Second-Quantized One-Body Operator

Purpose

Let $A$ act on a one-particle Hilbert space $\mathcal H_1$, and let $\{\lvert\varphi_i\rangle\}$ be an orthonormal one-particle basis. Define

$$A_{ij} = \langle\varphi_i\vert A \vert\varphi_j\rangle.$$

The number-conserving Fock-space lift is

$$d\Gamma(A) \equiv \widehat A = \sum_{ij} A_{ij} d_i^\dagger d_j.$$

The visible formula is the same for bosons and fermions. Statistics enters through the algebra and number-state action of $d_i,d_i^\dagger$.

On the $N$-particle sector,

$$\widehat A = \sum_{\alpha=1}^{N} A^{(\alpha)},$$

where $A^{(\alpha)}$ acts as $A$ on particle slot $\alpha$ and as the identity on the others. The canonical proof is at One-Body Operators.

At a glance

One-particle object	Fock-space lift
$A=\sum_{ij}A_{ij}\lvert i\rangle\langle j\rvert$	$d\Gamma(A)=\sum_{ij}A_{ij}d_i^\dagger d_j$
Identity $I_1$	Total number $\widehat N=\sum_i d_i^\dagger d_i$
Projector $\lvert u\rangle\langle u\rvert$	Mode number $d_u^\dagger d_u$
Diagonal $A=\sum_i a_i\lvert i\rangle\langle i\rvert$	$\sum_i a_i\widehat n_i$
Position kernel $A(x,y)$	$\int dx\,dy\,\psi^\dagger(x)A(x,y)\psi(y)$
Local operator $A_x$	$\int dx\,\psi^\dagger(x)A_x\psi(x)$
One-body expectation	$\langle\widehat A\rangle=\operatorname{Tr}_1(A\gamma)$
Matrix commutator	$[d\Gamma(A),d\Gamma(B)]=d\Gamma([A,B])$

The sums and integrals include all spatial, spin, band, species, and other internal labels needed to specify a complete one-particle mode.

Matrix-to-operator dictionary

The one-particle operator has the resolution

$$A = \sum_{ij} A_{ij} \lvert\varphi_i\rangle \langle\varphi_j\rvert.$$

The lift replaces each one-particle dyad by a mode-transfer bilinear:

$$\lvert\varphi_i\rangle \langle\varphi_j\rvert \quad\longmapsto\quad d_i^\dagger d_j.$$

The operator $d_i^\dagger d_j$ annihilates a particle in mode $j$ and creates one in mode $i$. It changes at most one occupied mode and preserves total particle number.

Diagonal terms measure occupations:

$$\widehat A_{\mathrm{diag}} = \sum_i A_{ii} \widehat n_i, \qquad \widehat n_i = d_i^\dagger d_i.$$

Off-diagonal terms transfer one particle:

$$\widehat A_{\mathrm{off}} = \sum_{i\neq j} A_{ij} d_i^\dagger d_j.$$

Dropping the off-diagonal terms is valid only when $A$ is diagonal in the chosen basis or when a stated approximation or symmetry makes their matrix elements irrelevant.

Identity, projectors, and diagonal operators

The one-particle identity has

$$(I_1)_{ij} = \delta_{ij},$$

so

$$d\Gamma(I_1) = \sum_i d_i^\dagger d_i = \widehat N.$$

On a fixed-$N$ sector this acts as

$$d\Gamma(I_1) = N I_N,$$

not as $I_N$. The lift applies one copy of the identity to each particle and adds the results.

For a normalized mode $\lvert u\rangle$,

$$A = \lvert u\rangle\langle u\rvert \quad\Longrightarrow\quad d\Gamma(A) = d_u^\dagger d_u.$$

If $A$ has eigenvectors $\lvert a_i\rangle$ and eigenvalues $a_i$,

$$A = \sum_i a_i \lvert a_i\rangle\langle a_i\rvert,$$

then in the eigenmode basis

$$d\Gamma(A) = \sum_i a_i\widehat n_i.$$

This is why a purely one-body Hamiltonian becomes a sum of independent mode energies after diagonalizing its one-particle matrix.

Hermiticity and adjoints

Taking the adjoint gives

$$\widehat A^\dagger = \sum_{ij} A_{ij}^* d_j^\dagger d_i = \sum_{ij} (A^\dagger)_{ij} d_i^\dagger d_j = d\Gamma(A^\dagger).$$

Therefore

$$A^\dagger=A \quad\Longrightarrow\quad \widehat A^\dagger=\widehat A.$$

In a complete faithful Fock representation containing the one-particle sector, the converse also holds. A quick matrix check is

$$A_{ij} = A_{ji}^*.$$

For a Hamiltonian, complex hopping coefficients must occur with their Hermitian-conjugate partners.

Number conservation

Let

$$\widehat N = \sum_k d_k^\dagger d_k.$$

For bosonic or fermionic canonical modes, ordinary commutators satisfy

$$[\widehat N,d_i^\dagger] = d_i^\dagger, \qquad [\widehat N,d_j] = - d_j.$$

Hence

$$\begin{aligned} [\widehat N,d_i^\dagger d_j] &= [\widehat N,d_i^\dagger]d_j + d_i^\dagger[\widehat N,d_j] \\ &= 0, \end{aligned}$$

and every additive one-body lift obeys

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

A one-body operator may change spin, momentum, site, band, or species labels, but the lift contains one creation and one annihilation and therefore preserves the total number.

Action on creation and annihilation operators

For either bosonic commutators or fermionic anticommutators in the underlying mode algebra, the even bilinear has ordinary commutators

$$[\widehat A,d_k^\dagger] = \sum_i A_{ik} d_i^\dagger,$$

and

$$[\widehat A,d_k] = - \sum_j A_{kj} d_j.$$

These identities are often the quickest way to compute transformations and Heisenberg equations. If

$$\widehat H = d\Gamma(h),$$

then

$$i\hbar \frac{d d_k}{dt} = [d_k,\widehat H] = \sum_j h_{kj}d_j.$$

The mode operators therefore evolve under the same one-particle matrix $h$ that evolves one-particle amplitudes.

Bilinear commutator algebra

Canonical bosonic and fermionic bilinears obey

$$[d_i^\dagger d_j,d_k^\dagger d_l] = \delta_{jk} d_i^\dagger d_l - \delta_{il} d_k^\dagger d_j.$$

The statistics-dependent terms cancel in this ordinary commutator. Therefore

$$[d\Gamma(A),d\Gamma(B)] = d\Gamma([A,B]).$$

Useful consequences include:

$$[A,B]=0 \quad\Longrightarrow\quad [d\Gamma(A),d\Gamma(B)]=0,$$

and, for a one-body Hamiltonian,

$$\frac{d}{dt} d\Gamma(A) = d\Gamma \left( \frac{\partial A}{\partial t} + \frac{i}{\hbar}[h,A] \right)$$

in the Heisenberg picture, with the displayed sign convention following $\dot O=\partial_tO+(i/\hbar)[H,O]$.

Bosonic and fermionic number-state action

For two distinct bosonic modes $i\neq j$,

$$d_i^\dagger d_j \lvert\ldots,n_i,\ldots,n_j,\ldots\rangle = \sqrt{n_j} \sqrt{n_i+1} \lvert\ldots,n_i+1,\ldots,n_j-1,\ldots\rangle.$$

The term vanishes when $n_j=0$.

For fermions, the transfer is nonzero only if

$$n_j=1, \qquad n_i=0.$$

In the ordered convention

$$\lvert n_1,\ldots,n_M\rangle = (c_1^\dagger)^{n_1} \cdots (c_M^\dagger)^{n_M} \lvert0\rangle,$$

the nonzero transfer has sign

$$(-1)^{ \sum_{k=\min(i,j)+1}^{\max(i,j)-1} n_k }.$$

The compact lift formula does not display this sign because it is already encoded in the fermionic operator algebra. Do not guess signs from particle labels; fix a mode order and apply the operators in that convention.

Basis covariance

Let a new orthonormal basis be

$$\lvert\chi_p\rangle = \sum_i \lvert\varphi_i\rangle U_{ip}, \qquad U^\dagger U=I.$$

The corresponding creation operators and matrix transform as

$$e_p^\dagger = \sum_i U_{ip} d_i^\dagger, \qquad A' = U^\dagger A U.$$

Then

$$\sum_{pq} A'_{pq} e_p^\dagger e_q = \sum_{ij} A_{ij} d_i^\dagger d_j.$$

The matrix and ladder operators are basis dependent; the Fock-space operator is not. Rotating only the coefficients or only the mode operators changes the physical operator.

The simple formula assumes an orthonormal basis. In a nonorthogonal orbital set, overlap matrices and a dual basis enter. Inserting nonorthogonal matrix elements directly into the orthonormal formula generally gives the wrong operator.

Continuum field form

Let

$$\psi(x) = \sum_i \varphi_i(x)d_i, \qquad \psi^\dagger(x) = \sum_i \varphi_i^*(x)d_i^\dagger,$$

where $x$ can include position and internal labels. If

$$A(x,y) = \langle x\vert A\vert y\rangle,$$

then

$$d\Gamma(A) = \int dx\,dy\, \psi^\dagger(x) A(x,y) \psi(y).$$

For a local or differential one-particle operator $A_x$,

$$d\Gamma(A) = \int dx\, \psi^\dagger(x) A_x \psi(x).$$

Examples include the local potential

$$\widehat V = \int d^d x\, V(\mathbf x) \psi^\dagger(\mathbf x) \psi(\mathbf x),$$

and the nonrelativistic kinetic energy

$$\widehat T = \int d^d x\, \psi^\dagger(\mathbf x) \left( - \frac{\hbar^2}{2m} \nabla^2 \right) \psi(\mathbf x).$$

For differential operators, integration by parts, boundary conditions, and operator domains are part of the definition. Unsmeared fields are operator-valued distributions, so continuum products may also require regularization.

Lattice and internal-state examples

A one-particle tight-binding matrix $h_{ij}$ lifts to

$$\widehat H_1 = \sum_{ij} h_{ij} d_i^\dagger d_j.$$

Nearest-neighbor hopping is one-body:

$$\widehat H_t = - t \sum_{\langle ij\rangle} \left( d_i^\dagger d_j + d_j^\dagger d_i \right).$$

It couples two sites but transfers only one particle. By contrast, $\widehat n_i\widehat n_j$ is a two-body density interaction.

For spin-$1/2$ modes with site or orbital label $r$,

$$\widehat S^a = \frac{\hbar}{2} \sum_{r} \sum_{\sigma\sigma'} d_{r\sigma}^\dagger (\sigma^a)_{\sigma\sigma'} d_{r\sigma'}.$$

The spin indices are part of the complete one-particle label. Off-diagonal Pauli-matrix elements rotate spin while preserving total particle number.

Expectations from the one-body density matrix

Define the unnormalized one-body reduced density matrix by

$$\gamma_{ij} \equiv \langle d_j^\dagger d_i \rangle.$$

Then

$$\operatorname{Tr}_1\gamma = \sum_i \langle d_i^\dagger d_i\rangle = \langle\widehat N\rangle.$$

Every one-body expectation is

$$\begin{aligned} \langle\widehat A\rangle &= \sum_{ij} A_{ij} \langle d_i^\dagger d_j\rangle \\ &= \sum_{ij} A_{ij} \gamma_{ji} \\ &= \operatorname{Tr}_1(A\gamma). \end{aligned}$$

Index conventions vary across fields. If another source defines $\gamma_{ij}=\langle d_i^\dagger d_j\rangle$, its trace formula will contain the corresponding transpose or reordered indices. Declare the definition before contracting.

The one-body matrix determines means of all one-body observables, but not generally their variances:

$$\widehat A^2$$

contains a two-body contribution after the ladder operators are reordered. Two many-body states can share the same $\gamma$ while differing in pair correlations, entanglement, and higher moments.

What is not an additive one-body lift

The operator grading is determined by how many particles are acted on at once, not merely by polynomial degree:

Structure	Typical form	Classification
Scalar	$E_0 I$	Zero-body constant
Number-conserving bilinear	$A_{ij}d_i^\dagger d_j$	Additive one-body lift
Source	$\eta_i d_i^\dagger+\eta_i^*d_i$	Changes particle number; not a lift
Pairing	$\Delta_{ij}d_i^\dagger d_j^\dagger+\mathrm{h.c.}$	Quadratic but number nonconserving
Pair interaction	$V_{ij;kl}d_i^\dagger d_j^\dagger d_l d_k$	Two-body

A Hartree, Fock, Kohn–Sham, or other mean-field Hamiltonian can have bilinear form while representing an interacting problem approximately. Its form does not make the original microscopic interaction one-body.

Likewise, a Bogoliubov transformation can make a Hamiltonian diagonal in quasiparticle bilinears. The operator grading must state whether it refers to bare particles, projected orbitals, or quasiparticles.

Finite-basis construction checks

For a computational many-body basis:

1. Declare the complete ordered one-particle mode list.
2. Compute $A_{ij}$ in that exact basis.
3. Verify $A=A^\dagger$ when the observable should be Hermitian.
4. Construct each allowed transfer $d_i^\dagger d_j$ with the chosen bosonic or fermionic action rules.
5. Check $[\widehat N,\widehat A]=0$.
6. Restrict to the one-particle sector and verify that the matrix equals $A$.
7. For $A=I_1$, verify that the result is $\widehat N$.
8. Test one small unitary basis rotation and confirm covariance.

A finite one-particle projection $P_1$ produces

$$A_{\mathrm{proj}} = P_1 A P_1.$$

Its lift is exact for the projected model, not automatically for the full system. Effective transformations can induce two-body and higher pieces in an observable even when the original operator was one-body.

Common mistakes

• Reversing the matrix convention and using $\langle\varphi_j\vert A\vert\varphi_i\rangle$ as $A_{ij}$.
• Mixing coefficients from one mode basis with ladder operators from another.
• Dropping off-diagonal transfer terms because the state, rather than the operator, is diagonal in a chosen basis.
• Treating hopping as two-body merely because it connects two sites.
• Calling every quadratic expression an additive one-body lift.
• Forgetting the spin, band, or species part of a complete mode label.
• Omitting fermionic ordering signs in occupation-basis matrix elements.
• Using an orthonormal-basis formula for nonorthogonal orbitals without a dual basis or overlap metric.
• Assuming $\langle\widehat A\rangle$ and $\operatorname{Var}(\widehat A)$ require the same reduced information.
• Inferring that diagonalizing $h$ solves a Hamiltonian whose interaction remains quartic.
• Ignoring domains and boundary terms for unbounded differential operators.

Exercises

1. Identity and number conservation

Show that $d\Gamma(I_1)=\widehat N$ and that every $d\Gamma(A)$ commutes with $\widehat N$.

Solution

In an orthonormal basis,

$$(I_1)_{ij} = \delta_{ij}.$$

Therefore

$$d\Gamma(I_1) = \sum_{ij} \delta_{ij} d_i^\dagger d_j = \sum_i d_i^\dagger d_i = \widehat N.$$

Using

$$[\widehat N,d_i^\dagger] = d_i^\dagger, \qquad [\widehat N,d_j] = - d_j,$$

one finds

$$[\widehat N,d_i^\dagger d_j] = 0.$$

Linearity then gives

$$[\widehat N,d\Gamma(A)] = 0.$$

2. Two-mode hopping matrix

Let

$$A = \begin{pmatrix} \epsilon_1 & t\\ t^* & \epsilon_2 \end{pmatrix}.$$

Write its Fock-space lift and check Hermiticity.

Solution

Insert the four matrix elements:

$$\widehat A = \epsilon_1 d_1^\dagger d_1 + \epsilon_2 d_2^\dagger d_2 + t d_1^\dagger d_2 + t^* d_2^\dagger d_1.$$

The diagonal terms are Hermitian when $\epsilon_1,\epsilon_2$ are real. The last two terms are adjoints of one another, so the whole operator is Hermitian.

3. Basis covariance

Given $\lvert\chi_p\rangle=\sum_i\lvert\varphi_i\rangle U_{ip}$, show that $A'=U^\dagger A U$ and the transformed ladder operators reproduce the same Fock-space operator.

Solution

The new matrix is

$$\begin{aligned} A'_{pq} &= \langle\chi_p\vert A\vert\chi_q\rangle \\ &= \sum_{ij} U_{ip}^* A_{ij} U_{jq}, \end{aligned}$$

so $A'=U^\dagger A U$. The creation and annihilation operators are

$$e_p^\dagger = \sum_i U_{ip}d_i^\dagger, \qquad e_q = \sum_j U_{jq}^*d_j.$$

Substitution gives

$$\sum_{pq} A'_{pq} e_p^\dagger e_q = \sum_{ij} A_{ij} d_i^\dagger d_j,$$

using $UU^\dagger=I$.

4. One-body expectation

Using $\gamma_{ij}=\langle d_j^\dagger d_i\rangle$, derive $\langle\widehat A\rangle=\operatorname{Tr}_1(A\gamma)$ and evaluate the result for $A=I_1$.

Solution

Directly,

$$\langle\widehat A\rangle = \sum_{ij} A_{ij} \langle d_i^\dagger d_j\rangle = \sum_{ij} A_{ij} \gamma_{ji}.$$

The final sum is exactly

$$\operatorname{Tr}_1(A\gamma).$$

For $A=I_1$,

$$\langle d\Gamma(I_1)\rangle = \operatorname{Tr}_1\gamma = \langle\widehat N\rangle,$$

as required.

Canonical explanations

• One-Body Operators
• Many-Body One-Body Operators
• Creation and Annihilation Operators
• Number Operators
• Field Operators
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Two-Body Operators

Related lookup pages

• Correlation Functions
• Second Quantization Bridge
• Fock Space Bridge
• Common Operators Table

References

• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Westview Press, 1998.
• P. Ring and P. Schuck, The Nuclear Many-Body Problem, Springer, 1980.
• A. J. Coleman and V. I. Yukalov, Reduced Density Matrices: Coulson’s Challenge, Springer, 2000.