far defied categorization as conventional metals or insulators. This has led to speculation that they might involve a new, and as-yet unidentified, parent vacuum. Proposals for such a vacuum include the Mott insulator, the resonating valence bond, the non-Fermi liquid, and the loop-current insulator[4–7].

However, there is a much simpler potential explanation: the zero-temperature phases of the cuprates are conventional,and the strange behaviors are just critical phenomena and glassiness associated with transitions among two or more of these phases [8–10]. This view is supported by the absence of experimental evidence for new states of matter at lowest temperature scales. It is also supported by theory, in that none of the proposed theoretical alternatives to conventional metals and insulators can be (1) written down in a straight forward way at zero temperature or (2) shown to be stabilized by any simple Hamiltonian. There is no mathematical case that any of them actually exist.

I know Laughlin has a reputation of being strong headed (to say the least), but I want to understand the basis for these statements. Clearly he has reasons to say what he said in such absolute terms, but mainstream physicists treat the Mott insulator as if its existence is unquestionable. The "textbook" examples of Mott insulators are NiO and La2CuO4, which definitely seem to be insulators at zero temperature, but presumably these don't "count" to Laughlin.

So, my question is: does the Mott insulator exist as a distinct ground state?

If the Mott insulator does exist, why does Laughlin argue it does not? If it doesn't exist, why is this point seemingly ignored by most physicists?

An aside: I disagree with the comments by user:wcc which claim ultracold atom simulations have "solved" this issue long ago and that Laughlin is only referring to superconducting copper oxides. Putting aside the fact that ultracold atom simulators cannot reach the ground state of the Hubbard model (they are much too hot to be anywhere close), Laughlin claims the Mott insulator does not exist, and that is a much stronger, purely mathematical statement. Laughlin's PRL is also from 2014, which is many years after the start of cold atoms on optical lattices which began in the late 90's.

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  • $\begingroup$ For those without access to PRB, the preprint version of the first article mentioned is available at arxiv.org/pdf/1306.5359.pdf $\endgroup$ Apr 5 '20 at 19:40
  • $\begingroup$ @Clara Diaz Sanchez thank you! $\endgroup$
    – KF Gauss
    Apr 5 '20 at 20:59
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    $\begingroup$ It has been realized countless times with ultracold atoms in optical lattice...and the subject has been quite beaten to death for nearly twenty years since this seminal work (nature.com/articles/415039a). That started with bosonic atoms but people have realized ferminionc Mott insulators as well. In a deep lattice, there are only two terms, tunneling and on-site repulsion, so the Hubbard model is a good description for atoms in optical lattice. It seems Laughlin's comments are intended specifically for cuprates. $\endgroup$
    – wcc
    Apr 7 '20 at 0:30
  • $\begingroup$ @wcc I understand that sentiment, but it doesn't help answer the question. Laughlin should be quite aware of these things, his papers were published in 2014 after all. As devil's advocate, I guess he would say that cold atom experiments are irrelevant because they are very "hot" on the scale of their fundamental interactions $U$ and $t$, so they have no business making statements about zero temperature. $\endgroup$
    – KF Gauss
    Apr 7 '20 at 5:03
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    $\begingroup$ @KFGauss, what makes you say that cold atoms are very hot in (t,U) scale? They are pretty cold. It's not cold enough in the superexchange scale (t^2/U) though. $\endgroup$
    – wcc
    Apr 7 '20 at 18:03

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