Two-Body Operators

A two-body operator acts on pairs of particles; pair interactions and pair-counting observables are examples. 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 unsymmetrized two-particle matrix elements in an ordered product basis. We assume the same pair operator for every pair and $[v,P_{12}]=0$, so it preserves the identical-particle sector. 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.

Helpful background. One-Body Operators explains lifting from particle slots to modes; Creation and Annihilation Operators fixes ordering.

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.

Action on Number States

For four distinct modes, use the occupation vector $\mathbf n$ and unit vectors $\mathbf e_i$. Then

$$a_i^\dagger a_j^\dagger a_l a_k|\mathbf n\rangle =\sqrt{(n_i+1)(n_j+1)n_kn_l}\, |\mathbf n+\mathbf e_i+\mathbf e_j-\mathbf e_k-\mathbf e_l\rangle.$$

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, incoming modes must be occupied when annihilated, and outgoing modes must be available after those annihilations. An outgoing index may equal an incoming index: for $i\ne j$, $c_i^\dagger c_j^\dagger c_jc_i=N_iN_j$ is nonzero when both modes were initially occupied. 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

$$v(\mathbf x,\mathbf y) = g\,\delta^{(3)}(\mathbf x-\mathbf y),$$

the field-operator form becomes

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

In three spatial dimensions this is a regulated low-energy contact theory, not an unqualified multiplication operator by a delta potential. Matching the scattering length gives the leading dilute-gas coupling $g=4\pi\hbar^2a_s/m$; a cutoff-dependent bare coupling used in loop calculations must be matched consistently. Two-Body Operators in Many-Body Models discusses this effective interaction.

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.

Exercises

1. 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

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

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

2. 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

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

To check the factor, interchange dummy indices $k\leftrightarrow l$ in the exchange contribution. Since $c_kc_l=-c_lc_k$, its minus sign cancels the minus in $V_{ij;kl}-V_{ij;lk}$. The two contributions are equal, turning the prefactor $1/4$ into $1/2$.

3. 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.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• J. Hubbard, “Electron correlations in narrow energy bands”, Proceedings of the Royal Society A 276, 238-257, 1963.
• G. D. Mahan, Many-Particle Physics, 3rd ed., Kluwer Academic/Plenum, 2000.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Addison-Wesley, 1988.