Two-Body Operators

A two-body operator is an interaction term that acts on pairs of particles. In a fixed-$N$ first-quantized Hilbert space it has the form

$$\widehat V^{(N)} = \sum_{1\le \alpha<\beta\le N} v^{(\alpha\beta)},$$

where $v^{(\alpha\beta)}$ acts on the $\alpha$th and $\beta$th particle slots. In Fock-space notation the same number-conserving pair interaction is written

$$\widehat V = \frac12 \sum_{ijkl} V_{ij;kl} d_i^\dagger d_j^\dagger d_l d_k.$$

Here $d_i,d_i^\dagger$ are bosonic or fermionic mode operators, and

$$V_{ij;kl} = \langle\varphi_i\otimes\varphi_j\vert v \vert\varphi_k\otimes\varphi_l\rangle$$

are two-particle matrix elements in an ordered product basis. The order of the operator string matters: $d_l d_k$ annihilates the incoming modes $k,l$, while $d_i^\dagger d_j^\dagger$ creates the outgoing modes $i,j$.

The companion many-body application guide develops statistics-adapted tensors, two-body reduced density matrices, pair correlations, basis covariance, effective interactions, and computational validation.

First-Quantized Pair Interactions

For distinguishable particles, a pair interaction $v^{(\alpha\beta)}$ acts nontrivially on two tensor slots and as the identity on all the others. For three particles,

$$\widehat V^{(3)} = v^{(12)}+v^{(13)}+v^{(23)}.$$

For identical particles, the slots are still formal bookkeeping labels. A physical pair interaction must treat every unordered pair in the same way. That is why the fixed-$N$ interaction is a sum over unordered pairs:

$$\sum_{\alpha<\beta} v^{(\alpha\beta)}.$$

Equivalently,

$$\sum_{\alpha<\beta} v^{(\alpha\beta)} = \frac12 \sum_{\alpha\ne\beta} v^{(\alpha\beta)}.$$

The factor $1/2$ is already visible in first quantization: the ordered pair $(\alpha,\beta)$ and the ordered pair $(\beta,\alpha)$ describe the same physical pair.

Matrix Elements

Choose an orthonormal one-particle basis $\{\lvert\varphi_i\rangle\}$. The ordered product states

$$\lvert\varphi_k\rangle\otimes\lvert\varphi_l\rangle$$

span the two-particle product space before imposing bosonic or fermionic exchange symmetry. Define

$$V_{ij;kl} = \langle\varphi_i\otimes\varphi_j\vert v \vert\varphi_k\otimes\varphi_l\rangle.$$

If $v$ is Hermitian, then

$$V_{ij;kl} = V_{kl;ij}^*.$$

If the interaction is symmetric under exchange of its two arguments, then

$$V_{ij;kl} = V_{ji;lk}.$$

These symmetries are properties of the matrix elements and the interaction. They do not remove the need to keep a consistent operator ordering in the second-quantized expression.

Second-Quantized Form

The Fock-space lift of the pair interaction is

$$\widehat V = \frac12 \sum_{ijkl} V_{ij;kl} d_i^\dagger d_j^\dagger d_l d_k.$$

The operator $d_l d_k$ first annihilates one particle from mode $k$ and one from mode $l$. The operator $d_i^\dagger d_j^\dagger$ then creates one particle in mode $j$ and one in mode $i$, with the displayed ordering. The term therefore preserves total particle number:

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

It generally does not preserve individual mode occupations. Interactions can scatter particles between modes even when total particle number is fixed.

For bosons, the algebra of the $a_i$ operators supplies the square-root occupation factors. For fermions, the algebra of the $c_i$ operators supplies both Pauli exclusion and the signs from reordering modes.

Factor of One Half

The factor $1/2$ prevents double counting of unordered pairs. In field language, the same convention appears as

$$\widehat V = \frac12 \int d^3x\,d^3y\, \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf y) v(\mathbf x,\mathbf y) \psi(\mathbf y) \psi(\mathbf x).$$

If $v(\mathbf x,\mathbf y)=v(\mathbf y,\mathbf x)$, then the integration over $\mathbf x,\mathbf y$ counts each physical pair twice. The prefactor removes that double counting.

Expanding each field operator in a one-particle basis gives the mode-index expression above; the expansion itself is the topic of Mode Expansions.

The factor can change when the coefficient convention changes. For example, fermionic many-body theory often uses antisymmetrized matrix elements

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

With this convention, the same interaction is often written

$$\widehat V = \frac14 \sum_{ijkl} \langle ij\Vert kl\rangle c_i^\dagger c_j^\dagger c_l c_k.$$

Both formulas are standard. What is not allowed is mixing the prefactor from one convention with the matrix elements from the other.

Fermionic Signs

For fermions, the operator ordering is part of the definition. Interchanging two fermionic creation operators gives a minus sign:

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

Likewise,

$$c_l c_k = -c_k c_l.$$

The matrix-element convention and the operator ordering must be matched. With the unsymmetrized product-basis matrix elements $V_{ij;kl}$, the standard normal-ordered expression is

$$\widehat V = \frac12 \sum_{ijkl} V_{ij;kl} c_i^\dagger c_j^\dagger c_l c_k.$$

The signs produced by the anticommutation relations implement exchange antisymmetry. One should not add extra sign rules by hand after the operator ordering has been fixed.

Action on Number States

For bosons, and for four distinct modes $i,j,k,l$,

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

If some indices coincide, the same operator algebra gives the correct factorial factors. For example,

$$a^\dagger a^\dagger aa \lvert n\rangle_B = n(n-1)\lvert n\rangle_B.$$

For fermions, the corresponding operator is zero unless the incoming modes are occupied and the outgoing modes are available. The sign is determined by the fixed mode ordering used to define the occupation-number basis.

Examples

Contact Interaction for Bosons

For spinless bosons with

$$v(\mathbf x,\mathbf y) = g\,\delta(\mathbf x-\mathbf y),$$

the field-operator form becomes

$$\widehat V_{\mathrm{contact}} = \frac g2 \int d^3x\, \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf x) \psi(\mathbf x) \psi(\mathbf x).$$

This is the standard local interaction used in dilute Bose-gas models. The coupling $g$ is an effective low-energy parameter, not a universal microscopic constant.

Coulomb Interaction

For particles of charge $q$ interacting through the Coulomb potential,

$$v(\mathbf x,\mathbf y) = \frac{q^2} {4\pi\epsilon_0\lvert\mathbf x-\mathbf y\rvert},$$

one writes, suppressing spin labels,

$$\widehat V_C = \frac12 \int d^3x\,d^3y\, \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf y) \frac{q^2} {4\pi\epsilon_0\lvert\mathbf x-\mathbf y\rvert} \psi(\mathbf y) \psi(\mathbf x).$$

For electrons, the complete mode label includes spin, and the field expression normally includes sums over spin components.

Hubbard Interaction Preview

On a lattice with spinful fermions, the on-site Hubbard interaction is

$$\widehat V_H = U\sum_r n_{r\uparrow}n_{r\downarrow}.$$

In creation and annihilation operators,

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

This is a two-body term: it counts pairs of opposite-spin fermions occupying the same lattice site.

Relation to One-Body Operators

A number-conserving one-body operator contains one creation and one annihilation operator:

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

A number-conserving two-body operator contains two creation and two annihilation operators:

$$\widehat V = \frac12 \sum_{ijkl} V_{ij;kl} d_i^\dagger d_j^\dagger d_l d_k.$$

The distinction is physical. One-body terms describe independent motion, external potentials, spin rotations, or mode mixing. Two-body terms describe pair interactions, scattering, and correlations generated by interactions. They combine into the standard many-particle Hamiltonian.

Common Mistakes

• Dropping the factor $1/2$ when summing over ordered pairs or over both integration variables.
• Using the one-body bilinear $d_i^\dagger d_j$ for a genuine pair interaction.
• Mixing unsymmetrized matrix elements with the $\frac14$ prefactor used for antisymmetrized fermionic matrix elements.
• Reordering fermionic operators without tracking the sign.
• Forgetting that spin is part of the mode label in contact, Coulomb, and lattice examples.
• Assuming a two-body operator changes total particle number; the number-conserving form has two creations and two annihilations.
• Treating every interaction as a contact interaction. Coulomb, dipolar, exchange, and effective interactions have different matrix elements and domains of validity.

Cross-Links

• Creation and Annihilation Operators
• One-Body Operators
• Number Operators
• Mode Expansions
• Field Operators
• Interactions and Coupling Terms
• Many-Particle Hamiltonians
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Occupation-Number Basis
• Fermionic Fock Space
• 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.
• G. D. Mahan, Many-Particle Physics, 3rd ed., Kluwer Academic/Plenum, 2000.
• J. Hubbard, “Electron correlations in narrow energy bands”, Proceedings of the Royal Society A 276, 238-257, 1963.

Exercises

1. Explain why $\sum_{\alpha<\beta}v^{(\alpha\beta)}=\frac12\sum_{\alpha\ne\beta}v^{(\alpha\beta)}$ for a symmetric pair interaction.

Solution

The ordered sum contains both $(\alpha,\beta)$ and $(\beta,\alpha)$ for every unordered pair. If the interaction is symmetric, these two terms describe the same physical pair and have the same operator action. Therefore the ordered sum is twice the unordered sum, and the factor $1/2$ removes the double counting.

2. Show that the second-quantized two-body operator preserves total particle number.

Solution

For either bosons or fermions,

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

Thus

$$\begin{aligned} [N_{\mathrm{tot}},d_i^\dagger d_j^\dagger d_l d_k] &= d_i^\dagger d_j^\dagger d_l d_k +d_i^\dagger d_j^\dagger d_l d_k \\ &\quad -d_i^\dagger d_j^\dagger d_l d_k -d_i^\dagger d_j^\dagger d_l d_k \\ &=0. \end{aligned}$$

Every term has two creations and two annihilations, so $\widehat V$ commutes with $N_{\mathrm{tot}}$.

3. Derive the contact-interaction form for spinless bosons from $v(\mathbf x,\mathbf y)=g\delta(\mathbf x-\mathbf y)$.

Solution

Insert the potential into the field expression:

$$\widehat V = \frac12 \int d^3x\,d^3y\, \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf y) g\delta(\mathbf x-\mathbf y) \psi(\mathbf y) \psi(\mathbf x).$$

The delta function sets $\mathbf y=\mathbf x$, giving

$$\widehat V = \frac g2 \int d^3x\, \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf x) \psi(\mathbf x) \psi(\mathbf x).$$

4. For fermions, compare the unsymmetrized and antisymmetrized conventions.

Solution

With unsymmetrized matrix elements,

$$\widehat V = \frac12 \sum_{ijkl} V_{ij;kl} c_i^\dagger c_j^\dagger c_l c_k.$$

With antisymmetrized matrix elements

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

the same interaction is commonly written

$$\widehat V = \frac14 \sum_{ijkl} \langle ij\Vert kl\rangle c_i^\dagger c_j^\dagger c_l c_k.$$

The two forms agree when their prefactors and matrix-element definitions are used consistently.

5. Show that the Hubbard term $U n_{r\uparrow}n_{r\downarrow}$ is a two-body operator.

Solution

Using $n_{r\sigma}=c_{r\sigma}^\dagger c_{r\sigma}$,

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

Move $c_{r\uparrow}$ past $c_{r\downarrow}^\dagger$, picking up a minus sign, and then use the anticommutation of the two annihilation operators to put the result in normal order:

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

The result contains two creation and two annihilation operators, so it is a two-body term.