Field Operators and Many-Body Wavefunctions

Writing a many-electron wavefunction is an exercise in bookkeeping you would rather not do. Six electrons demand a $6\times6$ Slater determinant, seven demand $7\times7$ , and the antisymmetry has to be nailed on by hand every single time. Worse, the particle number is welded into the very shape of the Hilbert space, so a process that creates or destroys a particle — a photon absorbed, a Cooper pair broken — has no natural home. There is a formalism that dissolves all of this, and its central objects are the field operators $\hat{\psi}(\mathbf{r})$ and $\hat{\psi}^\dagger(\mathbf{r})$ . A field operator is nothing mystical: it is the creation/annihilation algebra rewritten in the position basis. Where a mode operator $\hat{a}_k^\dagger$ adds a particle to orbital $k$ , the field operator $\hat{\psi}^\dagger(\mathbf{r}) = \sum_k \phi_k^*(\mathbf{r})\,\hat{a}_k^\dagger$ adds one at the point $\mathbf{r}$ . The per-mode ladder algebra — how $\hat{a}_k$ and $\hat{a}_k^\dagger$ raise and lower occupation — is developed in Creation and Annihilation Operators; here we graft position onto it and watch what falls out. What falls out is remarkable. Because two creation operators (anti)commute, the amplitude $\langle 0|\hat{\psi}(\mathbf{r}_N)\cdots\hat{\psi}(\mathbf{r}_1)|\Psi\rangle$ is automatically symmetric (bosons) or antisymmetric (fermions) — the Slater determinant assembles itself out of the operator ordering, and the Pauli principle becomes the one-line identity $(\hat{\psi}^\dagger(\mathbf{r}))^2 = 0$ . Observables lose their $N$ -dependence: one kinetic-energy integral and one interaction integral govern a single atom or a whole metal. This article builds the definitions, derives the canonical algebra and the wavefunction bridge, and closes the loop back to the determinants and permanents you already know — showing they were the algebra's output all along.

Key takeaways

Field operators $\hat{\psi}^\dagger(\mathbf{r})$ and $\hat{\psi}(\mathbf{r})$ create and annihilate a particle at a point $\mathbf{r}$ , packaging the whole ladder algebra of Creation and Annihilation Operators into a continuous position label.
They obey $[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')]_\mp = \delta^3(\mathbf{r}-\mathbf{r}')$ — commutator for bosons, anticommutator for fermions — the position-space face of the spin–statistics split.
Acting on the vacuum, $N$ field operators build the totally (anti)symmetric $N$ -particle state with no explicit symmetrization: the algebra does the work.
The first-quantized wavefunction is an amplitude, $\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N) = \tfrac{1}{\sqrt{N!}}\langle 0|\hat{\psi}(\mathbf{r}_N)\cdots\hat{\psi}(\mathbf{r}_1)|\Psi\rangle$ , reproducing Slater Determinants and permanents exactly.
One- and two-body operators collapse to $N$ -independent integrals $\hat{H}=\int\hat{\psi}^\dagger\hat{h}\hat{\psi} + \tfrac12\int\hat{\psi}^\dagger\hat{\psi}^\dagger V\hat{\psi}\hat{\psi}$ .
Key idea: a field operator is the creation/annihilation algebra written in the position basis, $\hat{\psi}^\dagger(\mathbf{r})=\sum_k\phi_k^*(\mathbf{r})\hat{a}_k^\dagger$ ; from the single fact $[\hat{\psi},\hat{\psi}^\dagger]_\mp=\delta^3$ , every symmetric/antisymmetric many-body wavefunction follows automatically.

The core insight

A field operator is the creation/annihilation algebra written in the position basis: $\hat{\psi}^\dagger(\mathbf{r})=\sum_k\phi_k^*(\mathbf{r})\,\hat{a}_k^\dagger$ . Its (anti)commutation relation with $\hat{\psi}(\mathbf{r}')$ is the position-space delta function, and from that one algebraic fact the entire apparatus of (anti)symmetric many-body wavefunctions follows.

Where this lives today

Field operators are the working language of every modern many-body laboratory. Cold-atom experiments realize the Bose–Hubbard and fermionic Hubbard models by expanding $\hat{\psi}(\mathbf{r})=\sum_i w_i(\mathbf{r})\hat{a}_i$ in localized Wannier orbitals, so that the on-site interaction $U$ and hopping $t$ are just matrix elements of the second-quantized one- and two-body operators. Quantum-gas microscopes then image the density operator $\hat{n}(\mathbf{r})$ site by site, turning an abstract operator into a photograph of atoms on a lattice. The same machinery drives the numerics that predict materials: coupled-cluster and configuration-interaction methods in quantum chemistry, dynamical mean-field theory for correlated electrons, and tensor-network states all live in the occupation-number representation that field operators define. And the direct descendant is quantum field theory itself, where $\hat{\psi}$ is promoted to a relativistic field and the same creation/annihilation structure describes particle production at colliders.

The physical picture

Picture space threaded by a complete set of single-particle orbitals $\{\phi_k(\mathbf{r})\}$ — plane waves, harmonic-oscillator states, Wannier functions, whatever basis suits the problem. Each orbital $k$ has an occupation you can raise with $\hat{a}_k^\dagger$ or lower with $\hat{a}_k$ . The field operator $\hat{\psi}^\dagger(\mathbf{r})$ is the one superposition of these mode creators that deposits a particle sharply at the geometric point $\mathbf{r}$ . It is the exact operator analogue of the first-quantized identity $|\mathbf{r}\rangle = \sum_k \phi_k^*(\mathbf{r})\,|k\rangle$ : swap each ket $|k\rangle$ for $\hat{a}_k^\dagger|0\rangle$ and you have $\hat{\psi}^\dagger(\mathbf{r})|0\rangle = |\mathbf{r}\rangle$ .

Figure

Field operator as a superposition of mode creators

Why does antisymmetry arrive for free? Because for fermions $\hat{\psi}^\dagger(\mathbf{r}_1)\hat{\psi}^\dagger(\mathbf{r}_2) = -\hat{\psi}^\dagger(\mathbf{r}_2)\hat{\psi}^\dagger(\mathbf{r}_1)$ . Deciding which point to "fill first" costs a minus sign, and that single algebraic statement is the Pauli principle and is every Slater determinant. You never write a determinant down; the operator ordering writes it for you. Set $\mathbf{r}_1 = \mathbf{r}_2$ and the state vanishes — two fermions cannot sit at the same point, the content of $(\hat{\psi}^\dagger)^2 = 0$ .

Figure

Antisymmetry from operator ordering — the exchange node

The great economy shows up in observables. A two-body Coulomb interaction in first quantization is a sum $\sum_{i<j}V(\mathbf{r}_i-\mathbf{r}_j)$ whose very form depends on how many particles are present. In second quantization it is the single, $N$ -independent expression $\tfrac12\iint\hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r}')V(\mathbf{r}-\mathbf{r}')\hat{\psi}(\mathbf{r}')\hat{\psi}(\mathbf{r})$ . One operator governs a hydrogen atom, a block of copper, and a dilute gas. And because the number density $\hat{n}(\mathbf{r})=\hat{\psi}^\dagger(\mathbf{r})\hat{\psi}(\mathbf{r})$ integrates to a number operator with eigenvalues $0,1,2,\dots$ , Fock space holds all particle numbers at once — impossible in fixed- $N$ first quantization, indispensable the moment particles appear or vanish.

The core equations

The definition ties everything to the single-particle basis. The field operator is the mode-operator vector contracted against the orbitals, so that "acting at $\mathbf{r}$ " means "the superposition of mode actions that a point-localized state would demand."

\hat{\psi}(\mathbf{r})=\sum_k \phi_k(\mathbf{r})\,\hat{a}_k,\qquad \hat{\psi}^\dagger(\mathbf{r})=\sum_k \phi_k^*(\mathbf{r})\,\hat{a}_k^\dagger \tag{1}

where: $\hat{\psi}(\mathbf{r})$ removes a particle at $\mathbf{r}$ ; $\hat{\psi}^\dagger(\mathbf{r})$ adds one at $\mathbf{r}$ ; $\phi_k(\mathbf{r})=\langle\mathbf{r}|k\rangle$ are complete orthonormal single-particle orbitals; $\hat{a}_k,\hat{a}_k^\dagger$ are the mode operators whose ladder algebra $\hat{a}_k^\dagger|n_k\rangle=\sqrt{n_k+1}\,|n_k+1\rangle$ is owned by Creation and Annihilation Operators. Read $\hat{\psi}^\dagger(\mathbf{r})$ as "sum over all the ways to add a particle, weighted by how much orbital $k$ overlaps the point $\mathbf{r}$ ." The canonical algebra is the whole engine. It descends from the mode (anti)commutators of Creation and Annihilation Operators by one application of basis completeness, and it is where the boson/fermion choice lives.

[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')]_{\mp}=\delta^3(\mathbf{r}-\mathbf{r}'),\qquad [\hat{\psi}(\mathbf{r}),\hat{\psi}(\mathbf{r}')]_{\mp}=0 \tag{2}

where: $[A,B]_\mp$ is the commutator ( $-$ , bosons) or anticommutator ( $+$ , fermions); $\delta^3$ is the three-dimensional Dirac delta. The right-hand side is a delta and not a Kronecker $\delta_{kk'}$ precisely because the label is now continuous — this is the position-space image of the mode relation. The wavefunction is an amplitude, not an operator. This is the bridge that makes second quantization equivalent to first quantization rather than a new theory: the familiar $\Psi$ is recovered as the overlap of the abstract state with a string of point-localizing field operators.

\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N)=\frac{1}{\sqrt{N!}}\,\langle 0|\hat{\psi}(\mathbf{r}_N)\cdots\hat{\psi}(\mathbf{r}_1)|\Psi\rangle \tag{3}

where: the $1/\sqrt{N!}$ makes $\langle\Psi|\Psi\rangle=\int|\Psi|^2$ ; the raw overlap (without the prefactor) already delivers the un-normalized (anti)symmetric combination, and the $1/\sqrt{N!}$ is the normalization — a distinction worth pinning down, and one the pitfall box returns to. Observables lose their particle-number scaffolding. A one-body operator is a single spatial integral; the two-body interaction is a double integral with a specific "sandwich" ordering that automatically forbids self-interaction.

\hat{O}_1=\int\! d^3r\,\hat{\psi}^\dagger(\mathbf{r})\,o(\mathbf{r})\,\hat{\psi}(\mathbf{r})=\sum_{k,l}\langle k|o|l\rangle\,\hat{a}_k^\dagger\hat{a}_l,\qquad \hat{V}=\frac{1}{2}\iint d^3r\,d^3r'\,\hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r}')\,V\,\hat{\psi}(\mathbf{r}')\hat{\psi}(\mathbf{r}) \tag{4}

where: $o(\mathbf{r})$ is any one-body operator (kinetic $-\hbar^2\nabla^2/2m$ , external $U(\mathbf{r})$ ); $\langle k|o|l\rangle=\int\phi_k^*\,o\,\phi_l$ ; the factor $\tfrac12$ removes double-counting of the unordered pair; the reversed inner order $\hat{\psi}(\mathbf{r}')\hat{\psi}(\mathbf{r})$ (note the swap versus the outer $\hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r}')$ ) is what makes the term produce exactly $\sum_{i<j}V(\mathbf{r}_i-\mathbf{r}_j)$ with both direct and exchange pieces. Which relation to reach for depends on what you are doing, and the following table is the practical map.

you have	use	it gives
a definition of $\hat{\psi}$ in a chosen basis	Eq. (1)	the operator in terms of the mode ladder
two field operators to reorder	Eq. (2)	the (anti)commutator, with the delta or a sign
an abstract Fock state $\|\Psi\rangle$	Eq. (3)	the ordinary $\Psi(\mathbf{r}_1,\dots)$ , (anti)symmetric for free
a one- or two-body observable	Eq. (4)	an $N$ -independent integral of field operators

Derivation AFrom the mode algebra to the position-space delta

The goal is to prove Eq. (2): that the field operators inherit a Dirac-delta (anti)commutator from the Kronecker-delta mode algebra. The strategy is to substitute the definition (1) into the bracket, push the bracket onto the mode operators where the algebra of Creation and Annihilation Operators applies, and then collapse the double sum with basis completeness. The one thing to watch is that the sign convention ( $\mp$ ) rides along untouched from the mode operators to the fields, so the boson/fermion choice is made once, at the level of $\hat{a}_k$ , and never again.

Step 1. Write the bracket using the definitions, keeping the statistics sign $\mp$ symbolic. Because the coefficients $\phi_k(\mathbf{r})$ are ordinary complex numbers, they pull straight out of the bracket:

[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')]_\mp=\sum_{k,k'}\phi_k(\mathbf{r})\,\phi_{k'}^*(\mathbf{r}')\,[\hat{a}_k,\hat{a}_{k'}^\dagger]_\mp.

Why the coefficients factor out: $\phi_k(\mathbf{r})$ is a number, and brackets are bilinear, so only the operator part carries the algebra.

Step 2. Insert the mode relation $[\hat{a}_k,\hat{a}_{k'}^\dagger]_\mp=\delta_{kk'}$ — this is the result owned by Creation and Annihilation Operators. The Kronecker delta kills one sum:

=\sum_{k,k'}\phi_k(\mathbf{r})\,\phi_{k'}^*(\mathbf{r}')\,\delta_{kk'}=\sum_k\phi_k(\mathbf{r})\,\phi_k^*(\mathbf{r}').

Step 3. Recognize the surviving sum as the completeness relation of the single-particle basis. Because $\{\phi_k\}$ is complete, $\sum_k\phi_k(\mathbf{r})\phi_k^*(\mathbf{r}')=\langle\mathbf{r}|\mathbf{r}'\rangle=\delta^3(\mathbf{r}-\mathbf{r}')$ :

[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')]_\mp=\delta^3(\mathbf{r}-\mathbf{r}').

The same substitution with $[\hat{a}_k,\hat{a}_{k'}]_\mp=0$ gives $[\hat{\psi}(\mathbf{r}),\hat{\psi}(\mathbf{r}')]_\mp=0$ . Verified: a discrete-space check confirms both the orthonormality $\sum_x\phi_k(x)\phi_l(x)=\delta_{kl}$ and the completeness $\sum_k\phi_k(x)\phi_k(x')=\delta_{xx'}$ for a particle-in-a-box (DST) basis to machine precision; matrix representations of the fermionic modes reproduce $\{\hat{a}_k,\hat{a}_l^\dagger\}=\delta_{kl}$ and $(\hat{a}_k^\dagger)^2=0$ exactly.

Go deeper · Derivation B · 1 minThe two-body wavefunction emerges (anti)symmetric from the ordering

Derivation BThe two-body wavefunction emerges (anti)symmetric from the ordering

This is the structural heart of the formalism, and the result this article owns: the (anti)symmetry of a many-body wavefunction is not imposed — it is output by the operator ordering. The goal is to compute the raw amplitude $\langle 0|\hat{\psi}(\mathbf{r}_1)\hat{\psi}(\mathbf{r}_2)|1_a1_b\rangle$ for a two-particle state $|1_a1_b\rangle=\hat{a}_a^\dagger\hat{a}_b^\dagger|0\rangle$ occupying orthonormal orbitals $\phi_a,\phi_b$ , and to read off its symmetry. The strategy is to expand each field operator in modes, keep only the terms that survive against the two occupied creators, and let the (anti)commutation signs assemble the answer. Watch the normalization: the raw overlap carries no $1/\sqrt{2}$ — that lives in the wavefunction definition (3).

Step 1. Expand $\hat{\psi}(\mathbf{r}_i)=\phi_a(\mathbf{r}_i)\hat{a}_a+\phi_b(\mathbf{r}_i)\hat{a}_b+(\text{other modes})$ . Only the $a,b$ terms can survive against $\hat{a}_a^\dagger\hat{a}_b^\dagger|0\rangle$ ; any other annihilator hits the vacuum on the right. So four products of two annihilators contribute.

Step 2. Evaluate the surviving amplitudes with the algebra. For fermions, anticommuting $\hat{a}_a\hat{a}_b$ through $\hat{a}_a^\dagger\hat{a}_b^\dagger$ picks up a relative minus sign between the $(a\!:\!\mathbf{r}_1,\,b\!:\!\mathbf{r}_2)$ pairing and the swapped one. The result is

\langle 0|\hat{\psi}(\mathbf{r}_1)\hat{\psi}(\mathbf{r}_2)|1_a1_b\rangle=\pm\big[\phi_a(\mathbf{r}_1)\phi_b(\mathbf{r}_2)-\phi_a(\mathbf{r}_2)\phi_b(\mathbf{r}_1)\big]\quad(\text{fermions}),

where the overall $\pm$ is a fixed ordering convention.

Step 3. Repeat for bosons: the commutator gives no relative sign, so the two pairings add:

\langle 0|\hat{\psi}(\mathbf{r}_1)\hat{\psi}(\mathbf{r}_2)|1_a1_b\rangle=\phi_a(\mathbf{r}_1)\phi_b(\mathbf{r}_2)+\phi_a(\mathbf{r}_2)\phi_b(\mathbf{r}_1)\quad(\text{bosons}).

Step 4. Read off the symmetry and normalize. Swapping $\mathbf{r}_1\leftrightarrow\mathbf{r}_2$ flips the fermion amplitude ( $-1$ ) and fixes the boson amplitude ( $+1$ ) — antisymmetry and symmetry, produced entirely by the algebra. Multiplying by the definition's $1/\sqrt{2}$ gives the unit-normalized wavefunctions $\Psi=\tfrac{1}{\sqrt2}[\phi_a(\mathbf{r}_1)\phi_b(\mathbf{r}_2)\mp\phi_a(\mathbf{r}_2)\phi_b(\mathbf{r}_1)]$ — the two-fermion Slater determinant and the two-boson permanent, recovered without ever writing one. Setting $\mathbf{r}_1=\mathbf{r}_2$ in the fermion case gives $0$ : the exclusion node. Verified: an explicit two-mode Fock-space computation (Jordan–Wigner matrices for fermions, truncated occupation-number matrices for bosons) reproduces the raw overlaps above exactly for generic orthonormal $\phi_a,\phi_b$ ; the fermion amplitude is antisymmetric with a vanishing diagonal, the boson amplitude is symmetric, and the $1/\sqrt2$ -scaled wavefunction is unit-normalized to machine precision.

Go deeper · Derivation C · 1 minThe Heisenberg equation for the field

Derivation CThe Heisenberg equation for the field

The goal is to show that under a one-body Hamiltonian the field operator itself obeys the single-particle Schrödinger equation as an operator equation — the cleanest statement that second quantization contains first quantization. The strategy is to compute $i\hbar\,\partial_t\hat{\psi}=[\hat{\psi},\hat{H}]$ with $\hat{H}=\int\hat{\psi}^\dagger\hat{h}\hat{\psi}$ , using the algebra to collapse the commutator to a delta and integrate it away. The subtlety is that $[\hat{\psi},\hat{H}]$ is a genuine commutator regardless of statistics (because $\hat{H}$ is an even, number-conserving observable), even though the fields themselves anticommute for fermions.

Step 1. Set up the commutator with $\hat{h}(\mathbf{r}')=-\tfrac{\hbar^2}{2m}\nabla'^2+U(\mathbf{r}')$ :

[\hat{\psi}(\mathbf{r}),\hat{H}]=\int d^3r'\,\hat{h}(\mathbf{r}')\,[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')\hat{\psi}(\mathbf{r}')].

Step 2. Evaluate the inner commutator for both statistics. Using $[\hat{A},\hat{B}\hat{C}]=\{\hat{A},\hat{B}\}\hat{C}-\hat{B}\{\hat{A},\hat{C}\}$ for fermions (only the first anticommutator is nonzero) and the ordinary identity for bosons gives the same result:

[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')\hat{\psi}(\mathbf{r}')]=\delta^3(\mathbf{r}-\mathbf{r}')\,\hat{\psi}(\mathbf{r}').

Why statistics drops out here: the number-like combination $\hat{\psi}^\dagger\hat{\psi}$ commutes with $\hat{\psi}$ in the same way for either sign, because the two sign flips from the two anticommutations cancel.

Step 3. Integrate the delta and assemble the Heisenberg equation:

i\hbar\,\partial_t\hat{\psi}(\mathbf{r},t)=[\hat{\psi},\hat{H}]=\Big(-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf{r})\Big)\hat{\psi}(\mathbf{r},t).

The field obeys the one-particle Schrödinger equation. Adding the two-body term appends a nonlinear piece $\int V(\mathbf{r},\mathbf{r}')\hat{\psi}^\dagger(\mathbf{r}')\hat{\psi}(\mathbf{r}')\,d^3r'\;\hat{\psi}(\mathbf{r})$ , whose mean-field reduction is the Gross–Pitaevskii/Hartree equation. Verified: the matrix identity $[\hat{a}_j,\sum_{kl}h_{kl}\hat{a}_k^\dagger\hat{a}_l]=\sum_l h_{jl}\hat{a}_l$ is confirmed to machine precision for a random Hermitian $h$ in both the fermionic Fock space and (below the occupation-truncation ceiling) the bosonic one.

Limiting cases

The framework specializes cleanly, and each limit is a sanity check that it contains what it should.

case	result	meaning
single particle ( $N=1$ )	$\|\Psi\rangle=\int d^3r\,\Psi(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r})\|0\rangle$ , and $\Psi(\mathbf{r})=\langle 0\|\hat{\psi}(\mathbf{r})\|\Psi\rangle$	second quantization collapses to ordinary one-particle QM with no surplus structure
fermions	$\{\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')\}=\delta^3,\;(\hat{\psi}^\dagger)^2=0$	anticommutators enforce antisymmetry; nilpotency is the Pauli principle as an identity
bosons	$[\hat{\psi}(\mathbf{r}),\hat{\psi}^\dagger(\mathbf{r}')]=\delta^3$	commutators give symmetric states (permanents); repeated occupation allowed
mean field / condensate	$\hat{\psi}(\mathbf{r})\to\langle\hat{\psi}(\mathbf{r})\rangle=\Psi_0(\mathbf{r})$	the field acquires a classical value; $\hat{H}$ yields the Gross–Pitaevskii equation

Common trapThe field operator is not the wavefunction

The most common trap is to read $\hat{\psi}(\mathbf{r})$ as "the wavefunction at $\mathbf{r}$ ." It is not: $\hat{\psi}(\mathbf{r})$ is an operator on Fock space that removes a particle at $\mathbf{r}$ ; the wavefunction is the number $\Psi(\mathbf{r})=\langle 0|\hat{\psi}(\mathbf{r})|\Psi\rangle$ in the one-particle sector. A second, subtler trap concerns that $1/\sqrt{N!}$ in Eq. (3): the raw overlap $\langle 0|\hat{\psi}(\mathbf{r}_N)\cdots\hat{\psi}(\mathbf{r}_1)|\Psi\rangle$ already gives the un-normalized (anti)symmetric combination — e.g. $\phi_a(\mathbf{r}_1)\phi_b(\mathbf{r}_2)\mp\phi_a(\mathbf{r}_2)\phi_b(\mathbf{r}_1)$ for two particles — and it is the definition's $1/\sqrt{N!}$ (not any hidden factor in the operators) that turns it into the unit-normalized Slater determinant. Finally: never casually reorder two $\hat{\psi}^\dagger$ 's for fermions — the sign flip is the antisymmetry, and dropping it silently corrupts the physics.

Worked examples

Example 1 — The two-fermion amplitude, computed. Take two orthonormal orbitals and evaluate the raw amplitude and the normalized wavefunction at a concrete pair of points. Let $\phi_a,\phi_b$ be an orthonormal pair on a two-point space, $\phi_a=(\cos\theta,\sin\theta)$ and $\phi_b=(-\sin\theta,\cos\theta)$ with $\theta=0.3\,$ rad, so $\phi_a\cdot\phi_a=\phi_b\cdot\phi_b=1$ and $\phi_a\cdot\phi_b=0$ .

Step 1. Enumerate the two positions $x_1=0,\,x_2=1$ . Then $\phi_a(0)=\cos0.3=0.9553$ , $\phi_b(1)=\cos0.3=0.9553$ , $\phi_a(1)=\sin0.3=0.2955$ , $\phi_b(0)=-\sin0.3=-0.2955$ .

Step 2. Apply the fermion result: the raw overlap is $\phi_a(x_1)\phi_b(x_2)-\phi_a(x_2)\phi_b(x_1) = (0.9553)(0.9553)-(0.2955)(-0.2955)=0.9126+0.0873=1.0000$ .

Step 3. Compute the diagonal (exchange node). At $x_1=x_2=0$ : $\phi_a(0)\phi_b(0)-\phi_a(0)\phi_b(0)=0$ identically — the amplitude to find both fermions at the same point is exactly zero, the Pauli hole.

Step 4. Normalize. The wavefunction is $\Psi=\tfrac{1}{\sqrt2}\times(\text{raw})$ , so $\Psi(0,1)=1/\sqrt2=\mathbf{0.7071}$ and $\Psi(0,0)=\mathbf{0}$ . A direct Fock-space matrix computation reproduces exactly $\langle 0|\hat{\psi}(x_1)\hat{\psi}(x_2)|1_a1_b\rangle=\pm(\phi_a\phi_b-\text{swap})$ , antisymmetric, with the diagonal node.

Example 2 — Kinetic energy of a filled Fermi sea. Consider $N$ free fermions filling plane-wave modes $\mathbf{k}$ up to the Fermi momentum $k_F$ . The kinetic operator is $\hat{T}=\sum_{\mathbf{k}}\tfrac{\hbar^2 k^2}{2m}\hat{a}_{\mathbf{k}}^\dagger\hat{a}_{\mathbf{k}}$ , the mode form of Eq. (4) with $o=-\hbar^2\nabla^2/2m$ diagonal in $\mathbf{k}$ .

Step 1. The ground-state expectation is $\langle\hat{T}\rangle=\sum_{|\mathbf{k}|<k_F}\tfrac{\hbar^2k^2}{2m}$ , since $\langle\hat{a}_{\mathbf{k}}^\dagger\hat{a}_{\mathbf{k}}\rangle=1$ for occupied modes and $0$ otherwise.

Step 2. Convert the sum to an integral (spin- $\tfrac12$ , volume $V$ ): $\langle\hat{T}\rangle=\dfrac{2V}{(2\pi)^3}\displaystyle\int_0^{k_F}\dfrac{\hbar^2k^2}{2m}\,4\pi k^2\,dk=\dfrac{V\hbar^2}{2\pi^2 m}\dfrac{k_F^5}{5}$ .

Step 3. Express per particle using $N=\dfrac{2V}{(2\pi)^3}\dfrac{4\pi}{3}k_F^3=\dfrac{V k_F^3}{3\pi^2}$ . Dividing, $\dfrac{\langle\hat{T}\rangle}{N}=\dfrac{3}{5}\dfrac{\hbar^2 k_F^2}{2m}=\dfrac{3}{5}E_F$ .

Step 4. The average kinetic energy per electron is three-fifths of the Fermi energy — the standard free-electron-gas result, obtained here from a single $N$ -independent operator with the occupation supplied by $\langle\hat{a}_{\mathbf{k}}^\dagger\hat{a}_{\mathbf{k}}\rangle$ . No wavefunction was written.

Origins · 1 minField Operator and how this was discovered

Field Operator

1927–1928 · Göttingen

The quantized field entered physics from two directions at once. In 1927 Dirac quantized the electromagnetic field, promoting its Fourier amplitudes to creation and annihilation operators to explain spontaneous emission — the first field operators. That same year Jordan and Klein wrote down a second-quantized matter field for bosons with the commutator $[\hat{\psi},\hat{\psi}^\dagger]=\delta$ ; in 1928 Jordan and Wigner supplied the fermionic version with the anticommutator, the algebraic origin of the Pauli principle. Fock's 1932 construction of the direct-sum "Fock space" gave the states a permanent home. The formalism was the launch pad for quantum field theory, where $\hat{\psi}$ becomes a relativistic field.

History

1927

— Dirac quantizes the electromagnetic field (creation/annihilation operators), Proc. R. Soc. A 114, 243

1927

— Jordan and Klein give the second-quantized boson matter field, Z. Phys. 45, 751

1928

— Jordan and Wigner introduce anticommuting fermion field operators, Z. Phys. 47, 631

1932

— Fock constructs Fock space, the direct sum over particle number, Z. Phys. 75, 622

1957

— Bogoliubov's field-operator theory of the weakly interacting Bose gas

1961

— Gross and Pitaevskii obtain the mean-field condensate equation from the field Hamiltonian

Going further

References. Sakurai & Napolitano, Modern Quantum Mechanics, §7 (Identical Particles, Second Quantization) · Shankar, Principles of Quantum Mechanics, Ch. 10 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics Vol. II, Ch. XIV–XV · Fetter & Walecka, Quantum Theory of Many-Particle Systems, Ch. 1 · Bruus & Flensberg, Many-Body Quantum Theory in Condensed Matter Physics, Ch. 1 · Negele & Orland, Quantum Many-Particle Systems, Ch. 1.

Provenance. Definitions and the (anti)commutation algebra follow Fetter & Walecka Ch. 1, Sakurai & Napolitano §7, and Bruus & Flensberg Ch. 1. Every algebraic claim here is machine-checked: the field (anti)commutator from mode algebra plus completeness, the two-body amplitude emerging (anti)symmetric from operator ordering (explicit two-mode Fock spaces, both statistics), the Slater determinant from Wick contraction ( $N=2,3$ ), $(\hat{\psi}^\dagger)^2=0$ , the number-operator spectrum, the Heisenberg field equation, and the single-particle limit — 20 checks, all green. The per-mode ladder algebra is quoted from, not re-derived here — its owner is Creation and Annihilation Operators.

Builds on: Indistinguishability of Identical Particles · Symmetric and Antisymmetric States · Bosons and Fermions · Slater Determinants · Creation and Annihilation Operators. Leads to: Permanents and Symmetrized Boson States · Exchange Forces and Exchange Energy · Pauli Exclusion Principle · Spin–Statistics Theorem.

:::faq[Does $[\hat{\psi},\hat{H}]$ use a commutator even for fermions?] Yes. Although the fields anticommute, the equation of motion $i\hbar\,\partial_t\hat{\psi}=[\hat{\psi},\hat{H}]$ always uses a genuine commutator, because $\hat{H}$ is a number-conserving (even) observable. When you evaluate $[\hat{\psi},\hat{\psi}^\dagger\hat{\psi}]$ using $[\hat{A},\hat{B}\hat{C}]=\{\hat{A},\hat{B}\}\hat{C}-\hat{B}\{\hat{A},\hat{C}\}$ , the two sign flips from the two anticommutations cancel, and you get $\delta^3\hat{\psi}$ — the same as for bosons. That is why the field obeys the single-particle Schrödinger equation regardless of statistics.

Is "second quantization" quantizing the wavefunction a second time?

No — the name is historical and misleading. Nothing is quantized twice. Second quantization is a change of representation: from labelling states by particle coordinates $\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N)$ to labelling them by how many particles occupy each mode. It is exactly equivalent to ordinary ("first-quantized") quantum mechanics, as the wavefunction bridge $\Psi=\tfrac{1}{\sqrt{N!}}\langle 0|\hat{\psi}\cdots\hat{\psi}|\Psi\rangle$ shows. The field $\hat{\psi}$ is an operator, not a re-quantized $\Psi$ .

Why a Dirac delta on the right of the field (anti)commutator, not a Kronecker delta?

Because the label is continuous. The mode operators carry a discrete index $k$ and satisfy $[\hat{a}_k,\hat{a}_{k'}^\dagger]_\mp=\delta_{kk'}$ , a Kronecker delta. Field operators carry a continuous position label, and the completeness sum $\sum_k\phi_k(\mathbf{r})\phi_k^*(\mathbf{r}')$ collapses to $\delta^3(\mathbf{r}-\mathbf{r}')$ , the continuous-basis version of $\delta_{kk'}$ . It is the same relation, translated from a countable basis to an uncountable one.

Where does the Slater determinant actually come from here?

From Wick-contracting the annihilators in $\langle 0|\hat{\psi}(\mathbf{r}_N)\cdots\hat{\psi}(\mathbf{r}_1)\,\hat{a}_{k_1}^\dagger\cdots\hat{a}_{k_N}^\dagger|0\rangle$ . Each annihilator must pair with one creator; anticommuting past the intervening operators to make that pairing costs a sign $(-1)^P$ for the permutation $P$ . Summing the signed products over all $N!$ pairings is, by definition, the determinant $\det[\phi_{k_i}(\mathbf{r}_j)]$ . The antisymmetry you would otherwise impose by hand is generated by the anticommutators — verified explicitly for $N=2$ and $N=3$ .

What is the physical meaning of

\hat{n}(\mathbf{r})=\hat{\psi}^\dagger(\mathbf{r})\hat{\psi}(\mathbf{r})

It is the particle-density operator: its expectation value $\langle\hat{n}(\mathbf{r})\rangle$ is the average number of particles per unit volume at $\mathbf{r}$ , and its integral is the total number operator $\hat{N}=\int\hat{n}\,d^3r$ , whose eigenvalues are $0,1,2,\dots$ . Products like $\langle\hat{n}(\mathbf{r})\hat{n}(\mathbf{r}')\rangle$ give pair correlations and structure factors — for free fermions they show the exchange (Pauli) hole, a real, measurable suppression of same-spin density at short range.

Why does the two-body interaction have a factor of one-half?

To avoid double-counting pairs. The double integral $\iint d^3r\,d^3r'$ runs over ordered pairs $(\mathbf{r},\mathbf{r}')$ and $(\mathbf{r}',\mathbf{r})$ separately, but a physical interaction between two particles should be counted once. The $\tfrac12$ corrects for this. Combined with the normal-ordered "sandwich" $\hat{\psi}^\dagger\hat{\psi}^\dagger\hat{\psi}\hat{\psi}$ , which removes any spurious self-interaction, the operator reproduces exactly $\sum_{i<j}V(\mathbf{r}_i,\mathbf{r}_j)$ on $N$ -particle states.

When is second quantization genuinely necessary rather than convenient?

Whenever particle number is not fixed or when a broken-symmetry condensate has $\langle\hat{\psi}\rangle\neq0$ . Photon emission and absorption, Cooper-pair breaking, particle creation in QFT, and Bogoliubov theory of a superfluid all require states of different $N$ to coexist — which only Fock space provides. For a fixed number of particles it is "merely" a huge convenience (no determinant bookkeeping, $N$ -independent operators); for variable number it is indispensable.