# How to confirm whether a system is mott-insulator or not from DFT calculations?

I am using Quantum-Espresso and I need to study mott-insulator. Being a beginner, I am not sure, how do I confirm whether my system is a mott-insulator or not?

I am by no means a DFT expert, so another may be able to provide a better/more complete answer. Nevertheless...

The local density approximation (or its leading gradient extension) do not hold in the limit of a Mott insulator. This being because the Mott insulator is constructed from occupying maximally-localized electron orbitals, and therefore the gradients of density are not guaranteed to be small. The density does not slowly vary like an electron fluid. So you will never observe a Mott insulator using QEspresso.

Density Functional Theory (DFT) works within what I would refer to as the "weak coupling" paradigm. By this, I mean that the interactions are considered to be small relative to the kinetic energy, such that the interactions can be introduced as a perturbation of the non-interacting problem (i.e the electron gas). Spin isn't important to make this point, so let's consider a general (normal-ordered) Hamiltonian of the kind

$$H = \int d{\bf r}\; c^{\dagger}_{\bf r}\big(\frac{-\nabla^2}{2}\big)c_{\bf r} + \int d{\bf r}\; c^{\dagger}_{\bf r}U({\bf r})c_{\bf r} + \int d{\bf r}d{\bf r'} c^{\dagger}_{\bf r}c^{\dagger}_{\bf r'}\frac{1}{2}V({\bf r}-{\bf r'})c_{\bf r'}c_{\bf r} ,$$ where $$U$$ is the (one-body) static lattice potential and $$V$$ is the (two-body) electron-electron interaction. Now $$H$$ is an operator whose eigenstates we would like to solve. But because of the interactions, the full $$H$$ is a challenging problem to solve outright. However we know the solution in the case that we turn off the interactions (i.e $$V=0$$), which is just a 1-body problem of an electron in a lattice potential. It's solution being to fill up the available (non-occupied) single-particle states of the system until we run out of electrons -- the Fermi sea. This can either be a metal or insulator, depending on whether the valence electrons occupy a partially filled band or complete the occupation of a band.

Thus without interactions, the electron density is governed entirely by the crystal potential. If we can treat the effect of the interactions perturbatively, say by slowly turning up the size of $$V$$, then we could re-solve for the density perturbatively in terms of the original Fermi sea. Physically, we can imagine the electron density slightly readjusts from the original density into an energetically better state -- but not so much, and it remains a "Fermi sea" of electrons filling the lowest available states. But this Fermi sea is slightly modified from the one the crystal potential determined without interactions. In other words, the largest effect of the perturbation is to modify the single-particle states which the electrons occupy to form the sea -- but it is still a sea!

Let us imagine a new state $$|\Omega\rangle$$, which is just a Slater determinant whose single-particle wavefunctions are yet to be determined, and which has electron density $$n({\bf r})$$. Thus we should think of $$n$$ as being variational, such that the energy expectation value $$E[n]=\langle\Omega[n]|H|\Omega[n]\rangle$$ is a functional of the different possible $$n$$. The $$n=n^*$$ which minimizes $$E[n]$$ corresponds to the state $$|\Omega[n^*]\rangle$$ with the lowest energy -- our new solution. This energy functional has the form $$E[n] = \int d{\bf r}\; v({\bf r})n({\bf r}) + \frac{1}{2}\int d{\bf r}d{\bf r}'n({\bf r})V({\bf r}-{\bf r}')n({\bf r}') + T[n] + E_{xc}[n] ,$$ where $$T[n]$$ is the kinetic energy of a system of non-interacting electrons with density $$n$$, and $$E_{xc}[n]$$ is the exchange-correlation energy due to the two-body interaction.

The form of $$E[n]$$ above is general enough for any Slater determinant with total electron density $$n$$. For an arbitrary density, $$E_{xc}[n]$$ may depend on not only the density at any point in space $$n[r]$$ but also its gradients $$\nabla n[r]$$. However, DFT makes a few simplifying assumptions about the nature of $$E_{xc}[n]$$, in particular, that it can be expanded to leading order in the density (local density approximation) or to leading order in powers of the gradient [1].

Now the rest of this explanation is to make a point about why this approximation is bad for interaction-driven insulators. We will not consider the local density approximation (or gradient extension) in what follows. Let $$E[n]$$ be exact for a given $$|\Omega[n]\rangle$$.

Let us start with a system whose ground state $$|\Omega[n^*]\rangle$$ is a Fermi sea. For this to be true, the interaction effects must be small. This is only justified depending on the density of the electron fluid, which can be understood using some unit analysis. For our fluid, the kinetic energy $$\nabla^2$$ goes like units of $$1/x^2 \sim n^{2/3}$$. Considering the Coulomb interaction $$V({\bf r}-{\bf r'})=\frac{1}{|{\bf r}-{\bf r}'|} ,$$ scales like units of $$1/x \sim n^{1/3}$$, then the kinetic energy dominates at large densities.

In other words, we could say that our system is an electron gas, so long as the average separation of the electrons in the fluid is small. But as the density shrinks, we get pushed closer and closer into a regime in which the interactions dominate. At the extreme low-density end of this spectrum, the system is entirely dominated by interactions, and thus the ground state is one which minimizes the energy by separating the particles as far apart as possible. Rather than a smooth homogenous fluid, we have a sharp arrangement of localized particles with marginal-to-no density in between.

A cartoon of such a state would be delta-function-like orbitals arranged in a crystal. The density of this state varies sharply like the derivative of a delta-function. More physically, this may be either a Wigner crystal or a Mott insulator depending on the relative density of the electrons to that of the static background crystal lattice.

There is another way to think of a Mott insulator. In addition to the Fermi sea, there are other (effectively) single-particle problems hidden in $$H$$ as well. For example, let us consider that $$H$$ describes a metal when $$V=0$$, then turn on small repulsive $$V$$. While we turn up a repulsive interaction from zero, it is possible that the bandwidth of the partially occupied band becomes smaller than the interaction energy. In such a circumstance, it may be more energetically favorable for valence electrons to occupy localized orbitals. In other words, if the Hubbard model for the partially occupied band can minimize its energy by placing one particle per orbital, thus separating the particles as far apart as possible, it will do so. This particular example can only happen at special integer fillings of the band, and is what one would call a correlated (Mott-like) insulator. Why again is this a single-particle problem? Because it has a single-particle solution -- electrons occupying single-particle states, just not in the same way as the Fermi sea. This what I would refer to as a "strong coupling" paradigm, where kinetic energy is then taken as a perturbation.

Again, unlike the run-of-the-mill DFT based on slowly varying electron fluids, the gradients of the density are not small, owing to the localization of the electron orbitals needed to minimize the interaction energy.

you may be interested in:

There may be a way for you to study Mott-like (i.e interaction-driven) insulators starting from your DFT results. Essentially, one could use the bands outputted from QEspresso in order to construct maximally-localized Wannier functions for the valence bands of your interesting material [2]. This later step involves a program known as Wannier-90, which can be interfaced with QEspresso. These Wannier functions can be used to construct a Hubbard-like tight-binding model, which will permit a Mott insulator at special fillings, and for strong enough interactions.

There are a lot of details to Wanner construction which I'm glazing over, so check out Ref [2].