University Physics IV · Limits of Classical Physics · 1.8
What the Photon Evidence Establishes
Two of the most famous experiments in modern physics do not show what they are usually said to show. The move to learn is a sharper question — not whether the quantum model fits the data, but whether any classical rival is ruled out by it — and then the measurements that do the ruling out.
Build the model
Connect the measurement to the mechanism.
The photon was not established by the experiments used to teach it. Take the atom to be quantum mechanical and leave the electromagnetic field classical — the semiclassical theory Lamb and Scully pressed in 1969 — and first-order perturbation theory returns everything the photoelectric effect is famous for: a sharp threshold at ħω = φ, because a classical field can only drive a transition the atom's own levels allow; a current proportional to intensity, because the transition rate is; and prompt emission, because a rate is constant from the moment the field arrives rather than a bucket slowly filling. Compton's result stands no better.
The shift Δλ = (h/mₑ c)(1 − cos θ) is two-body relativistic kinematics, and ħ enters it through the electron's de Broglie relations as much as the field's; Schrödinger recovered the same formula in 1927 by Bragg-scattering a classical wave off an electron matter wave. What both experiments establish is that matter absorbs and recoils in quanta — a claim about the detector, not the light. Establishing the field itself needs a quantity for which classical optics proves a bound: a non-negative intensity distribution forces g⁽²⁾(0) ≥ 1 and Mandel's Q ≥ 0.
Antibunched light from a single emitter breaks the first, sub-Poissonian counting breaks the second, and the photon arrives late, only after you have said what would have refuted it.
- Simple definition
- An experiment establishes that the electromagnetic field is quantised only if its result lies outside what every classical field — any non-negative distribution over classical intensities — could have produced.
- Example
- Cauchy–Schwarz puts that boundary at g⁽²⁾(0) ≥ 1; a single atom gives g⁽²⁾(0) → 0, and Grangier's 1986 beam splitter returned α = 0.18 ± 0.06, some thirteen standard deviations outside anything classical.
Threshold, current ∝ intensity and prompt onset all follow with no quantised field in the calculation.
A is the amplitude of a classical vector potential; ρ(Ef) is the density of final states per joule; I in W m⁻².
Predicts over an hour's delay; none is seen. That refutes classical energy accretion, not the classical field.
φ = 2.28 eV for sodium, I = 10⁻² W m⁻² and σ = πa₀² = 8.8 × 10⁻²¹ m² give tacc = 4.15 × 10³ s.
Conservation alone gives it, and Schrödinger got it in 1927 from a classical wave on an electron matter wave.
h/mₑ c = 2.426 pm is the electron Compton wavelength; θ is the scattering angle; Δλ comes out in picometres.
One number separates the theories: thermal 2, coherent 1, ideal single emitter 0.
Dimensionless. Cauchy–Schwarz on a non-negative intensity also forces g⁽²⁾(0) ≥ g⁽²⁾(τ).
Q < 0 needs a negative quasi-probability, so no classical field fakes it — but detector loss dilutes it.
Counts per gate T, dimensionless. The second equality needs T short beside the g⁽²⁾ dip; a longer gate replaces g⁽²⁾(0) by its gate average g⁽²⁾T. Classical Q ≥ 0; coherent Q = 0; single-mode thermal Q = ⟨n⟩; loss maps Q → ηQ.
Grangier 1986 measured α = 0.18 ± 0.06: one detector fires at a time, which no wave split in two can do.
Pₜ, Pᵣ, Pc are transmit, reflect and coincidence probabilities per trigger; loss cancels in the ratio.
Consistency is cheap; exclusion is what counts
An experiment supports a model when the model predicts what happens. It establishes the model only when every rival predicts something else. That second bar is far higher, and the history of the photon is where students most often miss the difference. The rival here is not classical physics wholesale — nobody defends a classical atom after 1913 — but semiclassical theory: quantum atoms obeying the Schrödinger equation, driven by a Maxwell field containing no photons. It is a real theory with real predictions, and nothing excluded it until Clauser's 1974 coincidence test and the antibunching measurements that followed. To retire it you need a quantity for which classical optics proves a bound. Three bounds matter here: g⁽²⁾(0) ≥ 1, Q ≥ 0, and α ≥ 1. Each is a theorem about any field whose intensity is a non-negative random variable, so a measurement on the wrong side of one is not merely evidence for quantisation — it is a refutation of the alternative.
The photoelectric effect, done with a classical field
Put the atom in a classical field E₀ cos ωt. The perturbation −(e/m)A⋅p̂ connects a bound state to the continuum, and Fermi's golden rule gives a transition rate R = (2π/ħ)|⟨f|H′|i⟩|²ρ(Ef). The photoelectric signatures fall out at once. The golden rule places the electron at Eᵢ + ħω, and ρ(Ef) vanishes unless that energy lies in the continuum, so there is a sharp threshold at ħω = φ and the fastest electrons leave with Kₘₐₓ = ħω − φ: the energy per electron tracks frequency because the atom's level structure does, not because the light arrives in lumps. The rate goes as |E₀|², hence as intensity, so brighter light gives more electrons rather than faster ones. And R is a rate, constant once the field has been on for a cycle or two, so the first count from the surface is expected within nanoseconds however dim the beam. Lamb and Scully put it bluntly in 1969: every textbook feature of the photoelectric effect is a statement about a quantised detector.
Why the accumulation-time argument was never decisive
The argument that looked decisive runs like this. Let each sodium atom intercept only the light falling on its own geometric cross-section, and let that energy pile up until it reaches the work function. At a microwatt per square centimetre the formula card's estimate comes out at over an hour, whereas Lawrence and Beams in 1928 found the first electrons within 3 × 10⁻⁹ s of the light arriving. The conclusion drawn was "therefore photons". But look at what the estimate assumed: that energy accretes continuously into an atom-sized target until a threshold is crossed. The semiclassical calculation never says that. It gives a probability per unit time — small at low intensity, but a small rate still produces its first count almost immediately when you are watching some 10¹⁵ atoms per square centimetre. The null delay result kills classical accretion. It leaves the classical field standing.
Compton's shift, and what Bothe–Geiger added
Compton's 1923 result Δλ = (h/mₑ c)(1 − cos θ), with h/mₑ c = 2.426 pm, is derived by conserving relativistic energy and momentum between a lump carrying ħω and ħk and a free electron. It fits the data beautifully. But ħ enters that derivation through the electron's de Broglie relations as much as the field's, and in 1927 Schrödinger obtained the same shift by letting a classical wave Bragg-scatter off the moving density grating of an electron matter wave. A second experiment did more. Bohr, Kramers and Slater had proposed in 1924 that energy and momentum are conserved only statistically, so scattered photon and recoil electron need not appear together. Bothe and Geiger in 1925, and Compton and Simon the same year, found them in coincidence event by event, and the BKS model died. That is a real advance — but note precisely what it settled: strict conservation in individual events. The semiclassical account never gave up strict conservation, so the coincidence result left it untouched.
The inequality that light can break
Here is the bound in four lines. Let the light be a classical field whose intensity I at the detector is some non-negative random variable. Then g⁽²⁾(0) = ⟨I²⟩/⟨I⟩², and since ⟨I²⟩ − ⟨I⟩² = ⟨(I − ⟨I⟩)²⟩ ≥ 0, it follows that g⁽²⁾(0) ≥ 1. Cauchy–Schwarz applied at two times gives a second bound, g⁽²⁾(0) ≥ g⁽²⁾(τ), so a classical correlation function can never climb out of a dip. Thermal light sits at g⁽²⁾(0) = 2, bunched; a coherent laser sits exactly at 1; nothing classical goes below. Now take a single atom or quantum dot. Having emitted, it is in its ground state and cannot emit again until re-excited, so two clicks at zero delay have probability zero: g⁽²⁾(0) = 0, recovering to 1 over the excited-state lifetime. Kimble, Dagenais and Mandel saw this in resonance fluorescence in 1977; Grangier, Roger and Aspect made it unambiguous in 1986, sending heralded photons onto a beam splitter and finding coincidences at a fifth of the classical minimum.
Sub-Poissonian counting, and what it costs to see
The counting statistics carry the same verdict in a different currency. Count photons in a gate and compare the variance with the Poisson value: the Mandel parameter Q = (⟨Δn²⟩ − ⟨n⟩)/⟨n⟩ is zero for a laser, equal to ⟨n⟩ for single-mode thermal light, and non-negative for any classical field, because a fluctuating positive intensity can only add noise to a Poisson process. Q < 0 — quieter than Poisson — is forbidden classically, and Short and Mandel reported it in atomic resonance fluorescence in 1983. The catch is fragility. A detector of efficiency η thins the count binomially, and the thinning maps Q → ηQ, so a source seen through a 10% detector looks only a tenth as sub-Poissonian; stray Poissonian light dilutes it in the same way, by the emitter's share of the total. Normalised correlations do not suffer this, because loss multiplies numerator and denominator of g⁽²⁾ by the same η² — which is why antibunching is the workhorse and Q the connoisseur's measurement. Hong–Ou–Mandel interference and Bell tests now do the same job in other coinage.
Change one variable at a time
Make the relationship visible.
Raise the emitter share s from 0 and watch the solid curve sink through the classical floor at 1 — that crossing, not the photoelectric threshold, is what no classical field imitates. Then lengthen the lifetime: the dip widens without deepening, because τ₀ sets how long the emitter stays dark and s sets how dark.
g⁽²⁾(0)0.190
DIP FWHM5.55 ns
g⁽²⁾ AT τ = 10 ns0.934
MANDEL Q AT ⟨n⟩ = 0.10-0.081
Live interpretationg⁽²⁾(0): 0.190. DIP FWHM: 5.55 ns. g⁽²⁾ AT τ = 10 ns: 0.934. MANDEL Q AT ⟨n⟩ = 0.10: −0.081
Catch the common trap
Explain before calculating.
A heralded single-photon source feeds a 50:50 beam splitter with a detector in each output. Per trigger, the transmitted detector fires with probability Pₜ = 5.0 × 10⁻³, the reflected one with Pᵣ = 5.0 × 10⁻³, and both together with Pc = 5.0 × 10⁻⁶. The detectors are about 10% efficient. What does the run establish?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA sodium surface of work function φ = 2.28 eV is lit at I = 1.0 μW cm⁻². Treating each atom as intercepting only the light that falls on σ = πa₀² = 8.8 × 10⁻²¹ m², find how long classical energy accretion would need to free one electron, and say what the observed prompt emission actually refutes.
- Work function in joules: φ = 2.28 eV × 1.602 × 10⁻¹⁹ J eV⁻¹ = 3.65 × 10⁻¹⁹ J.
- Intensity in SI: 1.0 μW cm⁻² = 1.0 × 10⁻⁶ W ÷ 1.0 × 10⁻⁴ m² = 1.0 × 10⁻² W m⁻².
- Power intercepted by one atom: P = Iσ = 1.0 × 10⁻² × 8.8 × 10⁻²¹ = 8.8 × 10⁻²³ W.
- Accumulation time: t = φ/P = 3.65 × 10⁻¹⁹ ÷ 8.8 × 10⁻²³ = 4.15 × 10³ s, about 69 minutes. Lawrence and Beams measured the delay to be under 3 × 10⁻⁹ s — twelve orders of magnitude smaller.
- What fails is the bucket-filling picture. The semiclassical calculation returns a transition probability per unit time, not a stored energy, so the first electron leaves promptly however dim the beam; the intensity sets how many follow it per second.
Answert ≈ 4.15 × 10³ s against a measured bound of 3 × 10⁻⁹ s. The null result refutes continuous energy accretion into an atom-sized target — it says nothing against a classical description of the field.
MediumA photon-counting run on a sub-Poissonian source records a mean of ⟨n⟩ = 100 counts per gate with variance ⟨Δn²⟩ = 64. Find Q and the gate-averaged correlation g⁽²⁾T, state which classical bound is broken, then work out what the same source shows through detectors of efficiency η = 0.25.
- Q = (⟨Δn²⟩ − ⟨n⟩)/⟨n⟩ = (64 − 100)/100 = −0.36. Classical light requires Q ≥ 0, so this run sits outside everything a non-negative intensity distribution can produce.
- Convert with Q = ⟨n⟩[g⁽²⁾T − 1]: g⁽²⁾T = 1 + Q/⟨n⟩ = 1 − 0.36/100 = 0.9964 — below 1, but by only 0.36%. A gate that collects 100 counts is far longer than the antibunching dip, so the average sits near 1 even if g⁽²⁾(0) itself is close to 0; that is why Q, not the correlation, is the sensitive statistic when ⟨n⟩ is large.
- Loss scales the mean: ⟨n⟩′ = η⟨n⟩ = 0.25 × 100 = 25.
- Loss adds partition noise to the variance: ⟨Δn²⟩′ = η²⟨Δn²⟩ + η(1 − η)⟨n⟩ = 0.0625 × 64 + 0.1875 × 100 = 4.00 + 18.75 = 22.75.
- So Q′ = (22.75 − 25)/25 = −0.090, exactly ηQ — a quarter of what the source carries. But g⁽²⁾T′ = 1 + (−0.090)/25 = 0.9964, the same number as before: loss rescales the numerator and denominator of a normalised correlation alike.
AnswerQ = −0.36 and g⁽²⁾T = 0.9964; the bound broken is Q ≥ 0. Through 25% detection Q reads −0.090 while g⁽²⁾T is unchanged — efficiency dilutes the counting statistic, never a normalised correlation.
HardIn 3.0 × 10⁷ trigger gates a heralded source gives Nₜ = 1.20 × 10⁵ transmitted counts, Nᵣ = 1.20 × 10⁵ reflected counts and Nc = 87 coincidences. Find the accidental coincidences expected from independent channels, the anticorrelation parameter α, and how far the result sits below the classical bound.
- Single-channel probabilities per gate: Pₜ = 1.20 × 10⁵ ÷ 3.0 × 10⁷ = 4.00 × 10⁻³, and Pᵣ is the same.
- If the detectors fired independently the expected coincidences would be N Pₜ Pᵣ = 3.0 × 10⁷ × (4.00 × 10⁻³)² = 3.0 × 10⁷ × 1.60 × 10⁻⁵ = 480.
- Only 87 were seen, so α = Pc/(Pₜ Pᵣ) = 87/480 = 0.181. A classical field obeys α ≥ 1, and a coherent state sits exactly at 1.
- Uncertainty is dominated by Poisson counting on the coincidences: √87 = 9.3, a relative 10.7%. Each singles channel contributes √(1.2 × 10⁵) ÷ (1.2 × 10⁵) = 0.29%, negligible in quadrature. So uα = 0.107 × 0.181 = 0.0194.
- Distance below the bound: (1 − 0.181)/0.0194 = 42 standard deviations. One detector fires at a time; the photon does not split at the beam splitter.
- State the limit of the claim. This excludes every field with a non-negative intensity distribution, but it does not make the photon a classical particle: send the same heralded photons into a Mach–Zehnder interferometer and full-visibility fringes return.
Answer480 accidentals expected against 87 observed, so α = 0.181 ± 0.019, about 42 standard deviations below the classical floor of 1. No classical field reproduces it — yet the same photons still interfere.