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An annihilation operator lowers the occupation number operator eigenvalue by one and annihilates the Fock vacuum.

The BCS ground state is a product of empty and occupied pair states:The modulus squares are needed when the amplitudes are complex. Each pair block is a normalized two-state quantum superposition; multiplying independent blocks preserves normalization.

The positive-energy Bogoliubov quasiparticle annihilation operators annihilate this paired mean-field state, so it is their vacuum even though physical electron occupations are nonzero. The pair momentum is zero, but $k$ labels all relative momenta in the chosen pairing space; it is not restricted to $k=0$. ^[ref]

The state superposes different total particle numbers. A fixed-number approximation can project it onto one number sector. The utility of the unprojected state is that independent pair blocks make the BCS anomalous average, excitation spectrum and BCS gap equation directly calculable.

A Bogoliubov--de Gennes Hamiltonian is the matrix representation of a quadratic fermion Hamiltonian in a Nambu spinor combining annihilation operators and creation operators. It treats ordinary one-particle motion and fermionic pairing in one eigenvalue problem.

For a column $c$ of $N$ fermionic annihilation operators, let $h=h^{†}$ and $\Delta =-{\Delta }^T$. With $\Psi =(c,c^{†}{}^T{)}^T$, defineThe canonical anticommutation relations give the identityThe factor $1/2$ compensates for the full particle-hole doubling; the trace constant follows when the hole block is normal-ordered.

A two-by-two opposite-spin pairing blockhas eigenvalues $E_{\pm }=\pm \sqrt{{\xi }^2+|\Delta {|}^2}$, as follows from its characteristic polynomial. Pairing opens a spectral gap at $\xi =0$. Negative and positive partners in a fully doubled representation are related by Particle-hole symmetry of a Bogoliubov--de Gennes Hamiltonian; they are not independent species of physical quasiparticles. A reduced block and a full doubled basis therefore require consistent counting conventions.

A Bogoliubov quasiparticle is an excitation created by the diagonal mode operator obtained through a Bogoliubov transformation. Its vacuum can contain particles in the original modes because creation and annihilation operators mix. Examples include antiferromagnetic spin waves, Bose-gas phonons and paired-fermion excitations.

Expand an interacting Bose field about a macroscopically occupied condensate and retain terms quadratic in its fluctuations. A Bogoliubov transformation diagonalizes the stable quadratic Hamiltonian. For contact repulsion this gives the Bogoliubov quasiparticle dispersion $E_k=\sqrt{{ϵ}_k({ϵ}_k+2g_B{n}_0)}$, a linear sound mode at small momentum and particle-like behavior at large momentum. Mixing creation and annihilation operators also produces quantum depletion of a Bose condensate. The approximation requires controlled fluctuations; it does not give an exact strongly interacting liquid or the full fluctuation-dominated critical region.

A Bogoliubov transformation is a canonical transformation of mode operators mixing creation operators and annihilation operators. It diagonalizes quadratic Hamiltonians or relates different vacuum descriptions. Bosonic and fermionic canonical bracket constraints differ. A nonzero creation-mixing coefficient makes its vacuum contain particles relative to the original mode decomposition.

This is the symmetric-particle construction of Fock space.

For one-particle Hilbert space ${\mathcal{H}}_1$, the bosonic Fock space isCreation operators add one symmetrized particle and annihilation operators remove one.

For two equal-energy bosonic species with annihilation operators $a,b$, set $c_{-}=(a-b)/\sqrt{2}$ and let $N_{-}$ be its number operator. Since $[N_{-},c_{-}]=-c_{-}$, conjugation by $e^{i\pi N_{-}}$ reverses the antisymmetric combination while fixing $(a+b)/\sqrt{2}$, and therefore exchanges $a,b$ and their creation operators. For the particle and antiparticle modes of a complex scalar field this realizes charge conjugation, fixes the Fock vacuum, and squares to one. The normal-ordered generator $\int (a^{†}-b^{†})(a-b)$ equals $2N_{-}$. Writing the annihilation factors first adds an infinite constant in the continuum, which must be subtracted or regulated before defining its exponential.

The electromagnetic four-potential has a vanishing temporal momentum and a Gauss law constraint in gauge theory. Gauge reduction leaves two transverse canonical pairs. Their canonical commutation relations give the photon creation and annihilation operators. Covariant Gupta-Bleuler quantization instead retains auxiliary polarization modes and selects the same physical state space by a subsidiary condition and null-state quotient.

A linear change of creation operators and annihilation operators is canonical when it preserves their defining brackets. For finitely many bosonic modes, the canonical commutation relations require $U{U}^{†}-V{V}^{†}=I$ and $U{V}^T-V{U}^T=0$. For fermions, the canonical anticommutation relations instead require $U{U}^{†}+V{V}^{†}=I$ and $U{V}^T+V{U}^T=0$. A Bogoliubov transformation uses such mixing to diagonalize a quadratic Hamiltonian or relate different vacuum descriptions. Infinite-mode unitary implementation has further conditions, as in bosonic mode mixing implementability.

The free transverse photon field is a Hermitian sum of two transverse-polarization plane-wave modes with creation and annihilation operators. Their standard bosonic commutator gives the transverse equal-time commutator. The normal-ordered Hamiltonian is a sum of positive oscillator energies.

Choose the negative-frequency Dirac spinors as the charge conjugates of the positive-frequency spinors. Transposing $\bar{u}(/p-m)=0$ and using the charge-conjugation matrix identity gives $(/p+m)C{\bar{u}}^T=0$, so the displayed choice solves the correct antiparticle equation. In the mode expansion of a Dirac field, the unitary charge-conjugation operator then exchanges particle and antiparticle annihilation operators, with conjugate phases on their creation operators. A different spinor basis can insert conventional phases or a spin-index relabeling; the displayed equality specifies a compatible choice.

Suddenly shifting a quantum harmonic oscillator equilibrium from zero to a signed position $x_0$ leaves the old ground state unchanged at the instant of the quantum quench. The new annihilation operator is $a^{\prime }=a-\lambda$, with $\lambda =\sqrt{m\omega /(2\hslash )}{x}_0$, so $a^{\prime }|0⟩=-\lambda |0⟩$. In the translated energy eigenstates $|n^{\prime }⟩=e^{-ip{x}_0/\hslash }|n⟩$, the expansion isThe sign follows by translating the old vacuum by $-x_0$ before expanding it in the old Fock states. A Baker--Campbell--Hausdorff formula with central commutator separates the resulting exponential into creation and annihilation parts. Evolution with the shifted Hamiltonian preserves this coherent state, with $\alpha (t)={\alpha }_0{e}^{-i\omega t}$. Its mean occupation is $|{\alpha }_0{|}^2$ and its mean energy above the new potential minimum is $\hslash \omega (|{\alpha }_0{|}^2+1/2)$. Omitting $(-\lambda {)}^n$ in favour of $|\lambda {|}^n$ changes the state when $x_0>0$.

For a coherent state, write $a|\alpha ⟩=\alpha |\alpha ⟩$ and $|\alpha ⟩={\sum }_n{c}_n|n⟩$ in number states. The annihilation operator gives $c_{n+1}\sqrt{n+1}=\alpha c_n$, so $c_n=c_0{\alpha }^n/\sqrt{n!}$. Normalization yields $|c_0{|}^2=e^{-|\alpha {|}^2}$. Thus photon statistics follows a Poisson distribution with $\mu =|\alpha {|}^2$, and the normal ordering of every moment gives $⟨(a^{†}{)}^k{a}^k⟩={\mu }^k$. This factorization is the defining virtue of coherent states for optical calculations.^[ref]

As an original comparison, choose $\mu =2$. Then $p_0=e^{-2}$, $p_1=p_2=2e^{-2}$, $\mathrm{Var}N=2$, and $⟨N(N-1)⟩=4$. A two-photon Fock state also has mean two but has $\mathrm{Var}N=0$ and pair moment two. ConsequentlyIn particular, $|1⟩$ cannot supply a pair. Nevertheless, photon statistics alone does not determine a quantum state: averaging $|\alpha ⟩⟨\alpha |$ over a uniform quantum phase removes its off-diagonal entries while retaining exactly the same $p_n$ and these equal-time moments.

With $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, a free real scalar field has mode expansion of a free field $\varphi =\int d{\Pi }_p(a{e}^{-ipx}+a^{†}{e}^{ipx})$ and commutator $[a(p),a^{†}(q)]=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Substitution into the quadratic Hamiltonian density gives $H=\frac{1}{2}\int d{\Pi }_p{E}_p(a^{†}a+a{a}^{†})$. The terms with two annihilation operators or two creation operators cancel by $E_p^2={\mathbf{p}}^2+m^2$. Normal ordering removes the constant vacuum energy. The commutators $[H,a(p)]=-E_{p}a(p)$ and $[H,a^{†}(p)]=E_p{a}^{†}(p)$ give the boson energy without changing under this subtraction. Replacing $a(p)$ by $\sqrt{2E_p}c(p)$ recovers the noncovariant normalization $[c(p),c^{†}(q)]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$.

For a quantum harmonic oscillator, creation and annihilation operators move between energy eigenstates in steps of one oscillator quantum. Here $|n⟩$ is a normalized number state with nonnegative integer $n$.

The bosonic ladder operators satisfy

Writing $N=A^{†}A$ gives $N|n⟩=n|n⟩$ and $H=\hslash \omega (N+\frac{1}{2})$. Thus applying $A^{†}$ raises the energy by $\hslash \omega$. The annihilation operator obeys $A|0⟩=0$; it does not turn the ground state into a state below the zero-point energy. The normalization factors make repeated application consistent with the commutator. ^[ref]

Hermitian combinations of pair creation and annihilation operators. With $K_3=(N+1/2)/2$, their commutators are $[K_1,K_2]=-4i{K}_3$, $[K_2,K_3]=i{K}_1$ and $[K_3,K_1]=i{K}_2$. A coherent state has both variances $⟨N⟩+1/2$, saturating the corresponding uncertainty product.

Retain the first $N\ge 2$ number states of a quantum harmonic oscillator. Let ${\Pi }_N$ be the orthogonal projection onto their span and compress $X_N={\Pi }_{N}X{\Pi }_N$, $P_N={\Pi }_{N}P{\Pi }_N$, restricted to this span. The compressed annihilation operator $a_N$ satisfiesFor $n<N-1$, the two products give $n+1$ and $n$; on the top state they give $0$ and $N-1$. Thus, with $\ell =\sqrt{\hslash /(m\omega )}$,On every quantum superposition of $|0⟩,\dots ,|N-2⟩$, this commutator acts exactly as $i\hslash I_N$. For a normalized retained state with coefficients $c_n$,Small probability at the cutoff must therefore be judged relative to $1/N$ for this expectation. For $N=8$, the diagonal of $[X_N,P_N]/(i\hslash )$ is $(1,1,1,1,1,1,1,-7)$, with zero trace.

This is unavoidable for finite matrices: trace of a matrix commutator gives $\mathrm{tr}(XP-PX)=0$, whereas $\mathrm{tr}(i\hslash I_N)=i\hslash N\ne 0$. Compression does not preserve products: ${\Pi }_{N}XP{\Pi }_N$ can use intermediate states outside the retained span, while $X_N{P}_N$ cannot. Accordingly, compressing the full oscillator Hamiltonian operator gives ${\Pi }_{N}H{\Pi }_N=\hslash \omega (a_N^{†}{a}_N+I_N/2)$, but building it from the compressed squares givesThe two prescriptions agree on lower number states and disagree at the boundary. In the infinite oscillator, the canonical commutation relation holds on the common invariant dense subspace of finite number state combinations. Its unbounded position and momentum operators do not admit the finite trace argument; writing the formal identity does not assert that both products act on every vector of the Hilbert space.

For a one-particle Hilbert space $H$, the bosonic or fermionic Fock space is ${\mathcal{F}}_{\pm }(H)={⨁}_{n=0}^{\infty }{H}^{{\otimes }_{\pm }n}$, using symmetric or antisymmetric tensor powers. Its zeroth summand is the one-dimensional vacuum sector. Creation operators and annihilation operators change particle number. The bosonic Fock space and fermionic Fock space implement the corresponding exchange statistics.

The oscillator vacuum is annihilated by all annihilation operators. Acting with creation operators constructs Fock states. In single-string quantization this is the unexcited string, not the empty spacetime state.