Photon Number States

A photon number state, or optical Fock state, has a definite number of excitations in one or more specified electromagnetic modes. For one normalized mode $f$,

$$\hat N_f\lvert n_f\rangle = n\lvert n_f\rangle, \qquad n=0,1,2,\ldots .$$

The subscript matters. A photon is an excitation of a mode, and a mode includes its frequency content, spatial profile, polarization, and temporal envelope. The statement “this pulse is a one-photon state” is incomplete until the relevant mode or mode subspace has been identified.

Number states have exact excitation number, vanishing number variance, and no first-order field amplitude. Nonvacuum number states additionally have nonclassical counting correlations; the vacuum is a coherent state. Those facts do not imply that:

• a photon is a tiny localized pellet following a trajectory;
• every detector gate returns exactly the prepared photon number;
• a weak laser pulse is a one-photon state;
• a definite-number state possesses an ordinary but unknown optical phase;
• a small measured value of $g^{(2)}(0)$ by itself certifies every desirable property of a single-photon source.

Required background. Quantized Electromagnetic Modes defines optical mode operators, Number States gives oscillator eigenstates, Mode Occupations explains occupation bases, and Density Operators distinguishes mixtures from coherent superpositions.

Optical Occupation and Detector Counts

The phrase “photon number” can refer to several related but distinct objects:

Question	Mathematical object
How many excitations occupy mode $f$?	eigenvalue of $\hat N_f=\hat a_f^\dagger\hat a_f$
What number would an ideal measurement of that mode return?	distribution $P_f(n)=\langle n_f\vert\rho\vert n_f\rangle$
How many detector events occur in a gate?	a detector-dependent count distribution
How many photons occupy all modes in a set $\mathcal B$?	$\hat N_{\mathcal B}=\sum_{\mu\in\mathcal B}\hat N_\mu$
How many photons leave an open system per unit time?	output flux or counting intensity

Mode occupations and their sums are observables of the specified field state. An output flux likewise needs a defined propagating channel and time normalization. A raw count record additionally depends on loss, collection, mode-selective filtering, detector efficiency, dark counts, dead time, timing jitter, and number resolution.

The clean model below begins with lossless bosonic modes. Open propagation and detection are introduced as quantum channels and measurements, not silently folded into the definition of the state.

Number States of One Optical Mode

Let $\hat a_f$ annihilate one excitation in a normalized electromagnetic mode $f$, with

$$[\hat a_f,\hat a_f^\dagger]=1.$$

The mode number operator is

$$\hat N_f = \hat a_f^\dagger\hat a_f.$$

Starting from the mode vacuum,

$$\hat a_f\lvert0_f\rangle=0,$$

the normalized number states are

$$\lvert n_f\rangle = \frac{(\hat a_f^\dagger)^n}{\sqrt{n!}} \lvert0_f\rangle.$$

They obey

$$\hat a_f\lvert n_f\rangle = \sqrt n\,\lvert(n-1)_f\rangle,$$ $$\hat a_f^\dagger\lvert n_f\rangle = \sqrt{n+1}\,\lvert(n+1)_f\rangle,$$

and

$$\hat N_f\lvert n_f\rangle = n\lvert n_f\rangle.$$

For an ideal monochromatic normal mode of angular frequency $\omega_f$, the free Hamiltonian contribution is

$$\hat H_f = \hbar\omega_f \left( \hat N_f+\frac12 \right).$$

Thus $\lvert n_f\rangle$ is both a number eigenstate and an energy eigenstate. For a finite-bandwidth traveling wave packet, however, a state with one photon in the packet need not be an exact free-energy eigenstate. Definite excitation number and definite energy are different assertions.

The optical field in a number state

For one mode at a fixed spacetime point, a field quadrature has the form

$$\hat X_\theta = \frac{ \hat a_f e^{-i\theta} + \hat a_f^\dagger e^{i\theta} }{\sqrt2}.$$

Every number state has

$$\langle n_f\vert\hat a_f\vert n_f\rangle=0, \qquad \langle\hat X_\theta\rangle=0.$$

Its quadrature variance is independent of $\theta$:

$$(\Delta X_\theta)^2 = n+\frac12.$$

The mean electric field therefore vanishes, but its fluctuations do not. The state $\lvert n_f\rangle$ is not a classical wave of fixed amplitude whose phase has merely been hidden. Its field statistics differ from those of a phase-randomized classical amplitude.

The vacuum $n=0$ has variance $1/2$ in every quadrature, the smallest common variance of a phase-invariant state. A squeezed state can reduce one quadrature below $1/2$ at the expense of the other. Adding a definite photon number increases both conjugate quadrature variances equally rather than selecting a classical oscillation phase.

One Photon Means One Mode Excitation

A normalized wave-packet mode can be assembled from a continuum of frequency, wavevector, and polarization modes. In schematic notation,

$$\hat a_f^\dagger = \sum_\lambda \int d^3k\, f_\lambda(\mathbf k) \hat a_{\mathbf k\lambda}^\dagger,$$

with

$$\sum_\lambda \int d^3k\, \lvert f_\lambda(\mathbf k)\rvert^2 = 1.$$

The associated one-photon state is

$$\lvert1_f\rangle = \hat a_f^\dagger\lvert0\rangle.$$

Its label $f$ contains the full mode specification. Another normalized mode $g$ has overlap

$$[\hat a_f,\hat a_g^\dagger] = \langle f\vert g\rangle.$$

Consequently,

$$\langle1_f\vert1_g\rangle = \langle f\vert g\rangle.$$

Perfectly distinguishable photons occupy orthogonal modes; perfectly indistinguishable single photons occupy the same mode apart from degrees of freedom deliberately used to label paths or qubits.

A photon does not have to be monochromatic

Free propagation replaces the spectral amplitude by $f_\lambda(\mathbf k)e^{-i\omega_{\mathbf k}t}$. Total photon number stays one, while energy can have nonzero variance. The packet evolution and its energy moments are derived in Quantized Electromagnetic Modes. The fixed input packet need not remain an eigenmode of occupation at later times; its freely evolved packet does.

Mode basis dependence

Suppose one input mode is transformed into two orthogonal output modes:

$$\hat a_f^\dagger \longmapsto t\,\hat b^\dagger + r\,\hat c^\dagger, \qquad \lvert t\rvert^2+\lvert r\rvert^2=1.$$

Then

$$\lvert1_f\rangle \longmapsto t\lvert1_b,0_c\rangle + r\lvert0_b,1_c\rangle.$$

The total photon number remains exactly one, but neither output port has definite local occupation. This is ordinary unitary mode mixing, not creation or destruction of the photon.

The general basis-dependence of occupation is developed on Mode Occupations.

Many-Mode Optical Fock States

For orthonormal optical modes $f_1,f_2,\ldots$, a multimode number state is

$$\lvert n_1,n_2,\ldots\rangle = \prod_j \frac{ (\hat a_j^\dagger)^{n_j} }{ \sqrt{n_j!} } \lvert0\rangle.$$

It satisfies

$$\hat N_j \lvert n_1,n_2,\ldots\rangle = n_j \lvert n_1,n_2,\ldots\rangle.$$

The total number in the chosen mode set is

$$\hat N_{\mathrm{tot}} = \sum_j\hat N_j.$$

A state may have definite total number without definite occupation of each mode. For example,

$$\lvert\psi\rangle = \frac{ \lvert1_a,0_b\rangle + e^{i\phi}\lvert0_a,1_b\rangle }{\sqrt2}$$

has one photon in total and a meaningful relative phase $\phi$ between paths. This does not contradict the absence of an absolute phase for a single-mode number state. Relative phase resides in coherence between different mode occupations.

Mixed one-photon states

Not every source emitting at most one photon prepares a pure mode $\lvert1_f\rangle$. Conditioned on the one-photon sector, a general state is

$$\rho_1 = \sum_j q_j \lvert1_{f_j}\rangle \langle1_{f_j}\rvert, \qquad \sum_jq_j=1,$$

in its eigenmode decomposition. Its modal purity is

$$\mathcal P_1 = \operatorname{Tr}(\rho_1^2) = \sum_jq_j^2.$$

A source can therefore have nearly perfect photon-number purity and still emit photons in an incoherent mixture of temporal or spectral modes. This distinction is central to two-photon interference.

Photon Number Statistics

For a single selected mode in state $\rho$, the ideal photon-number distribution is

$$P(n) = \langle n\vert\rho\vert n\rangle, \qquad \sum_{n=0}^{\infty}P(n)=1.$$

The moments are

$$\langle \hat N^k\rangle = \sum_{n=0}^{\infty}n^kP(n).$$

For a number state $\lvert n_0\rangle$,

$$P(n)=\delta_{n,n_0},$$

so

$$\langle\hat N\rangle=n_0, \qquad (\Delta N)^2=0.$$

The probability-generating function

$$G(z) = \langle z^{\hat N}\rangle = \sum_{n=0}^{\infty}P(n)z^n$$

is useful because

$$\left. \frac{d^rG}{dz^r} \right|_{z=1} = \langle \hat N(\hat N-1)\cdots(\hat N-r+1) \rangle.$$

These falling-factorial moments are exactly the operator moments naturally selected by ideal absorptive photodetection:

$$\langle\hat N^{\underline r}\rangle = \left\langle (\hat a^\dagger)^r\hat a^r \right\rangle.$$

For $\lvert n_0\rangle$,

$$\langle\hat N^{\underline r}\rangle = \begin{cases} \dfrac{n_0!}{(n_0-r)!}, & r\le n_0, \\[6pt] 0, & r>n_0. \end{cases}$$

There cannot be an $(n_0+1)$-fold ideal absorption coincidence from an $n_0$-photon state.

Variance, Fano factor, and Mandel parameter

Three related statistics are

$$(\Delta N)^2 = \langle\hat N^2\rangle - \langle\hat N\rangle^2,$$ $$F = \frac{(\Delta N)^2}{\langle\hat N\rangle},$$

and

$$Q = \frac{ (\Delta N)^2-\langle\hat N\rangle }{ \langle\hat N\rangle } = F-1.$$

For an exact nonvacuum number state,

$$F=0, \qquad Q=-1.$$

For a Poisson distribution, $F=1$ and $Q=0$. Sub-Poissonian statistics means $F<1$, or equivalently $Q<0$. It is a sufficient signature of nonclassicality for the measured mode under the standard photodetection model.

Normalized intensity correlation

For a single selected mode with nonzero mean occupation, define the normalized factorial moment (often denoted $g^{(2)}(0)$) by

$$g^{(2)}(0) = \frac{ \langle \hat a^\dagger\hat a^\dagger\hat a\hat a \rangle }{ \langle\hat a^\dagger\hat a\rangle^2 } = \frac{ \langle\hat N(\hat N-1)\rangle }{ \langle\hat N\rangle^2 }.$$

For $\lvert n\rangle$ with $n\ge1$,

$$g^{(2)}(0) = 1-\frac1n.$$

In particular,

$$g^{(2)}(0)=0$$

for an ideal one-photon state. Two detections from the same single excitation are impossible. For large $n$, $g^{(2)}(0)$ approaches one even though the state remains an exact, highly nonclassical number state. Therefore $g^{(2)}(0)\approx1$ does not by itself prove that light is coherent or classical.

The relation

$$Q = \langle\hat N\rangle \left[ g^{(2)}(0)-1 \right]$$

follows directly from $\hat N^2=\hat N(\hat N-1)+\hat N$. It holds for the same number observable with finite nonzero mean. For a pulse this is a mode- or gate-integrated statistic, not automatically the pointwise two-time correlation of a stationary beam. In particular, an isolated monochromatic Fock state has constant $g^{(2)}(\tau)=1-1/n$; sub-Poissonian occupation alone does not establish a temporal rise away from zero delay. Antibunching explains that distinction.

An input $|3\rangle$ gives an exact occupation at $n=3$. Independent loss with $\eta=0.6$ produces probabilities $(0.064,0.288,0.432,0.216)$ for $k=0,1,2,3$ counts. It cannot produce $k>3$ without added light or detector noise. The count variance changes, while $g^{(2)}=2/3$ survives ideal uniform loss at nonzero efficiency.

Prepared Number Versus Detected Counts

An ideal number measurement projects onto

$$\Pi_n=\lvert n\rangle\langle n\rvert.$$

Real photodetection usually implements a different positive-operator-valued measure after a lossy channel. For a detector with independent efficiency $\eta$ and perfect number resolution, each of the $n$ incident photons is registered with probability $\eta$. The conditional count distribution is

$$\begin{aligned} P(k\mid n) &= \binom nk \eta^k (1-\eta)^{n-k}, \\ &\qquad 0\le k\le n. \end{aligned}$$

For an incident number distribution $P_{\mathrm{in}}(n)$,

$$P_{\mathrm{det}}(k) = \sum_{n=k}^{\infty} \binom nk \eta^k (1-\eta)^{n-k} P_{\mathrm{in}}(n).$$

This is Bernoulli thinning. For an input $\lvert n_0\rangle$ with $n_0>0$ and $\eta>0$,

$$\langle K\rangle = \eta n_0,$$ $$\operatorname{Var}(K) = \eta(1-\eta)n_0.$$

The observed Fano factor and Mandel parameter become

$$F_{\mathrm{det}}=1-\eta, \qquad Q_{\mathrm{det}}=-\eta.$$

Loss weakens sub-Poissonian number squeezing but does not make the detected count distribution a delta function at a smaller integer.

Why normalized correlations can survive loss

Under independent loss with $\eta>0$,

$$\langle K\rangle = \eta\langle N\rangle,$$

and

$$\langle K(K-1)\rangle = \eta^2 \langle N(N-1)\rangle.$$

Hence

$$g_{\mathrm{det}}^{(2)}(0) = g_{\mathrm{in}}^{(2)}(0)$$

when there are no dark counts, backgrounds, saturation effects, or mode-dependent efficiencies. This useful cancellation does not mean loss is irrelevant: brightness, heralding efficiency, count uncertainty, and higher-order inference all degrade.

On–off detectors

An on–off detector distinguishes no click from at least one click but does not resolve $k=1$ from $k=2$. For an $n$-photon input with no dark counts,

$$P(\text{no click}\mid n) = (1-\eta)^n,$$ $$P(\text{click}\mid n) = 1-(1-\eta)^n.$$

A click therefore does not certify that exactly one photon arrived. Detector tomography, calibrated attenuation, multiplexing, or coincidence measurements may be needed to infer the incident distribution. The number-resolving and threshold POVMs, dark counts, dead time, and detector calibration are developed on optical Photon Counting. Conditional time records and quantum-jump dynamics are treated on open-system Photon Counting.

Number and Optical Phase

A number state has no preferred absolute optical phase. The precise reason is not merely that some heuristic uncertainty product becomes large. The number spectrum is bounded below, and this obstructs an everywhere-defined self-adjoint phase operator satisfying an unrestricted canonical commutator with $\hat N$.

Suppose one naively assumed

$$[\hat N,\hat\Phi]=i\hat I$$

on the number states. Taking the diagonal matrix element in $\lvert n\rangle$ would give

$$\langle n\vert [\hat N,\hat\Phi] \vert n\rangle = 0,$$

whereas the proposed right-hand side gives $i$. The contradiction signals a domain and spectrum problem, not a failure of quantum mechanics.

Phase shifts are well defined

The unitary phase-shift operator is unproblematic:

$$\hat U(\varphi) = e^{-i\varphi\hat N}.$$

It acts on a number state as

$$\hat U(\varphi)\lvert n\rangle = e^{-in\varphi}\lvert n\rangle.$$

This is only a global phase for a single number state, so every measurement probability is invariant under $\varphi$. A superposition of different number states can acquire observable relative phases.

The exponential phase operator is not unitary

A useful number-lowering shift is

$$\hat E = \sum_{n=0}^{\infty} \lvert n\rangle\langle n+1\rvert = (\hat N+1)^{-1/2}\hat a.$$

It obeys

$$\hat E\hat E^\dagger=\hat I,$$

but

$$\hat E^\dagger\hat E = \hat I-\lvert0\rangle\langle0\rvert.$$

The missing inverse at the vacuum boundary prevents $\hat E$ from being a unitary exponential of an ordinary Hermitian phase operator.

Phase POVMs

Optical phase can nevertheless be described operationally by a positive-operator-valued measure. Introduce generalized phase kets

$$\lvert\theta\rangle = \sum_{n=0}^{\infty} e^{in\theta}\lvert n\rangle, \qquad 0\le\theta<2\pi.$$

They are not normalizable Hilbert-space vectors, but they define the canonical phase POVM

$$d\Pi(\theta) = \frac{d\theta}{2\pi} \lvert\theta\rangle\langle\theta\rvert.$$

For a state with number-basis elements $\rho_{mn}=\langle m\vert\rho\vert n\rangle$, the phase probability density is

$$p(\theta) = \frac{1}{2\pi} \sum_{m,n=0}^{\infty} \rho_{mn} e^{i(n-m)\theta}.$$

For $\rho=\lvert n_0\rangle\langle n_0\rvert$,

$$p(\theta)=\frac{1}{2\pi}.$$

This phase observable refers the signal to an external phase reference; there is no directly measurable reference-free absolute phase. Its outcomes are uniformly distributed for a number state. More generally, every number-diagonal state has a uniform canonical phase distribution because it contains no coherence between different number sectors.

Pegg–Barnett construction

Pegg and Barnett define phase first in an $(s+1)$-dimensional truncated number space. The orthonormal phase states are

$$\lvert\theta_m^{(s)}\rangle = \frac{1}{\sqrt{s+1}} \sum_{n=0}^{s} e^{in\theta_m} \lvert n\rangle,$$

where

$$\theta_m = \theta_0+\frac{2\pi m}{s+1}, \qquad m=0,\ldots,s.$$

Physical expectation values are calculated at finite $s$, and only then is the limit $s\to\infty$ taken. This produces a consistent Hermitian phase observable within the prescription. It should not be shortened to the claim that an unrestricted canonical $\hat\Phi$ exists on the full oscillator Hilbert space.

What “phase uncertainty” should mean

Phase is periodic, so an ordinary linear variance depends on the branch cut. Useful measures include:

• the canonical phase distribution $p(\theta)$;
• circular moments such as $\langle e^{i\theta}\rangle$;
• the Holevo variance where applicable;
• a specified interferometric phase-estimation error;
• quadrature statistics measured relative to a local oscillator.

For a number state, the first circular moment vanishes and the canonical phase distribution is uniform. Writing

$$\Delta N\,\Delta\Phi\ge\frac12$$

without defining $\Phi$, its domain, and the periodic uncertainty measure is not a reliable statement.

Absolute Phase, Relative Phase, and References

Optical phase is observed relative to a reference mode, path, clock, or local oscillator. Consider

$$\lvert\psi\rangle = \frac{ \lvert1_a,0_b\rangle + e^{i\phi}\lvert0_a,1_b\rangle }{\sqrt2}.$$

The total photon number is exactly one, yet a beam splitter can convert the relative phase $\phi$ into output probabilities. There is no contradiction: $\phi$ relates two amplitudes in a multimode state and is not an absolute single-mode phase.

Likewise, homodyne detection uses a strong local oscillator to define which field quadrature is measured. A phase estimate always inherits assumptions about the reference and mode matching. Removing the reference can turn an apparently phase-coherent description into a number-diagonal reduced state.

Single-Photon Sources

An ideal pulsed single-photon source would produce, on demand, the same pure wave-packet state on every trigger:

$$\rho_{\mathrm{ideal}} = \lvert1_f\rangle\langle1_f\rvert.$$

For number-insensitive diagnostics, phase averaging permits a block-diagonal description in photon-number sectors,

$$\rho = p_0\rho_0 + p_1\rho_1 + p_2\rho_2 + \cdots,$$

where $\rho_n$ is normalized within the $n$-photon sector and

$$\sum_{n=0}^{\infty}p_n=1.$$

Here $p_n=\operatorname{Tr}(\Pi_n\rho)$ and $p_n\rho_n=\Pi_n\rho\Pi_n$ for the total-number projector $\Pi_n$. A general source can also have coherences between different number sectors; the displayed sum omits them and does not describe all phase-sensitive measurements. Within each sector, $\rho_1$ may be mixed and $\rho_2$ may contain photons in distinguishable modes.

Source qualities are multidimensional

A useful source characterization reports several quantities rather than one headline number:

Property	Operational question
preparation probability	how often does a trigger create the desired state?
brightness	how often is a useful photon delivered at a stated reference plane?
multiphoton suppression	how small are $p_{n\ge2}$ or $g^{(2)}(0)$?
collection efficiency	how much emitted light enters the target mode?
heralding efficiency	given a herald, how often is its partner present?
purity	is the one-photon sector a pure spatiotemporal mode?
indistinguishability	do photons from separate trials interfere as identical modes?
repetition rate	how often can a source attempt emission?
stability	do these properties remain constant over the acquisition?

Numbers must be quoted at a stated plane. “Source efficiency” measured at the first lens, fiber output, chip facet, or detector are not interchangeable.

Deterministic emitters

An emitter restricted to one radiative two-level transition can emit at most one photon before re-excitation. An initially excited multilevel atom can instead produce several photons in a cascade; the one-photon claim requires excluding such channels. Selected transitions in atoms, ions, molecules, color centers, and quantum dots can approximate the two-level case. Under pulsed excitation, the ideal cycle is

$$\lvert g,0\rangle \longrightarrow \lvert e,0\rangle \longrightarrow \lvert g,1_f\rangle.$$

Antibunching follows because the emitter must be excited again before another photon can be emitted. Real imperfections include re-excitation within the same pump pulse, nonradiative decay, dephasing, spectral diffusion, blinking, imperfect collection, and emission into unwanted modes. Photon Antibunching develops the conditional reset dynamics and the measured continuous-wave and pulsed correlation signatures.

A resonator can enhance emission into a selected mode and shape the output wave packet. Cavity-assisted Raman protocols can transfer an atomic excitation into a propagating photon in a controlled temporal mode. These processes connect to Spontaneous Emission and later cavity-QED treatments.

Heralded pair sources

For one signal–idler pair of orthogonal Schmidt modes, an undepleted classical pump generates an ideal two-mode squeezed state. Absorb its fixed pair phase into the mode definitions:

$$\begin{aligned} \lvert\Psi\rangle &= \sqrt{1-\lambda^2} \sum_{n=0}^{\infty} \lambda^n \lvert n_s,n_i\rangle, \\ &\qquad 0\le\lambda<1. \end{aligned}$$

This is a single-mode-pair model, not a generic spectral description of a broadband source. Several Schmidt pairs generally participate, as explained in Parametric Down-Conversion. In the ideal no-background model, detecting an idler photon heralds the presence of a signal photon before signal-arm loss. In the weak pump limit, the dominant heralded term is $\lvert1_s\rangle$, but higher pair terms remain:

$$\lvert2_s,2_i\rangle, \quad \lvert3_s,3_i\rangle, \quad\ldots$$

Increasing the pump raises the herald rate and the multipair contamination. Loss and threshold herald detectors make the conditional state more subtle because a single herald click may have originated from several idler photons.

Spectral correlations can also leave the heralded signal in a mixed mode. Engineering the joint spectral amplitude, filtering, or mode-selective detection can improve purity, often at a cost in brightness or heralding efficiency.

Attenuated laser pulses are not single-photon states

A coherent pulse with mean photon number $\mu$ has Poisson statistics,

$$P(n) = e^{-\mu}\frac{\mu^n}{n!}.$$

When $\mu\ll1$,

$$\begin{gathered} P(0)\approx1-\mu, \qquad P(1)\approx\mu, \\ P(2)\approx\frac{\mu^2}{2}. \end{gathered}$$

Most pulses are vacuum, some contain one photon, and a nonzero fraction contain two or more. Attenuation changes $\mu$ but does not convert the coherent state into $\lvert1\rangle$. In particular,

$$g^{(2)}(0)=1$$

for an ideal coherent state at every nonzero amplitude, whereas $g^{(2)}(0)=0$ for $\lvert1\rangle$.

Weak coherent pulses are useful and often experimentally convenient. They should simply be named correctly.

Diagnosing a Single-Photon Source

For a weak source with negligible $p_{n\ge3}$,

$$\langle N\rangle \approx p_1+2p_2,$$

and

$$\langle N(N-1)\rangle \approx 2p_2.$$

Hence

$$g^{(2)}(0) \approx \frac{2p_2}{(p_1+2p_2)^2}.$$

If $p_2\ll p_1$, this is often simplified to

$$g^{(2)}(0) \approx \frac{2p_2}{p_1^2}.$$

The approximation must not be used when multiphoton components are sizable. Moreover, a small $g^{(2)}(0)$ says little about vacuum probability, brightness, spectral purity, or indistinguishability.

Hanbury Brown–Twiss test

Send the source into one input of a balanced beam splitter and place detectors at both outputs. An ideal single-photon input transforms as

$$\lvert1,0\rangle \longmapsto \frac{ \lvert1,0\rangle + i\lvert0,1\rangle }{\sqrt2},$$

for one common phase convention. There is no $\lvert1,1\rangle$ term, so an ideal trial cannot produce a true coincidence between the two outputs.

The ratio of zero-delay coincidences to an appropriate uncorrelated baseline estimates the appropriate $g^{(2)}$ after corrections and normalization. With threshold detectors, the simple coincidence/singles ratio needs small click probabilities so clicks are approximately linear in photon number; otherwise use the full click POVM. Background, afterpulsing, detector dead time, timing windows, and source intermittency can bias the estimate. Reporting only a fitted number without the acquisition and normalization model is incomplete.

Hanbury Brown–Twiss Interferometry develops the two-detector geometry, accidental baseline, timing convolution, and continuous-versus-pulsed normalization.

Indistinguishability is a separate test

Two sources with strongly suppressed multiphoton emission may still differ in frequency, temporal shape, polarization, or spatial mode. Hong–Ou–Mandel interference compares two photons and probes their modal overlap. For pure one-photon modes $f$ and $g$, ideal balanced-beam-splitter coincidence probability is

$$P_{\mathrm c} = \frac12 \left( 1-\lvert\langle f\vert g\rangle\rvert^2 \right).$$

Thus perfect mode overlap gives zero coincidences, while orthogonal modes give $1/2$. For independent mixed one-photon states, replace the squared overlap by $\operatorname{Tr}(\rho_f\rho_g)$. Identical independent copies give that state’s modal purity. Uniform mode-independent loss reduces the coincidence rate but cancels from the normalized ideal visibility; mode-dependent filtering, backgrounds, multiphoton terms, and detector response can change it. Number purity and indistinguishability answer different questions.

Nonclassicality of Number States

Quantum optics often calls a state classical if it can be written as a statistical mixture of coherent states with a nonnegative Glauber–Sudarshan probability measure, allowing delta measures:

$$\rho = \int d^2\alpha\, P(\alpha) \lvert\alpha\rangle\langle\alpha\rvert, \qquad P(\alpha)\ge0.$$

For such a state, normally ordered field moments can be reproduced by a classical random complex amplitude. Number states with $n\ge1$ do not admit such a nonnegative probability measure. Their generalized $P$ distributions involve derivatives of delta functions. Singularity alone is not sufficient for nonclassicality: a coherent state has a positive delta measure.

Sub-Poissonian fluctuations

For any positive mixture of coherent states, conditional counts are Poissonian and classical intensity fluctuations can only add variance. Consequently,

$$\operatorname{Var}(N) \ge \langle N\rangle$$

under the ideal single-mode photodetection model. An observed

$$\operatorname{Var}(N) < \langle N\rangle$$

therefore rules out that classical model. Exact number states attain the strongest possible suppression, $\operatorname{Var}(N)=0$.

Anticorrelation at a beam splitter

A single classical wave entering a beam splitter produces fields in both outputs. Shot noise from ideal semiclassical photodetection does not create the perfect conditional anticorrelation of an ideal one-photon state. The beam-splitter test made this distinction operational: one excitation is found in one output or the other, never both in the same trial.

For $\lvert n\rangle$, the result

$$g^{(2)}(0)=1-\frac1n<1$$

is likewise incompatible with a positive classical intensity ensemble for that mode.

Wigner-function negativity

In the convention where the vacuum Wigner function is

$$W_0(\alpha) = \frac{2}{\pi} e^{-2\lvert\alpha\rvert^2},$$

the number-state Wigner function is

$$W_n(\alpha) = \frac{2}{\pi} (-1)^n L_n\left(4\lvert\alpha\rvert^2\right) e^{-2\lvert\alpha\rvert^2},$$

where $L_n$ is a Laguerre polynomial. For every $n\ge1$, this function has negative regions. Wigner negativity is another sufficient nonclassicality signature, but it is not equivalent to every other criterion for arbitrary states.

Worked Example: A Lossy Three-Photon State

Send $\lvert3\rangle$ through a pure-loss channel of transmissivity $0<\eta\le1$, with vacuum in the unused environmental port. The output state of the retained mode is number diagonal:

$$\rho_{\mathrm{out}} = \sum_{k=0}^{3} \binom3k \eta^k(1-\eta)^{3-k} \lvert k\rangle\langle k\rvert.$$

The probabilities are

$$P(0)=(1-\eta)^3,$$ $$P(1)=3\eta(1-\eta)^2,$$ $$P(2)=3\eta^2(1-\eta),$$ $$P(3)=\eta^3.$$

The mean and variance are

$$\langle N\rangle=3\eta, \qquad (\Delta N)^2=3\eta(1-\eta).$$

Yet

$$g^{(2)}(0) = \frac{ 3\cdot2\,\eta^2 }{ (3\eta)^2 } = \frac23,$$

independent of $\eta$ in the ideal loss-only model. The output is no longer an exact number state, while its normalized second-order correlation retains the input value.

Worked Example: Heralding with a Threshold Detector

Take the pair state

$$\lvert\Psi\rangle = \sqrt{1-\lambda^2} \sum_{n=0}^{\infty} \lambda^n \lvert n_s,n_i\rangle.$$

An ideal threshold detector on the idler reports “click” for every $n\ge1$. Conditioned on a click, the signal state is

$$\rho_{s\mid\mathrm{click}} = (1-\lambda^2) \sum_{n=1}^{\infty} \lambda^{2(n-1)} \lvert n_s\rangle\langle n_s\rvert.$$

The conditional single-photon weight is

$$P(1\mid\mathrm{click}) = 1-\lambda^2,$$

and the multiphoton weight is

$$P(n\ge2\mid\mathrm{click}) = \lambda^2.$$

Weak pumping suppresses multipair events, but it also lowers the unconditional herald rate. Nonunit idler efficiency further biases the conditional mixture because higher-pair events are more likely to produce at least one detected idler photon.

Photonic Qubits applies these states to path and polarization encodings. Phase-Space Distributions compares the corresponding Wigner, $P$, and $Q$ descriptions.

Exercises

1. Distinguish a Fock state from a phase-randomized wave

Compare $|n\rangle$ with an equal-phase mixture of coherent states $|\sqrt n e^{i\phi}\rangle$, for $n>0$. Both have zero quadrature mean and $\langle X^2\rangle=n+1/2$. Use $X^4={:}X^4{:}+3{:}X^2{:}+3/4$ to find their fourth moments.

Solution

For $|n\rangle$, only number-conserving normally ordered terms survive: $\langle{:}X^4{:}\rangle=3n(n-1)/2$ and $\langle{:}X^2{:}\rangle=n$. Therefore

$$\langle X^4\rangle_{\rm Fock}=\frac32n^2+\frac32n+\frac34.$$

For the phase-averaged coherent state, normal ordering replaces operators by complex amplitudes. The classical displacement is $\sqrt{2n}\cos\phi$, whose fourth average is $3n^2/2$. Thus

$$\langle X^4\rangle_{\rm phase\ average}=\frac32n^2+3n+\frac34.$$

The difference $3n/2$ distinguishes states with identical first and second quadrature moments. Rotational symmetry alone does not distinguish a Fock state from a phase-randomized coherent wave.

2. Correlations of an exact number state

Derive

$$g^{(2)}(0)=1-\frac1n$$

for $\lvert n\rangle$, and evaluate it for $n=1$, $2$, and $10$.

Solution

Apply the annihilation operators:

$$\hat a\hat a\lvert n\rangle = \sqrt{n(n-1)} \lvert n-2\rangle.$$

Therefore

$$\langle n\vert \hat a^\dagger\hat a^\dagger\hat a\hat a \vert n\rangle = n(n-1).$$

Since $\langle\hat a^\dagger\hat a\rangle=n$,

$$g^{(2)}(0) = \frac{n(n-1)}{n^2} = 1-\frac1n.$$

The requested values are

$$\begin{gathered} g^{(2)}_{n=1}(0)=0, \qquad g^{(2)}_{n=2}(0)=\frac12, \\ g^{(2)}_{n=10}(0)=0.9. \end{gathered}$$

The approach to one at large $n$ does not make the exact number state a coherent state; higher moments and phase-space structure remain different.

3. Loss and number resolution

An ideal $\lvert2\rangle$ state reaches a number-resolving detector with efficiency $\eta=0.6$. Find the probabilities of zero, one, and two counts, the mean count, the variance, and the observed $g^{(2)}(0)$.

Solution

Bernoulli thinning gives

$$P(k\mid2) = \binom2k (0.6)^k(0.4)^{2-k}.$$

Thus

$$\begin{gathered} P(0)=0.16, \qquad P(1)=0.48, \\ P(2)=0.36. \end{gathered}$$

The binomial mean and variance are

$$\langle K\rangle=2(0.6)=1.2,$$ $$\operatorname{Var}(K) = 2(0.6)(0.4) = 0.48.$$

The factorial moment is

$$\langle K(K-1)\rangle = 2P(2) = 0.72.$$

Therefore

$$g^{(2)}(0) = \frac{0.72}{(1.2)^2} = \frac12.$$

It equals the input value, illustrating loss invariance for ideal uniform thinning without background.

4. See the information lost by number-sector averaging

Compare $|\psi\rangle=(|0\rangle+|1\rangle)/\sqrt2$ with $\rho_{\rm mix}=(|0\rangle\langle0|+|1\rangle\langle1|)/2$. Find their mean number, $g^{(2)}$, field amplitude, and canonical phase distribution relative to a common reference.

Solution

Both have $P(0)=P(1)=1/2$, $\langle N\rangle=1/2$, and $g^{(2)}=0$. The pure state has $\langle a\rangle=1/2$ and $p(\theta)=(1+\cos\theta)/(2\pi)$. The mixture has $\langle a\rangle=0$ and $p(\theta)=1/(2\pi)$. Number counting cannot distinguish them, but phase-referenced measurements can. Describing a source solely by $p_n\rho_n$ discards this information.

5. One photon at a beam splitter

A photon enters port $a$ of a balanced beam splitter while port $b$ is in vacuum. Use

$$\hat a^\dagger \longmapsto \frac{ \hat c^\dagger+i\hat d^\dagger }{\sqrt2}$$

to find the output state and the photon-number covariance $\operatorname{Cov}(N_c,N_d)$.

Solution

The input is

$$\lvert1_a,0_b\rangle = \hat a^\dagger\lvert0\rangle.$$

After the beam splitter,

$$\lvert\psi_{\mathrm{out}}\rangle = \frac{ \lvert1_c,0_d\rangle + i\lvert0_c,1_d\rangle }{\sqrt2}.$$

Each output has mean occupation

$$\langle N_c\rangle = \langle N_d\rangle = \frac12.$$

There is never one photon in both outputs, so

$$\langle N_cN_d\rangle=0.$$

Hence

$$\begin{aligned} \operatorname{Cov}(N_c,N_d) &= \langle N_cN_d\rangle - \langle N_c\rangle\langle N_d\rangle \\ &= -\frac14. \end{aligned}$$

The negative covariance expresses output anticorrelation while total number remains exactly one.

6. Weak coherent pulse versus imperfect single-photon source

Compare two pulsed sources with the same mean photon number $\langle N\rangle=0.1$:

1. a coherent state;
2. the mixture $\rho=0.9\lvert0\rangle\langle0\rvert +0.1\lvert1\rangle\langle1\rvert$.

Find $P(2)$ and $g^{(2)}(0)$ for each.

Solution

For the coherent state, $\mu=0.1$ and

$$P_{\mathrm{coh}}(2) = e^{-0.1}\frac{0.1^2}{2} \approx 4.52\times10^{-3}.$$

Every nonvacuum coherent state has

$$g_{\mathrm{coh}}^{(2)}(0)=1.$$

The vacuum–one-photon mixture has

$$P_{\mathrm{mix}}(2)=0.$$

Moreover,

$$\langle N(N-1)\rangle=0,$$

so

$$g_{\mathrm{mix}}^{(2)}(0)=0.$$

Equal mean energy does not imply equal photon statistics. Vacuum probability also does not prevent the conditional nonvacuum output from having perfect multiphoton suppression.

7. Modal purity of a one-photon mixture

A source emits exactly one photon, but its one-photon state is

$$\rho_1 = q\lvert1_f\rangle\langle1_f\rvert + (1-q)\lvert1_g\rangle\langle1_g\rvert,$$

where $\langle f\vert g\rangle=0$. Compute the modal purity and determine its minimum over $q$. Find the ideal Hong–Ou–Mandel visibility for two independently emitted copies of this state.

Solution

Orthogonality removes cross terms in $\rho_1^2$:

$$\rho_1^2 = q^2\lvert1_f\rangle\langle1_f\rvert + (1-q)^2 \lvert1_g\rangle\langle1_g\rvert.$$

Therefore

$$\mathcal P_1 = \operatorname{Tr}(\rho_1^2) = q^2+(1-q)^2.$$

Differentiating,

$$\frac{d\mathcal P_1}{dq} = 4q-2,$$

so the minimum occurs at $q=1/2$ and equals

$$\mathcal P_{1,\min} = \frac12.$$

The source has perfect photon-number purity but imperfect modal purity. Photon counting alone cannot distinguish this mixture from a pure one-photon mode. For independent copies the visibility is $V_{\rm HOM}=\operatorname{Tr}\rho_1^2=q^2+(1-q)^2$, and the minimum visibility is $1/2$. Correlations between successive emissions would require their joint state instead of this independent-copy formula.

8. Threshold-heralded pair source

For the ideal pair state on this page, condition on a threshold idler click. Compute the conditional mean signal photon number and $g_s^{(2)}(0)$ in terms of $x=\lambda^2$.

Solution

The conditional distribution is geometric on $n=1,2,\ldots$:

$$P(n\mid\mathrm{click}) = (1-x)x^{n-1}.$$

Its mean is

$$\langle N_s\rangle = \sum_{n=1}^{\infty} n(1-x)x^{n-1} = \frac{1}{1-x}.$$

Using the second derivative of the geometric series,

$$\langle N_s(N_s-1)\rangle = \frac{2x}{(1-x)^2}.$$

Therefore

$$g_s^{(2)}(0) = \frac{ \langle N_s(N_s-1)\rangle }{ \langle N_s\rangle^2 } = 2x = 2\lambda^2.$$

As $\lambda\to0$, the heralded state approaches $\lvert1\rangle$ and $g_s^{(2)}(0)\to0$, while the unconditional probability of obtaining a herald also tends to zero.

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