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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.

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.

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 $[c(p),c^{†}(q)]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$.

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.

For transverse coupling with equal mode amplitudes $\lambda$, Fermi's golden rule and the oscillator creation operator and annihilation operator givewhere the first index is the final spin and $g$ is the density of states per unit angular frequency. Decay creates a quantum and carries the factor $n_{\Omega }+1$; excitation annihilates one and carries $n_{\Omega }$. For nonuniform resonant occupations, $n_{\Omega }$ is their density-weighted mean.

With a thermal Bose-Einstein distribution, $n_{\Omega }/(n_{\Omega }+1)=e^{-\beta \hslash \Omega }$. The two-state master equation therefore has the Boltzmann distribution as its unique stationary populations when ${\Gamma }_0>0$. The population deviation relaxes at rate ${\Gamma }_0(2n_{\Omega }+1)$. If there is no resonant coupling, both rates vanish and stationarity does not determine the populations.

For a number operator eigenstate $|n⟩$, the annihilation operator and creation operator satisfy $⟨a⟩=⟨a^{†}⟩=⟨a^2⟩=⟨(a^{†}{)}^2⟩=0$, $⟨a^{†}a⟩=n$ and $⟨a{a}^{†}⟩=n+1$. Inserting $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ and $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$ gives the displayed second moments and zero means.

Thus the kinetic energy and potential energy each have mean $\hslash \omega (n+1/2)/2$, and the uncertainty product is $\Delta x\Delta p=\hslash (n+1/2)$. Among number states only the ground state saturates the Heisenberg uncertainty relation. A quantum superposition can have nonzero position and momentum means despite each constituent number state's zero means.

The Holstein–Primakoff transformation represents a spin-$S$ operator algebra with a bosonic creation operator and annihilation operator:The order of the square root and oscillator operator matters. On the physical Fock states $|n⟩$, $0\le n\le 2S$, the raising and lowering matrix elements are $\sqrt{n(2S-n+1)}$ and $\sqrt{(n+1)(2S-n)}$. Their squared difference gives $[S^{+},S^{-}]|n⟩=2(S-n)|n⟩$, proving the spin commutation relations. Expanding the square root yields the linear spin-wave approximation and its interaction corrections.

For a free real scalar field expanded with measure $d^{3}p/[(2\pi {)}^{3}2E_p]$ and $f_p=e^{-ip\cdot x}$, the displayed Klein-Gordon inner product extracts its annihilation operator. Indeed $(f_p,f_q)=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$ and $(f_p,f_q^{\ast })=0$; spatial Fourier inversion removes the measure. Taking the Hermitian adjoint gives $a^{†}(\mathbf{p})=-i\int d^{3}x{f}_p\overleftrightarrow{{\partial }_t}\varphi$. The equal-time canonical commutation relation then gives $[a(\mathbf{p}),a^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$.

The Lamb-Dicke regime means the wave-induced phase change across the occupied motional wavepacket is small. For one normal mode, let $x_0=\sqrt{\hslash /(2m\nu )}$ be the ground state's position standard deviation, and write $x=x_0(a+a^{†})$ using creation and annihilation operators. Here $m$ is mass and $\hslash$ the Planck constant divided by $2\pi$. The effective wave vector component $k_{\mathrm{eff}}$ is the single beam's component for a direct transition, or the difference of two beam components for a Raman transition. With dimensionless $\eta =|k_{\mathrm{eff}}|x_0$, the useful condition is $\eta \sqrt{2n+1}\ll 1$ over the populated number states, rather than merely $\eta \ll 1$.

Absorb the sign of $k_{\mathrm{eff}}$ into the pulse phase. Expanding the coupling factor givesThe constant term preserves $n$; the annihilation operator gives $a|n⟩=\sqrt{n}|n-1⟩$; the creation operator gives $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The different square roots explain why removing a motional quantum becomes impossible at $n=0$, while adding one remains possible. Smallness of $\eta$ controls this spatial expansion; resolving the resulting transitions is a separate frequency requirement.

In linear optics, the response is linear in the optical field and propagation or passive mixing maps mode annihilation operators to linear combinations of mode annihilation operators. Wave plates and beam splitters can rotate or mix photonic modes, but do not by themselves supply a deterministic interaction between independent photon-number qubits. Conditional measurements and auxiliary photons change that resource model.