research40 min studyLesson 61 in Physics

Second quantization and lattice Hamiltonians

Build many-particle lattice models from occupation states, fermionic operators, hopping, and local interactions.

Before you begin

On this page

The question for today

What you’ll understand

Construct a fermionic occupation basis with a fixed orbital ordering.
Translate hopping and density interactions into creation and annihilation operators.
Separate a change of representation from a physical approximation.

A solid contains too many electrons for a list of individually labeled particle trajectories to be useful. Quantum indistinguishability makes those labels inappropriate anyway. Second quantization organizes the same many-particle quantum mechanics by asking which one-particle orbitals are occupied. Lattice Hamiltonians then make an additional modeling choice: retain selected localized orbitals and the couplings relevant to a specified energy range. Keeping these two steps distinct prevents elegant notation from hiding an uncontrolled approximation.

Occupation replaces particle labels

A one-particle orbital includes every label needed to specify its state, including spin when spin is retained. For $M$ fermionic orbitals, each occupation is zero or one, giving $2^M$ basis states before imposing a fixed particle number. The sector with $N$ particles instead contains $\binom{M}{N}$ states. This rapid growth survives the compact notation: writing a Hamiltonian in one line does not make its eigenvectors inexpensive to find. Bosonic orbitals allow arbitrary nonnegative occupations, so a numerical calculation usually needs an explicitly checked occupation cutoff.

Choose an ordering of the orbitals once and define occupation states by applying creation operators in that order to the vacuum. Exchanging two fermionic creation operators changes the sign. This sign is part of antisymmetry, not an extra repulsive force between electrons. Two opposite-spin electrons may occupy the same spatial orbital because their complete spin-orbitals differ. Conversely, adding a second electron to an already occupied identical spin-orbital gives zero. A useful hand calculation always states its ordering before comparing off-diagonal matrix elements.

|n_1\cdots n_M\rangle=(c_1^\dagger)^{n_1}\cdots(c_M^\dagger)^{n_M}|0\rangle,\qquad n_i\in\{0,1\}

This convention fixes the fermionic signs. The vacuum is the state annihilated by every orbital annihilation operator.↔ Scroll wide equations horizontally.

Operators encode motion and exclusion

The fermionic canonical anticommutation relations say that annihilating and then creating in one orbital differs from doing the operations in the opposite order. Their sum is the identity. They imply that the number operator $n_i=c_i^\dagger c_i$ is a projection: its only eigenvalues are zero and one. A hopping operator $c_i^\dagger c_j$ removes an electron from orbital $j$ and adds it to orbital $i$ , returning zero if the move violates occupation constraints. Its many-particle sign depends on the declared ordering.

A physical Hamiltonian must be Hermitian. Consequently, a hopping amplitude from one orbital to another appears with the complex-conjugate reverse amplitude. A complex phase can encode a magnetic gauge potential, but a phase assigned to one link is not by itself gauge invariant. Products around loops can carry physical flux information. For number-conserving terms, every monomial contains equal numbers of creation and annihilation operators. Their commutator with the total number vanishes even though an individual site occupation need not remain constant during coherent motion.

\{c_i,c_j^\dagger\}=\delta_{ij},\quad \{c_i,c_j\}=0,\quad n_i^2=n_i,\quad \widehat N=\sum_i n_i

These relations are for fermions in an orthonormal orbital basis; bosonic operators instead satisfy canonical commutators.↔ Scroll wide equations horizontally.

Choose a model and audit what it leaves out

A minimal spinful lattice model combines nearest-neighbor hopping with an energy cost for opposite spins sharing a site. The hopping favors delocalization, whereas the interaction changes the relative energy of configurations containing double occupancy. Neither term represents a little electron choosing a classical jump time. Exact quantum evolution is a superposition of configurations. Which processes may be omitted depends on orbital separation, screening, filling, and the energy or temperature window. Longer-range hopping, nonlocal interactions, multiple orbitals, disorder, and lattice motion can each matter in a real material.

The exact two-site, one-particle example isolates hopping, without establishing thermodynamic many-body behavior. The attached two-level lab can illustrate probabilities and relative phase for a single particle distributed between two sites, using the two basis states as left and right. It cannot implement fermionic anticommutation, construct a many-electron Fock space, or solve the interacting Hamiltonian. This comparison identifies which many-body structure has been removed.

H=-t\sum_{\langle ij\rangle,\sigma}(c_{i\sigma}^\dagger c_{j\sigma}+c_{j\sigma}^\dagger c_{i\sigma})+U\sum_i n_{i\uparrow}n_{i\downarrow}

Here each undirected neighboring pair is counted once, t is real, and the retained orbitals form the specified lattice model.↔ Scroll wide equations horizontally.

Work it through

One particle on two sites

Use one spin species and a single link with $t=2\,\mathrm{eV}$ . In the ordered basis $|10\rangle,|01\rangle$ , find the two energies and the occupation of each site in the ground state.

The interaction term is absent for one particle. The hopping matrix is $H=\begin{pmatrix}0&-2\\-2&0\end{pmatrix}\mathrm{eV}$ . Both diagonal entries vanish because neither site has an on-site energy in this example.
The normalized symmetric vector $(1,1)/\sqrt2$ has energy $-2\,\mathrm{eV}$ ; the antisymmetric vector $(1,-1)/\sqrt2$ has energy $+2\,\mathrm{eV}$ . The level separation is therefore $4\,\mathrm{eV}$ .
In the lower state, each basis amplitude has squared magnitude $1/2$ , so $\langle n_1\rangle=\langle n_2\rangle=1/2$ . A measurement finds one complete particle on one site, not half a particle on each site.

The result

Coherent delocalization lowers the ground-state energy by 2 eV relative to a localized basis state with zero energy expectation.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

A one-particle occupation analogy

Interpret the two basis states as one particle on the left or right site. At a mixing angle of $90^\circ$ , vary relative phase while checking that both occupation probabilities remain one half. Playback precesses the phase of this single pure two-level state; it is not many-body lattice evolution.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTwo occupation probabilities alone do not determine the coherence needed for a hopping-energy expectation.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Anticommutation gives the square of a fermionic creation operator as zero. Opposite spin states are different orbitals and must be represented by different operators.

Your turn

Try it yourself

Set $t=0$ and $U=6\,\mathrm{eV}$ in the spinful two-site model with exactly two electrons. Compare one electron on each site with two opposite-spin electrons on the same site. Explain whether the separated configurations have a unique ground state.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Evaluate the product of spin occupations on each site. At zero hopping the Hamiltonian is diagonal in this occupation basis.

Compare with the solution

A separated configuration has no site with both spins occupied and therefore has energy zero. A doubly occupied site contributes one factor $n_{i\uparrow}n_{i\downarrow}=1$ , giving energy $6\,\mathrm{eV}$ . The separated states are lower, but their spin configurations are degenerate in this Hamiltonian at $t=0$ . No exchange splitting or preferred spin pattern can be inferred until an additional process is included.

Keep the idea

Take it with you

Second quantization is a representation of indistinguishable particles; selecting lattice orbitals is a separate approximation.
Fermionic signs require a fixed ordering and cannot be replaced by a classical exclusion rule alone.
A short Hamiltonian can still act on an exponentially large many-body state space.

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