Fermionic Operators in Many-Body Models

Fermionic creation and annihilation operators make antisymmetric many-particle states usable in a mode basis. A bilinear $c_i^\dagger c_j$ transfers a fermion from mode $j$ to mode $i$, a density product $n_{i\uparrow}n_{i\downarrow}$ detects double occupancy, and ordered strings of operators build hopping, interactions, currents, and pairing terms.

The canonical anticommutation relations and an ordered occupation basis determine every sign. Keeping that order consistent lets us translate between operator expressions and matrix elements without guessing from a diagram.

Helpful background. Fermionic anticommutation relations fixes the algebra; occupation-number representation supplies the ordered matrix basis.

Complete modes and their order

A complete mode includes spin, orbital, and other retained labels. Fix $1<2<\cdots<M$ and write number states with creators in ascending order; products act on kets from right to left. We use $n_i=c_i^\dagger c_i$ and count each unordered lattice bond once. Changing the mode order is a basis transformation, so states and operators must be transformed together.

Pauli exclusion for one mode

The canonical algebra gives $c^2=(c^\dagger)^2=0$, $n^2=n$, and $cc^\dagger=I-n$. Thus one complete mode is either empty or occupied. Two opposite-spin particles may occupy the same spatial orbital because they occupy distinct complete modes.

Ordered ladder actions

For $|\boldsymbol n\rangle=(c_1^\dagger)^{n_1}\cdots(c_M^\dagger)^{n_M}|0\rangle$, define $Q_j(\boldsymbol n)=\sum_{k<j}n_k$. The nonzero actions are

$$\begin{aligned} c_j|\boldsymbol n\rangle&=(-1)^{Q_j}|\boldsymbol n-\boldsymbol e_j\rangle &&(n_j=1),\\ c_j^\dagger|\boldsymbol n\rangle&=(-1)^{Q_j}|\boldsymbol n+\boldsymbol e_j\rangle &&(n_j=0). \end{aligned}$$

All other actions vanish. The sign counts occupied modes crossed while moving the operator to its ordered position. Occupation-number representation shows how to encode and index these states.

The sign $(-1)^{Q_j}$ follows the declared order. For a product of operators, update the occupations after each rightmost action before counting parity for the next one.

Why the Sign Cannot Be Dropped

Different fermionic creation operators anticommute:

$$c_i^\dagger c_j^\dagger = -c_j^\dagger c_i^\dagger, \qquad i\ne j.$$

The minus sign is the operator version of exchanging two columns in a Slater determinant. Removing it changes interference between many-body amplitudes, matrix elements around loops in configuration space, exchange energies, and correlation functions.

Rephasing a basis vector and transforming every coefficient consistently leaves predictions unchanged. Relative phases between interfering paths to the same final state can affect observables. A mode reorder is therefore a valid basis transformation, while deleting parity signs from only some operations is a change of theory.

Fermions Are Not Hard-Core Bosons

Both a fermionic mode and a hard-core bosonic mode have occupation $0$ or $1$. Their algebras differ between distinct modes:

$$\{c_i,c_j^\dagger\} = 0, \qquad i\ne j,$$

whereas distinct hard-core boson operators are usually chosen to commute.

Equal local dimensions do not imply equal many-body operator algebras. Mapping fermions to spins or hard-core bosons requires nonlocal parity strings, boundary-condition choices, or other structure that preserves the exchange signs.

Bilinear Transfer and Its Sign

For $i\ne j$, define the number of occupied modes strictly between $i$ and $j$:

$$P_{ij}(\boldsymbol n) = \sum_{ \min(i,j)<k<\max(i,j) } n_k.$$

Then

$$c_i^\dagger c_j \lvert\boldsymbol n\rangle ={} (-1)^{P_{ij}(\boldsymbol n)} n_j(1-n_i) \times \lvert \boldsymbol n +\boldsymbol e_i -\boldsymbol e_j \rangle.$$

The result is nonzero only when the source is occupied and the destination is empty. Unlike bosonic transfer, there is no square-root enhancement. The nontrivial factor is the parity sign.

Modes adjacent in the chosen order have $P_{ij}=0$, but neighboring lattice sites need not correspond to adjacent entries in a spin-orbital ordering.

Worked Sign Example

Use the site-major order

$$(1\uparrow,1\downarrow,2\uparrow,2\downarrow)$$

and the bitstring state

$$\lvert1,0,1,1\rangle.$$

Consider moving the $2\downarrow$ fermion into the empty $1\downarrow$ mode:

$$c_{1\downarrow}^\dagger c_{2\downarrow} \lvert1,0,1,1\rangle.$$

The modes strictly between $1\downarrow$ and $2\downarrow$ contain the occupied $2\uparrow$ mode, so

$$P_{1\downarrow,2\downarrow} = 1.$$

Therefore

$$c_{1\downarrow}^\dagger c_{2\downarrow} \lvert1,0,1,1\rangle = -\lvert1,1,1,0\rangle.$$

The minus sign would change if a different bit ordering were used, together with a corresponding phase change of the basis vectors. The Hamiltonian spectrum would not.

General One-Body Terms

A number-conserving one-body Hamiltonian is

$$H_1 = \sum_{i,j} h_{ij} c_i^\dagger c_j,$$

with

$$h_{ij} = h_{ji}^*$$

for Hermiticity. Diagonal terms are mode energies,

$$H_\epsilon = \sum_i \epsilon_i n_i,$$

while off-diagonal terms transfer fermions between modes.

The matrix $h$ can include kinetic energy, lattice hopping, spin-orbit coupling, orbital hybridization, Zeeman terms, or external potentials. Its indices must cover every label on which the one-particle operator acts.

Spinful Hopping

For spin-preserving lattice hopping,

$$\begin{aligned} H_t = -\sum_{\langle i,j\rangle,\sigma} \bigl( t_{ij}c_{i\sigma}^\dagger c_{j\sigma} + t_{ij}^*c_{j\sigma}^\dagger c_{i\sigma} \bigr). \end{aligned}$$

Each unordered bond is counted once. The Hermitian-conjugate term moves the fermion in the reverse direction.

Spin-dependent hopping is represented by a matrix,

$$H_t = -\sum_{i,j} \sum_{\sigma,\sigma'} c_{i\sigma}^\dagger t_{ij}^{\sigma\sigma'} c_{j\sigma'}.$$

Hermiticity requires

$$t_{ji} = t_{ij}^\dagger$$

in spin space. Off-diagonal spin entries describe spin flips or spin-orbit-coupled hopping and need not conserve $N_\uparrow$ and $N_\downarrow$ separately.

Complex Hopping and Gauge Choice

Under a local rephasing

$$c_i \longmapsto e^{i\chi_i}c_i,$$

the coefficient representation changes as

$$t_{ij} \longmapsto e^{i(\chi_i-\chi_j)}t_{ij}.$$

Individual bond phases depend on the orbital convention. The argument of the product of hopping amplitudes around a closed loop is gauge invariant modulo $2\pi$ and can represent magnetic or synthetic flux.

Fermionic parity signs from the Fock basis and Peierls phases from hopping coefficients are different structures. Both can appear in one matrix element, and neither should be silently absorbed into the other.

Bond Current

For one spin-preserving bond,

$$H_{t,ij} = -\sum_\sigma \left( t_{ij}c_{i\sigma}^\dagger c_{j\sigma} + t_{ij}^*c_{j\sigma}^\dagger c_{i\sigma} \right),$$

define current from $i$ to $j$ by

$$\frac{dn_i}{dt} = -I_{i\to j}$$

for the contribution of that bond. The Heisenberg equation gives

$$\begin{aligned} I_{i\to j} = -\frac{i}{\hbar} \sum_\sigma \bigl( t_{ij}c_{i\sigma}^\dagger c_{j\sigma} - t_{ij}^*c_{j\sigma}^\dagger c_{i\sigma} \bigr). \end{aligned}$$

Thus directed transport depends on the imaginary part of the hopping coherence. Density Operators and Current Operators derives the continuity equation and discusses spin-current conventions.

Complete Spin-Orbital Labels

A spinful lattice mode is labeled by the pair

$$\alpha=(i,\sigma).$$

The canonical algebra is

$$\{c_{i\sigma},c_{j\sigma'}^\dagger\} = \delta_{ij}\delta_{\sigma\sigma'}I.$$

At each site,

$$n_i = n_{i\uparrow}+n_{i\downarrow}.$$

The four local states are

$$\lvert0\rangle, \qquad \lvert\uparrow\rangle, \qquad \lvert\downarrow\rangle, \qquad \lvert\uparrow\downarrow\rangle.$$

The doubly occupied state contains two fermions in different complete modes. Pauli exclusion forbids two copies of $i\uparrow$ or two copies of $i\downarrow$, not one of each.

Local Spin Operators

For spin-$1/2$ fermions, define the dimensionless onsite spin generators

$$S_i^a = \frac12 \sum_{\sigma,\sigma'} c_{i\sigma}^\dagger (\sigma^a)_{\sigma\sigma'} c_{i\sigma'},$$

where $\sigma^a$ is a Pauli matrix. Explicitly,

$$S_i^+ = c_{i\uparrow}^\dagger c_{i\downarrow}, \qquad S_i^- = c_{i\downarrow}^\dagger c_{i\uparrow},$$

and

$$S_i^z = \frac12 \left( n_{i\uparrow}-n_{i\downarrow} \right).$$

Physical angular momentum is $\hbar\mathbf S_i$ in this convention. These operators are fermion bilinears and preserve total particle number. They vanish on the empty and doubly occupied single-orbital site, while acting as spin-$1/2$ generators on the singly occupied subspace.

The local moment obeys $\mathbf S_i^2=\tfrac34(n_i-2D_i)$, where $D_i=n_{i\uparrow}n_{i\downarrow}$: charge fluctuations reduce the probability of a spin-bearing site. Coupling a separate localized spin to one such conduction-electron bilinear gives the exchange term in the Kondo Model Preview.

Hubbard Onsite Interaction

The single-band Hubbard interaction is

$$H_U = U \sum_i n_{i\uparrow}n_{i\downarrow}.$$

The double-occupancy operator

$$D_i = n_{i\uparrow}n_{i\downarrow}$$

is a projector:

$$D_i^2 = D_i.$$

It has eigenvalue one only on $\lvert\uparrow\downarrow\rangle$. Therefore $U>0$ penalizes double occupancy and $U<0$ favors it relative to singly occupied configurations.

Penalizing a doublon with finite $U$ and removing it from the Hilbert space are different operations. The latter produces the projected, noncanonical fermion operators defined in the t–J Model Preview.

In normal-ordered ladder-operator form,

$$n_{i\uparrow}n_{i\downarrow} = c_{i\uparrow}^\dagger c_{i\downarrow}^\dagger c_{i\downarrow} c_{i\uparrow}.$$

The displayed order is important. Reordering one pair without its compensating minus sign produces the wrong operator.

The Hubbard model as an operator example

Combining spin-preserving hopping, the onsite double-occupancy projector and $-\mu N$ gives the single-band Hubbard model. The model is quadratic at $U=0$, diagonal in occupations at zero hopping, and has zero interaction in the one-particle sector. A uniform chemical potential shifts all energies equally inside one fixed-$N$ sector. These limits test signs without requiring a many-site calculation.

General Two-Body Interactions

With unsymmetrized product-basis matrix elements,

$$H_2 = \frac12 \sum_{i,j,k,l} V_{ij;kl} c_i^\dagger c_j^\dagger c_l c_k.$$

Hermiticity and exchange symmetry of the first-quantized interaction imply

$$V_{ij;kl} = V_{kl;ij}^*, \qquad V_{ij;kl} = V_{ji;lk}.$$

Fermionic theory often uses antisymmetrized matrix elements

$$\langle ij\Vert kl\rangle = V_{ij;kl} - V_{ij;lk}.$$

Then the same operator is commonly written

$$H_2 = \frac14 \sum_{i,j,k,l} \langle ij\Vert kl\rangle c_i^\dagger c_j^\dagger c_l c_k.$$

The factors $1/2$ and $1/4$ belong to different coefficient conventions. Combining an antisymmetrized coefficient with the unsymmetrized prefactor is a factor-of-two error. The foundational lift is derived in Two-Body Operators, while the many-body application guide explains coefficient conventions and reduced-density contractions.

Density, Exchange, and Pair-Hopping Terms

A density interaction between distinct modes, with real symmetric coefficients $V_{ij}=V_{ji}$, is

$$H_V = \frac12 \sum_{i\ne j} V_{ij}n_i n_j,$$

or $\sum_{\langle i,j\rangle}V_{ij}n_i n_j$ when unordered bonds are counted once.

Multi-orbital models can also contain exchange and pair-hopping structures. For two orbitals $a,b$, representative terms include

$$c_{a\uparrow}^\dagger c_{b\downarrow}^\dagger c_{a\downarrow} c_{b\uparrow}$$

and

$$c_{a\uparrow}^\dagger c_{a\downarrow}^\dagger c_{b\downarrow} c_{b\uparrow}.$$

Their signs depend on the displayed operator order. A shorthand such as “exchange plus Hermitian conjugate” is safe only when the reference ordering is explicit.

Particle Number and Fermion Parity

The total number operator satisfies

$$[N,c_i] = -c_i, \qquad [N,c_i^\dagger] = c_i^\dagger.$$

A monomial with equal numbers of creation and annihilation operators conserves $N$. The corresponding global transformation is

$$c_i \longmapsto e^{i\theta}c_i.$$

Fermion parity is

$$\mathcal P_F = (-1)^N.$$

It acts on odd fermionic operators as

$$\mathcal P_F c_i\mathcal P_F = -c_i, \qquad \mathcal P_F c_i^\dagger\mathcal P_F = -c_i^\dagger.$$

Hamiltonians built from even fermionic monomials preserve parity. They may still break number conservation by changing $N$ in steps of two.

Pairing Terms

A quadratic pairing Hamiltonian can be written

$$H_\Delta = \frac12 \sum_{i,j} \left( \Delta_{ij}c_i^\dagger c_j^\dagger + \Delta_{ij}^*c_jc_i \right).$$

Because

$$c_i^\dagger c_j^\dagger = -c_j^\dagger c_i^\dagger,$$

only the antisymmetric coefficient part contributes:

$$\Delta_{ij} = -\Delta_{ji}$$

when $i,j$ are complete spin-orbital labels.

This does not mean the spatial part of every Cooper-pair wavefunction is odd. For spin singlets, antisymmetry can reside in spin while the spatial or orbital factor is symmetric. The complete pair amplitude must be antisymmetric under exchange of all one-particle labels.

Pairing terms do not conserve $N$, but they preserve fermion parity. Their diagonalization requires a Bogoliubov transformation in Nambu space and careful handling of constant terms. Bogoliubov Theory gives the general canonical construction and boson–fermion contrast; BCS Mean-Field Theory derives the self-consistent fermionic pairing saddle.

Normal Ordering and Reordering

Moving one fermionic operator past another produces a minus sign unless their anticommutator contributes a delta term. For example,

$$c_i c_j^\dagger = \delta_{ij}I - c_j^\dagger c_i.$$

For $i\ne j$, this is pure sign change. For $i=j$, it also produces the identity.

Never reorder a long fermionic monomial by visual symmetry alone. Use adjacent swaps, canonical anticommutation relations, or a symbolic routine that tracks both parity and contractions. Normal Ordering defines the creation-left convention; Normal Ordering in Many-Body QM develops filled-reference contractions and Hamiltonian decompositions.

Even and Odd Operators

A monomial containing an even number of fermionic ladder operators has even parity, as does a sum of such monomials. Number operators, hopping bilinears, spin densities, and ordinary interaction quartics are even. A single $c_i$ or $c_i^\dagger$ is odd.

Even operators localized on disjoint mode sets commute. Odd operators on disjoint sets anticommute. This distinction is why observable Hamiltonian terms can remain local even though a qubit representation of an individual fermion operator carries a parity string.

When partitioning a fermionic system into subsystems, tensor-product intuition must be matched to the chosen parity structure. One should not treat odd operators in separated regions as ordinary commuting subsystem observables.

Jordan–Wigner Preview

For a qubit chain with occupation convention

$$n_j = \frac{1-Z_j}{2},$$

one Jordan–Wigner representation is

$$c_j = \left( \prod_{k<j}Z_k \right) \frac{X_j+iY_j}{2}.$$

The $Z$ string records the same parity $(-1)^{Q_j}$ that appears in occupation-space action. Dropping it turns fermions into commuting hard-core degrees of freedom.

The string length depends on mode ordering, which is why orbital ordering affects qubit locality and tensor-network cost even though it does not change the fermionic physics.

Jordan–Wigner Transformation derives this representation and explains nearest-neighbor cancellations and periodic-boundary parity sectors. Bravyi and Kitaev discuss alternative encodings that change qubit locality without changing fermionic physics.

Projection and effective interactions

Each retained fermionic mode is exactly empty or occupied. Approximation enters through the choice of bands, orbitals, or active space, and through any extra many-particle constraint. A plain projection $PHP$ retains matrix elements of the original Hamiltonian; it does not automatically account for virtual visits to discarded states.

Eliminating high-energy states to construct an effective Hamiltonian can instead shift one-body terms and induce longer-range or higher-body interactions. For example, virtual doublons in a repulsive Hubbard model generate spin exchange of order $t^2/U$. The retained operators may also need transformation. A finite mode space is therefore not automatically accurate simply because Pauli exclusion makes each mode two dimensional.

Further projection changes the algebra itself. In the no-double-occupancy local space, $\widetilde c_{\sigma}=c_\sigma(1-n_{\bar\sigma})$ satisfies $\{\widetilde c_\sigma,\widetilde c_\sigma^\dagger\}=1-n_{\bar\sigma}$, not the identity. This is the constrained operator used in the t–J model.

Mean-Field Decoupling Is Not an Identity

An interaction such as

$$Un_{i\uparrow}n_{i\downarrow}$$

can be approximated in density, magnetic, or pairing channels. For a density decoupling,

$$n_{i\uparrow}n_{i\downarrow} \approx{} \langle n_{i\uparrow}\rangle n_{i\downarrow} + n_{i\uparrow}\langle n_{i\downarrow}\rangle - \langle n_{i\uparrow}\rangle \langle n_{i\downarrow}\rangle.$$

This is an approximation tied to a chosen channel and self-consistency condition. Other channels can be essential, and Fock exchange terms appear when the interaction and indices permit them.

Constant subtraction terms prevent double counting in the mean-field energy. Omitting them can leave eigenvectors unchanged while giving incorrect thermodynamic potentials and phase comparisons.

Observable Operator Patterns

Common fermionic observables include:

Observable	Operator form	Comment
occupation	$\langle n_i\rangle$	probability that complete mode $i$ is occupied
double occupancy	$\langle n_{i\uparrow}n_{i\downarrow}\rangle$	onsite opposite-spin pair probability
one-body coherence	$\langle c_i^\dagger c_j\rangle$	one-body density matrix and hopping coherence
spin density	$\langle\mathbf S_i\rangle$	local spin polarization under stated normalization
pairing amplitude	$\langle c_i c_j\rangle$	number-breaking or phase-referenced anomalous average
bond current	$\langle I_{i\to j}\rangle$	oriented transport convention

The operator ordering in correlation functions must be retained. In general,

$$\langle c_i c_j^\dagger\rangle \ne \langle c_j^\dagger c_i\rangle.$$

They are related by the anticommutator, not by casual reordering.

Exercises

Sign of a transfer

With order $1<2<3<4<5$, evaluate

$$c_2^\dagger c_5 \lvert1,0,1,0,1\rangle.$$

Check the result both by sequential ladder actions and by counting occupied modes strictly between $2$ and $5$.

Solution

First apply $c_5$. The occupied preceding modes are $1$ and $3$, so the annihilation sign is positive:

$$c_5\lvert1,0,1,0,1\rangle = \lvert1,0,1,0,0\rangle.$$

Next apply $c_2^\dagger$. Mode $1$ is occupied, giving a minus sign:

$$c_2^\dagger \lvert1,0,1,0,0\rangle = -\lvert1,1,1,0,0\rangle.$$

Equivalently, the modes strictly between $2$ and $5$ are $3$ and $4$, with occupations $1$ and $0$. Thus

$$P_{2,5}=1,$$

which gives the same negative sign.

Antisymmetry of the pairing matrix

Show that the symmetric part of $\Delta_{ij}$ contributes nothing to

$$\sum_{i,j} \Delta_{ij}c_i^\dagger c_j^\dagger.$$

Solution

Decompose

$$\Delta_{ij} = \Delta_{ij}^{A} + \Delta_{ij}^{S},$$

where $\Delta^A_{ij}=-\Delta^A_{ji}$ and $\Delta^S_{ij}=\Delta^S_{ji}$. For the symmetric part, relabel $i\leftrightarrow j$:

$$\begin{aligned} \sum_{i,j} \Delta_{ij}^{S} c_i^\dagger c_j^\dagger &= \sum_{i,j} \Delta_{ji}^{S} c_j^\dagger c_i^\dagger \\ &= -\sum_{i,j} \Delta_{ij}^{S} c_i^\dagger c_j^\dagger. \end{aligned}$$

The expression equals its negative and is therefore zero. Only the antisymmetric part contributes.

One-site Hubbard spectrum

For the one-site grand Hamiltonian

$$K = U n_\uparrow n_\downarrow - \mu(n_\uparrow+n_\downarrow),$$

find the energies of $\lvert0\rangle$, $\lvert\uparrow\rangle$, $\lvert\downarrow\rangle$, and $\lvert\uparrow\downarrow\rangle$.

Solution

The empty state has no particles or double occupancy:

$$E_0=0.$$

Each singly occupied state has one particle and no double occupancy:

$$E_\uparrow = E_\downarrow = -\mu.$$

The doubly occupied state has two particles and $D=1$:

$$E_{\uparrow\downarrow} = U-2\mu.$$

This elementary spectrum is a useful sign and chemical-potential check for Hubbard implementations.

Local spin and double occupancy

Verify $\mathbf S^2=\tfrac34(n-2D)$ on the four single-orbital states. What local moment remains when empty and doubly occupied states each have probability one half?

Solution

On the empty, up, down and doubly occupied states, $n-2D$ has eigenvalues $0,1,1,0$. The spin Casimir has corresponding eigenvalues $0,3/4,3/4,0$. Both operators are diagonal in these sectors, proving the identity. The specified mixture has $\langle n\rangle=1$ but $\langle\mathbf S^2\rangle=0$: mean filling one does not guarantee a local spin one half.

References

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