University Physics V · Foundations of Quantum Mechanics · 2.3

The photoelectric effect and what it does not prove

Millikan's slope hands you h to within a percent, and then the real lesson starts: every feature of the effect survives a classical field driving a quantised atom. This topic is where you learn to ask what an experiment establishes, what it merely permits, and which measurement — photon statistics — finally closes the gap.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Shine light of frequency ν on a metal and electrons leave with kinetic energy capped at hν − φ, above a threshold φ/h, within nanoseconds, at a rate proportional to intensity — and the stopping potential's slope against ν is h/e for every metal ever tried. Einstein read this as light arriving in quanta, and the reading stuck. But run the calculation the other way: keep the field classical, A(t) = A₀ cos ωt, quantise only the atom, and treat the coupling (e/m) A⋅p̂ in first-order time-dependent perturbation theory.

The golden rule then delivers a transition rate that is constant from t = 0 (promptness), proportional to |A₀|² (intensity), and non-zero only into continuum states at E_f = Eᵢ + ħω — the threshold and the h/e slope — because that ħω is a beat between the drive's e^(−iωt) and the stationary states' phases e^(−iEt/ħ), not a property of the light. So the photoelectric effect establishes quantised energy exchange with the atom and merely permits a quantised field. The cost of that honesty is that photon evidence must come from somewhere else: from intensity correlations, where every classical field obeys g²(0) ≥ 1 and a one-photon state measures g²(0) = 0.

Simple definition: The photoelectric effect is the ejection of electrons from a surface by light above a threshold frequency; its threshold, promptness and h/e stopping-potential slope all follow from a quantised atom in a classical field, so they evidence atomic energy levels, not photons.
Example: For sodium, φ = 2.28 eV: 400 nm light (hν = 3.10 eV) ejects electrons stopped by 0.82 V, while 600 nm light (2.07 eV) ejects none at any intensity — the threshold sits at λ₀ = 1240/2.28 nm = 544 nm.

Einstein's energy ledgere Vₛ = hν − φ

One line predicts the threshold ν₀ = φ/h, a Vₛ blind to intensity, and a slope no metal can alter.

Vₛ the stopping potential in V; φ the metal's work function in eV; hν the energy of one exchange in eV

The slope every metal sharesdVₛ/dν = h/e = 4.136 × 10⁻¹⁵ V s

Two stopping potentials at two known lines hand you h. Millikan reached half a percent in 1916 while calling the hypothesis behind the equation reckless.

equivalently 0.414 V per 10¹⁴ Hz; the metal moves only the intercept −φ/e

Photon-scale bookkeepinghc = 1240 eV nm

Keeps threshold and stopping-potential arithmetic in eV, the unit work functions are tabulated in.

λ in nm gives the exchange energy 1240/λ in eV; 544 nm ↔ 2.28 eV, sodium's threshold

Semiclassical photoemission rateΓ = (2π/ħ) |⟨f|(e/2m) A₀⋅p̂|i⟩|² ρ(E_f), E_f = Eᵢ + ħω

Constant from t = 0, proportional to |A₀|², and gated at ħω ≥ φ: all three photon signatures from an unquantised field.

A₀ the amplitude of the classical vector potential A(t) = A₀ cos ωt, in V s m⁻¹; the ½ is the cosine's absorbing half e^(−iωt); ρ(E_f) the continuum density of states in J⁻¹

The classical floor on coincidencesg²(0) = ⟨I²⟩/⟨I⟩² ≥ 1

Cauchy–Schwarz: any classical wave, split at a mirror, fires both detectors at least as often as chance.

I(t) ≥ 0 the classical intensity in W m⁻²; averages taken over the field's fluctuations

The single-photon zerog²(0) = ⟨n̂(n̂−1)⟩/⟨n̂⟩² = 0 for |1⟩

n̂(n̂−1)|1⟩ = 0 — one photon cannot fire two detectors. A measured g²(0) < 1 is the field's quantisation certificate.

n̂ = â†â the number operator on the detected mode; |1⟩ the one-photon Fock state

Three facts, and the classical field's absurd timetable

From Lenard's 1902 experiments onward the facts hardened. A maximum kinetic energy that grows linearly with frequency and ignores intensity; a threshold frequency below which nothing leaves however bright the lamp; and emission that begins promptly — Lawrence and Beams bounded the delay below about 3 ns in 1928. A classical field feeding energy continuously into a classical electron fails on the timetable spectacularly. At 1.0 μW m⁻² on sodium, an atom of cross-section ~1 × 10⁻¹⁹ m² collects ~1 × 10⁻²⁵ W, and gathering φ = 2.28 eV = 3.65 × 10⁻¹⁹ J takes 3.7 × 10⁶ s — six weeks against the measured nanoseconds. Note what that estimate assumed, though: an electron absorbing continuously, which is to say a classical atom. The failure convicts the pairing; it does not yet say which partner must be quantised.

Einstein's ledger turns the plot into a measurement of h

Postulate energy exchanged in lumps hν and the bookkeeping is one line: the fastest electron carries Kₘₐₓ = hν − φ, so the voltage that just stops it obeys eVₛ = hν − φ. Plot Vₛ against ν and every metal gives a straight line of the same slope h/e = 4.136 × 10⁻¹⁵ V s; the work function only slides the intercept −φ/e. Two mercury lines suffice: caesium stopped by 0.32 V at 546.1 nm and by 1.45 V at 365.0 nm gives slope 1.13 V / 2.72 × 10¹⁴ Hz = 4.15 × 10⁻¹⁵ V s, hence h = 6.65 × 10⁻³⁴ J s, within a third of a percent. That is exactly Millikan's 1916 programme — he ran it hoping to break Einstein's equation and instead confirmed it to half a percent, while writing that the theory behind it seemed untenable. His instinct that the equation outran its interpretation is the actual subject of this topic.

Quantise the atom, keep the field classical

Now run the calculation Lamb and Scully made famous in 1969. Let the atom have a bound state |i⟩ at energy Eᵢ and a continuum of states |f⟩, and couple it to the unquantised wave A(t) = A₀ cos ωt through Ĥ′(t) = (e/m) A(t)⋅p̂ = (e/2m) A₀⋅p̂ (e^(−iωt) + e^(iωt)). First-order time-dependent perturbation theory gives an amplitude c_f(t) whose sinc² factor collapses, after a few optical periods, onto energy conservation E_f = Eᵢ + ħω — only the e^(−iωt) term can reach the continuum; its partner would need a state ħω below |i⟩ — and Fermi's golden rule turns it into the rate Γ = (2π/ħ)|⟨f|(e/2m) A₀⋅p̂|i⟩|² ρ(E_f). Read the three signatures off. The rate is constant from t = 0, so the first electrons appear immediately: promptness without accumulation. It is proportional to |A₀|², so the count rate tracks intensity while each electron's energy does not. And the continuum is only reachable when ħω exceeds the binding energy: threshold, and with it the h/e slope. The ħ entered through the atom's stationary-state phases e^(−iEt/ħ) beating against the drive e^(−iωt) — nowhere was the field cut into lumps.

What the effect establishes, and what it merely permits

Hold the two calculations side by side. Einstein's photon ledger predicts the facts; the semiclassical golden rule predicts the same facts. When two models agree on every measured number, the experiment cannot separate them — it establishes only what both require and merely permits what one of them adds. Both require quantised atomic energy exchange: stationary states, with transitions bookkept in units set by ħ. Only one requires a granular field, so the photoelectric effect is evidence for the quantum atom and silent on the photon. Keep this discipline for the rest of the course: Compton scattering meets the same audit in the next topic (a Doppler model of a recoiling electron reproduces the shift, and only Bothe–Geiger coincidences force quantum-by-quantum transfer), and the duality bounds of UPV-02.07 are theorems about a two-state space, not about clumsy kicks. To convict the field itself you must measure a quantity for which classical and quantum field theories give different answers — not the energy of one electron, but the statistics of many detections.

Antibunching: the number a classical field cannot reach

Split the beam at a 50:50 mirror, put a detector on each output, and count coincidences inside a short window: g²(0) is the coincidence rate divided by what chance alone would give. Model the light classically as a fluctuating intensity I(t) ≥ 0, each detector firing in proportion to I; then g²(0) = ⟨I²⟩/⟨I⟩², and Cauchy–Schwarz forces ⟨I²⟩ ≥ ⟨I⟩². Every classical field has g²(0) ≥ 1: thermal light gives 2, an ideal laser exactly 1 — even a perfectly steady wave splits, so both halves fire together at least at chance. Quantum mechanically the same measurement reads g²(0) = ⟨n̂(n̂−1)⟩/⟨n̂⟩², and the one-photon state gives zero, since n̂(n̂−1)|1⟩ = 0: a photon delivered whole to one output cannot also fire the other. Kimble, Dagenais and Mandel saw g²(0) < 1 in the resonance fluorescence of single sodium atoms in 1977; Grangier, Roger and Aspect measured 0.18 ± 0.06 on a heralded cascade source in 1986, thirteen standard deviations below the classical floor. That — not 1905 — is where field quantisation became an experimental fact.

From counts to g²(0): the estimator and its floors

In the laboratory g²(0) is an estimator built from three numbers: singles rates R₁ and R₂, a coincidence window τ, and the measured coincidence rate R_c. Uncorrelated detectors coincide by accident at R₁R₂τ, so g²(0) ≈ R_c/(R₁R₂τ) — chance sits at 1 by construction. With R₁ = 9000 s⁻¹, R₂ = 8000 s⁻¹ and τ = 10 ns, chance predicts 0.72 coincidences per second; a true single-photon emitter should deliver almost none. The systematic floor comes from whatever mimics classical light: background counts, detector afterpulsing, and a second emitter in the focal volume — two independent single photons give g²(0) = 1/2. The statistical floor is Poisson: with N_c recorded coincidences the fractional uncertainty is 1/√N_c, so quoting g²(0) = 0.10 ± 0.02 needs about 25 of them. And the claim worth making is never 'g² is small' but 'g² sits below 1 by k standard deviations' — the error propagation is two lines of NumPy; the physics is knowing which floor you are standing on.

Change one variable at a time

Make the relationship visible.

Interactive model

Drive frequency ν8.0 ×10¹⁴ Hz

Work function φ2.28 eV

Relative intensity I40 %

Push the intensity slider: the bar grows and the dot refuses to move — a classical wave feeding a classical electron would move both, while a classical wave driving a quantised atom moves only the bar, exactly as drawn. Then raise φ from caesium's 1.95 eV toward zinc's 4.3 eV: the line slides right without tilting, because the slope h/e belongs to nature and only the intercept belongs to the metal.

PHOTON ENERGY hν3.31 eV

STOPPING POTENTIAL Vₛ1.03 V

THRESHOLD ν₀ = φ/h5.51 ×10¹⁴ Hz

EMISSION RATE40 %

Live interpretationPHOTON ENERGY hν: 3.31 eV. STOPPING POTENTIAL Vₛ: 1.03 V. THRESHOLD ν₀ = φ/h: 5.51 ×10¹⁴ Hz. EMISSION RATE: 40 %

Catch the common trap

Explain before calculating.

Misconception check

"The photoelectric effect proves light comes in photons." It does not. Threshold, prompt emission, intensity-proportional rates and the h/e slope are all reproduced by first-order perturbation theory with the classical field A₀ cos ωt driving a quantised atom: the ħω in E_f = Eᵢ + ħω is contributed by the atom's stationary-state phases, so every "photon signature" is really a signature of atomic energy levels. Einstein's inference was reasonable in 1905, when there was no quantum atom to absorb the credit; after matrix mechanics the semiclassical calculation quietly took it. What no classical field can do — weak, noisy or otherwise — is push g²(0) below 1, because ⟨I²⟩ ≥ ⟨I⟩² always; only photon number, with ⟨n̂(n̂−1)⟩ < ⟨n̂⟩², reaches it, and that was first measured in 1977. Ask of every proof which quantisation it establishes and which it merely permits.

Four results are reported from a photoemission and photon-counting laboratory. Which one cannot be reproduced by any model in which the electromagnetic field is a classical wave?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium has work function φ = 2.28 eV. Light of wavelength 400 nm falls on it. Find the maximum photoelectron energy, the stopping potential, and the longest wavelength that ejects electrons at all.

Convert the wavelength with hc = 1240 eV nm: the energy of one exchange is hν = 1240/400 = 3.10 eV.
The ledger eVₛ = hν − φ gives Kₘₐₓ = 3.10 − 2.28 = 0.82 eV, so the stopping potential is Vₛ = 0.82 V. Intensity appears nowhere: a brighter lamp raises the count rate, never this number.
Threshold: the exchange must at least pay φ, so λ₀ = 1240/2.28 = 544 nm. At 600 nm the exchange is only 2.07 eV and nothing leaves at any intensity.
Audit the claim: every number here also follows from the golden rule with a classical field, because E_f = Eᵢ + ħω is enforced by the atom's phases. This arithmetic measures φ and h; it does not detect photons.

AnswerKₘₐₓ = 0.82 eV, Vₛ = 0.82 V, λ₀ = 544 nm — and none of the three is evidence of field quantisation.

MediumA caesium photocathode is illuminated with the mercury lines 546.1 nm and 365.0 nm; the stopping potentials are 0.32 V and 1.45 V. Extract h and the work function, then state which quantisation this measurement establishes and which it merely permits.

Frequencies: ν₁ = c/λ₁ = (2.998 × 10⁸ m s⁻¹)/(546.1 × 10⁻⁹ m) = 5.490 × 10¹⁴ Hz and ν₂ = (2.998 × 10⁸ m s⁻¹)/(365.0 × 10⁻⁹ m) = 8.214 × 10¹⁴ Hz.
Slope of the Vₛ–ν line: ΔVₛ/Δν = (1.45 − 0.32) V / (8.214 − 5.490) × 10¹⁴ Hz = 1.13/2.724 × 10⁻¹⁴ = 4.148 × 10⁻¹⁵ V s.
h = e × slope = 1.602 × 10⁻¹⁹ × 4.148 × 10⁻¹⁵ = 6.65 × 10⁻³⁴ J s, 0.3% above the accepted 6.626 × 10⁻³⁴ J s.
Intercept: φ/e = hν₁/e − Vₛ₁ = 4.148 × 10⁻¹⁵ × 5.490 × 10¹⁴ − 0.32 = 2.277 − 0.32 = 1.96 V, so φ = 1.96 eV — caesium's tabulated 1.95 eV within the rounding of two stopping potentials.
The audit: the slope establishes that energy reaches the atom in units ħω, and the golden rule with a classical field predicts the identical line, because E_f = Eᵢ + ħω is set by the atomic phases rather than by lumps in the light. Field quantisation is permitted, not established; h is measured either way.

Answerh = 6.65 × 10⁻³⁴ J s, φ = 1.96 eV. The experiment establishes quantised atomic energy exchange; it merely permits photons.

HardA candidate single-photon source feeds a 50:50 beamsplitter. Singles rates are R₁ = 9.0 × 10³ s⁻¹ and R₂ = 8.0 × 10³ s⁻¹, the coincidence window is τ = 10 ns, and 36 coincidences are recorded in 500 s. Compute g²(0) with its Poisson uncertainty, test the classical bound, and state what a second identical emitter in the focus would do.

Chance coincidences: uncorrelated detectors coincide at R₁R₂τ = 9.0 × 10³ × 8.0 × 10³ × 1.0 × 10⁻⁸ = 0.72 s⁻¹, so chance predicts 360 coincidences in 500 s.
Measured rate: R_c = 36/500 = 0.072 s⁻¹, so g²(0) = R_c/(R₁R₂τ) = 0.072/0.72 = 0.10.
Poisson statistics on 36 counts: fractional uncertainty 1/√36 = 1/6, so u = 0.100/6 = 0.0167 and g²(0) = 0.100 ± 0.017.
Classical test: every classical intensity obeys g²(0) = ⟨I²⟩/⟨I⟩² ≥ 1. The result sits (1 − 0.100)/0.0167 ≈ 54 standard deviations below the floor — no classical field model survives; the field is quantised.
Two identical independent emitters would give g²(0) = 1 − 1/2 = 0.50, because one photon from each can fire both detectors. Values below 0.5 are the practical certificate that a single emitter dominates.

Answerg²(0) = 0.100 ± 0.017, violating the classical bound g²(0) ≥ 1 by about 54σ. A second emitter would lift it to 0.5.