Creation, Annihilation, and Second Quantization

Second quantization is the operator language of Fock space. Creation and annihilation operators add or remove one quantum from a mode, their algebra encodes bosonic or fermionic statistics, and one- and two-body operators become sums of mode-operator products.

The historical name can mislead. The procedure does not quantize an already quantized particle a second time. It rewrites many-particle quantum mechanics in a representation where exchange symmetry and variable particle number are built into the state space and operator algebra.

Use the volume orientation to choose the physical subsystem or mode decomposition, and Fock Space and Occupation Number to construct the state space. This chapter owns the operator construction that acts on that space.

The construction sequence is

$$\begin{gathered} \text{choose one-particle modes} \\ \downarrow \\ \text{define creation and annihilation maps} \\ \downarrow \\ \text{impose the statistics algebra} \\ \downarrow \\ \text{lift one- and two-body operators}. \end{gathered}$$

Chapter Map

Task	Canonical page	Main output
add and remove mode quanta	Creation and Annihilation Operators	normalized action on bosonic and fermionic number states
encode bosonic statistics	Bosonic Commutation Relations	canonical commutators and symmetric Fock-space action
encode fermionic statistics	Fermionic Anticommutation Relations	canonical anticommutators, exclusion, and ordering signs
count particles and excitations	Number Operators	mode number, total number, and conservation commutators
change between mode and position language	Mode Expansions	completeness, basis changes, momentum modes, and spin labels
use position-space creation and annihilation fields	Field Operators	operator-valued distributions, density, and field algebra
lift single-particle observables	One-Body Operators	bilinear mode and field expressions
lift pair interactions	Two-Body Operators	quartic expressions, matrix elements, and double-counting factor
assemble physical models	Many-Particle Hamiltonians	gases, lattices, interactions, and pairing preview
reorder operator products	Normal Ordering	creation-left convention, fermionic signs, and reference vacuum
preview contraction bookkeeping	Wick’s Theorem Preview	Gaussian-state contractions and scope limits
identify what changes in relativistic theory	Bridge to QFT	free modes, fields, locality, antiparticles, and QFT boundaries

Creation and Annihilation Maps

Let $\lvert n_1,n_2,\ldots\rangle$ be an occupation-number state. A bosonic creation operator acts as

$$\begin{aligned} a_r^\dagger \lvert\ldots,n_r,\ldots\rangle &= \sqrt{n_r+1} \\ &\quad\times \lvert\ldots,n_r+1,\ldots\rangle, \end{aligned}$$

while annihilation acts as

$$a_r \lvert\ldots,n_r,\ldots\rangle = \sqrt{n_r}\, \lvert\ldots,n_r-1,\ldots\rangle.$$

The square-root factors preserve normalization and produce the canonical algebra. In particular, $a_r\lvert0\rangle=0$.

Fermionic operators change occupations between zero and one. With mode order $1,2,\ldots$ fixed, define

$$\eta_r = (-1)^{\sum_{s<r}n_s}.$$

Then

$$\begin{aligned} c_r^\dagger \lvert\ldots,n_r,\ldots\rangle &= \eta_r(1-n_r) \lvert\ldots,1,\ldots\rangle, \\ c_r \lvert\ldots,n_r,\ldots\rangle &= \eta_r n_r \lvert\ldots,0,\ldots\rangle. \end{aligned}$$

The sign records how many occupied modes the operator passes in the chosen ordering. Different consistent conventions represent the same physics, but mixing conventions produces errors.

Statistics Algebra

Bosonic mode operators satisfy

$$\begin{aligned} [a_r,a_s^\dagger]&=\delta_{rs}, \\ [a_r,a_s]&=0, \\ [a_r^\dagger,a_s^\dagger]&=0. \end{aligned}$$

Fermionic mode operators satisfy

$$\begin{aligned} \{c_r,c_s^\dagger\}&=\delta_{rs}, \\ \{c_r,c_s\}&=0, \\ \{c_r^\dagger,c_s^\dagger\}&=0. \end{aligned}$$

Setting $r=s$ in the final relation gives

$$(c_r^\dagger)^2=0,$$

which is Pauli exclusion in operator form. Bosonic creation operators commute, while fermionic creation operators anticommute and change sign under reordering.

Number Operators and Sector Changes

For either statistics, use the appropriate mode operators to define

$$\hat n_r=d_r^\dagger d_r, \qquad \hat N=\sum_r\hat n_r,$$

where $d_r=a_r$ for bosons and $d_r=c_r$ for fermions.

The universal number commutators are

$$\begin{aligned} [\hat N,d_r^\dagger]&=d_r^\dagger, \\ [\hat N,d_r]&=-d_r. \end{aligned}$$

Thus $d_r^\dagger$ raises total number by one and $d_r$ lowers it by one. Bilinears $d_r^\dagger d_s$ conserve total number because they remove one quantum and add one.

For fermions, $\hat n_r^2=\hat n_r$, so its eigenvalues are zero and one. Bosonic $\hat n_r$ has all nonnegative integer eigenvalues.

Mode Expansions

Let $\{\phi_r(\mathbf x)\}$ be a complete orthonormal basis of the one-particle Hilbert space. The nonrelativistic annihilation and creation fields are expanded as

$$\begin{aligned} \hat\psi(\mathbf x) &= \sum_r \phi_r(\mathbf x)d_r, \\ \hat\psi^\dagger(\mathbf x) &= \sum_r \phi_r^*(\mathbf x)d_r^\dagger. \end{aligned}$$

The inverse relation is

$$d_r = \int d^3x\, \phi_r^*(\mathbf x) \hat\psi(\mathbf x).$$

Completeness turns the discrete algebra into the equal-time field algebra. For bosons,

$$\left[ \hat\psi(\mathbf x), \hat\psi^\dagger(\mathbf y) \right] = \delta^{(3)}(\mathbf x-\mathbf y),$$

and for fermions the commutator is replaced by an anticommutator.

Field operators at sharp points are operator-valued distributions. Meaningful operators are obtained by smearing them with suitable wavefunctions or by placing the theory on a lattice or other regulated basis.

Basis Changes Preserve the Abstract Operator

Suppose the one-particle bases are related by

$$\lvert\alpha\rangle = \sum_r U_{\alpha r}\lvert r\rangle.$$

The mode operators transform contragrediently so that the field and physical observables remain unchanged. With the convention above,

$$b_\alpha^\dagger = \sum_r U_{\alpha r}d_r^\dagger, \qquad b_\alpha = \sum_r U_{\alpha r}^*d_r.$$

Matrix elements and operator coefficients must be transformed together. A diagonal one-body Hamiltonian in one basis may contain off-diagonal hopping or mixing terms in another.

One-Body Operators

Let $A$ be a one-particle operator with matrix elements

$$A_{rs} = \langle r|A|s\rangle.$$

Its number-conserving many-body lift is

$$\widehat A = \sum_{r,s} A_{rs}d_r^\dagger d_s.$$

In position language,

$$\widehat A = \int d^3x\, \hat\psi^\dagger(\mathbf x) A_{\mathbf x} \hat\psi(\mathbf x),$$

where $A_{\mathbf x}$ acts on the one-particle coordinate and any suppressed internal indices.

Examples include kinetic energy, external potentials, spin operators, mode energies, and hopping. The bilinear form is the canonical home of one-body physics in second quantization.

Two-Body Operators

Let a pair interaction have matrix elements

$$V_{rs;tu} = \langle r,s|V|t,u\rangle.$$

For an interaction summed over unordered particle pairs, the second-quantized operator is

$$\widehat V = \frac12 \sum_{r,s,t,u} V_{rs;tu} d_r^\dagger d_s^\dagger d_u d_t.$$

The factor $1/2$ removes double counting when the first-quantized interaction is $\sum_{i<j}V(i,j)$. Index order matters, especially for fermions; changing the convention for $V_{rs;tu}$ requires changing the operator order consistently.

For a position-space pair potential,

$$\begin{aligned} \widehat V =\frac12 \int d^3x\,d^3y\, &\hat\psi^\dagger(\mathbf x) \hat\psi^\dagger(\mathbf y) \\ &\times V(\mathbf x,\mathbf y) \hat\psi(\mathbf y) \hat\psi(\mathbf x). \end{aligned}$$

Contact interactions, Coulomb interactions, and Hubbard on-site terms are special cases after choosing modes and regularization.

Many-Particle Hamiltonians

A general number-conserving Hamiltonian with one- and two-body terms has the form

$$\begin{aligned} H &= \sum_{r,s}h_{rs}d_r^\dagger d_s \\ &\quad+ \frac12 \sum_{r,s,t,u} V_{rs;tu} d_r^\dagger d_s^\dagger d_u d_t. \end{aligned}$$

Every term contains equal numbers of creation and annihilation operators, so

$$[H,\hat N]=0.$$

Pairing terms such as $c_r^\dagger c_s^\dagger$ and $c_s c_r$ change particle number by two. They may appear in effective mean-field descriptions and preserve number parity even when they do not preserve $\hat N$. The physical interpretation and approximation must be stated.

Normal Ordering

Normal ordering moves creation operators to the left of annihilation operators. It is denoted by colons:

$$:d_r d_s^\dagger: = \begin{cases} d_s^\dagger d_r,&\text{bosons}, \\ -d_s^\dagger d_r,&\text{fermions}. \end{cases}$$

Commutators or anticommutators generate the difference between an original product and its normal-ordered form. For one bosonic mode,

$$a a^\dagger = a^\dagger a+1.$$

Normal ordering is defined relative to a chosen vacuum or reference state. In interacting many-body theory and QFT, changing the reference can change what “normal ordered” means.

Wick’s Theorem Is a Scoped Preview

Wick’s theorem reorganizes products or time-ordered products into normal-ordered terms plus contractions when the reference state has the required Gaussian structure. A contraction is bookkeeping for a two-point expectation value, not a new physical interaction.

The theorem is powerful because Gaussian states reduce higher correlators to two-point data. It does not apply unchanged to arbitrary interacting or non-Gaussian states. This volume provides only Wick’s Theorem Preview; full many-body and relativistic formulations belong in their canonical settings.

What Carries Over to QFT

Quantum mechanics already supplies:

• Fock space and number sectors;
• creation and annihilation operators;
• bosonic and fermionic algebras;
• mode expansions;
• operator-valued fields in position space;
• normal ordering and contraction language.

Relativistic QFT adds essential structure: Lorentz symmetry, local relativistic fields, antiparticles, microcausality, interacting vacua, renormalization, and particle concepts that may depend on background or asymptotic regime.

A nonrelativistic field $\hat\psi(\mathbf x)$ annihilates a particle at equal time within the many-body model. It should not be identified without qualification with a relativistic scalar, spinor, or gauge field. Use Bridge to QFT for the handoff.

A Construction Protocol

1. Choose and order modes

State the one-particle basis, internal labels, and fermionic ordering convention.

2. Select the statistics algebra

Use commutators for bosons and anticommutators for fermions. Do not mix the resulting reordering rules.

3. Define matrix elements

Compute $h_{rs}$ and $V_{rs;tu}$ from the one-particle basis before writing operator sums. Record index and complex-conjugation conventions.

4. Build operators by body order

One-body terms are bilinear. Two-body terms are quartic. Check the pair-counting factor and operator order.

5. Check Hermiticity and conservation laws

Verify $H^\dagger=H$, particle-number or parity commutators, spin or lattice symmetries, and the action on simple occupation states.

6. Change basis consistently

Transform both mode operators and coefficient matrices. The abstract observable must not depend on notation.

7. State the approximation boundary

Identify cutoffs, truncated modes, contact regularization, mean-field replacements, reference vacua, or nonrelativistic assumptions.

Common Mistakes

• Omitting bosonic square-root factors. Number-state normalization then fails.
• Ignoring fermionic mode ordering. Anticommutation signs depend on a consistent convention.
• Using commutators for fermions or anticommutators for bosons. Statistics is encoded in the algebra.
• Forgetting the pair factor $1/2$. This double counts unordered interactions under the stated convention.
• Reversing annihilation indices in a fermionic quartic term. Operator order and matrix-element convention must match.
• Treating $\hat\psi(\mathbf x)$ as an ordinary function. Sharp-point fields are operator-valued distributions.
• Calling every quartic term an interaction. Normal ordering and basis transformations can also produce quartic expressions in effective descriptions; inspect the model.
• Assuming normal ordering is reference independent. The vacuum or reference state matters.
• Calling second quantization a second physical quantization step. It is a many-body representation and operator formalism.
• Equating the nonrelativistic field language with full QFT. Relativistic locality and particle interpretation add essential content.

Reading Paths

Operator foundation: Creation and Annihilation Operators → Bosonic Commutation Relations or Fermionic Anticommutation Relations → Number Operators.

Mode and field language: Mode Expansions → Field Operators → Bridge to QFT.

Hamiltonian construction: One-Body Operators → Two-Body Operators → Many-Particle Hamiltonians.

Reordering and contractions: Normal Ordering → Wick’s Theorem Preview → Harmonic-Oscillator Ladder Operators.

Continuous-variable bridge: Mode Expansions → Continuous Variables and Modes → Gaussian States Preview.

Across-fields bridge: Many-Particle Hamiltonians → Entanglement Across Fields → Entanglement in Many-Body Physics.

Practice bridge: Reference, Problems, and Notebooks → Fock Space Exercises → Computational Notebooks.

References

• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Westview Press, 1998.
• A. Altland and B. Simons, Condensed Matter Field Theory, 3rd ed., Cambridge University Press, 2023.
• H. Bruus and K. Flensberg, Many-Body Quantum Theory in Condensed Matter Physics, Oxford University Press, 2004.
• A. F. G. Wyatt, Excitations in Quantum Fluids, Cambridge University Press, 1999.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Westview Press, 1995.

Exercises

Exercise 1: Recover the bosonic commutator

Act with $[a,a^\dagger]$ on a one-mode number state $\lvert n\rangle$ and show that the result is $\lvert n\rangle$.

Solution

First,

$$\begin{aligned} a a^\dagger\lvert n\rangle &= (n+1)\lvert n\rangle, \\ a^\dagger a\lvert n\rangle &= n\lvert n\rangle. \end{aligned}$$

Subtracting gives

$$[a,a^\dagger]\lvert n\rangle = \lvert n\rangle.$$

Because the number states span the Fock space on the natural dense domain, this realizes $[a,a^\dagger]=I$ there.

Exercise 2: Fermionic exclusion from the algebra

Use $\{c_r^\dagger,c_r^\dagger\}=0$ to show that the same fermionic mode cannot be created twice.

Solution

The anticommutator gives

$$2(c_r^\dagger)^2=0,$$

so $(c_r^\dagger)^2=0$. Acting twice on any state therefore gives the zero vector. In particular,

$$(c_r^\dagger)^2\lvert0\rangle=0.$$

This is the operator-algebra form of Pauli exclusion for a complete one-particle mode.

Exercise 3: A diagonal one-body Hamiltonian

Let the one-particle Hamiltonian satisfy $h\lvert r\rangle=\epsilon_r\lvert r\rangle$. Write its second-quantized form and find its action on $\lvert n_1,n_2,\ldots\rangle$.

Solution

The matrix elements are $h_{rs}=\epsilon_r\delta_{rs}$, so

$$H_0 = \sum_r\epsilon_r d_r^\dagger d_r = \sum_r\epsilon_r\hat n_r.$$

Therefore

$$\begin{aligned} H_0\lvert\mathbf n\rangle &= E_{\mathbf n}\lvert\mathbf n\rangle, \\ E_{\mathbf n} &= \sum_r\epsilon_r n_r, \end{aligned}$$

Here $\lvert\mathbf n\rangle=\lvert n_1,n_2,\ldots\rangle$. The result applies to bosons or fermions with the corresponding allowed occupations.

Exercise 4: Why the pair factor appears

Suppose a first-quantized interaction is $\sum_{i<j}V(i,j)$. Explain why a symmetric sum over ordered mode indices in second quantization carries a factor $1/2$.

Solution

The unordered pair sum counts each physical pair once. A fully expanded operator sum naturally includes both orderings associated with the two particle slots or field coordinates. For a symmetric pair potential, the contributions related by exchanging the two particles are equal.

Multiplying the ordered sum by $1/2$ removes this double counting. If the matrix elements are defined with antisymmetrization or another restricted convention already built in, the prefactor must be checked rather than copied mechanically.

Exercise 5: Number conservation of a quartic term

Show that

$$Q_{rstu} = d_r^\dagger d_s^\dagger d_u d_t$$

commutes with total number $\hat N$ for bosons or fermions.

Solution

Use $[\hat N,d_j^\dagger]=d_j^\dagger$, $[\hat N,d_j]=-d_j$, and the product rule for commutators. Each creation operator contributes $+1$ and each annihilation operator contributes $-1$:

$$\begin{aligned} [\hat N,Q_{rstu}] &= (1+1-1-1)Q_{rstu} \\ &=0. \end{aligned}$$

The argument uses ordinary commutators with the even operator $\hat N$ and works for either underlying statistics algebra.