Hubbard Dimer

The Hubbard dimer is the exactly diagonalizable two-site, two-spin fermion model in which hopping competes with onsite interaction and turns the half-filled ground state into a correlated mixture of a spin singlet and doublon–hole configurations.

The main example has two particles, one orbital per site, and a single hopping bond. Its six states reduce to a two-state mixing problem plus four uncoupled states, making charge fluctuations and antiferromagnetic exchange calculable by hand.

Helpful background. Hubbard Model explains the competition between hopping and onsite interaction; Fermionic Anticommutation Relations explains the signs in an ordered occupation basis.

Two Sites with Onsite Interaction

The reference dimer uses:

Field	Baseline choice
sites	two sites, labeled $1$ and $2$
physical modes	one spin-$\uparrow$ and one spin-$\downarrow$ fermionic mode per site
global mode order	$(1\uparrow,1\downarrow,2\uparrow,2\downarrow)$
full Hilbert space	four-mode Fock space, $\dim\mathcal F=16$
primary sector	total particle number $N=2$, corresponding to one particle per site on average
bond list	one undirected bond $\{1,2\}$
hopping	real $t>0$, with matrix element $-t$
interaction	onsite $U\in\mathbb R$
chemical potential	absent; $H$, not $H-\mu\widehat N$, is diagonalized
site offset and bias	absent
fields, flux, intersite interaction, and pairing	absent
benchmark point	$t=1$, $U=4$ in the $N=2$ sector

The primary reference is the repulsive half-filled problem $U>0$, but the exact formulas below remain valid for attractive $U<0$. The model has only one physical bond. Writing a two-site periodic sum with two directed site indices and then adding a Hermitian conjugate can double that bond accidentally.

Fermionic Modes and Ordered Fock Basis

The operators satisfy

$$\begin{aligned} \{c_{i\sigma},c_{j\sigma'}^\dagger\} &= \delta_{ij}\delta_{\sigma\sigma'}, \\ \{c_{i\sigma},c_{j\sigma'}\} &= 0. \end{aligned}$$

Define

$$n_{i\sigma} = c_{i\sigma}^\dagger c_{i\sigma}, \qquad n_i = n_{i\uparrow}+n_{i\downarrow}.$$

The declared mode order defines each bit state:

$$\lvert n_{1\uparrow} n_{1\downarrow} n_{2\uparrow} n_{2\downarrow} \rangle ={} \left(c_{1\uparrow}^\dagger\right)^{n_{1\uparrow}} \left(c_{1\downarrow}^\dagger\right)^{n_{1\downarrow}} \times \left(c_{2\uparrow}^\dagger\right)^{n_{2\uparrow}} \left(c_{2\downarrow}^\dagger\right)^{n_{2\downarrow}} \lvert0\rangle.$$

This ordering is not physical, but it fixes the signs of matrix elements and coordinate representations. A different ordering is equally valid only if basis states, operators, and observables are transformed consistently.

Full and fixed-number dimensions

With four fermionic modes,

$$\dim\mathcal F = 2^4 = 16.$$

The sector with total particle number $N$ has dimension

$$\dim\mathcal H_N = \binom4N.$$

Thus the dimensions for $N=0,1,2,3,4$ are

$$1,\ 4,\ 6,\ 4,\ 1.$$

Because spin-conserving hopping preserves $N_\uparrow$ and $N_\downarrow$ separately,

$$\dim\mathcal H_{N_\uparrow,N_\downarrow} = \binom2{N_\uparrow} \binom2{N_\downarrow}.$$

At $N=2$, the sector decomposes as

$$\begin{aligned} \dim\mathcal H_{2,0}&=1,\\ \dim\mathcal H_{1,1}&=4,\\ \dim\mathcal H_{0,2}&=1. \end{aligned}$$

Half-filled six-state basis

In the baseline order, a convenient $N=2$ basis is:

Bit state	Local-state notation	$S_{\mathrm{tot}}^z$
1100	$\lvert\uparrow\downarrow,0\rangle$	$0$
1010	$\lvert\uparrow,\uparrow\rangle$	$1$
1001	$\lvert\uparrow,\downarrow\rangle$	$0$
0110	$\lvert\downarrow,\uparrow\rangle$	$0$
0101	$\lvert\downarrow,\downarrow\rangle$	$-1$
0011	$\lvert0,\uparrow\downarrow\rangle$	$0$

Local-state notation is compact; the ordered creation-operator definition fixes its signs.

Hamiltonian and Conventions

The fixed-number Hamiltonian is

$$H ={} -t \sum_{\sigma=\uparrow,\downarrow} \left( c_{1\sigma}^\dagger c_{2\sigma} + c_{2\sigma}^\dagger c_{1\sigma} \right) + U \left( n_{1\uparrow}n_{1\downarrow} + n_{2\uparrow}n_{2\downarrow} \right).$$

Write

$$H = H_t+H_U.$$

The kinetic term moves one fermion without changing its spin. The interaction counts doublons:

$$D = n_{1\uparrow}n_{1\downarrow} + n_{2\uparrow}n_{2\downarrow}, \qquad H_U = UD.$$

Repulsive $U>0$ raises configurations with a doubly occupied site. Attractive $U<0$ lowers them.

Sign of the hopping

The site rephasing

$$c_{2\sigma} \longmapsto -c_{2\sigma}$$

maps $t\mapsto-t$ for both spins. Because the dimer has no closed loop, the spectrum depends on $t^2$. Fixing $t>0$ chooses a basis convention, not a distinct physical phase.

Chemical potential and energy offsets

In a grand-canonical calculation,

$$K = H-\mu\widehat N, \qquad \widehat N = \sum_{i,\sigma}n_{i\sigma}.$$

Every level in sector $N$ then shifts by $-\mu N$. A uniform onsite energy $\epsilon_0$ similarly adds $\epsilon_0N$. Neither changes eigenvectors within one fixed-$N$ sector, but both matter when comparing particle sectors.

A common particle–hole-centered form is

$$H_{\mathrm c} ={} H_t + U \sum_{i=1}^{2} \left( n_{i\uparrow}-\frac12 \right) \left( n_{i\downarrow}-\frac12 \right).$$

For the two-site dimer,

$$H_{\mathrm c} = H - \frac U2\widehat N + \frac U2.$$

In the $N=2$ sector this is simply

$$H_{\mathrm c} = H-\frac U2.$$

Therefore centered and uncentered spectra differ by an energy shift even when they describe the same fixed-sector eigenstates.

Matrix Representation and Fermionic Signs

Focus first on the $N_\uparrow=N_\downarrow=1$ block. Define

$$\begin{aligned} \lvert D_1\rangle &= \lvert\uparrow\downarrow,0\rangle, \\ \lvert X_1\rangle &= \lvert\uparrow,\downarrow\rangle, \\ \lvert X_2\rangle &= \lvert\downarrow,\uparrow\rangle, \\ \lvert D_2\rangle &= \lvert0,\uparrow\downarrow\rangle. \end{aligned}$$

In the ordered basis

$$\left( \lvert D_1\rangle, \lvert X_1\rangle, \lvert X_2\rangle, \lvert D_2\rangle \right),$$

the block is

$$H_{1,1} = \begin{pmatrix} U&-t&t&0\\ -t&0&0&-t\\ t&0&0&t\\ 0&-t&t&U \end{pmatrix}.$$

These signs follow from the declared creation-operator order. In this particular block, replacing the basis vector $\lvert X_2\rangle$ by $-\lvert X_2\rangle$ makes all four independent nonzero hopping entries negative. That is a valid basis rephasing if every operator and state coordinate is transformed too. The singlet and triplet then have different coordinate formulas.

Consequently, a correct spectrum alone cannot certify the fermionic signs. Check matrix elements in the declared basis and spin observables as well. Arbitrarily replacing entries by absolute values in a general many-fermion matrix has no such guarantee.

The two remaining $N=2$ states,

$$\lvert T_+\rangle = \lvert\uparrow,\uparrow\rangle, \qquad \lvert T_-\rangle = \lvert\downarrow,\downarrow\rangle,$$

are one-dimensional $S_{\mathrm{tot}}^z=\pm1$ blocks.

Spin-Adapted Basis

Define

$$\begin{aligned} \lvert S\rangle &= \frac{ \lvert X_1\rangle-\lvert X_2\rangle }{\sqrt2}, \\ \lvert T_0\rangle &= \frac{ \lvert X_1\rangle+\lvert X_2\rangle }{\sqrt2}, \\ \lvert D_+\rangle &= \frac{ \lvert D_1\rangle+\lvert D_2\rangle }{\sqrt2}, \\ \lvert D_-\rangle &= \frac{ \lvert D_1\rangle-\lvert D_2\rangle }{\sqrt2}. \end{aligned}$$

The three states

$$\lvert T_+\rangle, \qquad \lvert T_0\rangle, \qquad \lvert T_-\rangle$$

form a spin-$1$ triplet. The states $\lvert S\rangle$, $\lvert D_+\rangle$, and $\lvert D_-\rangle$ are spin singlets.

With the phases above,

$$H_{S,D_+,D_-} = \begin{pmatrix} 0&-2t&0\\ -2t&U&0\\ 0&0&U \end{pmatrix}.$$

The hopping couples the singly occupied singlet to the symmetric doublon combination with amplitude $-2t$. It annihilates every triplet and does not mix the antisymmetric doublon. For the two parallel-spin triplets, Pauli exclusion blocks each hop separately. For the zero-magnetization triplet, individual hopping paths exist but their amplitudes cancel. Rephasing one basis vector can reverse both entries $\pm2t$, but cannot change an eigenvalue or observable.

Symmetries and Conserved Quantities

Charge conservation

The total particle number is conserved:

$$[H,\widehat N] = 0.$$

Spin-conserving hopping also gives

$$[H,N_\uparrow] = [H,N_\downarrow] = 0.$$

The latter pair resolves $S_{\mathrm{tot}}^z$ sectors but does not replace the full spin symmetry.

Spin rotations

Define dimensionless local spin operators

$$\mathbf s_i = \frac12 \sum_{\alpha,\beta} c_{i\alpha}^\dagger \boldsymbol\sigma_{\alpha\beta} c_{i\beta}.$$

For spin-independent hopping and interaction,

$$[H,\mathbf S_{\mathrm{tot}}] = 0, \qquad \mathbf S_{\mathrm{tot}} = \mathbf s_1+\mathbf s_2.$$

Hence

$$[H,\mathbf S_{\mathrm{tot}}^2] = [H,S_{\mathrm{tot}}^z] = 0.$$

The threefold zero-energy triplet degeneracy is enforced by spin $SU(2)$ symmetry. A split triplet in the baseline model is an implementation error, not finite-size physics.

Site exchange and time reversal

The symmetric dimer is invariant under exchanging sites $1$ and $2$, with the induced fermionic signs handled by the unitary mode permutation. Real spin-independent hopping is also time-reversal invariant. A site bias breaks site exchange, while a magnetic field or suitable spin-dependent hopping can break time reversal or spin rotations. A single spin-independent complex hopping phase on this isolated one-bond dimer is removable by a site rephasing; it is not a gauge-invariant magnetic flux.

Particle–hole structure

On the bipartite dimer, the transformation

$$c_{1\sigma} \longmapsto c_{1\sigma}^\dagger, \qquad c_{2\sigma} \longmapsto -c_{2\sigma}^\dagger$$

leaves the centered Hamiltonian invariant. It exchanges $N$ with $4-N$. In the uncentered convention this symmetry relates levels with sector-dependent energy shifts; it does not mean the raw $H$ spectrum is symmetric about zero.

The Two-State Singlet Problem

The dimer is a finite exact-diagonalization problem. Every eigenvalue and eigenstate in all 16 Fock states can be obtained analytically. This is exact for the declared two-site Hamiltonian, not an exact solution of a thermodynamic Hubbard lattice.

The nontrivial half-filled calculation reduces to a $2\times2$ matrix:

$$H_{S,D_+} = \begin{pmatrix} 0&-2t\\ -2t&U \end{pmatrix}.$$

Define

$$R = \sqrt{U^2+16t^2}.$$

The two mixed singlet energies are

$$E_\pm = \frac{U\pm R}{2}.$$

The remaining half-filled levels are

$$E_T = 0 \quad \text{with degeneracy }3,$$

and

$$E_D = U$$

for $\lvert D_-\rangle$.

For $t\ne0$,

$$E_- < \min(0,U),$$

so the $N=2$ ground state is a unique spin singlet for every finite real $U$. A finite dimer has correlation crossovers, not a thermodynamic Mott or magnetic phase transition.

Complete Fock-Space Spectrum

In the uncentered Hamiltonian with $t>0$ and no chemical potential, the complete spectrum is:

Particle number	Sector dimension	Energies including multiplicity
$0$	$1$	$0$
$1$	$4$	$-t$ twice, $+t$ twice
$2$	$6$	$E_-$; $0$ three times; $U$; $E_+$
$3$	$4$	$U-t$ twice, $U+t$ twice
$4$	$1$	$2U$

The dimensions sum to

$$1+4+6+4+1 = 16.$$

At special parameter values, levels in different rows or entries can coincide and increase the total degeneracy. A grand-canonical spectrum follows by subtracting $\mu N$ from every level in row $N$.

The $N=1$ sector is the spin-degenerate tight-binding dimer. The $N=3$ sector is its one-hole counterpart shifted by $U$, and the fully occupied state contains one doublon on each site.

The dimer retains two copies of the four-state local Hubbard space and one hopping bond. At half filling, hopping connects the singly occupied singlet to doublon–hole configurations, while spin conservation prevents coupling of the triplet to onsite pairs.

Exact Ground State

For $t>0$, write

$$\lvert\psi_-\rangle = \cos\vartheta \lvert S\rangle + \sin\vartheta \lvert D_+\rangle,$$

with $0<\vartheta<\pi/2$ and

$$\cos(2\vartheta) = \frac U R, \qquad \sin(2\vartheta) = \frac{4t}{R}.$$

Equivalently,

$$\tan\vartheta = \frac{R-U}{4t} = \frac{-E_-}{2t}.$$

For repulsive $U$, $\vartheta<\pi/4$ and the singly occupied singlet has greater weight. For attractive $U$, $\vartheta>\pi/4$ and the even doublon has greater weight. At $U=0$, both weights are equal.

The state is not the classical product

$$\lvert\uparrow,\downarrow\rangle.$$

It is a spin eigenstate with coherent charge fluctuations. Suppressing doublon weight at large positive $U$ does not remove quantum entanglement between the two spins.

Exact Ground-State Observables

Double occupancy

The total doublon operator is $D=\partial H/\partial U$. Feynman–Hellmann gives

$$\begin{aligned} \langle D\rangle_- &= \frac{\partial E_-}{\partial U} \\ &= \frac12 \left( 1-\frac U R \right) \\ &= \sin^2\vartheta. \end{aligned}$$

Site symmetry gives

$$\left\langle n_{1\uparrow}n_{1\downarrow} \right\rangle = \left\langle n_{2\uparrow}n_{2\downarrow} \right\rangle = \frac12\langle D\rangle_-.$$

Kinetic and interaction energies

Because $H_t=t\,\partial H/\partial t$,

$$\langle H_t\rangle_- = t\frac{\partial E_-}{\partial t} = -\frac{8t^2}{R}.$$

The interaction energy is

$$\langle H_U\rangle_- = \frac U2 \left( 1-\frac U R \right).$$

They add to $E_-$:

$$\langle H_t\rangle_- + \langle H_U\rangle_- = \frac{U-R}{2}.$$

Spin correlation and local moment

Only the $\lvert S\rangle$ component carries one spin on each site. Therefore

$$\left\langle \mathbf s_1\mathbin{\cdot}\mathbf s_2 \right\rangle_- = -\frac34\cos^2\vartheta = -\frac38 \left( 1+\frac U R \right).$$

Define the local moment diagnostic

$$m_i^2 = \left( n_{i\uparrow}-n_{i\downarrow} \right)^2.$$

At symmetric half filling,

$$\langle n_i\rangle = 1,$$

and

$$\langle m_i^2\rangle_- = 1-\langle D\rangle_- = \frac12 \left( 1+\frac U R \right).$$

This local moment approaches one at strong repulsion, but a local moment is not by itself long-range magnetic order.

Charge imbalance

The exact charge-fluctuation identity is

$$\left\langle (n_1-n_2)^2 \right\rangle_- = 4\langle D\rangle_-.$$

The singly occupied singlet has zero imbalance; either doublon–hole configuration has squared imbalance four.

Excitation and Addition-Energy Diagnostics

For repulsive $U\ge0$, the first half-filled excitation is the triplet. The singlet–triplet gap is

$$\Delta_{S-T} = E_T-E_- = \frac{R-U}{2}.$$

The conventional finite-cluster charge gap is

$$\Delta_c = E_0(N+1) + E_0(N-1) - 2E_0(N).$$

Using

$$E_0(1) = -t, \qquad E_0(2) = E_-, \qquad E_0(3) = U-t,$$

gives

$$\Delta_c^{\mathrm{dimer}} = R-2t.$$

At $U=0$, this quantity is $2t$, the finite bonding–antibonding level spacing. It does not establish an interaction-driven bulk insulating phase. A Mott claim requires a thermodynamic sequence and charge-response analysis.

For attractive $U<0$, the odd doublon state at energy $U$ lies below the triplet. The lowest excitation then probes coherent pair motion rather than spin exchange.

Exact Two-Level Dynamics

The singlet sector is a detuned two-level system. Starting from the singly occupied singlet $\lvert S\rangle$, the probability of occupying $\lvert D_+\rangle$ at time $\tau$ is

$$P_{S\to D_+}(\tau) = \frac{16t^2}{R^2} \sin^2 \left( \frac{R\tau}{2\hbar} \right).$$

Repulsion suppresses the maximum doublon probability through the factor $16t^2/R^2$. The triplets remain stationary up to an overall phase because hopping cannot create an onsite spin triplet in one orbital.

This coherent oscillation is an exact finite-system result. Coupling to more sites, a bath, disorder, or time-dependent fields introduces additional frequencies or damping mechanisms.

Important Limits

Noninteracting limit

At $U=0$,

$$E_- = -2t, \qquad E_+ = 2t.$$

The triplets and $\lvert D_-\rangle$ form a fourfold zero-energy manifold. The ground state contains two opposite-spin fermions in the bonding orbital and has

$$\langle D\rangle_- = \frac12.$$

Atomic limit

At $t=0$, singly occupied configurations have energy zero and doublon configurations have energy $U$. For $U>0$, the four states with one particle per site are degenerate. There is no singlet–triplet exchange splitting until virtual hopping is restored.

Strong repulsion

For $U\gg t>0$,

$$E_- = -\frac{4t^2}{U} + O\!\left( \frac{t^4}{U^3} \right),$$

and

$$\Delta_{S-T} = \frac{4t^2}{U} + O\!\left( \frac{t^4}{U^3} \right).$$

The doublon weight is

$$\langle D\rangle_- = \frac{4t^2}{U^2} + O\!\left( \frac{t^4}{U^4} \right).$$

Within the singly occupied subspace, the effective operator is

$$H_{\mathrm{eff}} = J \left( \mathbf s_1\mathbin{\cdot}\mathbf s_2 - \frac14 \right), \qquad J = \frac{4t^2}{U}.$$

It assigns energy $-J$ to the singlet and $0$ to the triplet, matching the leading exact dimer spectrum. Effective Hamiltonians in Quantum Matter derives the controlled projection.

Strong attraction

For $U=-\lvert U\rvert$ with $\lvert U\rvert\gg t$,

$$E_- = U - \frac{4t^2}{\lvert U\rvert} + O\!\left( \frac{t^4}{\lvert U\rvert^3} \right).$$

The ground state approaches $\lvert D_+\rangle$, and

$$\langle D\rangle_- \longrightarrow 1.$$

The splitting between even and odd doublon combinations is generated by second-order pair motion. This paired limit is not described by the repulsive Heisenberg reduction.

Minimal Worked Example: Two Fermions at U = 0

Define bonding and antibonding operators

$$b_\sigma^\dagger = \frac{ c_{1\sigma}^\dagger+c_{2\sigma}^\dagger }{\sqrt2}, \qquad a_\sigma^\dagger = \frac{ c_{1\sigma}^\dagger-c_{2\sigma}^\dagger }{\sqrt2}.$$

The kinetic Hamiltonian is

$$H_t = -t \sum_\sigma b_\sigma^\dagger b_\sigma + t \sum_\sigma a_\sigma^\dagger a_\sigma.$$

At $N=2$, the ground state is

$$\lvert\psi_0\rangle = b_\uparrow^\dagger b_\downarrow^\dagger \lvert0\rangle,$$

with energy $-2t$. Expanding in the ordered site basis gives

$$\begin{aligned} \lvert\psi_0\rangle ={}& \frac12 \big( \lvert D_1\rangle + \lvert X_1\rangle - \lvert X_2\rangle + \lvert D_2\rangle \big) \\ ={}& \frac{ \lvert S\rangle+\lvert D_+\rangle }{\sqrt2}. \end{aligned}$$

The minus sign multiplying $\lvert X_2\rangle$ comes from restoring canonical creation-operator order. The result is a spin singlet with equal singly occupied and doublon weights, exactly as $\vartheta=\pi/4$ predicts.

Numerical Benchmark: MB-B003

MB-B003 collects the same exact comparison values. Use the mode order above, the full $N=2$ sector, and

$$t=1, \qquad U=4.$$

Exact targets

At this point,

$$R = \sqrt{32} = 4\sqrt2.$$

The ordered six-state spectrum is

$$\begin{gathered} E_0 = 2-2\sqrt2 = -0.828427124746190, \\ E_1=E_2=E_3=0, \\ E_4=4, \\ E_5 = 2+2\sqrt2 = 4.828427124746190. \end{gathered}$$

The singlet–triplet gap is

$$\Delta_{S-T} = 0.828427124746190.$$

The ground-state total double occupancy is

$$\begin{aligned} \langle D\rangle_0 &= \frac12 \left( 1-\frac1{\sqrt2} \right) \\ &= 0.146446609406726. \end{aligned}$$

Useful additional targets are

$$\begin{aligned} \langle H_t\rangle_0 &= -\sqrt2, \\ \left\langle \mathbf s_1\mathbin{\cdot}\mathbf s_2 \right\rangle_0 &= -\frac38 \left( 1+\frac1{\sqrt2} \right), \\ \langle m_i^2\rangle_0 &= \frac12 \left( 1+\frac1{\sqrt2} \right). \end{aligned}$$

Basis-independent matrix checks

In the full $N=2$ sector,

$$\operatorname{Tr}H = 2U,$$

and

$$\operatorname{Tr}H^2 = 2U^2+8t^2.$$

At the benchmark point,

$$\operatorname{Tr}H = 8, \qquad \operatorname{Tr}H^2 = 40.$$

The implementation must also satisfy

$$[H,\widehat N] = 0, \qquad [H,\mathbf S_{\mathrm{tot}}^2] = 0, \qquad [H,S_{\mathrm{tot}}^z] = 0.$$

Comparing a Matrix Calculation

1. Enumerate exactly six $N=2$ states and four $N_\uparrow=N_\downarrow=1$ states.
2. Reproduce the complete six-level spectrum with multiplicities.
3. Recover one spin triplet of dimension three at energy zero.
4. Compare direct $\langle D\rangle_0$ with $\partial E_-/\partial U$.
5. Check the trace and second moment.
6. Change the global mode order, transform all operators consistently, and recover the same spectrum and observables.

Use the double-precision reference profile on Benchmark Problems: eigenvalues within $10^{-11}$ in units of $t$, listed ground-state observables within $10^{-10}$, and normalized residuals within $10^{-11}$.

Incorrect singlet mixing, a split triplet, or a missing factor of two inside $R$ usually indicates a fermionic-sign error, an omitted spin species, or a doubled bond. An unchanged spectrum after changing matrix-element signs can be a basis rephasing, so it does not establish that the original sign convention was implemented correctly.

Changing the Dimer

Variant	What changes	Consequence or further calculation
site bias	add $\frac{\Delta}{2}(n_1-n_2)$	inversion is broken and $\lvert D_+\rangle$, $\lvert D_-\rangle$ mix
Zeeman field	add $-hS_{\mathrm{tot}}^z$	triplet components split
intersite interaction	add $Vn_1n_2$	extended Hubbard dimer
spin-dependent hopping	replace $t$ by a spin matrix	spin $SU(2)$ can be reduced
pair hopping	add explicit doublon transfer	changes the paired-sector splitting
coupling to leads	attach bath modes	route toward impurity and transport models
more sites	extend the bond graph	Hubbard Model and the Hubbard Chain dossier
large positive $U$	project to low-energy singly occupied states	Effective Hamiltonians in Many-Body Systems
variational doublon suppression	multiply doublon amplitudes by a Gutzwiller factor	Variational Many-Body States

The dimer can test local algebra and short-range correlation, but it cannot establish a thermodynamic phase, spontaneous symmetry breaking, a bulk transport coefficient, or dimensional scaling.

Exercises

1. Count and classify the half-filled basis

Show that the $N=2$ sector has dimension six and decomposes into one spin triplet and three spin singlets.

Solution

Choosing two occupied modes from four gives

$$\dim\mathcal H_2 = \binom42 = 6.$$

The states with one fermion on each site form the tensor product of two spin-$1/2$ spaces:

$$\frac12\otimes\frac12 = 1\oplus0.$$

They therefore contain three triplet states and one singlet. The two doublon–hole states have both fermions in the same spatial orbital and must be spin singlets. Thus

$$6 = 3_{\mathrm{triplet}} + 3_{\mathrm{singlet}}.$$

2. Recover the signed four-state block

Using the site-major mode order, act with the hopping Hamiltonian on $\lvert D_1\rangle$. Explain the opposite signs of the two resulting separated-spin states.

Solution

The spin-down hop from site $1$ to site $2$ gives

$$-t c_{2\downarrow}^\dagger c_{1\downarrow} \lvert D_1\rangle = -t\lvert X_1\rangle.$$

The spin-up hop gives

$$-t c_{2\uparrow}^\dagger c_{1\uparrow} \lvert D_1\rangle = t\lvert X_2\rangle.$$

The second sign changes because the annihilation and creation operators cross a different number of occupied lower-order modes. Therefore

$$H_t\lvert D_1\rangle = -t\lvert X_1\rangle + t\lvert X_2\rangle = -\sqrt2t\lvert S\rangle.$$

Repeating the calculation for $\lvert D_2\rangle$ gives the same singlet coupling. Their symmetric combination couples with amplitude $-2t$, while their antisymmetric combination decouples.

3. Derive the mixed singlet energies

Diagonalize the $\{\lvert S\rangle,\lvert D_+\rangle\}$ block and prove that $E_-<\min(0,U)$ for $t\ne0$.

Solution

The characteristic polynomial is

$$\det \begin{pmatrix} -E&-2t\\ -2t&U-E \end{pmatrix} = E^2-UE-4t^2.$$

Its roots are

$$E_\pm = \frac{ U\pm\sqrt{U^2+16t^2} }{2}.$$

For $t\ne0$,

$$\sqrt{U^2+16t^2} > \lvert U\rvert.$$

Hence $E_-<0$. Also

$$E_--U = \frac{ -U-\sqrt{U^2+16t^2} }{2} < 0.$$

Thus $E_-<\min(0,U)$ and the mixed singlet is the unique half-filled ground state.

4. Connect double occupancy and spin correlation

Starting from the mixing-angle ground state, derive $\langle D\rangle_-$ and $\langle\mathbf s_1\cdot\mathbf s_2\rangle_-$. Check the limits $U=0$ and $U\to+\infty$.

Solution

The doublon operator has eigenvalue zero on $\lvert S\rangle$ and one on $\lvert D_+\rangle$. Therefore

$$\langle D\rangle_- = \sin^2\vartheta = \frac12 \left( 1-\frac U R \right).$$

The spin correlation is $-3/4$ on the singly occupied singlet and zero on a doublon–hole state, so

$$\left\langle \mathbf s_1\cdot\mathbf s_2 \right\rangle_- = -\frac34\cos^2\vartheta = -\frac38 \left( 1+\frac U R \right).$$

At $U=0$, the two weights are equal:

$$\langle D\rangle_- = \frac12, \qquad \left\langle \mathbf s_1\cdot\mathbf s_2 \right\rangle_- = -\frac38.$$

As $U\to+\infty$,

$$\langle D\rangle_- \to0, \qquad \left\langle \mathbf s_1\cdot\mathbf s_2 \right\rangle_- \to-\frac34.$$

5. Match the strong-coupling spin Hamiltonian

Show that

$$H_{\mathrm{eff}} = J \left( \mathbf s_1\cdot\mathbf s_2-\frac14 \right)$$

with $J=4t^2/U$ reproduces the leading repulsive dimer energies.

Solution

For two spin-$1/2$ degrees of freedom,

$$\mathbf s_1\cdot\mathbf s_2 = \begin{cases} -3/4,&S=0,\\ +1/4,&S=1. \end{cases}$$

Therefore

$$E_S^{\mathrm{eff}} = J \left( -\frac34-\frac14 \right) = -J,$$

while

$$E_T^{\mathrm{eff}} = J \left( \frac14-\frac14 \right) = 0.$$

With $J=4t^2/U$, these match

$$E_- = -\frac{4t^2}{U} +\cdots, \qquad E_T=0.$$

Using $J\mathbf s_1\cdot\mathbf s_2$ without the constant gives the correct splitting but not the energy zero of the projected Hubbard dimer.

6. Derive the doublon oscillation probability

Start in $\lvert S\rangle$ at $\tau=0$ and derive $P_{S\to D_+}(\tau)$.

Solution

Subtract the irrelevant constant $U/2$ from the two-state Hamiltonian:

$$H' = \begin{pmatrix} -U/2&-2t\\ -2t&U/2 \end{pmatrix}.$$

Its eigenvalue splitting is

$$R = \sqrt{U^2+16t^2}.$$

For a two-level system with off-diagonal magnitude $2t$ and detuning $U$, the transition probability is

$$P_{S\to D_+}(\tau) = \frac{ 4(2t)^2 }{ U^2+4(2t)^2 } \sin^2 \left( \frac{R\tau}{2\hbar} \right).$$

Thus

$$P_{S\to D_+}(\tau) = \frac{16t^2}{R^2} \sin^2 \left( \frac{R\tau}{2\hbar} \right).$$

At $U=0$, complete transfer is possible. For $\lvert U\rvert\gg t$, the maximum probability is suppressed as $16t^2/U^2$.

References

• D. P. Arovas, E. Berg, S. A. Kivelson, and S. Raghu, “The Hubbard Model,” Annual Review of Condensed Matter Physics 13, 239–274 (2022), doi:10.1146/annurev-conmatphys-031620-102024.
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