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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.

For an unnormalized bosonic coherent state, the measure is $d^2\psi /\pi$ over the complex plane. Gaussian moments $\int d^2\psi e^{-|\psi {|}^2}{\psi }^n{\bar{\psi }}^m/\pi =n!{\delta }_{nm}$ recover the identity in the Fock state basis. For a fermionic coherent state, use independent Grassmann variables and an ordered Berezin integral with $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$. The coefficient of $-\bar{\psi }\psi$ in the weighted projector is the identity. Products of these one-mode measures give the finite-mode result.

Bose-Einstein condensation concerns occupation of a one-particle state. Matter-wave coherence concerns field correlations between positions or times. Superfluidity concerns a flow response, including suppression of dissipation under appropriate conditions. A density peak diagnoses neither a relative quantum phase between two separately prepared clouds nor a critical flow speed. These observables require different experiments.

For example, in a single-mode fixed-number Fock state, $⟨\hat{a}⟩=0$ but $⟨{\hat{a}}^{†}\hat{a}⟩=N$. Writing $\hat{\psi }(\mathbf{r})=\varphi (\mathbf{r})\hat{a}$ for that occupied mode givesThus vanishing mean field does not imply vanishing first-order matter-wave coherence. For two independent fixed-number modes, $⟨{\hat{a}}^{†}\hat{b}⟩=0$: there is no ensemble mean interference term establishing a fixed relative quantum phase. Interpreting fringes in an individual detection record requires considering correlations and the measurement; assigning a pre-existing definite relative quantum phase is unnecessary.

Wolfgang Ketterle's lecture describes a separate coherence test: two condensates expanded into one another and produced fringes. Selectively removing one condensate made the fringes disappear on alternating preparation cycles, testing against diffraction from the confinement. This compared two-cloud and one-cloud signals; it did not measure dissipationless flow.^[ref]

Two distinct normalized plane waves on a periodic one-dimensional ring of length $L$ give the displayed projected contact interaction. Momentum conservation excludes pair conversion between these two modes. Four mixed operator orderings give the coefficient of $n_1{n}_2$. At fixed particle number $N$, the energy is $g[N(N-1)+2n_1(N-n_1)]/(2L)$. For repulsive coupling, its minima are the two endpoint Fock states with all particles in one mode; a one-particle energy difference selects the lower mode. This ground-state statement concerns the two-mode projection.

A creation operator raises the occupation number operator eigenvalue of a quantum harmonic oscillator by one. Repeated application builds Fock states.

A Fock state has specified occupation numbers. Applying creation operators to the vacuum constructs multiparticle momentum eigenstates.

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.

In the 2007 cavity quantum electrodynamics experiment of Christine Guerlin and collaborators, including Sébastien Gleyzes, Michel Brune, Jean-Michel Raimond and Serge Haroche, an injected coherent state occupied a 51.1 GHz microwave cavity with niobium-coated mirrors exhibiting superconductivity. Velocity-selected rubidium chemical atoms were prepared as circular Rydberg atoms in levels with principal quantum numbers 50 and 51. Repeated atomic readouts progressively narrowed the inferred photon number to a Fock state; later records revealed jumps. The cavity damping time was $T_c=0.130\mathrm{s}$. Empty-cavity Ramsey interferometry scans fixed the analysis-phase origin; weak injected fields refined the response calibration. Finite fringe contrast required calibrated likelihoods; photon losses could interrupt convergence.^[ref]

The functional sequence is preparation, first Ramsey interferometry zone $R_1$, dispersive cavity $C$, second zone $R_2$, and state-selective detection. The first $\pi /2$ microwave pulse prepares an atomic quantum superposition; dispersive atom-cavity coupling adds a photon number-dependent relative quantum phase while suppressing energy exchange. The second pulse converts that quantum phase into atomic populations; state-selective field ionization reads $e$ or $g$. Successive chemical atoms carry the meter records away while the same field remains available.^[ref]

This arrangement makes quantum nondemolition measurement useful for following a field repeatedly and preparing a Fock state by conditioning. The photon number is inferred from a sequence of physical records; it is not a directly visible label attached to a photon. Each atomic probe makes a weak, partial measurement in quantum mechanics of photon number, and the accumulated records can approach a resolved count. A single binary meter need not identify one among many possible Fock states, as perfect discrimination of pure states requires orthogonality.

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.

A coherent state is a normalized eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding in the Fock state eigenbasis gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and wavefunction normalization fixesThe displacement operator therefore shifts the oscillator ground state in phase space. Distinct coherent states are not orthogonal:

The occupation probabilities form a Poisson distribution,For the quantum harmonic oscillator, the expectation values of position and momentum areThe fluctuations equal those of the ground state,Thus the position wavefunction is a displaced Gaussian wave packet that saturates the Heisenberg uncertainty principle.

Time evolution under $H=\hslash \omega (a^{†}a+1/2)$ preserves this form:The center follows the classical oscillator trajectory while the packet width stays constant. Its mean energy is $\hslash \omega (N+1/2)$; it is generally a superposition of energy eigenstates, rather than a state of definite energy. For $N\gg 1$, the relative occupation fluctuation is $1/\sqrt{N}$, explaining the classical appearance of a highly excited coherent state.

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis: ${\tilde{S}}^z=S-n$ with $n=a^{†}a$. The square root in ${\tilde{S}}^{+}$ supplies the correct spin-ladder matrix elements. In particular,Their commutator is $2(S-n)$ on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is onlyThe order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in $n/(2S)$. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of $S$ about an ordered classical configuration. Large $S$ alone is insufficient: the average depletion must remain small compared with $S$. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

Use the standard canonical anticommutation relations for the two distinct fermionic orbitals. In particular $\{a,b^{†}\}=\{a^{†},b\}=0$, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital fermionic Fock space interpretation. Normalize the original vacuum state by $⟨0|0⟩=1$ and fix the orderingThe canonical anticommutation relations give $a^2=(a^{†}{)}^2=0$ and $a{a}^{†}=1-a^{†}a$. Consequently the number operator is an orthogonal projector:The same holds for $N_b$. Hence each eigenvalue obeys $n^2=n$ and is either $0$ or $1$, expressing the Pauli exclusion principle. Also $[N_a,N_b]=0$, so the two number operators have a simultaneous orthonormal basis of Fock states:Each number operator has eigenvalues $0,1$, each with state degeneracy two. For $N_a$, the zero eigenspace is spanned by $|00⟩,|01⟩$ and the unit eigenspace by $|10⟩,|11⟩$; for $N_b$, these are respectively $|00⟩,|10⟩$ and $|01⟩,|11⟩$. Any normalized linear combination within one such eigenspace is also an eigenstate of that number operator.

The ordering convention will matter below: $b^{†}{a}^{†}|0⟩=-|11⟩$ and

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere is one fermionic Bogoliubov quasiparticle of energy $E$ per occupied transformed orbital. The two Nambu spinor matrix eigenvalues $\pm E$ should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.

The inverse Bogoliubov transformation gives $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$. Because $(a^{†}{b}^{†}{)}^2=0$ by the canonical anticommutation relations, the exponential truncates:The fermionic signs are fixed byIt follows thatBoth annihilation conditions hold precisely when $B=v/u$. Normalizing the paired fermionic vacuum, using the orthogonality of its two Fock states, gives $A^2(1+B^2)=1$. Choose the irrelevant overall phase so that $A>0$. The result isEquivalently,Its positive pair coefficient follows from the specified order $ab+b^{†}{a}^{†}$ in the Hamiltonian operator. The state has even fermion parity and is a quantum superposition of zero and two original particles. In the limit $\Delta \to 0^{+}$, $A\to 1$ and $B\to 0$, so it reduces to the original vacuum state, as expected for $\epsilon >0$.

Use the two-pair component of polarization-entangled down-conversion to account for identical bosons in the same collected modes. Write $a_H^{†}$ for a creation operator in mode $a$ with horizontal polarization, and similarly for the other modes. For the plus Bell state, the pair-creation operator is $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The normalized four-photon state isThese are normalized Fock states. In particular, $(a_H^{†}{)}^2|0⟩=\sqrt{2}|2H_a⟩$; omitting this factor would misnormalize the four-photon input.

The first polarizing beam splitter sends $a_H^{†}$ to $T_H^{†}$ and $a_V^{†}$ to $c_V^{†}$. Only the middle term can give exactly one trigger photon and leave another photon in $c$. In the first term both $a$ photons reach $T$ and none reaches $c$; in the last there is no trigger. Thus the relevant, still unnormalized component is

For the selected paths of the half-wave plate and beam splitter, takeHere $e$ labels the output towards $D_3$. An immaterial common reflection phase has been absorbed in the mode definitions. Keeping one photon at $D_3$ leavesAt the final polarizing beam splitter, the $H_{c^{\prime }}{H}_d$ alternative yields one $H$ at each of $D_1,D_2$, while $V_{c^{\prime }}{V}_d$ yields one $V$ at each. The two opposite-polarization alternatives send both photons to the same detector and fail fourfold coincidence counting. The accepted component is thereforeAfter conditioning and normalization, the three detected output modes haveThe optical mode matching must erase any timing, frequency or other record that distinguishes the two alternatives. Otherwise the same polarization populations could describe an incoherent mixed quantum state, rather than this quantum superposition. The Bouwmeester three-photon GHZ experiment uses this principle. The photonic GHZ preparation by fourfold coincidence is a measurement in the coincidence basis: detecting all four photons projects and consumes them. A click at $T$ by itself does not herald a guaranteed freely available three-photon state.

Put $b=\mathcal{E}{e}^{i(kr-\omega t)}$, allowing $\mathcal{E}$ to be complex. The Hermitian electric field is $\hat{E}=b\hat{a}+b^{\ast }{\hat{a}}^{†}$. The annihilation operator and creation operator change photon number, so their diagonal matrix elements in a Fock state vanish. Since $\hat{a}{\hat{a}}^{†}=\hat{n}+1$ and ${\hat{a}}^{†}\hat{a}=\hat{n}$,The terms containing ${\hat{a}}^2$ or ${\hat{a}}^{†2}$ connect orthogonal number states and contribute zero. Even as $n$ grows, there is no definite oscillating mean electric field; its quadrature fluctuations grow instead. A Fock state can represent fixed field energy, but not a classical sinusoidal wave of prescribed wave phase. This is distinct from representing a classical field with a randomly averaged phase.

Choose the normalization constant real and positive; its otherwise arbitrary phase is physically immaterial. Orthogonality of the Fock states givesActing with the annihilation operator and shifting the summation index givesThus a coherent state is an annihilation-operator eigenket. It is not also an eigenket of the creation operator: a common eigenket would make $[\hat{a},{\hat{a}}^{†}]$ annihilate it, contradicting $[\hat{a},{\hat{a}}^{†}]=I$. Taking the adjoint of the annihilation-operator equation instead gives the distinct, valid bra relationWith $\hat{n}={\hat{a}}^{†}\hat{a}$, the photon number expected value is $\mu =|\alpha {|}^2$. Normal ordering ${\hat{n}}^2=({\hat{a}}^{†}{)}^2{\hat{a}}^2+{\hat{a}}^{†}\hat{a}$ gives $⟨n^2⟩={\mu }^2+\mu$. ThereforeAt large mean photon number the relative fluctuations vanish, giving a classical-field limit even though quantum fluctuations remain. For the vacuum $\mu =0$, the relative ratio itself is undefined.

The Born rule applied to the number-state coefficient gives the Poisson distributionIt depends on the single parameter $\mu =⟨n⟩$, and has equal expected value and variance, consistent with the preceding moments.

For a Fock state $|n⟩$, raising or lowering the number makes an orthogonal state, so the optical quadrature has $⟨\hat{X}⟩=0$. Expanding its square givesThe first two terms have zero diagonal matrix elements, while the last two contribute $n+1$ and $n$. In particular, the vacuum retains quadrature variance $1/2$.

Denote the normalized initial field by $|F⟩$, and its shifted packets byHere the photon-number kets are orthogonal Fock states. Counting shared number labels givesAfter the interaction, linearity gives the joint stateTake the partial trace over the field. Put $d=|\alpha {|}^2-|\beta {|}^2$ and $c=\alpha {\beta }^{\ast }+{\alpha }^{\ast }\beta$. In the ordered qubit basis $(|0⟩,|1⟩)$, the reduced density matrix isFor example, its upper off-diagonal entry is half the overlap of the field accompanying $|1⟩$ with the field accompanying $|0⟩$, fixing the conjugation order. The trace is one, and positivity follows because this is the partial trace of a normalized pure state. For $k\ge 2$, $s_k=1-2/k$; at $k=1$ it is zero, not the continuation $-1$. This is the photon-number reference for a Hadamard gate channel, not an actual coherent-state packet.