Many-Particle Hamiltonians

A many-particle Hamiltonian describes how many identical particles or excitations move, interact, and exchange energy among modes. In first quantization, a standard fixed-$N$ Hamiltonian has the form

$$H_N = \sum_{\alpha=1}^N h^{(\alpha)} + \sum_{\alpha<\beta} v^{(\alpha\beta)}.$$

The first sum contains one-body terms: kinetic energy, external potentials, spin couplings, or hopping in a single-particle basis. The second sum contains pair interactions. In second quantization, the same number-conserving Hamiltonian is written

$$H = \sum_{ij} h_{ij}d_i^\dagger d_j + \frac12 \sum_{ijkl} V_{ij;kl} d_i^\dagger d_j^\dagger d_l d_k.$$

Here $d_i,d_i^\dagger$ denote bosonic or fermionic mode operators with the appropriate algebra. The visible structure is simple: one-body terms have one creation and one annihilation operator; two-body terms have two creations and two annihilations. In the displayed form the operator strings are already normal ordered: creation operators stand to the left of annihilation operators.

General Number-Conserving Hamiltonian

Let $\{\lvert\varphi_i\rangle\}$ be an orthonormal one-particle mode basis. A general number-conserving Hamiltonian with up to two-body interactions is

$$H = H_1+V+E_0,$$

with

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

and

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

The constant $E_0$ shifts all energies by the same amount. In nonrelativistic quantum mechanics it is often omitted unless absolute energy references, thermodynamics, or comparisons between sectors matter.

The one-body matrix elements are

$$h_{ij} = \langle\varphi_i\vert h\vert\varphi_j\rangle,$$

and the two-body matrix elements are

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

The Hamiltonian preserves total particle number when every term contains the same number of creation and annihilation operators:

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

It may still change individual mode occupations. Interactions and off-diagonal one-body terms scatter particles between modes even when total number is fixed.

Field-Operator Form

For continuum problems, the same Hamiltonian is often written in field-operator notation:

$$H = \int d^3x\, \psi^\dagger(\mathbf x) h_{\mathbf x} \psi(\mathbf x) + \frac12 \int d^3x\,d^3y\, \psi^\dagger(\mathbf x) \psi^\dagger(\mathbf y) v(\mathbf x,\mathbf y) \psi(\mathbf y) \psi(\mathbf x).$$

For spinful particles, spin labels are included in the fields and summed. For example, a spin-independent interaction has the form

$$V = \frac12 \sum_{st} \int d^3x\,d^3y\, \psi_s^\dagger(\mathbf x) \psi_t^\dagger(\mathbf y) v(\mathbf x,\mathbf y) \psi_t(\mathbf y) \psi_s(\mathbf x).$$

The field form is compact and basis-independent in appearance. The mode form is better for computations in a chosen basis. The mode expansion connects the two.

Noninteracting Particles

If the Hamiltonian contains only a one-body term, then

$$H_0 = \sum_{ij} h_{ij}d_i^\dagger d_j.$$

Choose the mode basis to diagonalize $h$:

$$h\lvert\epsilon_i\rangle = \epsilon_i\lvert\epsilon_i\rangle.$$

Then

$$H_0 = \sum_i \epsilon_i N_i.$$

This formula describes noninteracting bosons, noninteracting fermions, oscillator modes, trap eigenmodes, and single-particle band modes. The statistics enter through the allowed occupations:

$$n_i=0,1,2,\ldots \quad \text{for bosons}, \qquad n_i\in\{0,1\} \quad \text{for fermions}.$$

A noninteracting Bose gas and a noninteracting Fermi gas can therefore share the same one-particle spectrum $\epsilon_i$ while having very different many-particle spectra because their occupation rules differ.

Interacting Particles

Interactions add terms with two creation and two annihilation operators:

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

The operator annihilates particles in modes $k,l$ and creates particles in modes $i,j$. In a momentum basis, this is naturally interpreted as scattering between incoming and outgoing momenta. In a spatial orbital basis, it encodes Coulomb, contact, dipolar, exchange, or effective interactions through the matrix elements.

For a translationally invariant two-body interaction, momentum conservation appears in the matrix elements. A common schematic form is

$$V = \frac{1}{2\Omega} \sum_{\mathbf k,\mathbf p,\mathbf q} \widetilde v(\mathbf q)\, d_{\mathbf k+\mathbf q}^\dagger d_{\mathbf p-\mathbf q}^\dagger d_{\mathbf p} d_{\mathbf k},$$

where $\Omega$ is the volume and $\widetilde v(\mathbf q)$ is the Fourier transform of the potential. The precise normalization depends on the Fourier convention.

Interactions generally make occupation of the noninteracting modes nonconserved:

$$[H,N_i]\ne 0$$

for individual $i$, even when $[H,N_{\mathrm{tot}}]=0$.

Lattice Hamiltonians

In lattice models, modes are often localized orbitals on sites. Lattice Models Overview owns the complete graph, local-space, support, constraint, and effective-model context. The Tight-Binding Model owns the diagonalization and band interpretation of a typical one-body Hamiltonian

$$H_t = \sum_{rs} t_{rs}d_r^\dagger d_s.$$

The one-particle two-site version is Tight-Binding Dimer, where this one-body term reduces to a two-by-two Hamiltonian.

For spinful fermions, the Hubbard model is

$$H_{\mathrm{Hub}} = -t \sum_{\langle r,s\rangle,\sigma} \left( c_{r\sigma}^\dagger c_{s\sigma} + c_{s\sigma}^\dagger c_{r\sigma} \right) + U\sum_r n_{r\uparrow}n_{r\downarrow}.$$

The hopping term is one-body: it moves a fermion between neighboring sites. The on-site interaction is two-body: it counts opposite-spin pairs on the same site.

For bosons on a lattice, the Bose-Hubbard Hamiltonian is often written

$$H_{\mathrm{BH}} = -J \sum_{\langle r,s\rangle} \left( a_r^\dagger a_s + a_s^\dagger a_r \right) + \frac U2 \sum_r n_r(n_r-1) + \sum_r \epsilon_r n_r.$$

The factor $n_r(n_r-1)$ counts unordered pairs of bosons occupying the same site. This is the lattice analog of a local contact interaction. The full conventions, limits, and optical-lattice reduction live in Bose–Hubbard Model.

Pairing Terms Preview

Some effective Hamiltonians include terms that do not conserve particle number:

$$H_{\Delta} = \sum_{\mathbf k} \left( \Delta_{\mathbf k} c_{\mathbf k\uparrow}^\dagger c_{-\mathbf k\downarrow}^\dagger + \Delta_{\mathbf k}^* c_{-\mathbf k\downarrow} c_{\mathbf k\uparrow} \right).$$

These terms create or annihilate pairs, so

$$[H_{\Delta},N_{\mathrm{tot}}]\ne 0.$$

They appear in mean-field descriptions of superconductivity and paired superfluids. This does not mean microscopic electric charge conservation has disappeared. Rather, the effective mean-field Hamiltonian describes a subsystem coupled to a condensate or uses an approximation in which the phase of the pair field is treated as a classical parameter.

Pairing terms often preserve number parity:

$$(-1)^{N_{\mathrm{tot}}},$$

even though they do not preserve $N_{\mathrm{tot}}$ itself.

Mean-Field Approximation Preview

Interacting many-particle Hamiltonians are usually hard. Mean-field approximations replace some operator products by self-consistent averages. A schematic Hartree-type replacement is

$$\psi^\dagger\psi^\dagger\psi\psi \quad\longrightarrow\quad \langle\psi^\dagger\psi\rangle \psi^\dagger\psi + \text{constant terms},$$

with exchange or pairing channels added when required by the problem. The result is an effective one-body Hamiltonian whose coefficients depend on the state being solved for.

Mean-field theory is not exact in general. It is useful when fluctuations around the chosen average are controlled or when it gives a qualitatively correct starting point for perturbation theory. It can also fail badly in low-dimensional systems, near critical points, or when correlations are the main physics.

How to Read a Hamiltonian

When confronted with a second-quantized Hamiltonian, ask:

• What are the modes?
• Are the operators bosonic or fermionic?
• Which terms are one-body, two-body, or effective pairing terms?
• Does the Hamiltonian conserve total particle number?
• Which symmetries are visible: spin, translation, parity, time reversal, particle-hole, or lattice symmetries?
• Is the Hamiltonian microscopic, effective, or mean-field?
• Which basis makes the one-body part simple, and which basis makes the interaction simple?

This habit prevents many mistakes. A formula such as $c_i^\dagger c_j$ is not meaningful until the mode labels $i,j$ and the fermionic ordering convention are specified.

Common Mistakes

• Treating every second-quantized Hamiltonian as if it conserves particle number.
• Confusing individual mode-occupation conservation with total particle-number conservation.
• Dropping the factor $1/2$ in two-body interactions.
• Forgetting spin labels in Hubbard, Coulomb, and pairing terms.
• Treating a mean-field Hamiltonian as an exact microscopic Hamiltonian.
• Mixing Fourier normalization conventions in momentum-space interactions.
• Assuming a diagonal one-body Hamiltonian remains diagonal after interactions are added.

Cross-Links

• Mode Occupations
• Composite Hamiltonians
• Interactions and Coupling Terms
• Fock Space Examples
• Number Operators
• Mode Expansions
• Field Operators
• One-Body Operators
• Two-Body Operators
• Two-Body Operators in Many-Body Models
• Normal Ordering
• Wick’s Theorem Preview
• Second Quantization: Bridge to QFT
• Entanglement in Quantum Chemistry
• Formula Sheet
• Reference Bridge: Second Quantization
• Hubbard Model
• Bose–Hubbard Model
• VQE maps finite many-particle Hamiltonians to measured qubit operators and separates model, representation, optimization, and estimator errors.
• Simulation of Quantum Chemistry applies the operator language to molecular integrals, active spaces, fermion-to-qubit encodings, eigensolvers, and chemistry-specific validation.

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.
• G. D. Mahan, Many-Particle Physics, 3rd ed., Kluwer Academic/Plenum, 2000.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press, 2015.
• J. Hubbard, “Electron correlations in narrow energy bands”, Proceedings of the Royal Society A 276, 238-257, 1963.
• J. Bardeen, L. N. Cooper, and J. R. Schrieffer, “Theory of superconductivity”, Physical Review 108, 1175-1204, 1957.

Exercises

1. Number conservation. Show that

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

commutes with $N_{\mathrm{tot}}=\sum_k d_k^\dagger d_k$.

Solution

For either bosons or fermions,

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

Therefore

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

Every term in $H_1$ commutes with $N_{\mathrm{tot}}$, so $[H_1,N_{\mathrm{tot}}]=0$.

2. Noninteracting spectrum. If $H_0=\sum_i\epsilon_iN_i$, what is the energy of the occupation state $\lvert n_1,n_2,\ldots\rangle$?

Solution

Since $N_i\lvert n_1,n_2,\ldots\rangle=n_i\lvert n_1,n_2,\ldots\rangle$,

$$H_0\lvert n_1,n_2,\ldots\rangle = \left( \sum_i \epsilon_i n_i \right) \lvert n_1,n_2,\ldots\rangle.$$

The allowed values of $n_i$ depend on the statistics.

3. Hubbard terms. In $H_{\mathrm{Hub}}$, identify the one-body and two-body terms.

Solution

The hopping term

$$-t \sum_{\langle r,s\rangle,\sigma} \left( c_{r\sigma}^\dagger c_{s\sigma} + c_{s\sigma}^\dagger c_{r\sigma} \right)$$

is one-body because each term has one creation and one annihilation operator. The interaction

$$U\sum_r n_{r\uparrow}n_{r\downarrow}$$

is two-body because

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

4. Pairing and number. Show that $c_i^\dagger c_j^\dagger$ does not conserve total particle number.

Solution

Use

$$[N_{\mathrm{tot}},c_i^\dagger] = c_i^\dagger, \qquad [N_{\mathrm{tot}},c_j^\dagger] = c_j^\dagger.$$

Then

$$\begin{aligned} [N_{\mathrm{tot}},c_i^\dagger c_j^\dagger] &= [N_{\mathrm{tot}},c_i^\dagger]c_j^\dagger + c_i^\dagger[N_{\mathrm{tot}},c_j^\dagger] \\ &= 2c_i^\dagger c_j^\dagger. \end{aligned}$$

The operator creates two particles, so it changes total number by two.

5. Bose-Hubbard pair counting. Show that $n_r(n_r-1)/2$ counts unordered boson pairs on one site.

Solution

On a number state with $n_r$ bosons at site $r$,

$$\frac12 n_r(n_r-1)$$

is the number of unordered pairs that can be chosen from $n_r$ objects. For example, it gives $0$ for $n_r=0,1$, $1$ for $n_r=2$, and $3$ for $n_r=3$.