Fock States and the Photon Number Basis

A single mode of the quantized electromagnetic field is, mathematically, nothing but a quantum harmonic oscillator, and this article builds on two upstream results: the ladder algebra $[\hat a,\hat a^{\dagger}]=1$ with its discrete spectrum from aq-optics-field-quantization, and the unique vacuum $|0\rangle$ carrying zero-point energy $\tfrac12\hbar\omega$ from aq-optics-vacuum-zero-point-energy. Feed those into the number operator $\hat n=\hat a^{\dagger}\hat a$ and its eigenstates fall out: the Fock states, or photon-number states, $|n\rangle$ , each holding an exact integer count of quanta. Because $\hat n$ commutes with the free Hamiltonian, these are also the stationary energy eigenstates of the mode. Everything rests on the vacuum. The creation operator $\hat a^{\dagger}$ climbs the ladder one photon at a time; $\hat a$ descends and terminates at $|0\rangle$ ; and the normalized rungs $|n\rangle=(\hat a^{\dagger})^{n}/\sqrt{n!}\,|0\rangle$ form a complete orthonormal basis. Any field state whatsoever — a laser's coherent state, thermal light, a squeezed beam — is a superposition or mixture written in this basis. That is why the number basis is the universal ledger of quantum optics. Physically, a Fock state is the most non-classical light imaginable. A definite photon number forces a random phase, so the mean field $\langle\hat E\rangle$ is exactly zero for every $|n\rangle$ — even one holding a million photons — unlike the smooth oscillating wave of a classical source. What is sharp is the intensity: the distribution is a spike, $\Delta n=0$ , giving sub-Poissonian, antibunched light with Mandel $Q=-1$ . Those properties, plus fragility under loss, make a clean single photon a frontier resource, not a commonplace.

Key takeaways

Fock states $|n\rangle$ are eigenstates of the number operator $\hat n=\hat a^{\dagger}\hat a$ with integer eigenvalue $n$ — a single field mode holding exactly $n$ photons.
They are grown from the vacuum by the creation operator, $|n\rangle=(\hat a^{\dagger})^{n}/\sqrt{n!}\,|0\rangle$ , and form a complete orthonormal basis for any field state.
Each is an energy eigenstate, $E_n=\hbar\omega(n+\tfrac12)$ ; the extra $\tfrac12$ is the zero-point energy recalled from the vacuum article.
The phase is totally undefined: $\langle n|\hat E|n\rangle=0$ for all $n$ while the field variance grows as $n+\tfrac12$ , so number states are maximally non-classical.
Their statistics are sub-Poissonian ( $\Delta n=0$ , $Q=-1$ ) with $g^{(2)}(0)=1-1/n$ , so $|1\rangle$ is perfectly antibunched — the target of single-photon sources, but ruined by loss.
Key idea: a Fock state $|n\rangle$ carries a precisely defined photon number ( $\Delta n=0$ ) but a totally undefined phase; the ladder operators act as $\hat a^{\dagger}|n\rangle=\sqrt{n+1}\,|n+1\rangle$ and $\hat a|n\rangle=\sqrt{n}\,|n-1\rangle$ , generating the whole number basis from the vacuum.

The core insight

A Fock state $|n\rangle$ carries a precisely defined number of photons ( $\Delta n=0$ ) but a totally undefined phase; the creation and annihilation operators act as a ladder, $\hat a^{\dagger}|n\rangle=\sqrt{n+1}\,|n+1\rangle$ and $\hat a|n\rangle=\sqrt{n}\,|n-1\rangle$ , generating the entire number basis from the vacuum.

Where this lives today

The number basis became more than a blackboard tool once individual Fock states could be built and measured. In cavity and circuit QED, a single microwave mode is coupled to a superconducting qubit or Rydberg atom, and driving that coupling pumps the field up the ladder to prepare $|1\rangle,|2\rangle,\dots$ and even superpositions of them; the $\sqrt n$ scaling of the Jaynes–Cummings coupling — the factor from equation (3) — is read straight off the oscillation frequencies, confirming the ladder is real. Optically, single-photon sources based on semiconductor quantum dots, nitrogen-vacancy centres, and heralded parametric down-conversion emit near-ideal $|1\rangle$ states with measured $g^{(2)}(0)$ well below $0.05$ , verified in the Hanbury Brown–Twiss geometry of Example 2. These photons are the flying qubits of photonic quantum computing and key distribution, while higher number and NOON states push interferometry toward the Heisenberg limit $\Delta\phi\sim1/N$ , where $\Delta n=0$ beats the shot-noise floor $1/\sqrt N$ — limited in practice only by fragility to loss.

The physical picture

Picture a single mode — one colour, one direction, one polarization — as a quantum harmonic oscillator whose "position" and "momentum" are the two field quadratures. Its energy comes in equally spaced rungs separated by $\hbar\omega$ , and the rung is the photon: a state on the $n$ -th rung holds exactly $n$ quanta at energy $\hbar\omega(n+\tfrac12)$ . Climbing one rung with $\hat a^{\dagger}$ adds a photon, stepping down with $\hat a$ removes one, and the ladder cannot descend below the vacuum since $\hat a|0\rangle=0$ . The first figure draws this ladder with the real energies of a helium–neon laser mode.

Figure

The photon-number ladder for a 633 nm cavity mode: rungs are equally spaced by ħω = 1.96 eV, â† climbs one rung with amplitude √(n+1) while â descends with √n, and the vacuum sits at zero-point energy E₀ = 0.98 eV

A number state is perfectly counted but phase-blind. Number and phase are conjugate, so pinning $n$ to an exact integer smears the phase uniformly over $[0,2\pi)$ . In phase space it is not a blob but a ring: the Wigner function of $|n\rangle$ is rotationally symmetric and, for $n\ge1$ , dips negative — a rigorous certificate of non-classicality no classical mixture can fake. It is the deepest sense in which $|1\rangle$ is "one photon" yet not a little classical pulse. The ladder factors carry physics. The $\sqrt{n+1}$ in emission says an excited atom radiates even into an empty mode ( $n=0$ gives $\sqrt1=1$ ): that "+1" is spontaneous emission, the surviving $\sqrt n$ the stimulated part. Detecting or losing a photon applies $\hat a$ , dropping $n$ by one — which is why any loss erodes a number state into a mixture of lower ones. The second figure compares $|n\rangle$ with the two commonest field states at the same mean photon number.

Figure

Photon-number distribution P(n) at mean n̄ = 4: the Fock state is a single spike at n = 4 with Δn = 0 and P = 1 off scale, the coherent state is Poissonian peaking near P ≈ 0.195, and thermal light is geometric peaking at n = 0 with P₀ = 0.20

The core equations

The whole structure is generated by one operator. The number operator $\hat n=\hat a^{\dagger}\hat a$ is Hermitian and positive, so its eigenvalues are real and non-negative; those eigenvalues turn out to be non-negative integers, a fact derived below rather than assumed.

\hat n=\hat a^{\dagger}\hat a,\qquad \hat n|n\rangle=n\,|n\rangle,\quad n=0,1,2,\dots \qquad (1)

where: $\hat n$ — photon-number operator (Hermitian, positive); $\hat a,\hat a^{\dagger}$ — annihilation and creation operators from field quantization; $n$ — non-negative integer; $|n\rangle$ — the Fock state carrying exactly $n$ quanta. Measuring $\hat n$ on $|n\rangle$ returns $n$ with certainty and zero spread. The counting behaviour is fixed by a single commutator imported from field quantization, from which follow the two that make $\hat a,\hat a^{\dagger}$ raising and lowering operators.

[\hat a,\hat a^{\dagger}]=1,\qquad [\hat n,\hat a]=-\hat a,\qquad [\hat n,\hat a^{\dagger}]=+\hat a^{\dagger} \qquad (2)

where: $[\hat A,\hat B]=\hat A\hat B-\hat B\hat A$ . Commuting either operator through $\hat n$ shifts its eigenvalue by exactly $\mp1$ , so $\hat a$ steps a number eigenstate down and $\hat a^{\dagger}$ steps it up. Requiring each rung to stay unit-normalized fixes the amplitudes on the ladder: $\sqrt n$ for lowering (which kills the vacuum) and $\sqrt{n+1}$ for raising (whose surviving "+1" seeds spontaneous emission).

\hat a|n\rangle=\sqrt{n}\,|n-1\rangle,\qquad \hat a^{\dagger}|n\rangle=\sqrt{n+1}\,|n+1\rangle \qquad (3)

where: $\sqrt n$ — lowering amplitude, zero on $|0\rangle$ ; $\sqrt{n+1}$ — raising amplitude, never zero. The asymmetry between them is the quantum-optical origin of the split between stimulated ( $\propto n$ ) and spontaneous ( $\propto1$ ) emission. Iterating the raising relation from the vacuum builds every Fock state explicitly, with a factorial holding the norm at one.

|n\rangle=\frac{(\hat a^{\dagger})^{n}}{\sqrt{n!}}\,|0\rangle \qquad (4)

where: $|0\rangle$ — the unique normalizable state with $\hat a|0\rangle=0$ ; $\sqrt{n!}$ — normalization accumulated from $\sqrt1\cdot\sqrt2\cdots\sqrt n$ . The entire infinite tower is manufactured from one object, the vacuum, by repeated creation. The resulting states are orthonormal and complete — a genuine basis — so any state or operator of the mode can be written as a sum over number states.

\langle m|n\rangle=\delta_{mn},\qquad \sum_{n=0}^{\infty}|n\rangle\langle n|=\hat 1 \qquad (5)

where: $\delta_{mn}$ — Kronecker delta (distinct photon numbers are perfectly distinguishable); $\hat 1$ — identity on the single-mode Hilbert space. Completeness reduces any expectation value to a sum of number-state matrix elements. Since $\hat H$ for a free mode is $\hbar\omega(\hat n+\tfrac12)$ , the Fock states are automatically the energy eigenstates, with the uniform spectrum of Figure 1.

\hat H=\hbar\omega\!\left(\hat n+\tfrac12\right),\qquad E_n=\hbar\omega\!\left(n+\tfrac12\right) \qquad (6)

where: $\omega$ — angular frequency of the mode; $\tfrac12\hbar\omega$ — zero-point energy from aq-optics-vacuum-zero-point-energy, present even in $|0\rangle$ . Adding a photon costs exactly $\hbar\omega$ regardless of how many are already there. Now the surprise: despite definite energy, a number state has no field. The mean of the field operator vanishes for every $n$ while its mean square grows linearly with $n$ — definite intensity, undefined phase.

\langle n|\hat E|n\rangle=0,\qquad \langle n|\hat E^{2}|n\rangle\propto\left(n+\tfrac12\right) \qquad (7)

where: $\hat E\propto(\hat a\,e^{-i\omega t}+\hat a^{\dagger}e^{+i\omega t})$ — the single-mode field operator. The mean is zero because $\hat a,\hat a^{\dagger}$ are purely off-diagonal in the number basis, so a Fock state is a ring of all phases at once — never a tilted classical wave. Finally, the statistical signatures that make number states a technology: a delta-function distribution, extreme sub-Poissonian variance, and antibunching that is perfect at $n=1$ .

\Delta n=0,\qquad Q=\frac{(\Delta n)^{2}-\bar n}{\bar n}=-1,\qquad g^{(2)}(0)=\frac{\langle\hat a^{\dagger}\hat a^{\dagger}\hat a\hat a\rangle}{\langle\hat a^{\dagger}\hat a\rangle^{2}}=1-\frac1n\ \ (n\ge1) \qquad (8)

where: $\Delta n$ — photon-number standard deviation; $\bar n=n$ — mean; $Q$ — Mandel parameter ( $Q<0$ is classically impossible); $g^{(2)}(0)$ — normalized zero-delay intensity correlation. It starts at $0$ for a single photon (no two ever coincide) and climbs toward the coherent value $1$ as $n\to\infty$ .

Derivation AFrom the commutators to the integer ladder and its amplitudes

The goal is to prove, using only $[\hat a,\hat a^{\dagger}]=1$ and the positivity of the number operator, that the eigenvalues of $\hat n$ can only be non-negative integers, that the tower terminates at a vacuum annihilated by $\hat a$ , and that unit normalization forces exactly $\sqrt n$ and $\sqrt{n+1}$ . Nothing about "photons" is assumed — the whole discrete structure comes from one algebraic relation plus non-negative norms.

Step 1. Suppose $\hat n|\nu\rangle=\nu|\nu\rangle$ for real $\nu$ and normalized $|\nu\rangle$ . Then $\nu=\langle\nu|\hat a^{\dagger}\hat a|\nu\rangle=\lVert\hat a|\nu\rangle\rVert^{2}\ge0$ . Why: the eigenvalue is literally the squared length of $\hat a|\nu\rangle$ , and no vector has negative length — so $\hat n$ has no negative eigenvalue.

Step 2. Using $[\hat n,\hat a]=-\hat a$ , $\hat n(\hat a|\nu\rangle)=(\nu-1)(\hat a|\nu\rangle)$ , so $\hat a|\nu\rangle$ is an eigenstate one rung lower; repeating gives a descending sequence $\nu,\nu-1,\nu-2,\dots$ Why: this is what "lowering operator" means — each $\hat a$ steps one rung down the ladder.

Step 3. A strictly descending sequence would eventually pass below zero, contradicting Step 1, unless it stops — which requires $\hat a|\nu_{\min}\rangle=0$ , forcing $\nu_{\min}=0$ . Hence $\nu$ was reached from $0$ by integer steps, so $\nu\in\{0,1,2,\dots\}$ . Why: positivity plus a unit-step ladder is the pincer that quantizes the spectrum and manufactures the vacuum.

Step 4. Fix the amplitudes: writing $\hat a^{\dagger}|n\rangle=c_n|n+1\rangle$ , unit norm gives $|c_n|^{2}=\langle n|\hat a\hat a^{\dagger}|n\rangle=\langle n|(\hat n+1)|n\rangle=n+1$ , so $c_n=\sqrt{n+1}$ , and the adjoint relation gives $\hat a|n\rangle=\sqrt n\,|n-1\rangle$ . Why: the "+1" here comes from commuting $\hat a$ past $\hat a^{\dagger}$ — the same "+1" that survives on the vacuum and drives spontaneous emission.

Go deeper · Derivation B · 1 minWhy a number state has zero mean field but growing fluctuations

Derivation BWhy a number state has zero mean field but growing fluctuations

The goal is to show that the paradox — a state of definite energy has no field — follows algebraically from orthonormality of the number basis, and to compute how the field variance grows with $n$ . This is the calculation behind "maximally non-classical": the field mean is exactly zero for every $n$ while its spread keeps growing, so the state is a ring of all phases.

Step 1. With $\hat E=\mathcal E_0(\hat a\,e^{-i\omega t}+\hat a^{\dagger}e^{+i\omega t})$ , the expectation needs $\langle n|\hat a|n\rangle=\sqrt n\,\langle n|n-1\rangle=0$ and likewise $\langle n|\hat a^{\dagger}|n\rangle=0$ . Why: $\hat a$ and $\hat a^{\dagger}$ each change the photon number by one, so they connect only different Fock states — their diagonal elements vanish.

Step 2. Therefore $\langle n|\hat E|n\rangle=0$ for every $n$ , including huge $n$ : no mean field, no phase, the state points nowhere in phase space. Why: definite photon number and definite phase are complementary, so an exact integer $n$ must completely randomize the phase.

Step 3. For the variance, only the number-conserving terms of $\hat E^{2}$ survive on the diagonal: $\langle n|\hat a^{\dagger}\hat a+\hat a\hat a^{\dagger}|n\rangle=n+(n+1)=2n+1$ , so $\langle\hat E^{2}\rangle=\mathcal E_0^{2}(2n+1)\propto(n+\tfrac12)$ . Why: the $\hat a^2$ and $(\hat a^{\dagger})^2$ terms shift the number by two and average to zero, leaving the symmetric product — the same one that gives the zero-point variance at $n=0$ .

Step 4. Since the mean is zero, $2n+1$ is the full field variance: nonzero even for the vacuum ( $n=0$ gives $1$ ) and growing without bound. Why: this is the quantitative sense in which a number state has ever-larger fluctuations wrapped around a vanishing mean — the phase-space ring just grows in radius.

Limiting cases

Case	Result	Meaning
Vacuum, $n=0$	$\hat a\vert0\rangle=0,\ E_0=\tfrac12\hbar\omega$	Bottom of the ladder; nonzero zero-point energy despite "no photons".
Single photon, $n=1$	$\vert1\rangle=\hat a^{\dagger}\vert0\rangle,\ g^{(2)}(0)=0$	The prototypical antibunched state; $g^{(2)}(0)=0$ signals a single-photon source.
Two photons, $n=2$	$g^{(2)}(0)=1-\tfrac12=\tfrac12$	Sub-Poissonian but not perfectly antibunched — the pair can be detected together.
Large $n$	$E_n\approx n\hbar\omega$	Energy nears the classical value, yet the phase stays random and $\langle\hat E\rangle=0$ — never a classical wave.
Coherent-state limit	$\vert\alpha\rangle=e^{-\vert\alpha\vert^{2}/2}\sum_n\tfrac{\alpha^{n}}{\sqrt{n!}}\vert n\rangle$	A phase-locked Fock superposition; a classical field for $\vert\alpha\vert\gg1$ , with $Q=0$ .

Worked examples

1 · The energy ladder of a real laser mode (confidence builder)

A helium–neon laser mode has vacuum wavelength $\lambda=633\ \mathrm{nm}$ . Find the photon energy $\hbar\omega$ , the zero-point energy of $|0\rangle$ , the energy of $|1\rangle$ , and check how close $|10\rangle$ sits to the "classical" energy $10\hbar\omega$ .

Step 1. The mode frequency is $\nu=c/\lambda=(2.998\times10^{8})/(633\times10^{-9})=4.74\times10^{14}\ \mathrm{Hz}$ , so one quantum carries $\hbar\omega=h\nu=(6.626\times10^{-34})(4.74\times10^{14})=3.14\times10^{-19}\ \mathrm{J}=\mathbf{1.96\ eV}$ — the uniform rung spacing of Figure 1.

Step 2. The vacuum is not empty of energy: $E_0=\tfrac12\hbar\omega=\mathbf{0.98\ eV}$ , the zero-point energy from the vacuum article, present even at $\bar n=0$ .

Step 3. One rung up, $E_1=\tfrac32\hbar\omega=\mathbf{2.94\ eV}$ ; note $E_1-E_0=\hbar\omega=1.96\ \mathrm{eV}$ exactly, as a uniform ladder requires.

Step 4. For $|10\rangle$ , $E_{10}=10.5\,\hbar\omega=\mathbf{20.6\ eV}$ versus the naive $10\hbar\omega=19.6\ \mathrm{eV}$ ; the ratio $1.05$ shows the zero-point offset is still a $5\%$ correction at ten photons. Spacing $\hbar\omega=1.96$ eV, $E_0=0.98$ eV, $E_1=2.94$ eV, and $|10\rangle$ lies $5\%$ above $10\hbar\omega$ .

2 · Antibunching and reading a Hanbury Brown–Twiss measurement

Compute $g^{(2)}(0)$ for $|1\rangle$ , $|2\rangle$ and $|5\rangle$ , and interpret a lab measurement of $g^{(2)}(0)=0.05$ from a candidate single-photon source.

Step 1. From $g^{(2)}(0)=1-1/n$ : the single photon gives $1-1/1=\mathbf{0}$ , since the pair count $\langle\hat a^{\dagger}\hat a^{\dagger}\hat a\hat a\rangle=n(n-1)=0$ — no two photons can ever be detected at once.

Step 2. For two photons $g^{(2)}(0)=1-1/2=\mathbf{0.5}$ , for five $\mathbf{0.8}$ . All lie below the coherent value $1$ (and far below thermal $2$ ), so every Fock state is antibunched — but only $|1\rangle$ perfectly, since $n(n-1)\ne0$ for $n\ge2$ .

Step 3. A source is certified by the standard threshold $g^{(2)}(0)<0.5$ . A measured $0.05$ means the zero-delay coincidence rate is only $5\%$ of the accidental rate a laser of the same power would give — the residual quantifies multi-photon contamination. $g^{(2)}(0)=0,\,0.5,\,0.8$ for $|1\rangle,|2\rangle,|5\rangle$ ; a measured $0.05$ is a near-ideal single-photon source.

Physics of the portrait

The number states are named for the physicist who gave the many-particle field its natural home — a Hilbert space built rung by rung on the vacuum.

Origins · 1 minVladimir Fock and how this was discovered

Vladimir Fock

1898–1974 · second quantization and the many-particle Hilbert space

Vladimir Aleksandrovich Fock was a Soviet theoretical physicist whose name is stamped across quantum mechanics — the Hartree–Fock method, the Fock–Krylov theorem, the Fock–Schwinger gauge — but nowhere more permanently than on the space of number states. Shaped by the turmoil of revolution and two world wars (he was arrested more than once, and a wartime injury left him partly deaf), Fock produced in 1932 the paper Konfigurationsraum und zweite Quantelung, recasting quantum field theory in the now-universal language of a Hilbert space spanned by states of definite particle number, with creation and annihilation operators moving between them. That construction — Fock space — is precisely the tower of $|n\rangle$ this article is about, and "second quantization" is the act of promoting the classical field amplitude to the operators $\hat a,\hat a^{\dagger}$ that climb it. The elegance of the number basis — every field state written as balances in one universal ledger — is his lasting gift to quantum optics, even though the optical single-photon states that most vividly realize it came only decades after his death.

History

1900

— Planck quantizes the oscillator, allowing energy only in units of hν

1925

— Born, Heisenberg and Jordan derive the oscillator ladder operators within matrix mechanics

1927

— Dirac quantizes the radiation field, introducing photon creation and annihilation operators

1932

— Fock formalizes second quantization and the number-state basis now bearing his name

1963

— Glauber's quantum theory of optical coherence makes the Fock basis the reference for all field states

1977

— Kimble, Dagenais and Mandel measure photon antibunching, the first observation of sub-Poissonian light

Going further

Texts. Loudon, The Quantum Theory of Light, 3rd ed., ch. 4 (single-mode quantization and number states) · Gerry and Knight, Introductory Quantum Optics, ch. 2–3 (Fock states, coherent states, photon statistics) · Saleh and Teich, Fundamentals of Photonics, 2nd ed., ch. 12 · Foot, Atomic Physics, ch. 1 and the field-quantization appendix · Demtröder, Laser Spectroscopy, vol. 1. The key results collected here are: the integer spectrum and vacuum follow from positivity of $\hat n$ plus $[\hat a,\hat a^{\dagger}]=1$ ; the ladder amplitudes $\sqrt n,\sqrt{n+1}$ follow from unit normalization; $E_n=\hbar\omega(n+\tfrac12)$ ; $\langle\hat E\rangle=0$ with $\langle\hat E^{2}\rangle\propto(n+\tfrac12)$ ; Mandel $Q=-1$ and $g^{(2)}(0)=1-1/n$ (giving $0,\,0.5,\,0.8$ for $n=1,2,5$ ); and the coherent overlap $|\langle n|\alpha\rangle|^{2}=e^{-|\alpha|^{2}}|\alpha|^{2n}/n!$ with $\bar n=|\alpha|^{2}$ , $Q=0$ , $P(3)=P(4)=0.1954$ at $\bar n=4$ . The laser-mode numbers ( $\lambda=633$ nm $\Rightarrow\hbar\omega=1.96$ eV, $E_0=0.98$ eV, $E_1=2.94$ eV, $E_{10}=20.6$ eV) use $c=2.998\times10^{8}\ \mathrm{m/s}$ and $h=6.626\times10^{-34}\ \mathrm{J\,s}$ .

Prerequisites. Quantization of the Electromagnetic Field · Vacuum State and Zero-Point Energy · Quantum Harmonic Oscillator. Leads to. Coherent States · Photon Statistics and the Mandel Q Parameter · Single-Photon Sources · Squeezed States of Light.

Ask the tutor: Ask me to derive the $\sqrt n$ Jaynes–Cummings scaling that lets a qubit climb the Fock ladder, or why detecting one photon degrades a coherent state differently from a true $|1\rangle$ .

If a number state has definite energy, why does it have no electric field?

Because energy (photon number) and optical phase are complementary, like position and momentum: pinning the number to an exact integer maximally randomizes the phase, so the field averages to zero. Algebraically the field operator is built from $\hat a$ and $\hat a^{\dagger}$ , each changing the number by one, so their diagonal elements $\langle n|\hat a|n\rangle$ vanish — while the intensity, built from the number-conserving $\hat a^{\dagger}\hat a$ , stays sharp. That is why $\langle n|\hat E|n\rangle=0$ even for millions of photons — the field is a phase-space ring with no preferred direction.

What is the difference between a Fock state and a coherent state?

They are opposite extremes. A Fock state $|n\rangle$ has a perfectly definite photon number ( $\Delta n=0$ , $Q=-1$ ) and no phase or mean field. A coherent state $|\alpha\rangle$ , the best description of ideal laser light, has a well-defined phase and oscillating mean field but a Poisson-distributed photon number, so $\Delta n=\sqrt{\bar n}$ and $Q=0$ . And $|\alpha\rangle$ is itself a superposition of Fock states with weights $e^{-|\alpha|^{2}}|\alpha|^{2n}/n!$ , so the number basis underlies both: a laser is a broad, phase-locked comb of Fock states, $|n\rangle$ a single rung.

Why is only the single-photon state perfectly antibunched?

Antibunching is measured by $g^{(2)}(0)$ , proportional to the pair count $\langle\hat a^{\dagger}\hat a^{\dagger}\hat a\hat a\rangle=n(n-1)$ — the ways to pick two photons from $n$ . For $|1\rangle$ there is only one photon, so $n(n-1)=0$ and $g^{(2)}(0)=0$ : two can never arrive together. For $|2\rangle$ the pair count is $2$ , giving $1/2$ ; in general $1-1/n$ . Every Fock state is thus sub-Poissonian (below the coherent value $1$ ), but only $n=1$ empties the two-photon channel — which is why a $g^{(2)}(0)$ dip toward zero certifies a genuine single-photon source.

Are Fock states stationary, and what evolves in time?

Yes — a number state is stationary. Because $|n\rangle$ is an eigenstate of $\hat H=\hbar\omega(\hat n+\tfrac12)$ , time evolution only multiplies it by the global phase $e^{-i\omega(n+\frac12)t}$ ; every expectation value — energy, photon number, the (already zero) mean field, $g^{(2)}(0)$ — is time-independent. A coherent state is different: its mean field genuinely oscillates at $\omega$ to trace the classical wave. So the visible oscillation of a light field comes from superpositions of different- $n$ Fock states beating together, not from any single number state, which just sits still up to an unobservable phase.

Why are high-number Fock states so hard to make and keep?

Two reasons, both from the ladder. First, generation: no simple linear-optics recipe deterministically deposits exactly $n$ quanta for $n\ge2$ ; one needs strong nonlinear coupling (cavity/circuit QED) or heralding, with efficiencies falling fast in $n$ . Second, fragility: any loss applies $\hat a$ , mapping $|n\rangle$ to $|n-1\rangle$ and, averaged over unknown loss, into a mixture of lower Fock states, so a single lost photon destroys the definite number. This is why coherent states, being eigenstates of $\hat a$ and unchanged in character by loss, describe ordinary light while clean Fock states remain a laboratory achievement.

In what sense does "n photons" not mean "n particles in a box"?

The photon number counts elementary excitations of one delocalized field mode, not localized billiard balls. The state $|n\rangle$ fills whatever profile the mode function defines — spread over the whole beam or cavity, not sitting at $n$ points. Detecting "a photon" means applying $\hat a$ to that entire mode, lowering its excitation by one quantum, not plucking a pre-existing particle from a collection. This is why interference, indistinguishability, and the bosonic $\sqrt{n+1}$ enhancement all make sense: the quanta have no individual identities. The particle language is shorthand for the rung number on the ladder, and taking it too literally causes most confusion about single-photon experiments.