# Why is the distinction between Mott Insulators and Charge Transfer Insulators important?

Strongly-correlated metals often become insulators due to the repulsive Coulomb interaction, and the basic model here is the Mott-Hubbard Model:

$$H=-t\sum(\hat{c}_{i,\sigma}^{\dagger}\hat{c}_{j,\sigma}+\hat{c}_{j,\sigma}^{\dagger}\hat{c}_{i,\sigma})+U\sum\hat{n}_i^{\uparrow}\hat{n}_i^{\downarrow}$$

Where $U$ represents the Coulomb energy cost of having two electrons on the same site/state.

A very influential paper by Jaan, Allen, Sawatzky makes a distinction between the Mott insulator and the Charge-transfer insulator (J Zaanen, GA Sawatzky, JW Allen - Physical Review Letters, 1985).

For the charge transfer insulator, charges can move between individual sites within a unit cell (i.e. there are at least 2 orbital states for each unit cell $i$) with an energy cost $\Delta$. The charge transfer gap then represents the cost of moving an electron between the anion and cation within the unit cell. I assume this introduces another term in the Hubbard Hamiltonian that looks like: $$H_{CT}\propto\Delta\sum(\hat{c}_{C}^{\dagger}\hat{c}_{A}+\mathrm{h.c.})$$ Where $C$ denotes the cation, and $A$ the anion.

Often phase diagrams of $U$ and $\Delta$ are drawn like the one at the bottom of this post.

My question:

Why is the differentiation between the Charge transfer insulator and Mott insulator important? Sure, the physical origin of the gap $U$ and $\Delta$ require two different orbitals, but what difference does it make with regards to superconductivity, antiferromagnetism, etc.?

In other words, the Mott and Charge-transfer insulators are microscopically different, but who cares and why?

• If the cuprates were pure Mott insulators, a doped hole should make a triplet state according to Hund's rule. Instead, with a hole in the oxygen band, one can gets things like a Zhang-Rice singlet and other possibilities. – Pieter Jan 25 '18 at 22:24
• @Pieter, I am not terribly familiar with the cuprates (or many transition metal oxides), would you mind elaborating? What would the doped hole make a triplet state with? And what is a Zhang-Rice singlet? – KF Gauss Jan 26 '18 at 1:03

If the cuprates would be pure Mott insulators, a doped hole would be on a copper ion, making this in a $3d^8$ configuration. According to Hund's rule, this would be a triplet, parallel spin. Instead, with a hole mostly in the ligands, a singlet can be the state with lowest binding energy. A singlet charge carrier, as Zhang and Rice wrote in 1987: journals.aps.org/prb/abstract/10.1103/PhysRevB.37.3759

• This is the same as your comment, which was interesting but did not answer my question at all. My question is about the general idea behind the JAS paper, which actually predates the discovery of cuprates. – KF Gauss Jan 27 '18 at 0:43
• Well, you were asking about superconductivity. Which I cannot explain, but the character of the charge carriers is expected to be important. – Pieter Jan 27 '18 at 6:25
• So now there is a bounty on this question, which I had tried to answer, especially the superconductivity part. Yes, there are also consequences for magnetism, for example in lithium-doped nickeloxide Li$_x$Ni$_{1-x}$O. But of course it always depends on the specific compound. I do not have a general answer and that seems to be what this question requires. – Pieter Feb 1 at 17:29