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Bose–Hubbard Dimer

The Bose–Hubbard dimer is the two-mode bosonic model in which tunneling competes with onsite interactions; every fixed-particle-number sector is finite, and the N=2N=2 sector gives a complete analytic laboratory for bosonic enhancement, number squeezing, fragmentation, pair formation, and coherent tunneling.

This dossier owns the finite two-mode problem: its basis, matrix elements, spin representation, exact low-particle spectra, observables, dynamics, asymptotic limits, and reproducible checks. The Bose–Hubbard Model article owns the general lattice model, atomic-limit Mott physics, optical-lattice reduction, and thermodynamic phase boundary. MB-B005 remains the stable numerical contract.

Unless a variant is stated explicitly, this page uses:

FieldBaseline choice
modestwo bosonic modes, labeled 11 and 22
local occupationni=0,1,2,…n_i=0,1,2,\ldots
primary sectorfixed total particle number NN
bond listone undirected bond {1,2}\{1,2\}
hoppingreal J>0J>0, with one-particle matrix element −J-J
interactiononsite U∈RU\in\mathbb R
biasabsent
chemical potentialabsent because one fixed-NN sector is diagonalized
exchange symmetryexact interchange 1↔21\leftrightarrow2
benchmark sectorN=2N=2
benchmark pointJ=1J=1, U=2U=2

The dimer has one physical link. Writing a two-site periodic nearest-neighbor sum and then adding its Hermitian conjugate can count that link twice. A calculation is reproducible only after the bond convention and the meaning of JJ are stated.

The mode operators satisfy

[bi,bj†]=δijI,[bi,bj]=[bi†,bj†]=0,\begin{aligned} [b_i,b_j^\dagger] &= \delta_{ij}I, \\ [b_i,b_j] &= [b_i^\dagger,b_j^\dagger] =0, \end{aligned}

with number operators ni=bi†bin_i=b_i^\dagger b_i. Their action on a normalized occupation state is

b1∣n1,n2⟩=n1 ∣n1−1,n2⟩,b1†∣n1,n2⟩=n1+1 ∣n1+1,n2⟩,\begin{aligned} b_1\lvert n_1,n_2\rangle &= \sqrt{n_1}\, \lvert n_1-1,n_2\rangle, \\ b_1^\dagger\lvert n_1,n_2\rangle &= \sqrt{n_1+1}\, \lvert n_1+1,n_2\rangle, \end{aligned}

and similarly for mode 22.

The unrestricted two-mode Fock space is infinite dimensional. The Hamiltonian conserves

N^=n1+n2,\widehat N=n_1+n_2,

so a fixed-NN basis is

BN={∣n,N−n⟩:n=0,1,…,N}.\mathcal B_N = \bigl\{ \lvert n,N-n\rangle: n=0,1,\ldots,N \bigr\}.

Its dimension is

dim⁡HN=N+1.\dim\mathcal H_N=N+1.

No additional onsite cutoff is needed in this sector: number conservation already implies 0≤ni≤N0\le n_i\le N. A cutoff smaller than NN changes the model by deleting valid states.

The symmetric dimer Hamiltonian is

H=−J(b1†b2+b2†b1)+U2[n1(n1−1)+n2(n2−1)].\begin{aligned} H ={}& -J \left( b_1^\dagger b_2 + b_2^\dagger b_1 \right) \\ &+ \frac{U}{2} \left[ n_1(n_1-1) + n_2(n_2-1) \right]. \end{aligned}

The hopping term delocalizes particles. Repulsive U>0U>0 suppresses occupation imbalance and onsite pairs, while attractive U<0U<0 favors clustering. The interaction counts unordered pairs: a mode with nn bosons contributes Un(n−1)/2Un(n-1)/2.

At fixed NN, adding −μN^-\mu\widehat N shifts every eigenvalue by −μN-\mu N without changing eigenvectors or level spacings. The chemical potential matters only when sectors with different NN compete.

Write

∣n⟩N≡∣n,N−n⟩.\lvert n\rangle_N \equiv \lvert n,N-n\rangle.

The diagonal element is

dn=U2[n(n−1)+(N−n)(N−n−1)]=U[(n−N2)2+N(N−2)4].\begin{aligned} d_n &= \frac{U}{2} \left[ n(n-1) + (N-n)(N-n-1) \right] \\ &= U \left[ \left(n-\frac N2\right)^2 + \frac{N(N-2)}{4} \right]. \end{aligned}

The only nonzero off-diagonal element connecting neighboring occupations is

N⟨n+1∣H∣n⟩N=−J(n+1)(N−n).{}_N\langle n+1\rvert H\lvert n\rangle_N = -J\sqrt{(n+1)(N-n)}.

The factor under the square root is the bosonic enhancement. Replacing it by a constant JJ turns the calculation into a different tight-binding problem in occupation space.

If

∣ψ⟩=∑n=0Ncn∣n⟩N,\lvert\psi\rangle = \sum_{n=0}^{N}c_n\lvert n\rangle_N,

then the eigenvalue equation is the finite recurrence

Ecn=dncn−Jn(N−n+1) cn−1−J(n+1)(N−n) cn+1,\begin{aligned} E c_n ={}& d_n c_n - J\sqrt{n(N-n+1)}\,c_{n-1} \\ &- J\sqrt{(n+1)(N-n)}\,c_{n+1}, \end{aligned}

with c−1=cN+1=0c_{-1}=c_{N+1}=0. Thus every fixed-NN problem is a real symmetric tridiagonal matrix of order N+1N+1.

Define

Sx=12(b1†b2+b2†b1),Sy=12i(b1†b2−b2†b1),Sz=12(n1−n2).\begin{aligned} S_x &= \frac12 \left( b_1^\dagger b_2+b_2^\dagger b_1 \right), \\ S_y &= \frac{1}{2i} \left( b_1^\dagger b_2-b_2^\dagger b_1 \right), \\ S_z &= \frac12(n_1-n_2). \end{aligned}

These operators obey [Sa,Sb]=iϵabcSc[S_a,S_b]=i\epsilon_{abc}S_c. In the fixed-NN sector,

S=N2,S2=S(S+1)I.S=\frac N2, \qquad \mathbf S^2=S(S+1)I.

The Hamiltonian becomes

H=−2JSx+USz2+UN(N−2)4I.H = -2J S_x + U S_z^2 + \frac{U N(N-2)}{4}I.

The dimer is therefore a single spin S=N/2S=N/2 in a transverse field with quadratic, one-axis anisotropy. The final term is constant at fixed NN; dropping it is harmless for eigenvectors and gaps but changes absolute energies. Any comparison must say whether that offset was retained.

The occupation state ∣n,N−n⟩\lvert n,N-n\rangle corresponds to the spin state

∣S,m⟩,m=n−N2.\lvert S,m\rangle, \qquad m=n-\frac N2.

This map is exact. It is not a large-NN or mean-field approximation.

The global phase symmetry

bi⟼eiαbib_i\longmapsto e^{i\alpha}b_i

implies

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

It is the reason the infinite Fock space decomposes into finite blocks.

Let PP interchange the two modes:

P∣n,N−n⟩=∣N−n,n⟩.P\lvert n,N-n\rangle = \lvert N-n,n\rangle.

For the unbiased dimer,

[H,P]=0,P2=I.[H,P]=0, \qquad P^2=I.

Eigenstates may be assigned reflection parity p=±1p=\pm1. In spin language, the interchange sends Sz↦−SzS_z\mapsto-S_z and leaves SxS_x unchanged. A bias proportional to n1−n2n_1-n_2 breaks this symmetry.

In the declared occupation basis, HH is real symmetric. Complex conjugation is therefore a time-reversal symmetry for this spinless two-mode model. A tunneling phase can be removed on one isolated link by rephasing a mode; unlike a loop, a single dimer carries no gauge-invariant flux.

For J>0J>0, all off-diagonal entries of the irreducible fixed-NN matrix are negative. The Perron–Frobenius argument permits a nondegenerate ground state with strictly positive occupation-basis coefficients. Since mode exchange preserves positivity and commutes with HH, that ground state is even under PP for every finite NN and finite UU.

This statement matters in the attractive regime: the finite-system ground state is an even cat-like superposition, not a state localized permanently in one well. An arbitrarily small bias can nevertheless select one side once it exceeds the exponentially small parity splitting.

ObjectStatusQualification
fixed-NN Hilbert spaceexactdimension N+1N+1
matrix representationexactreal symmetric tridiagonal recurrence
N=0,1,2N=0,1,2 spectrumelementary closed formall eigenvectors and observables are analytic
arbitrary finite NNfinite exact diagonalizationcomplete after the basis and arithmetic precision are declared
Bethe-ansatz descriptionavailablealgebraic and differential-equation formulations exist, but roots still require solving coupled algebraic equations
large-NN phase-space dynamicscontrolled semiclassical limitfinite-NN tunneling, revivals, and parity restoration are quantum corrections
thermodynamic phase transitionnot defined for one fixed dimersharp transitions require an additional scaling limit

Calling the dimer “exactly solved” is useful only with an object attached. The (N+1)×(N+1)(N+1)\times(N+1) matrix is exact for every finite NN, while a compact formula in elementary functions for every eigenvalue is unavailable at generic NN.

For N=0N=0, the only state is ∣0,0⟩\lvert0,0\rangle and E=0E=0.

In the basis {∣1,0⟩,∣0,1⟩}\{\lvert1,0\rangle,\lvert0,1\rangle\},

HN=1=(0−J−J0).H_{N=1} = \begin{pmatrix} 0 & -J\\ -J & 0 \end{pmatrix}.

The bonding and antibonding states are

∣±⟩=∣1,0⟩±∣0,1⟩2,\lvert\pm\rangle = \frac{ \lvert1,0\rangle \pm \lvert0,1\rangle }{\sqrt2},

with energies E+=−JE_+=-J and E−=+JE_-=+J, where the subscript here denotes exchange parity. The interaction is invisible because one particle cannot form an onsite pair.

For N=2N=2, use the ordered basis

B2={∣2,0⟩,∣1,1⟩,∣0,2⟩}.\mathcal B_2 = \bigl\{ \lvert2,0\rangle, \lvert1,1\rangle, \lvert0,2\rangle \bigr\}.

The Hamiltonian is

HN=2=(U−2J0−2J0−2J0−2JU).H_{N=2} = \begin{pmatrix} U & -\sqrt2J & 0\\ -\sqrt2J & 0 & -\sqrt2J\\ 0 & -\sqrt2J & U \end{pmatrix}.

The matrix element −2J-\sqrt2J follows, for example, from

⟨1,1∣b2†b1∣2,0⟩=2,⟨1,1∣H∣2,0⟩=−2J.\begin{aligned} \langle1,1\rvert b_2^\dagger b_1 \lvert2,0\rangle &= \sqrt2, \\ \langle1,1\rvert H\lvert2,0\rangle &= -\sqrt2J. \end{aligned}

Introduce even and odd pair states

∣D+⟩=∣2,0⟩+∣0,2⟩2,∣D−⟩=∣2,0⟩−∣0,2⟩2.\begin{aligned} \lvert D_+\rangle &= \frac{ \lvert2,0\rangle+\lvert0,2\rangle }{\sqrt2}, \\ \lvert D_-\rangle &= \frac{ \lvert2,0\rangle-\lvert0,2\rangle }{\sqrt2}. \end{aligned}

The odd state is dark to ∣1,1⟩\lvert1,1\rangle:

H∣D−⟩=U∣D−⟩.H\lvert D_-\rangle = U\lvert D_-\rangle.

In the even basis {∣D+⟩,∣1,1⟩}\{\lvert D_+\rangle,\lvert1,1\rangle\},

Heven=(U−2J−2J0).H_{\mathrm{even}} = \begin{pmatrix} U & -2J\\ -2J & 0 \end{pmatrix}.

The two paths from ∣1,1⟩\lvert1,1\rangle into the left and right doublon states add coherently, changing the coupling from 2J\sqrt2J to 2J2J.

Define

Δ=U2+16J2.\Delta = \sqrt{U^2+16J^2}.

The three eigenvalues are

E−=U−Δ2,ED=U,E+=U+Δ2.\begin{aligned} E_- &= \frac{U-\Delta}{2}, \\ E_D &=U, \\ E_+ &= \frac{U+\Delta}{2}. \end{aligned}

For J>0J>0 their ordering is always

E−<ED<E+.E_-<E_D<E_+.

Choose 0<θ<π/20<\theta<\pi/2 through

cos⁡(2θ)=UΔ,sin⁡(2θ)=4JΔ.\cos(2\theta) = \frac{U}{\Delta}, \qquad \sin(2\theta) = \frac{4J}{\Delta}.

Normalized eigenstates are

∣ψ−⟩=cos⁡θ ∣1,1⟩+sin⁡θ ∣D+⟩,∣ψD⟩=∣D−⟩,∣ψ+⟩=cos⁡θ ∣D+⟩−sin⁡θ ∣1,1⟩.\begin{aligned} \lvert\psi_-\rangle &= \cos\theta\, \lvert1,1\rangle + \sin\theta\, \lvert D_+\rangle, \\ \lvert\psi_D\rangle &= \lvert D_-\rangle, \\ \lvert\psi_+\rangle &= \cos\theta\, \lvert D_+\rangle - \sin\theta\, \lvert1,1\rangle. \end{aligned}

Both mixed levels have even parity. Their avoided crossing has minimum separation 4J4J at U=0U=0. The odd level crosses neither even level because its energy remains between them.

Exact two-boson Bose–Hubbard dimer energies as functions of interaction strength.

Exact N=2N=2 spectrum in units of JJ. The even levels E−E_- and E+E_+ undergo an avoided crossing, while the odd pair state has ED=UE_D=U. For large negative U/JU/J, E−E_- and EDE_D form the nearly degenerate cat-state doublet.

Define the total onsite-pair operator

DB=12∑i=12ni(ni−1).D_B = \frac12 \sum_{i=1}^{2} n_i(n_i-1).

It has eigenvalue 11 on ∣D±⟩\lvert D_\pm\rangle and 00 on ∣1,1⟩\lvert1,1\rangle. Therefore

⟨DB⟩0=sin⁡2θ=12(1−UΔ).\begin{aligned} \langle D_B\rangle_0 &= \sin^2\theta \\ &= \frac12 \left( 1- \frac{U}{\Delta} \right). \end{aligned}

The same result follows from Feynman–Hellmann:

⟨DB⟩0=∂E−∂U.\langle D_B\rangle_0 = \frac{\partial E_-}{\partial U}.

For strong repulsion this probability vanishes as 4J2/U24J^2/U^2. For strong attraction it tends to 11.

Let

T=b1†b2+b2†b1.T = b_1^\dagger b_2+b_2^\dagger b_1.

Since ∂H/∂J=−T\partial H/\partial J=-T,

⟨T⟩0=−∂E−∂J=8JΔ.\langle T\rangle_0 = -\frac{\partial E_-}{\partial J} = \frac{8J}{\Delta}.

The coherence per particle is

C=⟨T⟩0N=4JΔ\mathcal C = \frac{\langle T\rangle_0}{N} = \frac{4J}{\Delta}

for N=2N=2. It equals 11 at U=0U=0 and tends to zero in both strong-interaction limits.

Define the population imbalance

I=n1−n2,z=IN.I=n_1-n_2, \qquad z=\frac{I}{N}.

Exchange symmetry gives ⟨I⟩0=0\langle I\rangle_0=0. For two particles,

⟨I2⟩0=4sin⁡2θ=2(1−UΔ),⟨z2⟩0=sin⁡2θ.\begin{aligned} \langle I^2\rangle_0 &= 4\sin^2\theta = 2 \left( 1- \frac{U}{\Delta} \right), \\ \langle z^2\rangle_0 &= \sin^2\theta. \end{aligned}

Each site has mean occupation 11 and variance

Var⁡(n1)=Var⁡(n2)=sin⁡2θ.\operatorname{Var}(n_1) = \operatorname{Var}(n_2) = \sin^2\theta.

Repulsion squeezes number fluctuations, whereas attraction turns them into macroscopic left-versus-right fluctuations.

The one-body density matrix is

ρij(1)=⟨bi†bj⟩0=(14J/Δ4J/Δ1).\rho^{(1)}_{ij} = \langle b_i^\dagger b_j\rangle_0 = \begin{pmatrix} 1 & 4J/\Delta\\ 4J/\Delta & 1 \end{pmatrix}.

Its natural occupations are

λ±=1±4JΔ,λ++λ−=2.\lambda_\pm = 1\pm\frac{4J}{\Delta}, \qquad \lambda_++\lambda_-=2.

At U=0U=0, λ+=2\lambda_+=2 and both particles occupy the bonding orbital. In either strong-interaction limit the two natural occupations approach 11 and 11. That same limiting spectrum has different many-body interpretations: a number-squeezed product ∣1,1⟩\lvert1,1\rangle for repulsion and a cat-like superposition for attraction.

Tracing out either mode leaves occupation probabilities

p0=p2=sin⁡2θ2,p1=cos⁡2θ.p_0=p_2=\frac{\sin^2\theta}{2}, \qquad p_1=\cos^2\theta.

The mode entropy is

S1∣2=−cos⁡2θln⁡(cos⁡2θ)−sin⁡2θln⁡(sin⁡2θ2).S_{1|2} = -\cos^2\theta\ln(\cos^2\theta) - \sin^2\theta \ln\left(\frac{\sin^2\theta}{2}\right).

This is entanglement between the two mode algebras at fixed number. It should not be confused with an entropy obtained by labeling identical particles.

Prepare one boson in each mode:

∣ψ(0)⟩=∣1,1⟩.\lvert\psi(0)\rangle = \lvert1,1\rangle.

Parity confines the evolution to the even two-level block. The probability of finding both bosons on the same site at time tt is

PD(t)=∣⟨D+∣ψ(t)⟩∣2=16J2Δ2sin⁡2(Δt2ℏ).\begin{aligned} P_D(t) &= \lvert\langle D_+\rvert\psi(t)\rangle\rvert^2 \\ &= \frac{16J^2}{\Delta^2} \sin^2\left( \frac{\Delta t}{2\hbar} \right). \end{aligned}

This is also ⟨DB(t)⟩\langle D_B(t)\rangle. At U=0U=0, the system oscillates completely between ∣1,1⟩\lvert1,1\rangle and ∣D+⟩\lvert D_+\rangle with angular frequency 4J/ℏ4J/\hbar. Repulsion or attraction detunes the pair sector, reduces the maximum transfer probability to 16J2/Δ216J^2/\Delta^2, and raises the oscillation frequency to Δ/ℏ\Delta/\hbar.

The exact finite-NN dynamics is quasiperiodic in the many-level case. Mean-field Josephson trajectories can describe a large-NN envelope, but they do not reproduce finite-NN collapse, revival, parity restoration, or tunneling between semiclassical self-trapped regions.

For U=0U=0, introduce bonding and antibonding modes

a+=b1+b22,a−=b1−b22.a_+ = \frac{b_1+b_2}{\sqrt2}, \qquad a_- = \frac{b_1-b_2}{\sqrt2}.

Then

H=−J(n+−n−).H = -J \left( n_+-n_- \right).

If mm bosons occupy the antibonding mode, the exact energy and parity are

Em=−J(N−2m),pm=(−1)m,m=0,1,…,N.E_m = -J(N-2m), \qquad p_m=(-1)^m, \qquad m=0,1,\ldots,N.

The ground state is the number-conserving condensate

∣bond;N⟩=(a+†)NN!∣0⟩,\lvert\mathrm{bond};N\rangle = \frac{(a_+^\dagger)^N}{\sqrt{N!}} \lvert0\rangle,

whose site occupations follow a binomial distribution.

For J=0J=0, every occupation state is an eigenstate with energy dnd_n. Repulsive U>0U>0 minimizes occupation imbalance:

  • if NN is even, the unique ground state is ∣N/2,N/2⟩\lvert N/2,N/2\rangle;
  • if NN is odd, the two least-imbalanced states are degenerate.

Attractive U<0U<0 instead minimizes the energy with the degenerate endpoint states ∣N,0⟩\lvert N,0\rangle and ∣0,N⟩\lvert0,N\rangle.

Strong repulsion and the even–odd effect

Section titled “Strong repulsion and the even–odd effect”

For even NN and U≫J>0U\gg J>0, nondegenerate perturbation theory gives

E0=UN(N−2)4−J2N(N+2)2U+O(J4U3).\begin{aligned} E_0 ={}& \frac{U N(N-2)}{4} \\ &- \frac{J^2N(N+2)}{2U} + O\left(\frac{J^4}{U^3}\right). \end{aligned}

The first correction is second order because one hop leaves the balanced state.

For odd NN, the two least-imbalanced states couple directly. Their leading even and odd energies are

Eeven/odd=U(N−1)24∓J(N+1)2+O(J2U).E_{\mathrm{even/odd}} = \frac{U(N-1)^2}{4} \mp \frac{J(N+1)}{2} + O\left(\frac{J^2}{U}\right).

Thus odd filling retains a first-order tunneling scale even when U/JU/J is large. The distinction is a finite-dimer parity effect, not a bulk Mott phase transition.

For U<0U<0 and ∣U∣≫J|U|\gg J, the endpoint states are connected only after all NN bosons tunnel. Degenerate perturbation theory yields the leading parity splitting

δEcat=2NJN(N−1)! ∣U∣N−1[1+O(J2U2)].\delta E_{\mathrm{cat}} = \frac{2N J^N} {(N-1)!\,|U|^{N-1}} \left[ 1+O\left(\frac{J^2}{U^2}\right) \right].

The even superposition is the finite-system ground state. Because the splitting falls rapidly with NN and ∣U∣/J|U|/J, a small bias, loss process, or dephasing perturbation can destroy coherent left–right superposition physics long before it appreciably changes the larger excitation scales.

Taking U/J→+∞U/J\to+\infty at fixed N≤2N\le2 suppresses double occupancy. At N=2N=2, only ∣1,1⟩\lvert1,1\rangle remains in the strict projected subspace and hopping has no first-order action. This is not the same as replacing the bosonic ladder factors by spin-1/21/2 matrix elements before taking the limit; projection and perturbative elimination must be performed consistently.

A number-conserving product state can be written

∣α,β;N⟩=(αb1†+βb2†)NN!∣0⟩,∣α∣2+∣β∣2=1.\lvert\alpha,\beta;N\rangle = \frac{ (\alpha b_1^\dagger+\beta b_2^\dagger)^N }{\sqrt{N!}} \lvert0\rangle, \qquad |\alpha|^2+|\beta|^2=1.

Define the classical imbalance and relative phase by

z=∣α∣2−∣β∣2,ϕ=arg⁡β−arg⁡α.z=|\alpha|^2-|\beta|^2, \qquad \phi=\arg\beta-\arg\alpha.

The variational energy, apart from a constant, is

Ecl(z,ϕ)=UN(N−1)4z2−JN1−z2cos⁡ϕ.E_{\mathrm{cl}}(z,\phi) = \frac{U N(N-1)}{4}z^2 - JN\sqrt{1-z^2}\cos\phi.

After division by JNJN, the interaction control is

Λ=U(N−1)2J.\Lambda = \frac{U(N-1)}{2J}.

Many references replace N−1N-1 by NN in the large-NN limit. That approximation is harmless asymptotically but should not be inserted into the exact N=2N=2 formulas. The phase-space separatrix organizes Josephson oscillations and mean-field self-trapping, with the threshold depending on the initial zz and ϕ\phi. A finite quantum dimer has crossovers and long tunneling times rather than permanently disconnected phase-space regions.

ObservableDefinitionWhat it diagnoses
imbalanceI=n1−n2=2SzI=n_1-n_2=2S_zleft–right localization and symmetry breaking
coherenceC=⟨b1†b2+b2†b1⟩/N\mathcal C=\langle b_1^\dagger b_2+b_2^\dagger b_1\rangle/Nbonding-orbital coherence
onsite pairsDB=12∑ini(ni−1)D_B=\frac12\sum_i n_i(n_i-1)interaction energy and clustering
number squeezingξn2=4Var⁡(Sz)/N\xi_n^2=4\operatorname{Var}(S_z)/Nfluctuations relative to a binomial coherent state
one-body density matrixρij(1)=⟨bi†bj⟩\rho^{(1)}_{ij}=\langle b_i^\dagger b_j\ranglenatural occupations and fragmentation
parity⟨P⟩\langle P\rangleexchange sector and cat-state coherence
mode entropyS1∣2=−Tr⁡(ρ1ln⁡ρ1)S_{1\vert2}=-\operatorname{Tr}(\rho_1\ln\rho_1)entanglement between mode algebras
current from 1 to 2I1→2=iJℏ(b2†b1−b1†b2)I_{1\to2}=\frac{iJ}{\hbar}(b_2^\dagger b_1-b_1^\dagger b_2)coherent particle transfer

The continuity equation is

dn1dt=−I1→2,dn2dt=I1→2.\frac{d n_1}{dt} = -I_{1\to2}, \qquad \frac{d n_2}{dt} = I_{1\to2}.

Every observable should be reported with its normalization. In particular, ⟨T⟩\langle T\rangle, ⟨T⟩/N\langle T\rangle/N, the largest natural occupation, and the condensate fraction are related but numerically different quantities.

Minimal Worked Example: Two Bosons at U = 0

Section titled “Minimal Worked Example: Two Bosons at U = 0”

At U=0U=0, the two-boson ground state is

∣bond;2⟩=(a+†)22∣0⟩=12(∣2,0⟩+2∣1,1⟩+∣0,2⟩).\begin{aligned} \lvert\mathrm{bond};2\rangle &= \frac{(a_+^\dagger)^2}{\sqrt2} \lvert0\rangle \\ &= \frac12 \left( \lvert2,0\rangle + \sqrt2\lvert1,1\rangle + \lvert0,2\rangle \right). \end{aligned}

Its energy is −2J-2J. The probability distribution is

P(2,0)=14,P(1,1)=12,P(0,2)=14.P(2,0)=\frac14, \qquad P(1,1)=\frac12, \qquad P(0,2)=\frac14.

Hence

⟨DB⟩=12,C=1,Var⁡(n1)=12.\langle D_B\rangle=\frac12, \qquad \mathcal C=1, \qquad \operatorname{Var}(n_1)=\frac12.

The appearance of double occupancy does not signal attraction. It is the binomial fluctuation of two independent bosons condensed into a delocalized orbital.

The canonical contract is MB-B005: Two-Site Bose–Hubbard Model. Use the ordered basis

{∣2,0⟩,∣1,1⟩,∣0,2⟩}\bigl\{ \lvert2,0\rangle, \lvert1,1\rangle, \lvert0,2\rangle \bigr\}

with J=1J=1 and U=2U=2. Then

E−=1−5=−1.236067977499790…,ED=2,E+=1+5=3.236067977499790….\begin{aligned} E_- &= 1-\sqrt5 = -1.236067977499790\ldots, \\ E_D &=2, \\ E_+ &= 1+\sqrt5 = 3.236067977499790\ldots. \end{aligned}

The reflection parities are respectively even, odd, and even. At this point,

⟨DB⟩0=12(1−15)=0.276393202250021…,C0=25=0.894427190999916….\begin{aligned} \langle D_B\rangle_0 &= \frac12 \left( 1-\frac1{\sqrt5} \right) \\ &= 0.276393202250021\ldots, \\ \mathcal C_0 &= \frac{2}{\sqrt5} = 0.894427190999916\ldots. \end{aligned}

The normalized ground state can be audited as

∣ψ0⟩=1+1/52∣1,1⟩+1−1/52∣D+⟩.\begin{aligned} \lvert\psi_0\rangle ={}& \sqrt{ \frac{1+1/\sqrt5}{2} } \lvert1,1\rangle \\ &+ \sqrt{ \frac{1-1/\sqrt5}{2} } \lvert D_+\rangle. \end{aligned}

Before diagonalization, verify

Tr⁡H=2U,Tr⁡H2=2U2+8J2,det⁡H=−4UJ2.\begin{aligned} \operatorname{Tr}H &=2U, \\ \operatorname{Tr}H^2 &=2U^2+8J^2, \\ \det H &=-4UJ^2. \end{aligned}

At the benchmark point these become 44, 1616, and −8-8. Also verify Hermiticity, the vanishing commutator [H,P][H,P], and the number-sector dimension 33.

  1. Enumerate the fixed-N=2N=2 basis rather than imposing an independent local cutoff.
  2. Generate −2J-\sqrt2J from ladder-operator actions.
  3. Reproduce all three energies and parities.
  4. Check ⟨DB⟩0=∂E−/∂U\langle D_B\rangle_0=\partial E_-/\partial U.
  5. Check ⟨T⟩0=−∂E−/∂J\langle T\rangle_0=-\partial E_-/\partial J.
  6. Set U=0U=0 and recover {−2J,0,2J}\{-2J,0,2J\}.

For a dense 3×33\times3 double-precision calculation, absolute residuals near 10−1210^{-12} are reasonable. A larger tolerance must be justified by the arithmetic or solver. Agreement with eigenvalues alone is insufficient if observables are evaluated in a differently ordered basis.

The planned notebook bose_hubbard_dimer_exact.ipynb is indexed in Reproducible Notebooks. Its admission gate is exact agreement with MB-B005, including parity and both Feynman–Hellmann checks.

An asymmetric dimer adds

Hϵ=ϵ2(n1−n2)=ϵSz.H_\epsilon = \frac{\epsilon}{2}(n_1-n_2) = \epsilon S_z.

This breaks exchange parity, produces ⟨I⟩≠0\langle I\rangle\ne0, and can localize an attractive cat doublet when ∣ϵ∣N|\epsilon|N exceeds its tiny tunnel splitting.

Extended dimers may contain

Vn1n2,P[(b1†)2b22+(b2†)2b12].V n_1n_2, \qquad P \left[ (b_1^\dagger)^2b_2^2 + (b_2^\dagger)^2b_1^2 \right].

These terms change the spin anisotropy and selection rules. Their parameters cannot be folded into UU and JJ without an explicit derivation.

Time-dependent J(t)J(t) or ϵ(t)\epsilon(t) produces driven tunneling, while particle loss and dephasing require an open-system description. The closed Hamiltonian here remains the baseline against which those additions should be measured.

The Bose–Hubbard Chain dossier owns the one-dimensional many-site specialization. The broader Bose–Hubbard Model article owns general lattices, Mott lobes, and optical-lattice validity.

The N=2N=2 hopping matrix element is −2J-\sqrt2J, not −J-J. The occupation-dependent square root is part of the operator algebra.

There is one undirected bond. A periodic two-site sum can silently produce −2J(b1†b2+h.c.)-2J(b_1^\dagger b_2+\mathrm{h.c.}).

The onsite interaction counts pairs. The states ∣1,0⟩\lvert1,0\rangle and ∣0,1⟩\lvert0,1\rangle must have zero interaction energy.

Forgetting the fixed-N constant in the spin map

Section titled “Forgetting the fixed-N constant in the spin map”

H=−2JSx+USz2H=-2JS_x+US_z^2 differs from the declared Bose–Hubbard Hamiltonian by UN(N−2)/4UN(N-2)/4. Gaps agree; absolute energies generally do not.

Treating a finite crossover as a phase transition

Section titled “Treating a finite crossover as a phase transition”

A fixed dimer has a finite Hilbert space and analytic avoided crossings for J>0J>0. Mean-field bifurcations and large-NN scaling limits require their own qualifications.

Reading zero imbalance as absence of clustering

Section titled “Reading zero imbalance as absence of clustering”

An even cat state has ⟨n1−n2⟩=0\langle n_1-n_2\rangle=0 but a large imbalance variance. First moments alone cannot distinguish a balanced state from a symmetric superposition of imbalanced states.

Calling every two-mode condensate a Bose–Hubbard dimer

Section titled “Calling every two-mode condensate a Bose–Hubbard dimer”

The two-mode approximation assumes a controlled truncation to two orbitals with parameters that do not change appreciably as the state evolves. Strong deformation, higher-band population, density-assisted hopping, or appreciable mode overlap can invalidate the minimal Hamiltonian.

For arbitrary NN, derive the diagonal element dnd_n and the hopping element between ∣n,N−n⟩\lvert n,N-n\rangle and ∣n+1,N−n−1⟩\lvert n+1,N-n-1\rangle. Explain why the matrix is tridiagonal.

Solution

The interaction is diagonal in the occupation basis:

dn=U2[n(n−1)+(N−n)(N−n−1)].d_n = \frac U2 \left[ n(n-1)+(N-n)(N-n-1) \right].

The operator b1†b2b_1^\dagger b_2 moves one boson from mode 22 to mode 11:

b1†b2∣n,N−n⟩=(n+1)(N−n)∣n+1,N−n−1⟩.b_1^\dagger b_2 \lvert n,N-n\rangle = \sqrt{(n+1)(N-n)} \lvert n+1,N-n-1\rangle.

Its Hermitian conjugate moves one boson in the opposite direction. A single hop changes nn only by ±1\pm1, so no matrix elements occur beyond the first off-diagonals.

Use n1=N/2+Szn_1=N/2+S_z and n2=N/2−Szn_2=N/2-S_z to derive the Schwinger-spin form, including its constant term.

Solution

The hopping operator is 2Sx2S_x. For the interaction,

n1(n1−1)+n2(n2−1)=n12+n22−N=2Sz2+N22−N.\begin{aligned} n_1(n_1-1)+n_2(n_2-1) &= n_1^2+n_2^2-N \\ &= 2S_z^2+\frac{N^2}{2}-N. \end{aligned}

Multiplication by U/2U/2 gives

H=−2JSx+USz2+UN(N−2)4I.H = -2JS_x +US_z^2 +\frac{UN(N-2)}{4}I.

Construct ∣D±⟩\lvert D_\pm\rangle, show that ∣D−⟩\lvert D_-\rangle is an eigenstate, and diagonalize the even block.

Solution

The hopping amplitudes from ∣1,1⟩\lvert1,1\rangle to the two doublon states are equal. They cancel in the odd combination and add in the even combination:

H∣D−⟩=U∣D−⟩,H\lvert D_-\rangle =U\lvert D_-\rangle,

while

Heven=(U−2J−2J0).H_{\mathrm{even}} = \begin{pmatrix} U&-2J\\ -2J&0 \end{pmatrix}.

Its characteristic polynomial is

E2−UE−4J2=0,E^2-UE-4J^2=0,

so

E±=U±U2+16J22.E_\pm = \frac{U\pm\sqrt{U^2+16J^2}}{2}.

Together with ED=UE_D=U, these are all three eigenvalues.

4. Obtain observables without differentiating eigenvectors

Section titled “4. Obtain observables without differentiating eigenvectors”

Use the ground-state energy to derive ⟨DB⟩0\langle D_B\rangle_0 and ⟨T⟩0\langle T\rangle_0.

Solution

Because

∂H∂U=DB,∂H∂J=−T,\frac{\partial H}{\partial U}=D_B, \qquad \frac{\partial H}{\partial J}=-T,

Feynman–Hellmann gives

⟨DB⟩0=∂E−∂U=12(1−UU2+16J2),⟨T⟩0=−∂E−∂J=8JU2+16J2.\begin{aligned} \langle D_B\rangle_0 &= \frac{\partial E_-}{\partial U} = \frac12 \left( 1-\frac{U}{\sqrt{U^2+16J^2}} \right), \\ \langle T\rangle_0 &= -\frac{\partial E_-}{\partial J} = \frac{8J}{\sqrt{U^2+16J^2}}. \end{aligned}

5. Explain the repulsive even–odd effect

Section titled “5. Explain the repulsive even–odd effect”

At J=0J=0 and U>0U>0, identify the ground manifold for even and odd NN. Show why the leading tunneling correction is second order for even NN but first order for odd NN.

Solution

For even NN, the unique minimum is ∣N/2,N/2⟩\lvert N/2,N/2\rangle. One hop leaves this state, so its first-order diagonal correction vanishes. The two neighboring virtual states each lie an energy UU higher, yielding

ΔE(2)=−J2N(N+2)2U.\Delta E^{(2)} = -\frac{J^2N(N+2)}{2U}.

For odd NN, the states

∣N+12,N−12⟩,∣N−12,N+12⟩\left\lvert\frac{N+1}{2},\frac{N-1}{2}\right\rangle, \qquad \left\lvert\frac{N-1}{2},\frac{N+1}{2}\right\rangle

are degenerate and differ by one hop. Their off-diagonal matrix element is

−JN+12.-J\frac{N+1}{2}.

Diagonalizing this two-state manifold produces a first-order even–odd splitting J(N+1)J(N+1).

6. Derive the pair-oscillation probability

Section titled “6. Derive the pair-oscillation probability”

Starting from ∣1,1⟩\lvert1,1\rangle, derive the probability of occupying either doublon state at time tt.

Solution

Only the even block participates. Subtracting U/2U/2 times the identity gives the two-level Hamiltonian

Heven−U2I=(U/2−2J−2J−U/2),H_{\mathrm{even}}-\frac U2I = \begin{pmatrix} U/2&-2J\\ -2J&-U/2 \end{pmatrix},

whose level separation is Δ=U2+16J2\Delta=\sqrt{U^2+16J^2}. The standard detuned two-level transition probability is therefore

PD(t)=(2⋅2J)2Δ2sin⁡2(Δt2ℏ)=16J2Δ2sin⁡2(Δt2ℏ).P_D(t) = \frac{(2\cdot2J)^2}{\Delta^2} \sin^2\left( \frac{\Delta t}{2\hbar} \right) = \frac{16J^2}{\Delta^2} \sin^2\left( \frac{\Delta t}{2\hbar} \right).

Because DBD_B projects onto the two-boson pair subspace, PD(t)=⟨DB(t)⟩P_D(t)=\langle D_B(t)\rangle.

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