Mott Insulators in the Atomic Limit of the Hubbard Model

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Hello! Let's continue our exploration of strongly correlated electron systems.

In the previous lesson, we derived the Hubbard model, the quintessential model for interacting electrons on a lattice. We established that its physics is dictated by a fundamental competition: the kinetic energy term ( $- t$ ), which encourages electrons to delocalize and move freely, and the on-site Coulomb repulsion ( $U$ ), which penalizes double occupancy and favors localization. The full Hamiltonian is:
$H = - t \sum_{⟨ i, j ⟩, σ} (c_{j σ}^{†} c_{iσ} + H . c .) + U \sum_{i} n_{i ↑} n_{i ↓} - μ \sum_{i, σ} n_{iσ}$
where $μ$ is the chemical potential that controls the average number of electrons.

Today, we will analyze this model in its most transparent, albeit extreme, limit: the atomic limit, where we assume the electrons are so localized that they cannot hop at all ( $t = 0$ ). This simplification will allow us to uncover a profound piece of physics: the formation of a Mott insulator. This is an insulating state driven purely by electron-electron interactions, a phenomenon entirely outside the scope of conventional band theory.

1. The Atomic Limit: A Collection of Independent Sites

When we set the hopping parameter $t = 0$ , the kinetic term in the Hubbard Hamiltonian vanishes. The sites on the lattice become completely decoupled from one another.
$H_{t = 0} = U \sum_{i} n_{i ↑} n_{i ↓} - μ \sum_{i, σ} n_{iσ} = \sum_{i} (U n_{i ↑} n_{i ↓} - μ (n_{i ↑} + n_{i ↓}))$
Since the Hamiltonian is now a sum of independent terms for each site $i$ , we can solve the problem for a single site, and the solution for the entire lattice will just be a combination of these single-site solutions. The Hamiltonian for a single, isolated site is:
$H_{site} = U n_{↑} n_{↓} - μ (n_{↑} + n_{↓})$

2. Solving the Single-Site Problem

The Hilbert space for a single site is small, consisting of only four possible states. Since the number operators $n_{↑}$ and $n_{↓}$ are diagonal in this basis, these states are the eigenstates of our Hamiltonian. Let's find their energies:

Empty site: $∣0 ⟩$ . The number of electrons is zero, so $n_{↑} = 0$ and $n_{↓} = 0$ .
$E_{0} = U (0) (0) - μ (0 + 0) = 0$
Singly occupied, spin up: $∣ ↑ ⟩$ . Here $n_{↑} = 1$ and $n_{↓} = 0$ .
$E_{↑} = U (1) (0) - μ (1 + 0) = - μ$
Singly occupied, spin down: $∣ ↓ ⟩$ . Here $n_{↑} = 0$ and $n_{↓} = 1$ .
$E_{↓} = U (0) (1) - μ (0 + 1) = - μ$
Note the spin degeneracy of the singly occupied state.
Doubly occupied: $∣ ↑↓ ⟩$ . Here $n_{↑} = 1$ and $n_{↓} = 1$ .
$E_{↑↓} = U (1) (1) - μ (1 + 1) = U - 2 μ$

The following reading provides a concise, formal summary of this single-site analysis and explores its consequences for the system's ground state and particle density.

The Limits of the Hubbard Model

The paper 'The Limits of the Hubbard Model' by David Grabovsky provides a clear and formal analysis of the Hubbard model in the atomic limit.

Please read section 6.1, 'Single Site: Strong Interactions' (pages 17-18). Focus on the diagonalization of the single-site Hamiltonian and the table of ground states. Pay close attention to Figure 1, which plots the average occupation number N versus the chemical potential µ. We will analyze this plot in detail later in the lesson.

3. The Mott Gap: The Energy Cost of Conduction

Now we can address the central question: why is this system an insulator? Let's consider the case of half-filling, where there is, on average, one electron per site. In the atomic limit with large $U$ , the ground state will consist of exactly one electron on every site to avoid the energy penalty $U$ .

For electrical conduction to occur, an electron must be able to move from one site to another. Imagine trying to move an electron from site $j$ to an adjacent site $i$ .

1D Mott-Hubbard Chain: Hopping and On-site Repulsion — This diagram illustrates the fundamental process for charge transport in a half-filled Hubbard chain. For an electron to move, it must hop onto a site that is already occupied, creating a doubly occupied site (a "doublon") and leaving behind an empty site (a "holon"). This process has an associated energy cost.

Let's calculate this cost.

Initial state: Two adjacent sites, $i$ and $j$ , are singly occupied. For simplicity, let's assume they have opposite spins to allow for the final state. The total energy from these two sites is $E_{initial} = E_{↑} + E_{↓} = - μ - μ = - 2 μ$ .
Final state: The electron from site $j$ has hopped to site $i$ . Site $i$ is now doubly occupied, and site $j$ is empty. The total energy is $E_{final} = E_{↑↓} + E_{0} = (U - 2 μ) + 0 = U - 2 μ$ .

The energy required to create this mobile charge excitation is the difference:
$Δ E = E_{final} - E_{initial} = (U - 2 μ) - (- 2 μ) = U$
This is the Mott gap. It represents the energy cost to create a pair of charge carriers (a doubly-occupied site and an empty site). If $U$ is large compared to the thermal energy $k_{B} T$ , such excitations are suppressed. The electrons are "locked" in place by the strong repulsion, and the system cannot conduct electricity.

This is fundamentally different from a band insulator, which is insulating because it has a completely filled valence band separated from an empty conduction band. Here, the band would be half-filled and thus predicted to be a metal by band theory. The insulating behavior is a pure correlation effect.

4. Upper and Lower Hubbard Bands

The discrete energy levels we found can be reinterpreted as the foundation for the electronic band structure in a correlated system.

The energy level for a singly occupied site ( $- μ$ ) corresponds to what will become the Lower Hubbard Band. Excitations out of this band correspond to removing an electron, leaving an empty site.
The energy level for adding an electron to a singly occupied site to create a double occupancy ( $U - μ$ ) corresponds to the Upper Hubbard Band.

The Mott gap $U$ is the energy separation between these two bands. When we reintroduce a small amount of hopping ( $t > 0$ ), these discrete levels broaden into bands, but a gap of order $U$ remains as long as $U ≫ t$ .

The following video provides an excellent visual explanation of how these atomic levels evolve into the Mott-Hubbard band structure.

Introduction to Correlated Materials. Lecture 2: The Mott-Hubbard Band Structure

In this segment from her lecture series, Prof. Nicola Spaldin explains how the discrete atomic energy levels broaden into bands as atoms are brought closer together (equivalent to increasing the hopping t).

Please watch the clip from 10:55 to 15:33. Focus on the diagram of energy versus bandwidth. This clip clearly explains how the singly occupied level forms the lower filled band and the doubly occupied level forms the upper empty band, with the Mott-Hubbard gap separating them.

In the strict atomic limit ( $t = 0$ ), the density of states (DOS) consists of just two delta-function peaks separated by $U$ . This is shown in another clip from the same lecture series.

Introduction to Correlated Materials. Lecture 4: The Hubbard Hamiltonian

This short clip visualizes the density of states in the two extreme limits of the Hubbard model.

Watch the segment from 21:09 to 23:01. The key part is the diagram shown around 22:00, which depicts the DOS in the atomic limit as two sharp peaks separated by U, representing the lower and upper Hubbard bands before they acquire any bandwidth from hopping.

5. A Thermodynamic View of the Gap

The existence of the Mott gap can also be understood from a more formal, thermodynamic perspective by examining the relationship between particle number and chemical potential, as shown in Figure 1 of the Grabovsky reading.

At zero temperature, the chemical potential is the energy required to add one particle to the system: $μ = \partial E / \partial N$ .

To fill the system from $N = 0$ to $N = 1$ (one electron per site), we add electrons to empty sites. The energy cost is associated with the $- μ$ level. The chemical potential will be around $μ \approx 0$ .
Once the system is at half-filling ( $N = 1$ ), every site is occupied by one electron. To add the next electron, we are forced to create a double occupancy. The energy cost for this jumps from $- μ$ to $U - 2 μ$ .
This means to increase the filling from just below 1 to just above 1, the chemical potential $μ$ must abruptly jump from a value near 0 to a value near $U$ .

A jump or discontinuity in the chemical potential as a function of particle density is the thermodynamic signature of an energy gap. Here, the jump of size $U$ at half-filling is precisely the Mott gap. The plateau at $⟨ N ⟩ = 1$ in the plot of $⟨ N ⟩$ vs $μ$ is the manifestation of this insulating phase.

Test your understanding!

The partition function for the single site is $Z = 1 + 2 e^{β μ} + e^{β (2 μ - U)}$ . Using the statistical mechanics relation $⟨ N ⟩ = \frac{1}{β Z} \frac{\partial Z}{\partial μ}$ , derive the expression for the average occupation number $⟨ N ⟩$ and verify that at half-filling ( $⟨ N ⟩ = 1$ ), the chemical potential must be $μ = U /2$ .

Show answer

First, let's calculate the derivative of the partition function with respect to $μ$ :
$\frac{\partial Z}{\partial μ} = \frac{\partial}{\partial μ} (1 + 2 e^{β μ} + e^{2 β μ - β U}) = 2 β e^{β μ} + 2 β e^{2 β μ - β U}$
Now, we can find the average occupation number:
$⟨ N ⟩ = \frac{1}{β Z} (2 β e^{β μ} + 2 β e^{2 β μ - β U}) = \frac{2 ( e ^{β μ} + e ^{β (2 μ - U)} )}{1 + 2 e ^{β μ} + e ^{β (2 μ - U)}}$
This matches the expression given in the resources (e.g., Appendix C.1 of Grabovsky, eq. C.4, or Exercise 12 in Scalettar).

Now, let's set $⟨ N ⟩ = 1$ and solve for $μ$ :
$1 = \frac{2 ( e ^{β μ} + e ^{β (2 μ - U)} )}{1 + 2 e ^{β μ} + e ^{β (2 μ - U)}}$
$1 + 2 e^{β μ} + e^{2 β μ - β U} = 2 e^{β μ} + 2 e^{2 β μ - β U}$
Subtracting $2 e^{β μ}$ from both sides:
$1 + e^{2 β μ - β U} = 2 e^{2 β μ - β U}$
$1 = e^{2 β μ - β U}$
Taking the natural logarithm of both sides:
$ln (1) = 0 = 2 β μ - β U$
$β U = 2 β μ$
$μ = \frac{U}{2}$
This confirms that the chemical potential required to achieve exactly half-filling is precisely half the interaction energy. This is the center of the Mott gap.

Conclusion

In this lesson, we have dissected the Hubbard model in its simplest form—the atomic limit. By neglecting kinetic energy, we were able to see with great clarity how strong Coulomb repulsion fundamentally alters the electronic properties of a system.

Key Takeaways:

In the atomic limit ( $t = 0$ ), the Hubbard model describes a set of independent sites, each with four possible occupation states and corresponding energies.
At half-filling, creating a mobile charge carrier (a doublon-holon pair) costs a finite energy $U$ .
This energy cost is the Mott gap, which makes the system an electrical insulator despite having a half-filled band. This is the defining characteristic of a Mott insulator.
The single-site energy levels associated with removing or adding an electron to a half-filled system form the basis of the Lower and Upper Hubbard Bands, respectively.
The Mott gap also manifests as a discontinuity of size $U$ in the chemical potential as a function of filling at $⟨ N ⟩ = 1$ .

Preview of the Next Lesson

The atomic limit provides the essential concept of the Mott insulator. But what happens in a more realistic scenario where hopping is present, even if it's small compared to $U$ ? The interplay between $t$ and $U$ gives rise to a rich variety of phenomena, including magnetism. In our next lesson, we will move away from the atomic limit and apply a mean-field approximation to the Hubbard model to understand how itinerant electron interactions can lead to ferromagnetism, deriving the famous Stoner criterion.