# 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
• 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 '19 at 17:29