Lecture 8: Hamiltonians, Observables, and Correlations

Fock space supplies a language of particle occupations and their changes, but algebra alone does not say which processes occur or what a probe reads. This lecture turns that language into many-body models, observables, currents, and correlations, thereby creating the operator layer used by equilibrium theory, response, spectroscopy, strong correlation, and pairing. It also establishes a discipline that those later subjects repeatedly need: a nonzero correlation is a property of an operator pair in a declared state, not by itself a mechanism or phase diagnosis. Many-body algebra acquires physical meaning only when Hamiltonian, preparation, observable, and comparison are specified together.

A two-electron double quantum dot and a two-site optical-lattice well turn the operator grammar into measurements. In either realization, tunneling competes with an energetic cost for double occupation. Two prepared states can have the same total particle number and the same average population on each site while having different probabilities for joint occupation and different spin correlations. Mean site populations therefore do not fully characterize the many-particle state. A charge sensor or site-resolved microscope can distinguish the occupation statistics; singlet–triplet spectroscopy and spin-resolved correlations ask complementary questions of the same state.

Lecture 7’s canonical operator algebra established how creation and annihilation operators act. We now use them to construct Hamiltonians and calculate measurable quantities. The algebra alone does not specify the Hamiltonian, the prepared state, or the observable needed to predict the double-dot measurements.

We will lift declared one- and two-particle matrix elements to many-body operators and use the Hubbard dimer to calculate observables, currents, and correlations without confusing an abstract operator with its basis matrix.

The Hubbard dimer below is trusted precisely at this local scale. It can test conservation sectors, singlet/triplet structure, double occupation, and density or spin correlations. It cannot establish a bulk phase, long-range order, or a thermodynamic transition. Keeping that boundary visible prevents an exact two-site diagonalization from being overinterpreted.

Given one- and two-particle matrix elements, how are the corresponding Fock-space operators constructed, and which algebraic tests distinguish a conserved observable, an interaction, and a time-dependent correlation?

The construction is operationally important. A missing factor of one half in an interaction changes an energy; a wrong mode order changes a hopping sign; and confusing Hermiticity with number conservation changes which sectors a Hamiltonian connects. Density correlations measured in scattering and current correlations used in response theory both inherit these choices.

From Lecture 7’s ordered-sign calculation, apply a fermionic hopping monomial to an ordered occupation state, state the canonical algebra, and give the creator and annihilator transformations under a unitary change of one-particle basis. The discrete Fourier map is one specialization; its chain calculation is optional. From quantum mechanics, recover the matrix representation of an additive one-particle observable and the Heisenberg equation. The correlations below are evaluated in declared pure states.

From one- and two-particle contributions to Fock-space operators

The same numerical matrix can participate in several descriptions only after its domain is named. Here it acts first on a one-particle orbital space and is then lifted to a Fock-space operator; the many-body state on which that operator is evaluated is another object. The lifting identities and dimer calculations are exact within the declared orbital model. Treating that model as a material description is phenomenological unless higher orbitals and interaction channels have been downfolded with an error estimate. A nonzero correlation records structure of the state relative to chosen operators; it does not by itself establish a causal mechanism.

Lifting a one-particle matrix to Fock space

Three levels of description will be kept distinct. An operator is an abstract linear map on the many-particle Hilbert space; a matrix is its representation in a declared basis; and an expectation or correlator also requires a state. A unitary basis change alters matrix entries and occupation labels while leaving operator identities, spectra, and consistently transformed expectations invariant. Connected correlation measures statistical dependence relative to the chosen state; by itself it neither proves causation nor identifies a propagating excitation.

Retain the ${M}$ -dimensional one-particle Hilbert space ${\mathcal H_1}$ and ordered orthonormal one-particle basis ${\mathcal B=(|1\rangle,\ldots,|M\rangle)}$ from Lecture 7. Let ${\hat o:\mathcal H_1\to\mathcal H_1}$ be a one-particle operator with coordinate matrix ${{\bigl[\hat o\bigr]}_{\mathcal B}=:o}$ and entries ${o_{ij}=\langle i|\hat o|j\rangle}$ , where ${i,j\in\{1,\ldots,M\}}$ . The array ${o=(o_{ij})}$ is a coefficient matrix and therefore carries no hat; ${\hat o}$ and its Fock-space lift are operators. We use fermions: ${\hat c_i}$ annihilates resolved mode ${i}$ . For finite ${M}$ , the finite-particle domain of Lecture 7 is all of fermion Fock space, ${\mathcal D_-=\mathcal F_-(\mathcal H_1)}$ , since ${0\leq N\leq M}$ . In the next definition, ${\hat o^{(a)}}$ acts as ${\hat o}$ on tensor slot ${a\in\{1,\ldots,N\}}$ and as the identity on every other slot. A slot position is not a mode label or a persistent particle identity. The sum over all slots preserves antisymmetry; an individual slot operator need not.

Let ${\hat o:\mathcal H_1\to\mathcal H_1}$ be a linear operator with matrix ${{\bigl[\hat o\bigr]}_{\mathcal B}=:o}$ , where ${o_{ij}=\langle i|\hat o|j\rangle}$ , in the chosen orthonormal one-particle basis. Its additive fermionic Fock-space lift is the operator ${\hat O_1}$ on the algebraic domain ${\mathcal D_-}$ defined sectorwise by

$\left.\hat O_1\right|_{\mathcal H_-^{(N)}} :=\sum_{a=1}^{N}\hat o^{(a)}, \qquad \left.\hat O_1\right|_{\mathcal H_-^{(0)}}=0.$

Equivalently, in this basis,

$\hat O_1=\sum_{ij}o_{ij}\hat c_i^\dagger\hat c_j.$

The lift preserves each particle-number sector and ${[\hat N,\hat O_1]=0}$ on ${\mathcal D_-}$ . If ${\hat o}$ is self-adjoint on the declared one-particle domain, the corresponding lift is self-adjoint on its natural Fock-space domain; for the finite-mode setting used here every matrix operator is bounded and the formula acts on all of ${\mathcal F_-}$ .

An annihilation operator removes occupation from mode ${j}$ , the coefficient ${o_{ij}}$ applies the one-particle map, and a creation operator restores the particle in mode ${i}$ . This order is forced by the matrix element; it is not a mnemonic to reverse casually.

The lift can be checked from matrix elements rather than accepted by analogy. Fock’s 1932 construction is the primary historical source for the operator representation across particle-number sectors; the precise additive lift and domain statement here are derived in the lecture’s declared finite-mode convention [Foc32, printed pp. 623–633, Eqs. (7)–(39)]. On a one-particle state ${|j\rangle=\hat c_j^\dagger|0\rangle}$ ,

$\begin{aligned} \hat O_1|j\rangle &=\sum_{ab}o_{ab}\hat c_a^\dagger\hat c_b \hat c_j^\dagger|0\rangle\\ &=\sum_{ab}o_{ab}\hat c_a^\dagger (\delta_{bj}-\hat c_j^\dagger\hat c_b)|0\rangle =\sum_a o_{aj}|a\rangle. \end{aligned}$

because ${\hat c_b\hat c_j^\dagger|0\rangle=\delta_{bj}|0\rangle}$ . This is exactly the column action of ${o}$ . On an occupation state the same bilinear removes one occupied ${j}$ , including its ordering sign, and inserts ${i}$ ; summing ${o_{ij}}$ therefore applies ${\hat o}$ to every occupied one-particle slot. If ${o=o^\dagger}$ , relabeling ${i\leftrightarrow j}$ in the adjoint proves ${\hat O_1=\hat O_1^\dagger}$ .

For the general sector action, CAR gives a commutator with a single creator:

(8.1)

$\begin{aligned} [\hat c_a^\dagger\hat c_b,\hat c_j^\dagger] &=\hat c_a^\dagger\{\hat c_b,\hat c_j^\dagger\} -\{\hat c_a^\dagger,\hat c_j^\dagger\}\hat c_b =\delta_{bj}\hat c_a^\dagger,\\ [\hat O_1,\hat c_j^\dagger]&=\sum_a o_{aj}\hat c_a^\dagger. \end{aligned}$

Let ${j_1<\cdots<j_N}$ . Apply the commutator product rule and ${\hat O_1|0\rangle=0}$ to the ordered creator product:

(8.2)

$\hat O_1\hat c_{j_1}^\dagger\cdots\hat c_{j_N}^\dagger|0\rangle =\sum_{r=1}^{N}\sum_a o_{a j_r} \hat c_{j_1}^\dagger\cdots\hat c_{j_{r-1}}^\dagger \hat c_a^\dagger \hat c_{j_{r+1}}^\dagger\cdots\hat c_{j_N}^\dagger|0\rangle.$

Each term replaces exactly one factor by its one-particle image while retaining its slot. By Lecture 7’s creator-product/exterior-product identity this is ${\sum_r\hat o^{(r)}}$ on the antisymmetric state. Sorting the result into mode order supplies any exchange sign; a repeated mode vanishes. These ordered states span each sector, so the identity proves the lift on all finite-mode Fock space, including the vacuum.

Use the one-body lift to calculate a specified two-fermion transition, then construct its Hermitian hopping partner. Choose three distinct modes ${p<q<r}$ , let ${|pq\rangle:=\hat c_p^\dagger\hat c_q^\dagger|0\rangle}$ , and take the one-particle map ${\hat o=|r\rangle\langle p|}$ . Find the lifted action on ${|pq\rangle}$ . For a supplied complex hopping energy ${\gamma}$ , also construct the Hermitian two-state matrix in ${(|pq\rangle,|qr\rangle)}$ .

Reuse the ordered-action ledger of the ordered-fermion sign example; the two occupied slots, rather than distinguishable particle labels, determine the operation. The lift is ${\hat O_1=\hat c_r^\dagger\hat c_p}$ , so

$\hat O_1|pq\rangle =\hat c_r^\dagger\hat c_p \hat c_p^\dagger\hat c_q^\dagger|0\rangle =\hat c_r^\dagger\hat c_q^\dagger|0\rangle =-|qr\rangle.$

In first-quantized language, ${(\hat o^{(1)}+\hat o^{(2)})|p\wedge q\rangle =|r\wedge q\rangle=-|q\wedge r\rangle}$ , the same result. The removal sign and insertion sign are therefore not extra physics: together they encode the antisymmetric wedge order while the operator remains within the two-particle sector.

To make a Hamiltonian, pair the supplied hopping amplitude ${\gamma}$ with its conjugate: $\hat h=\gamma|r\rangle\langle p|+\gamma^*|p\rangle\langle r|, \qquad \hat H_1=\gamma\hat c_r^\dagger\hat c_p +\gamma^*\hat c_p^\dagger\hat c_r.$ On ${\mathcal B=(|pq\rangle,|qr\rangle)}$ , the two column actions give ${[\hat H_1]}_{\mathcal B} =\begin{pmatrix}0&-\gamma^*\\-\gamma&0\end{pmatrix}.$ The conjugate off-diagonal entries check Hermiticity independently of the wedge calculation. Both terms have one creator and one annihilator, so ${[\hat N,\hat H_1]=0}$ ; both final states still have two occupied modes. Its energies ${\pm|\gamma|}$ have the same gap under rephasing of either basis state. This is a one-body lift even though the example used two fermions. The following covariance argument explains that invariance for an arbitrary legal one-particle basis change.

Basis covariance of the lift

Here body rank refers to a number-conserving operator after normal ordering (moving creation operators to the left of annihilation operators, including the terms generated by the canonical algebra) and algebraic simplification: a term with ${k}$ creation maps and ${k}$ annihilation maps is a ${k}$ -body term, and a sum has the largest nonzero rank present. A scalar multiple of the identity has rank zero. This convention distinguishes the number of particles on which an operator acts from the number of particles in the state. The normal-ordered product ${\hat c_i^\dagger\hat c_j}$ is one-body: it applies the one-particle map ${|i\rangle\langle j|}$ additively to all particles. Its body rank does not depend on how many particles the state contains. Operator order matters: for fermions, ${\hat c_i\hat c_i^\dagger=\hat I-\hat n_i}$ contains a scalar and a one-body term. For distinct resolved modes, a product of two number operators is two-body; for the same fermion mode, ${\hat n_i^2=\hat n_i}$ is one-body after simplification. A pair source with two creators or two annihilators changes particle number and has no body-rank label under this convention. Its creation/annihilation counts and sector change must instead be stated explicitly.

Let ${\mathcal E}$ and ${\mathcal E'}$ be orthonormal bases of the same finite one-particle Hilbert space, with ${|\alpha'\rangle=\sum_i|i\rangle U_{i\alpha}}$ for a unitary matrix ${U}$ . If ${o'=U^\dagger oU}$ and the corresponding creation and annihilation operators transform as

$\hat c_\alpha^{\prime\dagger}=\sum_iU_{i\alpha}\hat c_i^\dagger, \qquad \hat c'_\beta=\sum_jU_{j\beta}^*\hat c_j,$

then

$\sum_{\alpha\beta}o'_{\alpha\beta} \hat c_\alpha^{\prime\dagger}\hat c'_\beta =\sum_{ij}o_{ij}\hat c_i^\dagger\hat c_j.$

Thus the Fock-space operator and its body rank are basis independent when coefficients and creation/annihilation operators are transformed together.

To prove the statement, expand the left-hand side. Its coefficient of ${\hat c_p^\dagger\hat c_q}$ is

$\sum_{\alpha\beta ij} U_{i\alpha}^*o_{ij}U_{j\beta} U_{p\alpha}U_{q\beta}^* =\sum_{ij}o_{ij}\delta_{ip}\delta_{jq} =o_{pq},$

where unitarity was used twice. A one-body operator therefore does not become an interaction merely because its coefficient matrix becomes dense.

The one-body lift adds a contribution for each particle. For an interaction, we must specify what a pair contributes and whether the many-particle model is a sum of those pair contributions. This is physical input, not a consequence of Fock space or CAR.

Pairwise additivity before coefficient conventions

A pair interaction may assign an energy to a joint configuration or connect one pair state to another. For example, the Hubbard interaction assigns an energy cost when two opposite-spin fermions occupy the same site. Pairwise additivity assumes that the same interaction acts on every unordered particle pair, with no independent three-body contribution in this term. In particle-slot notation the one- and two-body contributions are

(8.3)

$\hat H_1^{(N)}=\sum_{a=1}^{N}\hat h^{(a)}, \qquad \hat H_2^{(N)}=\sum_{1\le a<b\le N}\hat v^{(a,b)}.$

Here ${a,b}$ are tensor-slot positions used to describe the action, not persistent particle identities or one-particle basis labels. The complete sums act on the antisymmetric sector ${\mathcal H_-^{(N)}}$ . The pair term is distinct from the complete two-particle Hamiltonian: ${\hat H^{(2)}=\hat h^{(1)}+\hat h^{(2)}+\hat v^{(1,2)}}$ . We now construct the pairwise action with Lecture 7’s fermionic states, remaining in the same Fock space.

In the finite one-particle basis fixed above, let ${\hat v_2:\wedge^2\mathcal H_1\to\wedge^2\mathcal H_1}$ be a self-adjoint pair interaction, with units of energy. Use the normalized ordered pair basis

$|i\wedge j\rangle:=\hat c_i^\dagger\hat c_j^\dagger|0\rangle, \qquad \overline V_{ij;kl}:=\langle i\wedge j|\hat v_2|k\wedge l\rangle, \quad i<j,\ k<l.$

Its pairwise additive lift ${\hat H_2}$ applies this same pair map to every unordered pair in each sector, with all other particles retained, and is zero on the vacuum and one-particle sectors. Hermiticity gives ${\overline V_{ij;kl}=\overline V_{kl;ij}^*}$ . The lift acts on all of the finite-dimensional fermion Fock space.

The coefficient ${\overline V_{ij;kl}}$ is the amplitude for the incoming fermion pair ${k,l}$ to become ${i,j}$ . Specifying this two-particle matrix does not determine an arbitrary many-particle Hamiltonian: the pairwise additivity assumption fixes the extension used here. A three-body term, for example, would vanish in the two-particle sector but could act in higher sectors.

Deriving the operator by removing and reinserting a pair

Let ${K=(k_1<\cdots<k_N)}$ and ${|K\rangle=\hat c_{k_1}^\dagger\cdots\hat c_{k_N}^\dagger|0\rangle}$ . For ${a<b}$ , write ${|K\setminus\{k_a,k_b\}\rangle}$ for the ordered occupation state with those two occupied labels removed. Here the indices mark positions in the ordered list of occupied orbitals, not tensor slots. Removing the earlier label first leaves ${a-1}$ crossings for the first annihilator and ${b-2}$ for the second. CAR therefore gives

(8.4)

$\hat c_{k_b}\hat c_{k_a}|K\rangle ={(-1)}^{a+b-3}|K\setminus\{k_a,k_b\}\rangle.$

Apply the pair matrix to the removed pair, create each possible output pair, and sum over the occupied pairs. The state left after removal carries the spectators. This is the action required by pairwise additivity:

(8.5)

$\hat H_2|K\rangle =\sum_{a<b}\sum_{i<j} {(-1)}^{a+b-3}\overline V_{ij;k_a k_b}\, \hat c_i^\dagger\hat c_j^\dagger |K\setminus\{k_a,k_b\}\rangle.$

The following quartic operator has exactly that action. Its annihilators select an occupied incoming pair; its creators insert the outgoing pair. Creation into an already occupied spectator orbital gives zero, and reordering any surviving state supplies its fermionic sign.

(8.6)

$\boxed{\displaystyle \hat H_2 =\sum_{\substack{i<j\\k<l}} \overline V_{ij;kl}\, \hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k.}$

For an ordered input pair ${p<q}$ , the contraction ${\hat c_l\hat c_k|p\wedge q\rangle =\delta_{kp}\delta_{lq}|0\rangle}$ for ${k<l}$ gives the immediate check ${\hat H_2|p\wedge q\rangle =\sum_{i<j}\overline V_{ij;pq}|i\wedge j\rangle}$ . The general- ${N}$ action above establishes the lift on every occupation state, not only on this two-particle check. There is no counting prefactor because each incoming and outgoing pair appears once. Every term removes two particles and creates two, so ${[\hat N,\hat H_2]=0}$ . Taking the adjoint and exchanging incoming and outgoing pairs proves ${\hat H_2^\dagger=\hat H_2}$ . For infinitely many one-particle states, the corresponding kernel sums and operator domains require separate hypotheses; the finite basis remains the working setting here.

The same operator in unrestricted-index conventions

To sum over all indices, extend ${\overline V}$ by antisymmetry in each pair: ${\overline V_{ji;kl}=-\overline V_{ij;kl}}$ and ${\overline V_{ij;lk}=-\overline V_{ij;kl}}$ , with repeated-index pairs zero. This is the sign change of the pair basis, not a new physical assumption. There are two orders for the incoming pair and two for the outgoing pair, so the unrestricted sum counts four equal contributions.

(8.7)

$\hat H_2=\frac14\sum_{ijkl}\overline V_{ij;kl}\, \hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k.$

A microscopic interaction is often supplied in an ordinary product basis. Write ${|i,j\rangle:=|i\rangle\otimes|j\rangle}$ , which is distinct from the physical fermionic state ${|i\wedge j\rangle}$ . Choose a self-adjoint representative ${\hat v}$ on ${\mathcal H_1\otimes\mathcal H_1}$ whose restriction is ${\hat v_2}$ and which commutes with the swap ${\hat S_{12}|i,j\rangle=|j,i\rangle}$ . This expresses independence from the labels assigned to identical particles. An extension exists, for example by setting it to zero on the symmetric subspace. Its symmetric-sector action does not affect the fermionic lift. The representative also makes the slot notation ${\sum_{a<b}\hat v^{(a,b)}|_{\mathcal H_-^{(N)}}}$ precise.

The raw coefficients use product states before antisymmetrization. They obey simultaneous-exchange symmetry and Hermiticity, but need not be antisymmetric in either index pair separately:

(8.8)

$V_{ij;kl}:=(\langle i|\otimes\langle j|)\hat v (|k\rangle\otimes|l\rangle), \qquad V_{ij;kl}=V_{ji;lk}, \qquad V_{ij;kl}=V_{kl;ij}^*.$

Expanding the normalized fermionic pairs gives ${\overline V_{ij;kl}=V_{ij;kl}-V_{ij;lk}}$ . Substituting this difference into the unrestricted antisymmetrized sum and relabeling the second term gives the alternative raw-coefficient form:

(8.9)

$\hat H_2=\frac12\sum_{ijkl}V_{ij;kl}\, \hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k.$

These are representations of the same pairwise additive operator. Keep the coefficient convention and summation range with their prefactor: ordered fermionic pairs use one, unrestricted antisymmetrized coefficients use one quarter, and raw product-state coefficients use one half. The next calculation derives these factors explicitly; the subsequent worked example checks them against an occupation-state energy.

Detailed calculation of the two prefactors

For the unrestricted antisymmetrized sum, fix distinct ${i<j}$ and ${k<l}$ . The four orders give identical terms, because exchanging a pair changes both its coefficient sign and its operator sign:

(8.10)

$\begin{aligned} &\overline V_{ij;kl}\hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k +\overline V_{ji;kl}\hat c_j^\dagger\hat c_i^\dagger\hat c_l\hat c_k\\ &\quad+\overline V_{ij;lk}\hat c_i^\dagger\hat c_j^\dagger\hat c_k\hat c_l +\overline V_{ji;lk}\hat c_j^\dagger\hat c_i^\dagger\hat c_k\hat c_l\\ &=4\,\overline V_{ij;kl} \hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k. \end{aligned}$

For the product-space representative, expand both normalized pair states. Each contributes a factor ${1/\sqrt2}$ . Use ${V_{ji;lk}=V_{ij;kl}}$ and ${V_{ji;kl}=V_{ij;lk}}$ , both from simultaneous exchange of the two tensor factors:

(8.11)

$\begin{aligned} \overline V_{ij;kl} &=\frac12\left(V_{ij;kl}-V_{ij;lk}-V_{ji;kl}+V_{ji;lk}\right)\\ &=V_{ij;kl}-V_{ij;lk}. \end{aligned}$

To obtain the raw prefactor, interchange the dummy incoming indices ${k,l}$ in the second sum. CAR reverses the annihilator order with a minus sign. The two minus signs then cancel:

(8.12)

$\begin{aligned} \hat H_2 &=\frac14\sum_{ijkl}(V_{ij;kl}-V_{ij;lk}) \hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k\\ &=\frac14\sum_{ijkl}V_{ij;kl} \left(\hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k -\hat c_i^\dagger\hat c_j^\dagger\hat c_k\hat c_l\right)\\ &=\frac12\sum_{ijkl}V_{ij;kl} \hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k. \end{aligned}$

As a state-level check, for ${p<q}$ the two incoming contractions are ${\hat c_l\hat c_k|p\wedge q\rangle =(\delta_{kp}\delta_{lq}-\delta_{kq}\delta_{lp})|0\rangle}$ . The raw form therefore gives

(8.13)

$\begin{aligned} \hat H_2|p\wedge q\rangle &=\frac12\sum_{ij}(V_{ij;pq}-V_{ij;qp}) \hat c_i^\dagger\hat c_j^\dagger|0\rangle\\ &=\sum_{i<j}\overline V_{ij;pq}|i\wedge j\rangle. \end{aligned}$

The last equality combines both outgoing orders, whose coefficient and creator signs cancel. The preceding dummy-index derivation is an operator identity on all of the finite Fock space; this two-particle calculation is a check, not its sole justification. In particle-slot language the raw one-half also corresponds to ${\sum_{a<b}\hat v^{(a,b)}=\frac12\sum_{a\ne b}\hat v^{(a,b)}}$ . The two-body lift still preserves particle number and is Hermitian in all three conventions.

Test the raw-kernel convention on two resolved modes ${1<2}$ . Supply only ${V_{12;12}=V_{21;21}=U}$ , with real energy ${U}$ , and set the other raw entries to zero. The task is to reconstruct the interaction and its energy on ${|11\rangle}$ . Compare the raw-kernel and antisymmetrized conventions.

The supplied entries obey the exchange and Hermiticity conditions. Keep the ordered terms and their prefactor together rather than treating ${U}$ as an already-counted pair energy. Both incoming orders occur in the full sum: $\hat H_2=\frac U2\left( \hat c_1^\dagger\hat c_2^\dagger\hat c_2\hat c_1 +\hat c_2^\dagger\hat c_1^\dagger\hat c_1\hat c_2\right) =U\hat n_1\hat n_2.$ Two sign changes turn the second term into the first; the half cancels the two equal contributions. Direct action gives energy ${U}$ on ${|11\rangle}$ and zero on ${|00\rangle,|10\rangle,|01\rangle}$ . This checks physical pair counting by checking occupation-state energies directly. In the antisymmetrized convention, the four entries are ${U,-U,-U,U}$ for ${(12;12),(12;21),(21;12),(21;21)}$ . Their operator signs yield four equal terms, so the factor ${1/4}$ gives the same energy. Finally the operator is Hermitian and commutes with ${\hat N}$ , since it is a real product of commuting number operators. A factor ${1/2}$ multiplying those four antisymmetrized entries would give ${2U}$ , exposing the convention mismatch. This local pair penalty is the onsite term used in the dimer below.

Take six minutes and classify each monomial: ${\hat c_i^\dagger\hat c_j}$ , ${\hat c_i^\dagger\hat c_j^\dagger\hat c_l\hat c_k}$ , ${\hat c_i^\dagger\hat c_j^\dagger}$ , and ${\hat n_{i\uparrow}\hat n_{i\downarrow}}$ , filling four columns: body rank where defined, net particle-number change, whether the monomial is Hermitian as written or requires its conjugate partner, and the coefficient symmetry the term demands ( ${o_{ij}=o_{ji}^*}$ for the one-body term; ${V_{ij;kl}}$ consistent with the pair-counting convention just fixed). Check Hermiticity by taking the adjoint, and check coefficient symmetry against the declared pair-counting convention. These are separate tests.

The Hubbard dimer makes the conventions observable

Declaring the dimer and its number sectors

Model lineage

[ABKR22, secs. 1–2] — A modern review states the Hubbard Hamiltonian and its conserved quantities; the lecture’s two-site reduction remains course-owned.
[Gut63, pp. 159–162] — Early variational treatment of opposite-spin electronic correlations in a narrow-band setting.
[Hub63, pp. 238–257] — Hubbard’s 1963 narrow-band formulation, one member of the independently developed Gutzwiller–Hubbard–Kanamori lineage rather than the sole origin.
[Kan63, pp. 275–289] — Independent 1963 local-interaction formulation for correlated transition-metal electrons.

Return to Lecture 7’s spinful double dot. A resolved mode is now the pair ${(i,\sigma)}$ , with site ${i=1,2}$ and spin ${\sigma=\uparrow,\downarrow}$ . Keep the order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ and define ${\hat n_{i\sigma}:=\hat c_{i\sigma}^\dagger\hat c_{i\sigma}}$ . Each resolved mode has occupation zero or one; the site total ${\hat n_i:=\hat n_{i\uparrow}+\hat n_{i\downarrow}}$ can be zero, one or two. Opposite spins on one site are distinct modes, so Lecture 7’s distinct-mode density product gives the onsite pair energy below. The hopping energy ${t}$ and onsite pair energy ${U}$ are real model parameters. The two-site Hubbard Hamiltonian is

(8.14)

$\hat H_{\mathrm{Hub}} =-t\sum_{\sigma} \left( \hat c_{1\sigma}^\dagger\hat c_{2\sigma} +\hat c_{2\sigma}^\dagger\hat c_{1\sigma} \right) +U\sum_{i=1}^2\hat n_{i\uparrow}\hat n_{i\downarrow}.$

The hopping is one-body: it moves a particle between sites without changing the total count. The onsite term is two-body because it detects a pair of opposite spins on the same site. Since ${\hat n_{i\uparrow}\hat n_{i\downarrow}}$ has eigenvalue one only on a doubly occupied site, ${U>0}$ raises precisely those configurations.

The total number operator is

$\hat N=\sum_{i\sigma}\hat n_{i\sigma}.$

Using ${[\hat N,\hat c_i^\dagger]=\hat c_i^\dagger}$ and ${[\hat N,\hat c_j]=-\hat c_j}$ , each hopping bilinear commutes with ${\hat N}$ ; number operators commute as well. Therefore

(8.15)

$[\hat H_{\mathrm{Hub}},\hat N]=0.$

The same calculation shows ${[\hat H_{\mathrm{Hub}},\hat S^z]=0}$ for

$\hat S^z=\frac{\hbar}{2}\sum_i (\hat n_{i\uparrow}-\hat n_{i\downarrow}).$

This operator is the total spin angular momentum along the quantization axis. Each up-spin particle contributes ${+\hbar/2}$ , and each down-spin particle contributes ${-\hbar/2}$ . When comparing with a dimensionless-spin convention, use ${\hat S^z/\hbar}$ .

For Equation (8.14), with ${t>0}$ and ${U\geq0}$ , construct the two-electron matrix and identify the combination that couples to double occupancy. Retain the mode order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ . In the ${N=2}$ , ${S^z=0}$ sector, choose the ordered occupation basis ${\mathcal B_{\mathrm{occ}}=(|D_1\rangle,|A\rangle,|B\rangle,|D_2\rangle)}$ , where

Use the local penalty checked in Example 8.2 and direct ordered actions, rather than guessing matrix signs from symmetry. Construct the matrix column by column. For ${|D_1\rangle=|1100\rangle}$ ,

$\begin{aligned} \hat c_{2\uparrow}^\dagger\hat c_{1\uparrow}|D_1\rangle &=\hat c_{2\uparrow}^\dagger|0100\rangle=-|B\rangle,\\ \hat c_{2\downarrow}^\dagger\hat c_{1\downarrow}|D_1\rangle &=\hat c_{2\downarrow}^\dagger(-|1000\rangle)=|A\rangle. \end{aligned}$

For the singly occupied states, the active hops give

$\begin{aligned} \hat c_{1\downarrow}^\dagger\hat c_{2\downarrow}|A\rangle &=\hat c_{1\downarrow}^\dagger(-|1000\rangle)=|D_1\rangle,\\ \hat c_{2\uparrow}^\dagger\hat c_{1\uparrow}|A\rangle &=\hat c_{2\uparrow}^\dagger|0001\rangle=|D_2\rangle,\\ \hat c_{1\uparrow}^\dagger\hat c_{2\uparrow}|B\rangle &=\hat c_{1\uparrow}^\dagger(-|0100\rangle)=-|D_1\rangle,\\ \hat c_{2\downarrow}^\dagger\hat c_{1\downarrow}|B\rangle &=\hat c_{2\downarrow}^\dagger|0010\rangle=-|D_2\rangle. \end{aligned}$

Finally, from ${|D_2\rangle=|0011\rangle}$ ,

$\begin{aligned} \hat c_{1\uparrow}^\dagger\hat c_{2\uparrow}|D_2\rangle &=\hat c_{1\uparrow}^\dagger|0001\rangle=|A\rangle,\\ \hat c_{1\downarrow}^\dagger\hat c_{2\downarrow}|D_2\rangle &=\hat c_{1\downarrow}^\dagger(-|0010\rangle)=-|B\rangle. \end{aligned}$

Multiplying every hopping action by ${-t}$ and adding the onsite energy gives

$\begin{aligned} \hat H_{\mathrm{Hub}}|D_1\rangle&=U|D_1\rangle-t|A\rangle+t|B\rangle,\\ \hat H_{\mathrm{Hub}}|A\rangle&=-t|D_1\rangle-t|D_2\rangle,\\ \hat H_{\mathrm{Hub}}|B\rangle&=t|D_1\rangle+t|D_2\rangle,\\ \hat H_{\mathrm{Hub}}|D_2\rangle&=-t|A\rangle+t|B\rangle+U|D_2\rangle. \end{aligned}$

Thus the coefficient matrix is

${[\hat H_{\mathrm{Hub}}]}_{\mathcal B_{\mathrm{occ}}} {=} \begin{pmatrix} U&-t&t&0\\ -t&0&0&-t\\ t&0&0&t\\ 0&-t&t&U \end{pmatrix}.$

Hermiticity now checks the separately calculated columns rather than supplying them. The signs depend on the frozen basis phases: rephasing ${|A\rangle}$ or ${|B\rangle}$ changes a row and column together but not the spectrum. A sign error here would incorrectly couple the triplet rather than the singlet.

The combinations

$|S\rangle=\frac{|A\rangle-|B\rangle}{\sqrt2}, \qquad |T_0\rangle=\frac{|A\rangle+|B\rangle}{\sqrt2}$

are, with this occupation convention, the spin singlet and the ${m=0}$ triplet. The triplet does not couple to double occupancy, while the singlet does. This fact will later produce superexchange, but Lecture 16 owns that physical development. Here it is a check that fermionic signs and spin symmetry agree.

The trace is ${2U}$ , the sum of the two doubly occupied diagonal energies. At ${t=0}$ , the matrix is diagonal with entries ${(U,0,0,U)}$ , an independent check of which configurations pay the pair penalty. Every column remains in ${N=2,S^z=0}$ , and the separately obtained entries are Hermitian. These checks constrain both signs and factors. The symmetry-adapted figure organizes this exact matrix into smaller blocks; it makes no low-energy approximation.

For the symmetry reduction, carry forward the ordered basis and one signed column from Example 8.3. With ${t>0}$ , ${U\geq0}$ , mode order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ , and the conserved ${N=2,S^z=0}$ sector, the four configurations are

(8.16)

$\mathcal B_{\mathrm{occ}}=(|D_1\rangle,|A\rangle,|B\rangle,|D_2\rangle) =(|1100\rangle,|1001\rangle,|0110\rangle,|0011\rangle).$

An up-spin hop out of ${|D_1\rangle}$ crosses the remaining down-spin creator; a down-spin hop has a removal sign and an insertion sign that cancel. Multiplying by the hopping coefficient ${-t}$ and adding the pair energy gives the column to retain:

(8.17)

$\begin{aligned} \hat c_{2\uparrow}^{\dagger}\hat c_{1\uparrow}|D_1\rangle&=-|B\rangle, &\hat c_{2\downarrow}^{\dagger}\hat c_{1\downarrow}|D_1\rangle&=|A\rangle,\\ \hat H_{\mathrm{Hub}}|D_1\rangle&=U|D_1\rangle-t|A\rangle+t|B\rangle. \end{aligned}$

The remaining columns, calculated in the same way in Example 8.3, give the following matrix. Hermiticity and the ${t=0}$ diagonal ${(U,0,0,U)}$ check its entries. To test its symmetry, define ${|S\rangle=(|A\rangle-|B\rangle)/\sqrt2}$ and ${|T_0\rangle=(|A\rangle+|B\rangle)/\sqrt2}$ , the singlet and zero-spin-projection triplet in this convention.

(8.18)

${[\hat H_{\mathrm{Hub}}]}_{\mathcal B_{\mathrm{occ}}} =\begin{pmatrix}U&-t&t&0\\-t&0&0&-t\\t&0&0&t\\0&-t&t&U\end{pmatrix}.$

Define ${|D_\pm\rangle=(|D_1\rangle\pm|D_2\rangle)/\sqrt2}$ . In the ordered symmetry-adapted basis ${\mathcal B_{\mathrm{sym}}=(|D_+\rangle,|S\rangle, \vert{}D_-\rangle,\vert{}T_0\rangle)}$ , predict which states remain isolated and which two states form the nontrivial block in Figure 8.1.

Constructing the sector matrices

Figure 8.1. The exact symmetry-adapted basis converts the four-state Hubbard-dimer matrix into one coupled two-state singlet block and two isolated eigenstates.

Two distinct symmetries explain these blocks. The spin-independent hopping is unchanged when the same spin rotation is applied at both sites, and the onsite term can be written ${U\hat n_{i\uparrow}\hat n_{i\downarrow}=U\hat n_i(\hat n_i-\hat I)/2}$ , which depends only on total site occupation. Thus global spin rotations preserve the Hamiltonian, with ${[\hat H_{\mathrm{Hub}},\hat{\mathbf S}_{\mathrm{tot}}^2]=0}$ , where ${\hat{\mathbf S}_{\mathrm{tot}}}$ is the total physical spin angular momentum. The singlet and triplet have total-spin-squared eigenvalues ${0}$ and ${2\hbar^2}$ , respectively; a commuting Hamiltonian cannot mix these eigenspaces. A doubly occupied single orbital is a spin singlet, so it cannot couple to the triplet.

Exchange of the two identical sites gives a second symmetry. Choose its unitary operator ${\hat P}$ to fix the vacuum and interchange corresponding creators: ${\hat P|0\rangle=|0\rangle}$ , ${\hat P\hat c_{1\sigma}^\dagger\hat P^{-1}=\hat c_{2\sigma}^\dagger}$ , and conversely. The two-site Hamiltonian is unchanged, so ${[\hat H_{\mathrm{Hub}},\hat P]=0}$ . Reordering the exchanged creators back into the declared order gives ${\hat P|A\rangle=-|B\rangle}$ , ${\hat P|B\rangle=-|A\rangle}$ , and ${\hat P|D_1\rangle=|D_2\rangle}$ . Consequently ${\hat P|D_-\rangle=-|D_-\rangle}$ , whereas ${\hat P|D_+\rangle=|D_+\rangle}$ and ${\hat P|S\rangle=|S\rangle}$ . The odd singlet ${|D_-\rangle}$ cannot mix with the even singlets ${|D_+\rangle}$ and ${|S\rangle}$ . This site exchange is distinct from exchange of identical-particle tensor slots.

In a coupled-dot experiment, the remaining singlet mixing shifts the singlet–triplet energy difference. The dimer predicts that level structure once ${t}$ and ${U}$ are declared; lead broadening, pulse errors and a full device stability diagram require additional modeling.

Use the matrix from Example 8.3 and the displayed symmetry basis to find the lowest singlet energy and the probability of double occupancy. Keep ${t>0}$ , ${U\geq0}$ , and ${\mathcal B_{\mathrm{sym}}=(|D_+\rangle,|S\rangle,|D_-\rangle,|T_0\rangle)}$ , where ${|D_\pm\rangle=(|D_1\rangle\pm|D_2\rangle)/\sqrt2}$ . Double occupancy means that either site contains both electrons.

The figure is a basis reorganization, not an approximation. The amplitudes from ${|D_1\rangle}$ and ${|D_2\rangle}$ add for ${|D_+\rangle\leftrightarrow|S\rangle}$ and cancel in the orthogonal channels. Accordingly,

${[\hat H_{\mathrm{Hub}}]}_{\mathcal B_{\mathrm{sym}}} {=} \begin{pmatrix} U&-2t&0&0\\ -2t&0&0&0\\ 0&0&U&0\\ 0&0&0&0 \end{pmatrix}.$

The nontrivial ${2\times2}$ singlet block has characteristic equation ${E(E-U)-4t^2=0}$ . Solving this quadratic gives

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

At ${t\to0}$ they approach ${0}$ and ${U}$ ; at ${U=0}$ they become ${\pm2|t|}$ , matching two electrons placed in the bonding/antibonding one-particle orbitals. These two independent limits test both the off-diagonal factor of two and the interaction convention.

Writing the ground state as ${a|D_+\rangle+b|S\rangle}$ , its second matrix row gives ${-2ta=E_-b}$ . For ${t>0}$ , normalization therefore yields $D:=\left\langle\sum_i\hat n_{i\uparrow}\hat n_{i\downarrow}\right\rangle =|a|^2=\frac{E_-^2}{E_-^2+4t^2} =\frac12\left(1-\frac{U}{\sqrt{U^2+16t^2}}\right).$ The first equality follows because the double-occupancy operator has entries ${(1,0)}$ in the coupled ${(D_+,S)}$ block. Differentiating the explicit eigenvalue gives the same ${\partial E_-/\partial U=D}$ , checking the normalized-eigenvector route. At ${U=0}$ , ${D=1/2}$ ; at fixed ${U>0}$ and ${t\to0}$ , ${D\sim4t^2/U^2\to0}$ . The probability is dimensionless and lies between zero and one half for repulsive ${U}$ . The point ${t=U=0}$ is degenerate and has no unique ground-state probability. The double-occupancy probability measures the weight of charge configurations in the dimer ground state; it does not establish a bulk phase. The conservation laws established before the matrix construction justify the sector used throughout; the block-pattern comparison checks the same laws in a different representation.

Reading interaction effects from the spectrum

Before viewing Figure 8.2, predict which particle-number blocks can be nonzero for ${\hat c_i^\dagger\hat c_j}$ , ${\hat c_i^\dagger\hat c_j^\dagger}$ , their Hermitian sum, and the Hubbard Hamiltonian. Then translate each pattern into a commutator with ${\hat N}$ .

Figure 8.2. For the dimer’s four spin orbitals, both particle-number axes run from zero through four. The three block diagrams show allowed couplings within each sector, from one sector to the next under creation, and between sectors differing by two under a pair term plus its adjoint. Equal schematic cells do not represent equal sector dimensions.

The dimer has four spin orbitals, so its particle-number sectors are ${N=0,1,2,3,4}$ . The figure is a second representation of the same algebra. If the block pattern and commutator conclusion disagree, one of the operator counts or basis labels is wrong. An equal superposition of different ${N}$ sectors may still be written mathematically, but a number-conserving Hamiltonian evolves each block independently [ABKR22].

Historical background

[Foc32, pp. 622–647] — Early configuration-space and second-quantized construction of bosonic and fermionic occupation sectors.

Hermiticity does not imply this block pattern. For distinct modes ${i,j}$ , the source ${\hat\Delta_{ij}=\hat c_i^\dagger\hat c_j^\dagger+ \hat c_j\hat c_i}$ is Hermitian, but $[\hat N,\hat\Delta_{ij}] =2\hat c_i^\dagger\hat c_j^\dagger-2\hat c_j\hat c_i\neq0.$ The correct diagnostic is the commutator, not whether the operator can appear in a physical Hamiltonian. Mark its ${\Delta N=\pm2}$ blocks on the figure before proceeding.

Connecting the localized and bonding-orbital limits

The four-state calculation used ${N=2,S^z=0}$ . To display the full ${N=2}$ spectrum, include the parallel-spin states ${|T_+\rangle=|1010\rangle}$ and ${|T_-\rangle=|0101\rangle}$ . They have no double occupancy, and hopping is Pauli blocked, so both have energy zero, like ${|T_0\rangle}$ . The six states therefore occupy four distinct branches: the two singlets with ${E_\pm=(U\pm\sqrt{U^2+16t^2})/2}$ , the singlet with ${E_{D_-}=U}$ , and the threefold-degenerate triplet with ${E_T=0}$ .

In Figure 8.3 we display the entire repulsive range on finite axes. Set ${R=\sqrt{U^2+t^2}>0}$ and ${t=R\sin\theta,\ U=R\cos\theta}$ , with ${0\leq\theta\leq\pi/2}$ . At fixed ${R}$ this traces a quarter-circle in the hopping-interaction plane; both couplings vary, while the overall energy scale stays fixed. The bottom axis is linear in ${\theta}$ ; the top axis gives ${t/U=\tan\theta}$ . The special points are ${\theta=0}$ , the atomic limit ${t=0,U=R}$ ; ${\theta=\pi/4}$ , where ${t=U=R/\sqrt2}$ ; and ${\theta=\pi/2}$ , the noninteracting limit ${U=0,t=R}$ . Substituting in the exact energies gives ${E_\pm/R=(\cos\theta\pm\sqrt{\cos^2\theta+16\sin^2\theta})/2}$ , ${E_{D_-}/R=\cos\theta}$ , and ${E_T/R=0}$ . The dimensionless probability ${D=(1-\cos\theta/\sqrt{\cos^2\theta+16\sin^2\theta})/2}$ uses the right axis. Energy normalization changes the shapes of the curves, but not their ordering, degeneracies, or the states’ double occupancy at a given ratio. The omitted point ${t=U=0}$ has neither an angle nor a nonzero energy scale.

Alongside the same-site probability ${D}$ , we compare the dimensionless bond coherence ${B={\langle\hat B\rangle}_G}$ in the same two-electron ground state, where the Hermitian operator ${\hat B=\frac12\sum_\sigma(\hat c_{1\sigma}^\dagger\hat c_{2\sigma} +\hat c_{2\sigma}^\dagger\hat c_{1\sigma})}$ measures the real intersite coherence in the declared real-hopping convention. The hopping Hamiltonian is ${\hat H_{\mathrm{hop}}=-2t\hat B}$ , so ${B}$ directly determines its kinetic-energy contribution; it is not a particle current. This normalization gives ${B=1}$ when both electrons occupy the bonding orbital.

Figure 8.3. The exact two-electron spectrum connects localized spins to a filled bonding orbital. The bottom axis shows theta and the top shows t/U; energy in units of R uses the left axis. The right axis serves both dimensionless ground-state observables: green dash-dot D is same-site double occupancy; reddish-purple solid B is real bond coherence. B starts linearly while D starts quadratically in t/U. The atomic, equal-coupling and noninteracting points are marked; equal couplings do not define a phase boundary. The endpoint lists name the states or degenerate subspaces and give normalized occupation-basis states. Atomic kets span the named subspaces. Short level lines match the energy curves in color and style; adjoining segments mark coincident branches at the same energy.

To calculate ${B}$ , use the normalized ground state ${|G\rangle=a|D_+\rangle+b|S\rangle}$ from the preceding example. For ${t>0}$ , choose real ${a,b>0}$ : the negative off-diagonal hopping matrix element selects their equal sign. The same hopping matrix shows that ${\hat B|D_+\rangle=|S\rangle}$ and ${\hat B|S\rangle=|D_+\rangle}$ . Since ${a^2=D}$ and ${b^2=1-D}$ , taking the expectation value and inserting the known ${D}$ gives

(8.19)

$\begin{aligned} B&=(a\langle D_+|+b\langle S|)(a|S\rangle+b|D_+\rangle) =2ab=2\sqrt{D(1-D)}\\ &=\frac{4t}{\sqrt{U^2+16t^2}} =\frac{4\sin\theta}{\sqrt{\cos^2\theta+16\sin^2\theta}}. \end{aligned}$

For ${t/U\ll1}$ , the two curves have different leading orders: ${B\sim4t/U}$ , whereas ${D\sim4t^2/U^2}$ . Coherence between the covalent and doubly occupied components is first order in their small mixing amplitude; the probability of double occupancy is its square. At ${t=0,U>0}$ , ${B=0}$ throughout the ground-state subspace, since the bond operator maps singly occupied states into doubly occupied states. At ${U=0,t>0}$ , ${B=1}$ while ${D=1/2}$ . The relation between these two observables was derived for this ground state; it is not an operator identity for arbitrary preparations.

For ${t/U\to0^+}$ , the ground state approaches the covalent singlet ${|S\rangle}$ , with one electron on each site and ${D\to0}$ . At exactly ${t=0,U>0}$ , that singlet and all three triplets are degenerate; the double-occupancy probability is zero throughout this ground-state subspace. For small positive hopping, the exact singlet–triplet gap gives $\Delta_{\mathrm{ST}}=E_T-E_- =\frac{\sqrt{U^2+16t^2}-U}{2} \simeq\frac{4t^2}{U}\qquad(t/U\ll1).$ The left-hand level labels use the occupation order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ . At the atomic endpoint, ${\{|1010\rangle,|0101\rangle,|1001\rangle,|0110\rangle\}}$ spans the zero-energy eigenspace, while ${\{|1100\rangle,|0011\rangle\}}$ spans the level at ${E=U=R}$ . These are orthonormal bases of degenerate spaces; individual energy branches do not select a unique basis at exactly zero hopping.

At ${U=0,t>0}$ , the energies in units of ${t}$ are ${-2,0,0,+2}$ : the middle level contains the three triplets and ${|D_-\rangle}$ , hence four states. The ground state is ${|G\rangle=(|D_+\rangle+|S\rangle)/\sqrt2 =\hat b_\uparrow^\dagger\hat b_\downarrow^\dagger|0\rangle}$ , where ${\hat b_\sigma^\dagger=(\hat c_{1\sigma}^\dagger+ \hat c_{2\sigma}^\dagger)/\sqrt2}$ creates an electron in the bonding orbital. Two electrons occupy that orbital, while the probability that they occupy the same site is only ${D=1/2}$ . For ${t>0}$ , the exact ground state evolves smoothly between these limits; the marked ratios are reference values, not phase boundaries. The right-hand labels expand the two nondegenerate singlets explicitly: $\begin{aligned} |E=-2R\rangle&=\frac{|D_+\rangle+|S\rangle}{\sqrt2} =\frac{|1100\rangle+|0011\rangle+|1001\rangle-|0110\rangle}{2},\\ |E=+2R\rangle&=\frac{|D_+\rangle-|S\rangle}{\sqrt2} =\frac{|1100\rangle+|0011\rangle-|1001\rangle+|0110\rangle}{2}. \end{aligned}$ The upper state fills the antibonding orbital. The middle level at ${E=0}$ is spanned by the four orthonormal states ${|1010\rangle}$ , ${|0101\rangle}$ , ${(|1001\rangle+|0110\rangle)/\sqrt2}$ , and ${(|1100\rangle-|0011\rangle)/\sqrt2}$ . Thus the side panels label the states behind each endpoint level, with the same vertical energy scale as the curves. Having connected energies to charge occupation and coherence, we now construct density, spin, and current operators for further readouts.

Constructing density, spin, and current from fields

The Hamiltonian and its invariant sectors are now fixed, but a measurable question still requires a specified observable. For a finite orthonormal set of one-particle orbitals ${\phi_i:\Omega\to\mathbb C}$ , where ${\Omega\subseteq\mathbb R^d}$ , ${d\in\{1,2,3\}}$ , and ${\sigma\in\{\uparrow,\downarrow\}}$ , define the fermion field and observables at the reference time ${\tau=0}$

(8.20)

$\hat\psi_\sigma(\mathbf r) =\sum_i\phi_i(\mathbf r)\hat c_{i\sigma}, \qquad \hat n(\mathbf r) =\sum_\sigma \hat\psi_\sigma^\dagger(\mathbf r) \hat\psi_\sigma(\mathbf r),$

$\hat{\mathbf S}_i =\frac{\hbar}{2}\sum_{\alpha\beta} \hat c_{i\alpha}^\dagger {(\boldsymbol\sigma)}_{\alpha\beta} \hat c_{i\beta} =\frac{\hbar}{2} \begin{pmatrix} \hat c_{i\uparrow}^\dagger\hat c_{i\downarrow} +\hat c_{i\downarrow}^\dagger\hat c_{i\uparrow}\\[3pt] -\mathrm i\hat c_{i\uparrow}^\dagger\hat c_{i\downarrow} +\mathrm i\hat c_{i\downarrow}^\dagger\hat c_{i\uparrow}\\[3pt] \hat c_{i\uparrow}^\dagger\hat c_{i\uparrow} -\hat c_{i\downarrow}^\dagger\hat c_{i\downarrow} \end{pmatrix}.$

Here ${i}$ labels a normalized spatial orbital ${\phi_i(\mathbf r)}$ , and ${(i,\sigma)}$ labels a spin orbital; in the dimer the spatial orbitals are localized on the two sites. The operator ${\hat n(\mathbf r)}$ is number density, with units ${\mathrm{length}^{-d}}$ , while ${\hat n_i}$ counts particles in orbital ${i}$ and is dimensionless. Orthonormality gives ${\int_\Omega\mathrm d^d r\,\hat n(\mathbf r)=\hat N}$ . In the finite basis, for each ${\mathbf r\in\Omega}$ , ${\hat\psi_\sigma(\mathbf r)}$ is an operator-valued function of position. If the basis is completed to a continuum field, evaluation at a point is no longer an ordinary bounded operator and ${\hat\psi_\sigma}$ is interpreted as an operator-valued distribution acting after smearing with test functions, that is, integrating the field against a smooth spatial test function. Neither object is a section in the geometric sense used later for Bloch eigenspaces. Naming the kind of object prevents illegal pointwise products from being hidden by familiar notation.

Here ${\boldsymbol\sigma=(\sigma^x,\sigma^y,\sigma^z)}$ is the tuple of unhatted ${2\times2}$ Pauli coefficient matrices in the ordered spin basis ${(|\uparrow\rangle,|\downarrow\rangle)}$ . The bold upright ${\mathbf r}$ is an ordinary position vector; the hatted bold symbol denotes an operator tuple. The site-spin operator ${\hat{\mathbf S}_i}$ has angular-momentum units; its definition includes ${\hbar}$ explicitly. The subscript on the field labels an up or down spin, whereas the bold Pauli symbol denotes three matrices. Substituting the Pauli matrices gives the displayed column, whose rows are the x, y, and z components. Each Pauli matrix has two nonzero entries, giving six terms in total. The first two components flip spin; the third measures the difference between up-spin and down-spin populations.

Verify the angular-momentum commutation relations directly from the second-quantized site-spin operators

$\hat S_i^a=\frac{\hbar}{2}\sum_{\alpha\beta} \hat c_{i\alpha}^\dagger{(\sigma^a)}_{\alpha\beta}\hat c_{i\beta}.$

Here ${i,j}$ label sites, ${a,b\in\{x,y,z\}}$ label spin components, and ${\alpha,\beta\in\{\uparrow,\downarrow\}}$ label the ordered spin basis. Work on the full fermionic Fock space of the finite set of spin-orbitals. The input is Lecture 7’s canonical anticommutation relations (CAR):

$\{\hat c_{i\alpha},\hat c_{j\beta}^\dagger\} =\delta_{ij}\delta_{\alpha\beta}\hat I, \qquad \{\hat c_{i\alpha},\hat c_{j\beta}\} =\{\hat c_{i\alpha}^\dagger,\hat c_{j\beta}^\dagger\}=0.$

Compute the ordinary commutator of the spin operators, including the case of different sites. No single-occupancy constraint or Hamiltonian assumption is imposed.

First use composite spin-orbital labels ${p,q,r,s}$ , so that their Kronecker delta includes both the site and spin labels. Define a bilinear by

$\hat E_{pq}:=\hat c_p^\dagger\hat c_q.$

Move the middle annihilator past the creator in each ordered product, using ${\hat c_q\hat c_r^\dagger=\delta_{qr}\hat I-\hat c_r^\dagger\hat c_q}$ :

$\begin{aligned} \hat E_{pq}\hat E_{rs} &=\delta_{qr}\hat E_{ps} -\hat c_p^\dagger\hat c_r^\dagger\hat c_q\hat c_s,\\ \hat E_{rs}\hat E_{pq} &=\delta_{sp}\hat E_{rq} -\hat c_r^\dagger\hat c_p^\dagger\hat c_s\hat c_q. \end{aligned}$

The two four-operator terms are equal: exchanging the creators gives one minus sign, and exchanging the annihilators gives a second minus sign.

$\hat c_r^\dagger\hat c_p^\dagger\hat c_s\hat c_q ={(-1)}^2\hat c_p^\dagger\hat c_r^\dagger\hat c_q\hat c_s =\hat c_p^\dagger\hat c_r^\dagger\hat c_q\hat c_s.$

Subtract the two products. The four-operator terms cancel, leaving the bilinear identity

$[\hat E_{pq},\hat E_{rs}] =\delta_{qr}\hat E_{ps}-\delta_{sp}\hat E_{rq}.$

Now restore the separate site and spin labels and apply this identity to the two bilinears appearing in the spin operators:

$\begin{aligned} &[\hat c_{i\alpha}^\dagger\hat c_{i\beta}, \hat c_{j\gamma}^\dagger\hat c_{j\delta}]\\ &\qquad=\delta_{ij}\left( \delta_{\beta\gamma}\hat c_{i\alpha}^\dagger\hat c_{j\delta} -\delta_{\delta\alpha}\hat c_{j\gamma}^\dagger\hat c_{i\beta} \right). \end{aligned}$

Multiply by the two Pauli-matrix entries and sum the spin indices. The site delta sets ${j=i}$ in the surviving terms. The first contraction forms ${\sigma^a\sigma^b}$ ; relabeling the dummy indices in the second forms ${\sigma^b\sigma^a}$ :

$\begin{aligned} [\hat S_i^a,\hat S_j^b] &=\frac{\hbar^2}{4}\delta_{ij} \sum_{\alpha\delta} \hat c_{i\alpha}^\dagger \left[{(\sigma^a\sigma^b)}_{\alpha\delta} -{(\sigma^b\sigma^a)}_{\alpha\delta}\right] \hat c_{i\delta}\\ &=\frac{\hbar^2}{4}\delta_{ij} \sum_{\alpha\delta}\hat c_{i\alpha}^\dagger {([\sigma^a,\sigma^b])}_{\alpha\delta}\hat c_{i\delta}. \end{aligned}$

Use the numerical Pauli-matrix identity ${[\sigma^a,\sigma^b]=2\mathrm i\sum_c\epsilon_{abc}\sigma^c}$ , where ${\epsilon_{xyz}=+1}$ , exchanging two indices reverses its sign, and a repeated index makes it zero. Recognizing the spin definition gives

$\begin{aligned} [\hat S_i^a,\hat S_j^b] &=\mathrm i\hbar\,\delta_{ij}\sum_c\epsilon_{abc} \left(\frac{\hbar}{2}\sum_{\alpha\delta} \hat c_{i\alpha}^\dagger{(\sigma^c)}_{\alpha\delta}\hat c_{i\delta}\right)\\ &=\mathrm i\hbar\,\delta_{ij}\sum_c\epsilon_{abc}\hat S_i^c. \end{aligned}$

Check the sign and prefactor directly against the six-term expansion. At one fixed site, introduce the two spin-flip bilinears

$\hat A_i:=\hat c_{i\uparrow}^\dagger\hat c_{i\downarrow}, \qquad \hat B_i:=\hat c_{i\downarrow}^\dagger\hat c_{i\uparrow}.$

The bilinear identity and the displayed spin components give

$\begin{gathered} [\hat A_i,\hat B_i]=\hat n_{i\uparrow}-\hat n_{i\downarrow},\\ \hat S_i^x=\frac{\hbar}{2}(\hat A_i+\hat B_i),\\ \hat S_i^y=\frac{\hbar}{2}(-\mathrm i\hat A_i+\mathrm i\hat B_i),\\ \hat S_i^z=\frac{\hbar}{2}(\hat n_{i\uparrow}-\hat n_{i\downarrow}). \end{gathered}$

The same-operator commutators vanish, and reversing the order changes the sign. Expanding the remaining two cross terms therefore gives

$\begin{aligned} [\hat S_i^x,\hat S_i^y] &=\frac{\hbar^2}{4} \left(\mathrm i[\hat A_i,\hat B_i]-\mathrm i[\hat B_i,\hat A_i]\right)\\ &=\frac{\mathrm i\hbar^2}{2} (\hat n_{i\uparrow}-\hat n_{i\downarrow}) =\mathrm i\hbar\hat S_i^z. \end{aligned}$

Thus the cyclic commutators have the usual angular-momentum form; components at different sites commute because of ${\delta_{ij}}$ . These are operator identities on the full finite Fock space, including empty and doubly occupied sites, on which the local spin vanishes. Both sides have units of angular momentum squared. This algebra alone does not establish spin conservation: that requires a separate commutator with the chosen Hamiltonian. Return to the field and spin definitions with the same basis and physical-spin convention.

We use the Heisenberg picture for dynamics, with reference time ${\tau=0}$ . The normalized state ${|\Psi_0\rangle}$ prepared then stays fixed, while the operators evolve. The unitary propagator ${\hat U(\tau,0)}$ , abbreviated ${\hat U(\tau)}$ , obeys ${\hat U(0)=\hat I}$ . For a Schrödinger-picture operator ${\hat A_{\mathrm S}(\tau)}$ , which may have explicit time dependence, the corresponding operator, physical expectation and equation of motion are

(8.21)

$\begin{aligned} \hat A_{\mathrm H}(\tau)&:=\hat U^\dagger(\tau)\hat A_{\mathrm S}(\tau)\hat U(\tau),\\ \langle A\rangle(\tau)&:=\langle\Psi_0|\hat A_{\mathrm H}(\tau)|\Psi_0\rangle,\\ \frac{\mathrm d\hat A_{\mathrm H}}{\mathrm d\tau}& =\frac{\mathrm i}{\hbar} [\hat H_{\mathrm H},\hat A_{\mathrm H}] +\left(\frac{\partial\hat A}{\partial\tau}\right)_{\mathrm H}. \end{aligned}$

Untimed operators in the field, spin and current definitions denote their reference-time values, ${\hat A=\hat A_{\mathrm H}(0)}$ . Thus ${\langle A\rangle(0)=\langle\Psi_0|\hat A|\Psi_0\rangle}$ ; at a later time, keep the same preparation and use ${\hat A_{\mathrm H}(\tau)}$ . Fixing the state does not make its expectation values constant. If ${\hat A}$ has no explicit time dependence and ${[\hat H,\hat A]=0}$ , its Heisenberg representative is constant.

To define a two-time correlation, we must also specify the prepared state and the order of the two operators.

For the continuum comparison, specify nonrelativistic particles of mass ${m>0}$ and charge ${-e}$ , with kinetic energy ${-\hbar^2\nabla^2/(2m)}$ , a real local scalar potential ${V(\mathbf r)}$ , and no vector potential. In the following continuum equations, all fields and densities are Heisenberg operators at the same time ${\tau}$ ; we suppress the subscript H and common time argument. The one-body field equation is

(8.22)

$\mathrm i\hbar\,\partial_\tau\hat\psi_\sigma =\left[-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf r)\right] \hat\psi_\sigma.$

Differentiate the density and use this equation and its adjoint. The two local-potential terms cancel, leaving

(8.23)

$\begin{aligned} \partial_\tau\hat n &=\frac{\mathrm i\hbar}{2m}\sum_\sigma \left[\hat\psi_\sigma^\dagger\nabla^2\hat\psi_\sigma -(\nabla^2\hat\psi_\sigma^\dagger)\hat\psi_\sigma\right]\\ &=\frac{\mathrm i\hbar}{2m}\nabla\cdot\sum_\sigma \left[\hat\psi_\sigma^\dagger\nabla\hat\psi_\sigma -(\nabla\hat\psi_\sigma^\dagger)\hat\psi_\sigma\right]. \end{aligned}$

Reading this as ${\partial_\tau(-e\hat n)+\nabla\cdot \hat{\mathbf j}_{\mathrm{ch}}=0}$ identifies the usual current for this kinetic Hamiltonian (with no added divergence-free term):

$\hat{\mathbf j}_{\mathrm{ch}}(\mathbf r) =-\frac{e\hbar}{2m\mathrm i}\sum_\sigma \left[ \hat\psi_\sigma^\dagger\nabla\hat\psi_\sigma -(\nabla\hat\psi_\sigma^\dagger)\hat\psi_\sigma \right].$

The derivation is understood on smooth finite-particle wave functions, with field products defined through that action or smearing. A nonlocal potential or different kinetic Hamiltonian requires its own continuity calculation; absence of a vector potential alone does not guarantee this formula. On a lattice, the current across bond ${i\to j}$ follows by differentiating the Hamiltonian with respect to a Peierls phase or by evaluating the change of local density.

Find the particle current from site ${i}$ to ${j}$ , then evaluate it at ${\tau=0}$ for the one-particle state prepared at that time ${|\psi_\phi\rangle=(|i\rangle+\mathrm e^{\mathrm i\phi}|j\rangle)/\sqrt2}$ . Let real ${t>0}$ have energy units and take one hopping bond ${\hat H_{ij}=-t(\hat c_i^\dagger\hat c_j+ \hat c_j^\dagger\hat c_i)}$ . Relate the number current to the charge current for electrons of charge ${-e}$ .

The continuity route determines the operator first and leaves the preparation for a separate expectation, with elapsed time ${\tau}$ and hopping energy ${t}$ . The evolution rule follows from ${\hat U(\tau)=\mathrm e^{-\mathrm i\hat H_{ij}\tau/\hbar}}$ : differentiating ${\hat U^\dagger(\tau)\hat n_i\hat U(\tau)}$ gives ${(\mathrm i/\hbar)\hat U^\dagger(\tau)[\hat H_{ij},\hat n_i]\hat U(\tau)}$ . At ${\tau=0}$ , where ${\hat c_k=\hat c_{k,\mathrm H}(0)}$ , the CAR give ${[\hat c_i^\dagger\hat c_j,\hat n_i]=-\hat c_i^\dagger\hat c_j}$ and the opposite sign for its adjoint. Thus

$\left.\frac{\mathrm d\hat n_{i,\mathrm H}(\tau)}{\mathrm d\tau}\right|_{\tau=0} =\frac{\mathrm i}{\hbar}[\hat H_{ij},\hat n_i] =\frac{\mathrm i t}{\hbar} (\hat c_i^\dagger\hat c_j-\hat c_j^\dagger\hat c_i) =-\hat J^{(N)}_{i\to j}.$

Thus the particle-number and charge currents are

$\hat J^{(N)}_{i\to j} =\frac{\mathrm i t}{\hbar} (\hat c_j^\dagger\hat c_i-\hat c_i^\dagger\hat c_j), \qquad \hat J^{(\mathrm{ch})}_{i\to j}=-e\hat J^{(N)}_{i\to j}.$

Exchanging ${i}$ and ${j}$ reverses the current, and adding the equations for ${\dot{\hat n}_i}$ and ${\dot{\hat n}_j}$ gives zero. These are respectively the orientation and local-conservation checks.

At ${\tau=0}$ , taking expectations in the fixed preparation ${|\psi_\phi\rangle}$ gives ${\langle\hat c_j^\dagger\hat c_i\rangle=\mathrm e^{-\mathrm i\phi}/2}$ and ${\langle\hat c_i^\dagger\hat c_j\rangle=\mathrm e^{\mathrm i\phi}/2}$ . Thus ${\langle\psi_\phi|\hat J^{(N)}_{i\to j,\mathrm H}(0)|\psi_\phi\rangle=(t/\hbar)\sin\phi}$ . At ${\phi=0}$ the bonding eigenstate has zero current; at ${\phi=\pi/2}$ , ${\left.\mathrm d\langle\hat n_{i,\mathrm H}(\tau)\rangle/\mathrm d\tau\right|_{\tau=0}=-t/\hbar}$ . The latter also follows by applying the two-site Schrödinger equation to its amplitudes. Number current has units inverse time and charge current acquires ${-e}$ ; at later times use the evolved current operator in the same fixed preparation, while the operator conservation identity holds at every time. A pairing source would change that identity, as the criterion below makes explicit.

The phase-state calculation separates population from transport. At ${\tau=0}$ in the same preparation ${|\psi_\phi\rangle}$ , both mean site populations are ${1/2}$ for every ${\phi}$ , whereas ${\langle\psi_\phi|\hat J^{(N)}_{i\to j,\mathrm H}(0)|\psi_\phi\rangle=(t/\hbar)\sin\phi}$ . The phase is relative to the fixed real hopping convention, not an independently observable gauge choice. These are initial-time expectations at ${\tau=0}$ , not a claim of a stationary current.

Figure 8.4. At fixed real hopping and fixed site populations, relative phase changes the magnitude and direction of the initial particle current at tau=0.

For the number-conserving hopping Hamiltonian

$\hat H_t=-\sum_{\langle ij\rangle}t_{ij} (\hat c_i^\dagger\hat c_j+\hat c_j^\dagger\hat c_i), \qquad t_{ij}=t_{ji}\in\mathbb R,$

define a current for every ordered neighboring pair ${(i,j)}$ by

$\hat J^{(N)}_{i\to j} :=\frac{\mathrm i t_{ij}}{\hbar} (\hat c_j^\dagger\hat c_i-\hat c_i^\dagger\hat c_j).$

It obeys ${\hat J^{(N)}_{j\to i}=-\hat J^{(N)}_{i\to j}}$ . The Heisenberg equation therefore gives the operator identity at every time

$\frac{\mathrm d\hat n_{i,\mathrm H}(\tau)}{\mathrm d\tau} +\sum_{j\sim i}\hat J^{(N)}_{i\to j,\mathrm H}(\tau)=0, \qquad \frac{\mathrm d\hat N_{\mathrm H}(\tau)}{\mathrm d\tau}=0,$

where the sum contains the ordered currents leaving ${i}$ . A single global orientation for each undirected edge may instead be used as incidence-matrix bookkeeping, provided its sign is reversed at the edge’s other endpoint. If a pairing term ${\hat H_\Delta}$ is added, the right-hand side becomes the local source ${\hat{\mathcal S}_i=(\mathrm i/\hbar)[\hat H_\Delta,\hat n_i]}$ , and ${\sum_i\hat{\mathcal S}_i=(\mathrm i/\hbar)[\hat H_\Delta,\hat N]}$ is generally nonzero. This scalar source has units of inverse time; it is distinct from the spin angular momentum ${\hat{\mathbf S}_i}$ . Hermiticity does not remove this source; at time ${\tau}$ its Heisenberg representative ${\hat{\mathcal S}_{i,\mathrm H}(\tau)=\hat U^\dagger(\tau)\hat{\mathcal S}_i\hat U(\tau)}$ appears on the right-hand side.

Before reading Figure 8.5, choose an orientation for both bonds touching site ${j}$ . Predict which terms cancel when the equation is summed over a periodic chain and which term cannot cancel pairwise.

Figure 8.5. A local density changes by incoming current minus outgoing current plus a source; internal bond currents cancel pairwise in the periodic-system sum.

The arrows make orientation part of the equation rather than an afterthought. Every internal bond enters the global sum once with each sign, whereas a true source ${\hat{\mathcal S}_j}$ survives unless the source operators themselves sum to zero. This is why Hermiticity alone cannot imply number conservation.

The same grammar constructs couplings. A local electron–phonon model may be written

$\hat H_{\mathrm{e-ph}} =g\sum_i\hat n_i(\hat a_i+\hat a_i^\dagger).$

Here ${g}$ is a real coupling energy and ${\hat n_i=\sum_\sigma\hat n_{i\sigma}}$ counts electrons on site ${i}$ . It conserves electron number but not phonon occupation. This does not contradict Lecture 6: equilibrium phonon number was unconstrained. The model also shows why “number conservation” must name which number.

Dynamics turns operator structure into a measurable correlation

Mean site populations ${\langle\hat n_1\rangle}$ and ${\langle\hat n_2\rangle}$ describe the sites separately. The equal-time density correlation ${\langle\hat n_1\hat n_2\rangle}$ also records joint occupation. Comparing it with ${\langle\hat n_1\rangle\langle\hat n_2\rangle}$ reveals whether the occupations tend to fluctuate together or oppositely: the difference is their covariance. These expectations use the same preparation at the reference time ${\tau=0}$ . Correlations can therefore distinguish preparations with identical mean densities, answering the question posed at the start of the lecture.

To ask how this information evolves, average an ordered product with one observable at time ${\tau}$ and the other at ${0}$ , keeping the Heisenberg preparation fixed. A two-time correlation retains information that even a constant mean can miss. For a ground-state preparation and a time-independent Hamiltonian, its oscillation frequencies reveal excitation energies, while its weights show which excitations the chosen observable accesses. We will derive this statement after defining the correlation precisely; no thermal ensemble or applied perturbation is needed.

Derive the Heisenberg equation, including the term for explicit time dependence, from Schrödinger evolution. Work on the finite fermionic Fock space used in this lecture. Let ${\tau}$ denote time, choose the reference time ${\tau=0}$ , and let ${\hat H_{\mathrm S}(\tau)}$ be a continuous self-adjoint Hamiltonian. The subscript S denotes the Schrödinger picture.

The unitary propagator ${\hat U(\tau,0)}$ , abbreviated ${\hat U(\tau)}$ , obeys

$\mathrm i\hbar\frac{\mathrm d\hat U}{\mathrm d\tau} =\hat H_{\mathrm S}(\tau)\hat U, \qquad \hat U(0)=\hat I, \qquad \hat U^\dagger\hat U=\hat U\hat U^\dagger=\hat I.$

Allow a differentiable operator ${\hat A_{\mathrm S}(\tau)}$ to have explicit time dependence. Keep the prepared vector ${|\Psi_0\rangle}$ arbitrary and normalized. After deriving the general equation, specialize to a constant Hamiltonian and check the evolution of a single fermionic annihilation operator.

In the Schrödinger picture the state evolves as ${|\Psi_{\mathrm S}(\tau)\rangle=\hat U(\tau)|\Psi_0\rangle}$ . Move the two propagators from the state vectors onto the operator in an expectation value:

$\begin{aligned} \langle\Psi_{\mathrm S}(\tau)|\hat A_{\mathrm S}(\tau)|\Psi_{\mathrm S}(\tau)\rangle &=\langle\Psi_0|\hat U^\dagger\hat A_{\mathrm S}(\tau)\hat U|\Psi_0\rangle\\ &=\langle\Psi_0|\hat A_{\mathrm H}(\tau)|\Psi_0\rangle. \end{aligned}$

Thus the same physical expectation is represented by a fixed state and the Heisenberg operator. Define the Hamiltonian in that picture in the same way:

$\hat A_{\mathrm H}(\tau):=\hat U^\dagger\hat A_{\mathrm S}(\tau)\hat U, \qquad \hat H_{\mathrm H}(\tau):=\hat U^\dagger\hat H_{\mathrm S}(\tau)\hat U.$

First find the derivatives of both propagators. Taking the adjoint reverses the operator order and changes the sign of ${\mathrm i}$ :

$\frac{\mathrm d\hat U}{\mathrm d\tau} =-\frac{\mathrm i}{\hbar}\hat H_{\mathrm S}\hat U, \qquad \frac{\mathrm d\hat U^\dagger}{\mathrm d\tau} =\frac{\mathrm i}{\hbar}\hat U^\dagger\hat H_{\mathrm S}.$

The product rule has three terms. The middle one differentiates only the explicit dependence of ${\hat A_{\mathrm S}(\tau)}$ ; the other two come from the change of picture:

$\begin{aligned} \frac{\mathrm d\hat A_{\mathrm H}}{\mathrm d\tau} &=\frac{\mathrm d\hat U^\dagger}{\mathrm d\tau}\hat A_{\mathrm S}\hat U +\hat U^\dagger\frac{\partial\hat A_{\mathrm S}}{\partial\tau}\hat U +\hat U^\dagger\hat A_{\mathrm S}\frac{\mathrm d\hat U}{\mathrm d\tau}\\ &=\frac{\mathrm i}{\hbar}\hat U^\dagger\hat H_{\mathrm S}\hat A_{\mathrm S}\hat U +\hat U^\dagger\frac{\partial\hat A_{\mathrm S}}{\partial\tau}\hat U -\frac{\mathrm i}{\hbar}\hat U^\dagger\hat A_{\mathrm S}\hat H_{\mathrm S}\hat U\\ &=\frac{\mathrm i}{\hbar}\hat U^\dagger [\hat H_{\mathrm S},\hat A_{\mathrm S}]\hat U +\hat U^\dagger\frac{\partial\hat A_{\mathrm S}}{\partial\tau}\hat U. \end{aligned}$

To express the first term entirely in the Heisenberg picture, insert ${\hat U\hat U^\dagger=\hat I}$ between the two operators in each product:

$\begin{aligned} \hat U^\dagger[\hat H_{\mathrm S},\hat A_{\mathrm S}]\hat U &=(\hat U^\dagger\hat H_{\mathrm S}\hat U) (\hat U^\dagger\hat A_{\mathrm S}\hat U)\\ &\quad-(\hat U^\dagger\hat A_{\mathrm S}\hat U) (\hat U^\dagger\hat H_{\mathrm S}\hat U)\\ &=[\hat H_{\mathrm H},\hat A_{\mathrm H}]. \end{aligned}$

Combining the terms gives the displayed equation, with the explicit derivative defined by conjugating the Schrödinger-picture derivative:

$\begin{aligned} \frac{\mathrm d\hat A_{\mathrm H}}{\mathrm d\tau} &=\frac{\mathrm i}{\hbar}[\hat H_{\mathrm H},\hat A_{\mathrm H}] +\left(\frac{\partial\hat A}{\partial\tau}\right)_{\mathrm H},\\ \left(\frac{\partial\hat A}{\partial\tau}\right)_{\mathrm H} &:=\hat U^\dagger\frac{\partial\hat A_{\mathrm S}}{\partial\tau}\hat U. \end{aligned}$

The last term is not the total derivative of ${\hat A_{\mathrm H}}$ . For example, if ${\hat A_{\mathrm S}(\tau)=f(\tau)\hat B}$ with fixed ${\hat B}$ , that term is ${f'(\tau)\hat B_{\mathrm H}(\tau)}$ , even when ${\hat B}$ commutes with the Hamiltonian.

For a time-independent Hamiltonian ${\hat H}$ , the propagator equation has the exponential solution. Since ${\hat U}$ then commutes with ${\hat H}$ , ${\hat H_{\mathrm H}=\hat H}$ , and

$\hat U(\tau)=\mathrm e^{-\mathrm i\hat H\tau/\hbar}, \qquad \hat A_{\mathrm H}(\tau) =\mathrm e^{\mathrm i\hat H\tau/\hbar} \hat A_{\mathrm S}(\tau) \mathrm e^{-\mathrm i\hat H\tau/\hbar}.$

As a sign check, take one fermionic orbital with ${\hat H=\varepsilon\hat n}$ , ${\hat n=\hat c^\dagger\hat c}$ , and real energy ${\varepsilon}$ . The Lecture 7 identity ${[\hat n,\hat c]=-\hat c}$ gives

$\frac{\mathrm d\hat c_{\mathrm H}}{\mathrm d\tau} =-\frac{\mathrm i\varepsilon}{\hbar}\hat c_{\mathrm H}, \qquad \hat c_{\mathrm H}(\tau) =\mathrm e^{-\mathrm i\varepsilon\tau/\hbar}\hat c, \qquad \hat n_{\mathrm H}(\tau)=\hat n.$

In the basis ${(|0\rangle,|1\rangle)}$ , this can also be checked directly: ${\hat U=\operatorname{diag}(1,\mathrm e^{-\mathrm i\varepsilon\tau/\hbar})}$ and ${\hat c=|0\rangle\langle1|}$ , so multiplication gives exactly the same phase. The frequency ${\varepsilon/\hbar}$ has units inverse time.

The general product-rule derivation also allows a time-dependent Hamiltonian. Do not replace its propagator by ${\mathrm e^{-\mathrm i\hat H(\tau)\tau/\hbar}}$ ; use the propagator equation given at the start. In an infinite-dimensional setting, the differentiated products require a common invariant domain and sufficient differentiability; the finite setting here avoids that issue.

Return to the operator equation: with no explicit time dependence, commutation with the Hamiltonian makes the operator constant. A prepared state is still needed for an expectation value, and does not enter the operator evolution law.

Fix a Hamiltonian ${\hat H}$ , Heisenberg operators ${\hat A_{\mathrm H}(\tau)}$ and ${\hat B_{\mathrm H}(\tau)}$ , and a normalized preparation represented either by a vector ${|\Psi\rangle}$ or by a positive trace-class operator ${\hat\rho}$ with ${\operatorname{Tr}\hat\rho=1}$ . This preparation is specified at the reference time and held fixed in the Heisenberg picture; the displayed operator arguments identify the measurement times. The ordered two-time correlation is

$C_{AB}^{\Psi}:\mathbb R\to\mathbb C, \qquad C_{AB}^{\Psi}(\tau) :=\langle\Psi| \hat A_{\mathrm H}(\tau)\hat B_{\mathrm H}(0) |\Psi\rangle,$

or, for the density-operator preparation,

$C_{AB}^{\rho}(\tau) :=\operatorname{Tr}\!\left[ \hat\rho\,\hat A_{\mathrm H}(\tau)\hat B_{\mathrm H}(0) \right].$

For a finite system and bounded ${\hat A,\hat B}$ , this is an ordinary complex-valued function of ${\tau}$ . For unbounded observables the preparation must lie in a common domain making the displayed expectation finite. An infinite-volume or infinite-time spectral limit may exist only as a tempered distribution in ${\mathcal S'(\mathbb R)}$ , to be paired with a test function. Neither the function nor that distribution is a response function: response additionally requires a declared perturbing source, readout, and causal variation.

The density-operator form above does not assume a thermal ensemble; it is a general preparation. Lectures 9–10 will specialize it to thermal states and separate fluctuation from causal response. Here the point is algebraic: once density, spin, or current has been written in terms of creation and annihilation operators, the Hamiltonian determines its evolution. The correlation also depends on the chosen operator and preparation, as the spectral weights below make explicit.

For a time-independent Hamiltonian with a complete orthonormal energy eigenbasis ${\{|m\rangle\}}$ with energies ${E_m}$ , a stationary eigenstate ${|g\rangle}$ of energy ${E_g}$ , and a bounded self-adjoint operator ${\hat A=\hat A^\dagger}$ , insert ${\hat I=\sum_m|m\rangle\langle m|}$ between the two observables:

(8.24)

$\begin{aligned} C_{AA}^{g}(\tau) &=\langle g|\mathrm e^{\mathrm i\hat H\tau/\hbar} \hat A\,\mathrm e^{-\mathrm i\hat H\tau/\hbar}\hat A|g\rangle\\ &=\sum_m\mathrm e^{\mathrm iE_g\tau/\hbar} \langle g|\hat A|m\rangle \mathrm e^{-\mathrm iE_m\tau/\hbar}\langle m|\hat A|g\rangle. \end{aligned}$

Since the observable is self-adjoint, the two matrix elements are complex conjugates. Their product is therefore a nonnegative modulus squared:

(8.25)

$C_{AA}^{g}(\tau) =\sum_m \mathrm e^{-\mathrm i(E_m-E_g)\tau/\hbar} {|\langle m|\hat A|g\rangle|}^2.$

A stationary preparation has constant ${\langle\hat A_{\mathrm H}(\tau)\rangle}$ , but ${\hat A|g\rangle}$ can contain other energy eigenstates. Their relative phases make the ordered two-time correlation vary. It may be complex even for Hermitian ${\hat A}$ ; it is not the product of two mean measurements.

For a nondegenerate ground state, write ${\delta\hat A=\hat A-\langle g|\hat A|g\rangle\hat I}$ . The connected correlation removes the elastic ${m=g}$ term, the zero-frequency contribution from returning to the same state:

(8.26)

$\begin{aligned} C_{\delta A\delta A}^{g}(\tau) &=\sum_{m\ne g}\mathrm e^{-\mathrm i(E_m-E_g)\tau/\hbar} {|\langle m|\hat A|g\rangle|}^2,\\ S_{\delta A\delta A}^{g}(\omega) &=2\pi\sum_{m\ne g}{|\langle m|\hat A|g\rangle|}^2 \delta\!\left(\omega-\frac{E_m-E_g}{\hbar}\right). \end{aligned}$

Here the Fourier factor is ${\mathrm e^{+\mathrm i\omega\tau}}$ ; the infinite-time transform is distributional. Integration over ${\mathrm d\omega/(2\pi)}$ gives ${C_{\delta A\delta A}^{g}(0)={\langle{(\delta\hat A)}^2\rangle}_g}$ . Degenerate ground states can leave additional zero-frequency transitions even after subtracting the mean. No response or Gaussian assumption enters this energy-basis insertion.

The frequencies are transition-energy differences selected by matrix elements. Thus an energy spectrum alone does not determine a measured correlation; the operator and reference state decide which differences carry weight. The finite dimers used here satisfy the complete-eigenbasis hypothesis. With continuous spectrum, the sum is replaced by the corresponding spectral-measure integral and may define a distributional spectrum. This returns to Lecture 6’s probe boundary in operator language.

Determine the ordered density-imbalance correlation of a one-electron dimer in its ground state, then change only the readout to total density. Write the positive hopping scale as ${t_{\mathrm h}:=t>0}$ , let ${|+\rangle=(|1\rangle+|2\rangle)/\sqrt2}$ be the ground state with energy ${-t_{\mathrm h}}$ , with antibonding state ${|-\rangle=(|1\rangle-|2\rangle)/\sqrt2}$ at ${+t_{\mathrm h}}$ . Define ${\hat\Delta=\hat n_1-\hat n_2}$ and calculate ${\langle+|\hat\Delta_{\mathrm H}(\tau)\hat\Delta|+\rangle}$ , including its spectrum with Fourier factor ${\mathrm e^{+\mathrm i\omega\tau}}$ .

The spectral insertion above selects transitions by their matrix elements. Since ${\hat\Delta|+\rangle=|-\rangle}$ and the antibonding state has energy ${+t_{\mathrm h}}$ , the only nonzero spectral term gives

$C_{\Delta\Delta}(\tau) =\langle+|\hat\Delta_{\mathrm H}(\tau)\hat\Delta|+\rangle =\mathrm e^{-2\mathrm i t_{\mathrm h}\tau/\hbar}.$

At ${\tau=0}$ , the result is ${\langle+|\hat\Delta^2|+\rangle=1}$ ; its Fourier spectrum contains one line at ${2t_{\mathrm h}/\hbar}$ . Changing the probe to total density ${\hat n_1+\hat n_2}$ instead gives a constant correlation because that operator commutes with the Hamiltonian. In this fixed-one-electron sector, ${C_{NN}(\tau)=1}$ and ${S_{NN}(\omega)=2\pi\delta(\omega)}$ . The changed operator, not the energy levels alone, decides which transition is visible.

The time argument enters only through the dimensionless phase ${2t_{\mathrm h}\tau/\hbar}$ , and ${C_{\Delta\Delta}(-\tau)={C_{\Delta\Delta}(\tau)}^*}$ , as required for a Hermitian observable evaluated in a stationary state. Use the Fourier convention with a positive exponential sign: $S_{\Delta\Delta}(\omega) =\int_{-\infty}^{\infty}\mathrm d\tau\,\mathrm e^{+\mathrm i\omega\tau}C_{\Delta\Delta}(\tau) =2\pi\delta(\omega-2t_{\mathrm h}/\hbar).$ The ideal spectrum is a distribution. Its integrated weight divided by ${2\pi}$ is one, agreeing with the equal-time value independently of line location. No causal step function belongs in this ordered correlation. The next example returns to the two-electron sector of Example 8.3 and compares two preparations using joint site densities at equal time. Both the sector and the observable change; the one-electron transition line is not carried over.

Equal-time density correlation in the dimer

Fix a normalized state ${|\Psi_0\rangle}$ on the finite dimer Fock space, and write ${\langle\hat A\rangle:=\langle\Psi_0|\hat A|\Psi_0\rangle}$ . Equal-time means that both operators are evaluated at the same time; here choose ${\tau=0}$ , so ${\hat n_i=\hat n_{i\mathrm H}(0)}$ . Remove each site’s mean to define its fluctuation operator:

$\delta\hat n_i:=\hat n_i-\langle\hat n_i\rangle\hat I, \qquad \langle\delta\hat n_i\rangle=0.$

For the commuting site-number operators, the connected equal-time density correlation is the real, dimensionless covariance

$\operatorname{Cov}(\hat n_i,\hat n_j) :=\langle\delta\hat n_i\,\delta\hat n_j\rangle.$

Expanding the product explains the subtraction. The means are scalar numbers, and normalization gives ${\langle\hat I\rangle=1}$ , so

(8.27)

$\begin{aligned} \langle\delta\hat n_i\,\delta\hat n_j\rangle &=\langle\hat n_i\hat n_j\rangle -\langle\hat n_i\rangle\langle\hat n_j\rangle -\langle\hat n_i\rangle\langle\hat n_j\rangle +\langle\hat n_i\rangle\langle\hat n_j\rangle\\ &=\langle\hat n_i\hat n_j\rangle -\langle\hat n_i\rangle\langle\hat n_j\rangle. \end{aligned}$

“Connected” removes the contribution from the separate mean populations and retains their joint fluctuations. Positive covariance means that occupations tend to deviate from their means in the same direction; negative covariance means opposite directions. Statistically independent joint number outcomes give zero covariance, but zero covariance alone proves neither independence nor absence of entanglement. For one site, the same construction is the variance: ${\operatorname{Var}(\hat n_i):=\langle{(\delta\hat n_i)}^2\rangle}$ .

At another common time, replace both number operators by their Heisenberg operators at that time and subtract their corresponding means, still evaluated in the fixed preparation. Equal-time does not require a stationary state. The following example compares the supplied preparations at ${\tau=0}$ .

Compare joint density measurements in two supplied states. Calculate their mean site occupations and connected equal-time density correlations. Take the normalized covalent singlet

$|S\rangle =\frac{|1001\rangle-|0110\rangle}{\sqrt2}.$ Compare it with the symmetric double-occupancy state $|D_+\rangle =\frac{|1100\rangle+|0011\rangle}{\sqrt2}.$ Retain the mode order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ and the total site densities ${\hat n_i=\hat n_{i\uparrow}+\hat n_{i\downarrow}}$ .

Evaluate the diagonal occupation operators in each component before subtracting the product of the means; covariance and variance below refer to the supplied state.

Each component of ${|S\rangle}$ has one electron on each site. Therefore

The connected equal-time correlation is

$\langle\hat n_1\hat n_2\rangle -\langle\hat n_1\rangle\langle\hat n_2\rangle=0.$

The supplied ${|D_+\rangle}$ still has ${\langle\hat n_1\rangle =\langle\hat n_2\rangle=1}$ , but ${\langle\hat n_1\hat n_2\rangle=0}$ , so the connected correlation is ${-1}$ . Equal one-site means do not imply equal fluctuations or joint occupations. Correlations contain information that one-point densities lose.

The two states can therefore be distinguished by repeated joint charge measurements. A sensor that reports only the mean charge on each dot gives the same result for ${|S\rangle}$ and ${|D_+\rangle}$ . For ${\hat Q_i=-e\hat n_i}$ , the mean site charges are ${-e}$ and the charge covariance is ${e^2\operatorname{Cov}(\hat n_1,\hat n_2)}$ . Repeated joint charge measurements distinguish them because their double- and cross-occupation probabilities differ. Spin-resolved measurements can also distinguish the states, even though each has zero mean spin on either site. With the physical spin operator ${\hat S_i^z=\hbar(\hat n_{i\uparrow}-\hat n_{i\downarrow})/2}$ , each component of ${|S\rangle}$ has opposite site spins. Thus ${{\langle\hat S_1^z\rangle}_S={\langle\hat S_2^z\rangle}_S=0}$ but ${{\langle\hat S_1^z\hat S_2^z\rangle}_S=-\hbar^2/4}$ . In ${|D_+\rangle}$ , every occupied site has an up/down pair, so both local means and ${{\langle\hat S_1^z\hat S_2^z\rangle}_{D_+}}$ vanish. These spin correlations have units of ${\hbar^2}$ . Zero local magnetization therefore does not imply zero spin correlation.

The Hubbard ground-state singlet mixes ${|S\rangle}$ and ${|D_+\rangle}$ when ${t>0}$ . Write ${|g\rangle=a|D_+\rangle+b|S\rangle}$ , with ${|a|^2+|b|^2=1}$ , and recall the double occupancy ${D=|a|^2}$ calculated above. Density operators are diagonal in the occupation basis, so cross terms vanish: ${{\langle\hat n_1\rangle}_g={\langle\hat n_2\rangle}_g=1}$ and ${{\langle\hat n_1\hat n_2\rangle}_g=|b|^2=1-D}$ . Consequently ${\operatorname{Cov}_g(\hat n_1,\hat n_2)=-D}$ . At ${U=0}$ this is ${-1/2}$ ; for ${U/t\to+\infty}$ it tends to zero. Increasing repulsion suppresses charge fluctuations. Lectures 15–16 develop the many-site Mott and superexchange physics; these equalities describe the isolated dimer.

We can check the calculated covariances using total-number conservation and exchange of the two sites. Both normalized states have exact total number two. Consequently ${\operatorname{Cov}(\hat n_1,\hat n_2)=-\operatorname{Var}(\hat n_1)}$ : the covalent state has variance zero, while the double-occupancy state has variance one, recovering the two calculated covariances. Interchanging sites also leaves each result unchanged. These are statements about the specified preparations; neither endpoint is the general finite- ${t,U}$ ground state. The comparison uses direct expectations in the two supplied states, without a factorization assumption.

The joint number distribution makes the equal-density comparison visible. In ${|S\rangle}$ , the only site-number outcome is ${(n_1,n_2)=(1,1)}$ . In ${|D_+\rangle}$ , the outcomes ${(2,0)}$ and ${(0,2)}$ each have probability ${1/2}$ . Thus both means equal one, but the mean product is one or zero, giving connected correlations zero or minus one. The diagram shows probabilities of number measurements; it omits quantum phases and does not identify either coherent state with an incoherent mixture.

Figure 8.6. Equal site means conceal different joint outcomes: the singlet has zero number covariance, whereas the symmetric doublon state has covariance minus one.

What a doublon measurement counts

Which part of that correlation can an experiment access? In the optical-lattice experiment of Jördens and colleagues, tunneling was first frozen by deepening the lattice. An interaction-dependent energy shift then let a radio-frequency pulse transfer atoms on doubly occupied sites into an initially empty internal state; absorption imaging counted the transferred atoms [Jor+08, author version v2, p. 2].

A doubly occupied site is called a doublon. To count these sites and the fraction of atoms they contain, write $\hat N_{\mathrm d}=\sum_i\hat n_{i\uparrow}\hat n_{i\downarrow}, \qquad f_{\mathrm d}=\frac{2\langle\hat N_{\mathrm d}\rangle}{N}.$ Here ${N>0}$ is the prepared total atom number, and ${f_{\mathrm d}}$ is the fraction of atoms on doublons, not the fraction of sites that are doublons. With one transferred atom per doublon, the ideal transferred count is ${N_{\mathrm d}}$ ; the factor two restores the two atoms that occupied that site. Transfer and detection calibration belong to the measurement, not to the operator identity.

For the two states in Example 8.9, this observable is zero in the covalent singlet ${|S\rangle}$ and one in the double-occupancy state ${|D_+\rangle}$ , despite identical mean site densities. This gives a concrete readout for their different occupation fluctuations. The published experiment averages over a trapped many-site gas, however: its result is not the ground-state curve of an isolated dimer, and double occupancy alone is neither full state reconstruction nor a sufficient diagnosis of a Mott insulator. Lectures 15–16 develop the many-site interpretation.

Wick reduction expresses higher correlations in terms of two-operator expectation values under an additional assumption about the state. A number-conserving fermionic Gaussian state is a preparation whose higher correlations are sums of products of two-operator expectation values, with the fermionic exchange signs. Such two-operator averages are called contractions. In a number-conserving state, the anomalous pair averages (two creators or two annihilators) vanish. A Slater determinant is a pure-state example. For ${G_{ij}:=\langle\hat c_i^\dagger\hat c_j\rangle}$ and distinct resolved modes ${i\ne j}$ , the four-factor rule gives

(8.28)

$\begin{aligned} \langle\hat n_i\hat n_j\rangle &=\langle\hat c_i^\dagger\hat c_i\rangle \langle\hat c_j^\dagger\hat c_j\rangle +\langle\hat c_i^\dagger\hat c_j\rangle \langle\hat c_i\hat c_j^\dagger\rangle\\ &=G_{ii}G_{jj}-G_{ij}G_{ji},\\ \operatorname{Cov}(\hat n_i,\hat n_j)&=-{|G_{ij}|}^2. \end{aligned}$

The omitted creator-creator contraction vanishes by number conservation; CAR gives ${\langle\hat c_i\hat c_j^\dagger\rangle=-G_{ji}}$ for ${i\ne j}$ . This is a Gaussian four-factor rule, not a rule for arbitrary interacting preparations. The labels denote individual spin-orbitals, not spin-summed site densities. The general Wick theorem is imported as a preview and not proved here; none of the preceding dimer expectations uses it.

Construct the Fock-space operator for a one-particle observable with matrix elements ${o_{ij}}$ . Determine whether it commutes with ${\hat N}$ . Then classify ${\hat c_i^\dagger\hat c_j^\dagger+\hat c_j\hat c_i}$ by sector change and explain why a Hermitian operator need not conserve particle number.

From a supplied one-particle matrix and two-particle matrix element, construct the Fock-space terms, including prefactors and operator order. Test Hermiticity and number conservation separately. Derive a local observable and specify the state and ordering for its two-time correlation. A response measurement also requires an applied perturbation and a measured observable; an ordered correlation is not itself a causal response. Change the one-particle basis and identify what remains invariant: the operator, body rank, and spectrum.

From operator construction to measurable correlations

Lectures 7–8 have replaced recurring labeled-particle bookkeeping by one operator construction:

$\begin{aligned} \text{identical-particle symmetry} &\Longrightarrow\text{occupation sectors},\\ \text{occupation sectors} &\Longrightarrow\text{canonical operator algebra},\\ \text{canonical operator algebra} &\Longrightarrow\text{basis-covariant operators},\\ \text{basis-covariant operators} &\Longrightarrow\text{conservation and correlations}. \end{aligned}$

The construction supports several reusable diagnoses.

A fermionic sign is computed from a declared mode order and operator action. It is not a process-dependent convention.
Creation and annihilation operators map between Fock sectors, but the physical Hamiltonian conserves a number exactly when its commutator with the corresponding number operator vanishes.
A unitary one-particle basis change transforms coefficients and creation/annihilation operators together. It preserves canonical algebra and body rank.
One- and two-body operator forms follow from matrix elements and a declared symmetrization convention. Their numerical prefactors cannot be copied independently of that convention.
Density, spin, current, interactions, and correlations are constructed from the same creation and annihilation operators.

The larger course use of the dimer is methodological. Exact symmetry-sector reduction is not an approximation; choosing two sites is. Local correlations are evidence about the state; they are not by themselves evidence of macroscopic order. These separations will control how the same operators are used for Mott localization, virtual exchange, response, and transport in extended systems.

This is a hard mastery gate. Later lectures may assume that the reader can

apply bosonic and fermionic operators with correct factors and signs;
transform between site and momentum modes;
construct one- and two-body operators from matrix elements;
test number conservation by a commutator; and
interpret a declared pure-state correlation.

The examples used declared pure-state preparations. Lecture 9 now changes the preparation: it introduces a thermal density operator and asks how a weak time-dependent source changes an expectation value.

When using this construction, first identify what each expression represents: an operator, its matrix in a specified basis, or a correlation evaluated in a specified state. Consistent basis transformations preserve spectra and expectations, but changing the state can change connected correlations without changing the Hamiltonian. A nonzero correlator is evidence about joint statistics at the declared separation and ordering; it is not automatically a causal influence, propagating mode, or macroscopic order parameter.

Problems

1. From site hopping to momentum occupations

For a periodic spinless chain,

$\hat H=-t\sum_j (\hat c_{j+1}^\dagger\hat c_j+\hat c_j^\dagger\hat c_{j+1}) +\Delta\sum_j{(-1)}^j\hat n_j.$

Transform the hopping term to momentum space, including every normalization factor.
Show that the staggered term couples ${k}$ and ${k+\pi/a}$ .
Write the reduced-zone ${2\times2}$ coefficient matrix and its Fock-space one-body operator.
Explain why diagonalizing that matrix is a one-particle basis change and does not alter fermionic statistics or body rank.

2. Hubbard dimer: signs, symmetry, and correlation

Use the four-mode order declared above and the ${N=2,S^z=0}$ basis ${(|D_1\rangle,|A\rangle,|B\rangle,|D_2\rangle)}$ .

Reconstruct the Hubbard-dimer matrix by direct operator action.
Verify ${[\hat H,\hat N]=[\hat H,\hat S^z]=0}$ .
Identify the singlet and triplet combinations and show which couples to double occupancy.
For ${|\Psi(\theta)\rangle=\cos\theta|S\rangle+ \sin\theta|D_+\rangle}$ , calculate ${\langle\hat n_1\rangle}$ , ${\langle\hat n_1\hat n_2\rangle}$ , and the connected equal-time correlation.

3. A lattice continuity equation and its failure

For a periodic spinless chain, let

$\hat H_t=-t\sum_j (\hat c_{j+1}^\dagger\hat c_j+\hat c_j^\dagger\hat c_{j+1}), \qquad \hat n_j=\hat c_j^\dagger\hat c_j.$

Use the Heisenberg equation to derive

$\frac{\mathrm d\hat n_j}{\mathrm d\tau} =\hat J^{(N)}_{j-1\to j}-\hat J^{(N)}_{j\to j+1}$

and identify the particle-current operator on a bond.

Sum the local equation over the ring and verify total-number conservation.
Add the real pairing term

$\hat H_\Delta=\Delta\sum_j (\hat c_j^\dagger\hat c_{j+1}^\dagger+ \hat c_{j+1}\hat c_j).$

Calculate ${[\hat N,\hat H_\Delta]}$ and identify the local source/sink interpretation that replaces a number-only continuity equation.

Explain why Hermiticity of ${\hat H_\Delta}$ does not imply particle-number conservation.

Advanced Appendix: Constrained Local Hilbert Spaces and Hubbard Operators

Ordinary fermion operators act on the full local four-state space ${\{|0\rangle,|\uparrow\rangle,|\downarrow\rangle, \vert{}\uparrow\downarrow\rangle\}}$ . Some low-energy models deliberately remove one or more of these states—for example, by excluding double occupancy. It is then useful to write maps intrinsic to the retained local space rather than pretend that projected fermion operators still satisfy canonical algebra [ABKR22].

Let ${\mathcal H_i^{\mathrm{ret}}}$ be an ${r}$ -dimensional retained local Hilbert space with ordered orthonormal basis ${{\{{|a\rangle}_i\}}_{a=1}^{r}}$ . For ${a,b\in\{1,\ldots,r\}}$ , the Hubbard operator is the linear map

$\hat X_i^{ab}:\mathcal H_i^{\mathrm{ret}}\to \mathcal H_i^{\mathrm{ret}}, \qquad \hat X_i^{ab}:={|a\rangle}_i\,{}_i\!\langle b|.$

These maps obey the matrix-unit, adjoint, and completeness relations

$\hat X_i^{ab}\hat X_i^{cd}=\delta_{bc}\hat X_i^{ad}, \qquad {(\hat X_i^{ab})}^\dagger=\hat X_i^{ba}, \qquad \sum_{a=1}^{r}\hat X_i^{aa}=\hat I_i^{\mathrm{ret}}.$

When embedded into an ordinary tensor product of retained site spaces, operators acting on distinct site factors commute. If ${\hat X_i^{ab}}$ changes fermion parity and is to represent a physical fermion map across sites, a global site/mode order or graded tensor-product convention must be supplied separately; the local matrix-unit definition alone does not imply canonical anticommutation relations [Hub65].

The multiplication rule is matrix-unit algebra, not a canonical commutation or anticommutation relation.

For the no-double-occupancy space ${\{|0\rangle,|\uparrow\rangle,|\downarrow\rangle\}}$ , local number and spin may be written

$\hat n_i=\hat X_i^{\uparrow\uparrow} +\hat X_i^{\downarrow\downarrow}, \qquad \hat S_i^+=\hbar\hat X_i^{\uparrow\downarrow}, \qquad \hat S_i^z=\frac{\hbar}{2} (\hat X_i^{\uparrow\uparrow}-\hat X_i^{\downarrow\downarrow}).$

Let ${\hat P_i^{\mathrm{ret}}}$ project the full four-state local space onto this retained subspace. The projected electron-creation operator is

$\hat c_{i\uparrow,\mathrm{proj}}^\dagger :=\hat P_i^{\mathrm{ret}}\hat c_{i\uparrow}^\dagger \hat P_i^{\mathrm{ret}} =\hat X_i^{\uparrow0}.$

Its local anticommutator is

$\{\hat X_i^{0\uparrow},\hat X_i^{\uparrow0}\} =\hat X_i^{00}+\hat X_i^{\uparrow\uparrow} =\hat I_i^{\mathrm{ret}}-\hat X_i^{\downarrow\downarrow},$

not the identity. The missing term records two blocking conditions: an existing up spin and the excluded double occupancy when a down spin is present.

This algebra is useful precisely because it exposes the changed local Hilbert space. Replacing ${\hat c_{i\uparrow,\mathrm{proj}}^\dagger}$ by the canonical ${\hat c_{i\uparrow}^\dagger}$ without the projector would silently reintroduce the state the model removed. Conversely, Hubbard operators are not “better fermions”; they are matrix units adapted to a constrained local basis.

Verify the ${\mathfrak{su}(2)}$ commutators of ${\hat S_i^\pm,\hat S_i^z}$ using only the multiplication rule. Then calculate the anticommutator of the projected up- and down-spin creation maps and explain which part of canonical fermion algebra survives between different sites only after a global sign convention or graded tensor product is supplied.

References

[ABKR22] Arovas, D. P., Berg, E., Kivelson, S. A., and Raghu, S. (2022). “The hubbard model.” Annual Review of Condensed Matter Physics 13, 239–274. https://doi.org/10.1146/annurev-conmatphys-031620-102024.

[Foc32] Fock, V. (1932). “Konfigurationsraum und zweite quantelung.” Zeitschrift für Physik 75 (9–10), 622–647. https://doi.org/10.1007/BF01344458.

[Gut63] Gutzwiller, M. C. (1963). “Effect of correlation on the ferromagnetism of transition metals.” Physical Review Letters 10 (5), 159–162. https://doi.org/10.1103/PhysRevLett.10.159.

[Hub63] Hubbard, J. (1963). “Electron correlations in narrow energy bands.” Proceedings of the Royal Society of London. Series A 276 (1365), 238–257. https://doi.org/10.1098/rspa.1963.0204.

[Hub65] Hubbard, J. (1965). “Electron correlations in narrow energy bands—IV. The atomic representation.” Proceedings of the Royal Society of London. Series A 285 (1403), 542–560. https://doi.org/10.1098/rspa.1965.0124.

[Jor+08] Jördens, R., Strohmaier, N., Günter, K., Moritz, H., and Esslinger, T. (2008). “A mott insulator of fermionic atoms in an optical lattice.” Nature 455 (7210), 204–207. https://doi.org/10.1038/nature07244.

[Kan63] Kanamori, J. (1963). “Electron correlation and ferromagnetism of transition metals.” Progress of Theoretical Physics 30 (3), 275–289. https://doi.org/10.1143/PTP.30.275.

Lecture 8: Hamiltonians, Observables, and Correlations

8.1 From one- and two-particle contributions to Fock-space operators

8.1.1 Lifting a one-particle matrix to Fock space

8.1.2 Basis covariance of the lift

8.1.3 Pairwise additivity before coefficient conventions

8.1.4 Deriving the operator by removing and reinserting a pair

8.1.5 The same operator in unrestricted-index conventions

8.1.6 Detailed calculation of the two prefactors

8.2 The Hubbard dimer makes the conventions observable

8.2.1 Declaring the dimer and its number sectors