One-Body Operators

A one-body operator is an observable or Hamiltonian term obtained by applying the same single-particle operator to each particle. If $A$ is an operator on the one-particle Hilbert space $\mathcal h$, then its fixed-$N$ action is

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

where $A^{(\alpha)}$ acts as $A$ on the $\alpha$th tensor slot and as the identity on the others. For identical particles this sum is symmetric under relabeling of the formal slots, so it maps the bosonic and fermionic subspaces into themselves.

The companion many-body application guide develops reduced-density-matrix expectations, transition rules, dynamics, truncation, and effective-operator cautions.

In occupation-number language, the same operator has the compact Fock-space form

$$\widehat A = \sum_{ij} A_{ij}\,d_i^\dagger d_j, \qquad A_{ij} = \langle\varphi_i\vert A\vert\varphi_j\rangle.$$

Here $\{\lvert\varphi_i\rangle\}$ is an orthonormal one-particle mode basis, and $d_i,d_i^\dagger$ denote either bosonic mode operators or fermionic mode operators with the appropriate algebra. The bilinear $d_i^\dagger d_j$ removes one particle from mode $j$ and creates one particle in mode $i$, so the net particle number is unchanged.

First-Quantized One-Body Operators

For distinguishable particles, the tensor-slot notation is literal. For three particles, for example,

$$\widehat A^{(3)} = A\otimes I\otimes I + I\otimes A\otimes I + I\otimes I\otimes A.$$

For identical particles, the slots are bookkeeping devices rather than physical labels. The same sum is nevertheless meaningful because it treats every slot identically. This is the standard way to lift a single-particle observable to the $N$-particle sector.

If $A=h$ is a one-particle Hamiltonian, then $\widehat h^{(N)}$ is the part of a many-particle Hamiltonian in which each particle moves independently in the same external one-particle operator. Interactions between pairs of particles are not one-body terms; they require two annihilation operators and two creation operators.

Matrix Elements in a Mode Basis

Choose an orthonormal basis $\{\lvert\varphi_i\rangle\}$ of $\mathcal h$. The one-particle operator can be written as

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

with matrix elements

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

If $A$ is Hermitian, then

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

Diagonal matrix elements weight occupations. Off-diagonal matrix elements transfer occupation between modes. This distinction is one of the main reasons second quantization is useful: it separates “how many particles are in each mode” from “how the operator mixes modes.”

Second-Quantized Form

The Fock-space lift of $A$ is

$$\widehat A = \sum_{ij} A_{ij}\,d_i^\dagger d_j.$$

This formula is often denoted $d\Gamma(A)$ in mathematical treatments. It acts on every particle in a variable-particle-number state while preserving each fixed-$N$ sector.

On the one-particle sector,

$$d_i^\dagger d_j \lvert\varphi_k\rangle = \delta_{jk} \lvert\varphi_i\rangle,$$

so

$$\widehat A \lvert\varphi_k\rangle = \sum_i A_{ik} \lvert\varphi_i\rangle,$$

which is exactly the one-particle action of $A$.

On the vacuum,

$$\widehat A\lvert0\rangle=0,$$

because a one-body operator needs a particle to act on. On an $N$-particle sector, the same formula reproduces the first-quantized sum $\sum_{\alpha=1}^N A^{(\alpha)}$.

Action on Number States

For bosons and $i\ne j$,

$$a_i^\dagger a_j \lvert\ldots,n_i,\ldots,n_j,\ldots\rangle_B = \sqrt{(n_i+1)n_j}\, \lvert\ldots,n_i+1,\ldots,n_j-1,\ldots\rangle_B.$$

For $i=j$ the same bilinear is the number operator:

$$a_i^\dagger a_i=N_i.$$

For fermions, $c_i^\dagger c_j$ moves one fermion from mode $j$ to mode $i$ when mode $j$ is occupied and mode $i$ is empty. The result is zero if $j$ is empty or if $i$ is already occupied. A sign may appear from the chosen ordering of fermionic modes:

$$c_i^\dagger c_j \lvert n_1,n_2,\ldots\rangle_F = \pm \lvert n_1',n_2',\ldots\rangle_F,$$

where $n_j'=0$, $n_i'=1$, and all other occupations are unchanged. The sign is not a new physical rule; it is the occupation-number representation of fermionic antisymmetry.

Examples

Mode Number

Take

$$A = \lvert\varphi_r\rangle \langle\varphi_r\rvert.$$

Then $A_{ij}=\delta_{ir}\delta_{jr}$, so

$$\widehat A = d_r^\dagger d_r = N_r.$$

A mode-occupation operator is therefore a one-body operator associated with a one-particle projector.

Diagonal One-Particle Hamiltonian

If the one-particle Hamiltonian is diagonal,

$$h\lvert\varphi_i\rangle = \epsilon_i\lvert\varphi_i\rangle,$$

then

$$\widehat H_1 = \sum_i \epsilon_i\,N_i.$$

This is the standard free Hamiltonian for independent particles or independent excitations in modes with energies $\epsilon_i$.

Off-Diagonal Hopping or Mixing

For two modes with

$$h = \begin{pmatrix} \epsilon_1 & t \\ t^* & \epsilon_2 \end{pmatrix},$$

the second-quantized Hamiltonian is

$$\widehat H_1 = \epsilon_1 d_1^\dagger d_1 +\epsilon_2 d_2^\dagger d_2 +t\,d_1^\dagger d_2 +t^* d_2^\dagger d_1.$$

The diagonal terms count occupations. The off-diagonal terms transfer occupation between modes.

Spin Operators

If the one-particle basis includes spin states $\lvert s\rangle$, then a spin component $S^a$ is lifted as

$$\widehat S^a = \sum_{ss'} \langle s\vert S^a\vert s'\rangle d_s^\dagger d_{s'}.$$

If the particle also has orbital modes, the mode label must include both orbital and spin quantum numbers. For example, $d_{\mu s}^\dagger$ creates a particle in orbital mode $\mu$ and spin state $s$.

Field-Operator Form

For a continuum position basis, introduce nonrelativistic field operators through the mode expansion:

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

The operator meaning of $\psi(\mathbf x)$ and $\psi^\dagger(\mathbf x)$ is treated in Field Operators.

If $A_x$ is the position-space differential or multiplicative operator representing $A$, then

$$A_{ij} = \int d^3x\, \varphi_i^*(\mathbf x) A_x\varphi_j(\mathbf x).$$

Substituting the mode expansion into the bilinear gives

$$\widehat A = \int d^3x\, \psi^\dagger(\mathbf x) A_x \psi(\mathbf x),$$

with the usual domain and boundary-condition assumptions needed for differential operators.

For a single species of nonrelativistic particles in an external potential,

$$\widehat H_1 = \int d^3x\, \psi^\dagger(\mathbf x) \left( -\frac{\hbar^2}{2m}\nabla^2 +U(\mathbf x) \right) \psi(\mathbf x).$$

The potential part can also be written using the density operator $n(\mathbf x)=\psi^\dagger(\mathbf x)\psi(\mathbf x)$:

$$\widehat U = \int d^3x\, U(\mathbf x)n(\mathbf x).$$

Bosonic and Fermionic Cases

The formula

$$\widehat A = \sum_{ij} A_{ij}d_i^\dagger d_j$$

has the same visible shape for bosons and fermions. What changes is the algebra of the $d$ operators and the allowed occupation numbers.

For bosons,

$$[a_i,a_j^\dagger]=\delta_{ij}, \qquad [a_i,a_j]=0.$$

For fermions,

$$\{c_i,c_j^\dagger\}=\delta_{ij}, \qquad \{c_i,c_j\}=0.$$

In both cases, the ordinary commutator with the total number operator vanishes:

$$[N_{\mathrm{tot}},\widehat A]=0.$$

This is because every term $d_i^\dagger d_j$ creates and annihilates exactly one particle. A one-body operator can move particles among modes, rotate spin, or mix internal states, but it cannot change total particle number.

Basis Dependence and Invariance

The coefficients $A_{ij}$ depend on the chosen one-particle basis. The Fock-space operator $\widehat A$ does not.

If

$$\lvert\chi_\alpha\rangle = \sum_i \lvert\varphi_i\rangle U_{i\alpha},$$

then the transformed matrix is

$$\widetilde A_{\alpha\beta} = \sum_{ij} U_{i\alpha}^* A_{ij}U_{j\beta},$$

and the transformed annihilation operators satisfy

$$\widetilde d_\alpha = \sum_i U_{i\alpha}^*d_i.$$

The bilinear form is invariant:

$$\sum_{\alpha\beta} \widetilde A_{\alpha\beta} \widetilde d_\alpha^\dagger \widetilde d_\beta = \sum_{ij} A_{ij}d_i^\dagger d_j.$$

Thus a mode basis is needed to write the formula, but the physical operator is basis-independent.

Common Mistakes

• Writing only one tensor-slot term in first quantization instead of summing over all particles.
• Replacing a non-diagonal one-body operator by $\sum_i A_{ii}N_i$ and losing the mode-mixing terms.
• Thinking $d_i^\dagger d_j$ changes particle number because it contains a creation operator.
• Forgetting that fermionic bilinears carry signs through the chosen mode ordering.
• Treating spin as an extra decoration rather than part of the complete one-particle mode label.
• Confusing one-body terms with interactions; pair interactions require two-body operators.
• Assuming the field-operator formula is relativistic. Here $\psi(\mathbf x)$ is a nonrelativistic annihilation field.

Cross-Links

• Occupation-Number Basis
• Number States
• Creation and Annihilation Operators
• Number Operators
• Mode Expansions
• Field Operators
• Two-Body Operators
• Many-Particle Hamiltonians
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Second Quantization: Bridge to QFT
• Reference Bridge: Second Quantization
• Fock Space Examples
• Formula Sheet
• Fock Space Exercises

References

• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Addison-Wesley, 1988.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.

Exercises

1. Show that $\widehat A=\sum_{ij}A_{ij}d_i^\dagger d_j$ reproduces the one-particle action of $A$ on $\lvert\varphi_k\rangle$.

Solution

Use $d_j\lvert\varphi_k\rangle=\delta_{jk}\lvert0\rangle$ and $d_i^\dagger\lvert0\rangle=\lvert\varphi_i\rangle$. Then

$$\widehat A\lvert\varphi_k\rangle = \sum_{ij} A_{ij} d_i^\dagger d_j \lvert\varphi_k\rangle = \sum_i A_{ik} \lvert\varphi_i\rangle.$$

This is the expansion of $A\lvert\varphi_k\rangle$ in the same basis.

2. Let $h$ be a two-mode one-particle Hamiltonian with $h_{11}=\epsilon_1$, $h_{22}=\epsilon_2$, $h_{12}=t$, and $h_{21}=t^*$. Write $\widehat H_1$.

Solution

Insert the matrix elements into the one-body formula:

$$\widehat H_1 = \epsilon_1 d_1^\dagger d_1 +\epsilon_2 d_2^\dagger d_2 +t\,d_1^\dagger d_2 +t^*d_2^\dagger d_1.$$

The first two terms count occupations; the last two move occupation between the two modes.

3. Compute $a_1^\dagger a_2\lvert1,3\rangle_B$.

Solution

First annihilate one boson in mode $2$, then create one in mode $1$:

$$a_1^\dagger a_2 \lvert1,3\rangle_B = \sqrt{3}\sqrt{2}\, \lvert2,2\rangle_B = \sqrt6\, \lvert2,2\rangle_B.$$

4. Show that every one-body bilinear $d_i^\dagger d_j$ commutes with $N_{\mathrm{tot}}$.

Solution

For both bosons and fermions,

$$[N_{\mathrm{tot}},d_i^\dagger] = d_i^\dagger, \qquad [N_{\mathrm{tot}},d_j] = -d_j.$$

Therefore

$$[N_{\mathrm{tot}},d_i^\dagger d_j] = [N_{\mathrm{tot}},d_i^\dagger]d_j +d_i^\dagger[N_{\mathrm{tot}},d_j] = d_i^\dagger d_j -d_i^\dagger d_j = 0.$$

5. In position space, write the second-quantized form of a one-particle potential $U(\mathbf x)$.

Solution

The one-particle operator is multiplication by $U(\mathbf x)$, so

$$\widehat U = \int d^3x\, \psi^\dagger(\mathbf x) U(\mathbf x) \psi(\mathbf x).$$

Equivalently, using $n(\mathbf x)=\psi^\dagger(\mathbf x)\psi(\mathbf x)$,

$$\widehat U = \int d^3x\, U(\mathbf x)n(\mathbf x).$$