research35 min studyLesson 551 in Physics

Hubbard atomic limit and the Mott gap

Solve the repulsive one-orbital atomic limit while separating a charge gap from magnetic ordering and lattice transition claims.

Before you begin

On this page

The question for today

What you’ll understand

Specify hopping, repulsion, filling and bond-counting conventions.
Derive atomic addition and removal thresholds.
Distinguish an exact atomic gap from dimension-dependent lattice phase boundaries.

A partially filled noninteracting band suggests mobile carriers, yet electron repulsion can make changing local charge expensive. The one-orbital Hubbard model isolates that competition between hopping and double occupancy. Its atomic limit is exactly soluble and exposes the meaning of a charge gap without requiring an ordered spin pattern. The same calculation does not supply a universal critical interaction for every lattice, dimension, filling or constrained magnetic phase.

Fix the model before comparing energy scales

Take one spinful orbital per site, real nearest-neighbor hopping t and a positive on-site repulsion U. Each undirected bond is counted once together with its Hermitian reverse term. The filling n counts both spins and ranges from zero to two electrons per site; half filling means one, not one half. The chemical potential controls this density in the grand ensemble. Orbital choices, lattice geometry and longer-range terms are additional modeling decisions rather than consequences of the compact notation.

The relevant kinetic scale is a bandwidth determined by both hopping and geometry, so quoting a critical ratio without specifying the lattice is ambiguous. A chain, square lattice and large-coordination lattice do not share the same bandwidth-to-hopping factor or phase diagram. The interaction represents repulsion in retained orbitals after other degrees of freedom have been eliminated; it need not equal an unscreened bare Coulomb matrix element. This lesson first removes hopping entirely, avoiding any uncontrolled assumption about a lattice ground state.

K=H-\mu N=-t\sum_{\langle ij\rangle,\sigma}(c_{i\sigma}^\dagger c_{j\sigma}+c_{j\sigma}^\dagger c_{i\sigma})+U\sum_i n_{i\uparrow}n_{i\downarrow}-\mu\sum_i n_i,\qquad n=\frac{N_e}{N_s}

t, U and μ are energies; Ne is electron number and Ns site count. The displayed bond sum counts each undirected link once, and ni includes both spin occupations.↔ Scroll wide equations horizontally.

Compare four on-site states exactly

When t vanishes, each site is independent and has four states: empty, spin up, spin down and doubly occupied. Their grand energies follow by counting particles and double occupancy. At zero temperature a singly occupied site minimizes the grand energy when the chemical potential lies strictly between zero and U. The two spin states remain degenerate. An atomic sample in this interval is incompressible with respect to an infinitesimal chemical-potential change but has no exchange process selecting a spin arrangement.

Starting from one electron, removing it costs the chemical potential measured from the lower threshold, while adding a second costs U minus chemical potential. These two costs depend on the chosen position inside the plateau. Their sum is U and equals the separation between addition and removal thresholds. Confusing either individual cost with the whole charge gap produces a factor-of-two error at particle-hole symmetry, where each cost is U over two.

K_0=0,\qquad K_\uparrow=K_\downarrow=-\mu,\qquad K_2=U-2\mu,\qquad 0<\mu<U:\ n=1

These are exact grand energies for the isolated repulsive orbital. At μ = 0 or U there are additional charge degeneracies; finite temperature rounds the zero-temperature occupancy steps.↔ Scroll wide equations horizontally.

Separate insulating charge from spin physics

For the atomic half-filled many-site model, a canonical charge-gap definition compares the energies with one extra or one fewer electron. It yields U even though every spin configuration remains degenerate. Thus a charge gap does not logically require antiferromagnetic order. Small hopping broadens charge excitations and generates a much lower spin-exchange scale; calculating these effects requires additional approximations or a controlled expansion rather than reusing the isolated-site answer unchanged.

A lattice Mott transition depends on dimension, density, frustration, temperature and whether ordered phases are permitted. Paramagnetic coexistence boundaries in a specified approximation are not universal Hubbard constants. Likewise a gap caused by symmetry-breaking band reconstruction must be distinguished from the atomic interaction mechanism. The probability lab can illustrate the arithmetic of independent double occupancy in a reference ensemble, but that binomial distribution is not the correlated Hubbard ground state and cannot decide whether a lattice is insulating.

\Delta_c=E_0(N_s+1)+E_0(N_s-1)-2E_0(N_s)=U\quad(t=0),\qquad E_{\rm remove}=\mu,\quad E_{\rm add}=U-\mu

E0 denotes the canonical many-body ground energy without the −μN term. The equality Δc = U is exact only in the atomic half-filled model with the specified single orbital and no additional interactions.↔ Scroll wide equations horizontally.

Work it through

Unequal addition and removal costs

For an isolated Hubbard orbital let $U=6\,\mathrm{eV}$ and $\mu=2\,\mathrm{eV}$ at zero temperature. Identify the lowest grand-energy sector and compute its charge-removal cost, charge-addition cost and total threshold gap.

The empty, singly occupied and doubly occupied grand energies are $0$ , $-2$ and $6-4=2\,\mathrm{eV}$ . A singly occupied state is lowest, with two degenerate spin choices.
Removing its electron changes grand energy from $-2$ to zero, costing $2\,\mathrm{eV}$ . Adding an opposite-spin electron changes it from $-2$ to $+2$ , costing $4\,\mathrm{eV}$ .
The two costs add to $6\,\mathrm{eV}$ , equal to the interval between the chemical-potential thresholds zero and U. Neither the spin degeneracy nor an assumed magnetic ordering temperature enters this charge calculation.

The result

The site is singly occupied; removal costs 2 eV and addition costs 4 eV, with a 6 eV charge gap.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

Independent occupancy is only a reference

Reset and set BOTH Success probability p = 0.50 and Trials n = 2. Predict mean one and variance 0.5. Reset and set p = 1 and n = 2; predict mean two and zero variance. Interpret the two independent Bernoulli slots only as a reference counting exercise. The exact binomial bars do not include Hubbard repulsion, fermionic coherence, a Mott gap or correlated many-site occupations.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA mean occupancy alone does not specify its fluctuations or the interacting state.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The singly occupied grand energy −μ lies below both 0 and U − 2μ precisely when μ is positive and smaller than U.

Your turn

Try it yourself

Keep $U=6\,\mathrm{eV}$ but choose $\mu=7\,\mathrm{eV}$ at zero temperature in the atomic model. Find the occupancy and the cost of removing one electron from its lowest grand-energy state.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compare all three charge-sector grand energies before applying the removal-cost formula; the half-filled plateau assumption no longer holds.

Compare with the solution

The grand energies are $K_0=0$ , $K_1=-7\,\mathrm{eV}$ and $K_2=6-14=-8\,\mathrm{eV}$ . Double occupancy is now lowest, so n equals two. Removing one electron changes grand energy from −8 to −7 eV and costs $1\,\mathrm{eV}$ , equal to μ minus U. Using the earlier singly occupied removal cost μ would be incorrect because its initial sector is different. The occupancy change occurs at the upper threshold μ = U, not at a universal value of hopping or lattice bandwidth.

Keep the idea

Take it with you

Half filling means one electron per spinful orbital, with a stated bond convention.
The atomic charge gap is U, while addition and removal costs separately depend on μ.
An atomic insulating plateau does not determine a lattice critical interaction or spin order.

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