University Physics V · Foundations of Quantum Mechanics · 2.4
Compton Scattering & Photon Momentum
Collide a photon with an electron the way you would collide billiard balls. Hand the photon the null four-momentum ħk, conserve all four components at once, and a single contraction predicts a wavelength shift the Thomson picture cannot produce — then be honest about which quantisation the measurement actually forces.
Build the model
Connect the measurement to the mechanism.
Treat the X-ray beam not as a wave shaking an electron but as quanta, each carrying the null four-momentum q = ħk, and scattering becomes relativistic two-body kinematics. Conservation reads q + p = q′ + p′; nobody measures the recoil electron's final state, so isolate it, square, and the mass shell p′⋅p′ = mₑ²c² wipes it out, leaving p⋅(q − q′) = q⋅q′. In the electron's rest frame that is one line from λ′ − λ = (h/mₑc)(1 − cos θ): an absolute shift, at most 4.85 pm, independent of the incident wavelength and of the intensity — predictions the classical Thomson picture, where the electron reradiates at exactly the driving frequency, gets wrong.
What the model costs is precision about its own evidence. The shift measures the single ratio h/mₑc, and a semiclassical Doppler treatment of a classically recoiling electron reproduces the same formula, so the shifted line alone does not force light quanta; only the Bothe–Geiger coincidences, which caught each scattered photon paired with its own recoil electron, closed that gap. And the derivation buys its simplicity by assuming a free electron at rest — an assumption every real target violates, which is exactly what the unshifted line and the Doppler-broadened Compton profile record.
- Simple definition
- Compton scattering is the elastic two-body collision of a single photon of four-momentum ħk with a quasi-free electron, whose four-momentum balance lengthens the photon's wavelength by (h/mₑc)(1 − cos θ).
- Example
- A 71.1 pm Mo Kα X-ray scattered through 90° off graphite emerges at 73.5 pm — shifted by exactly one Compton wavelength, 2.43 pm, whatever the beam's intensity or exposure time.
The quantum input: one quantum carries momentum as well as energy, so scattering off an electron becomes two-body collision kinematics.
k = (ω/c, k) is the four-wavevector, so ħk is in kg m s⁻¹; the zero norm is E = pc, the mass shell of a massless quantum
One squaring deletes the recoil electron's unmeasured final state — the entire derivation is this line plus a choice of frame.
p the electron's initial four-momentum, q, q′ the photon's; from squaring p′ = p + q − q′ with p′⋅p′ = mₑ²c²
An absolute shift: 0 forward, 2λC = 4.85 pm at backscatter — percent-level on an X-ray, a part in 10⁵ on visible light.
λC = h/mₑc = 2.426 pm, the electron's Compton wavelength; θ is the photon's scattering angle
What a scintillator records: backscatter peak at E′(180°), Compton edge at K(180°), and a ceiling E′(180°) < mₑc²/2 = 256 keV.
mₑc² = 511 keV sets ε = E/mₑc²; the electron takes K = E − E′, largest at θ = 180°
Kinematics fixes where the shifted line sits; the QED cross-section fixes how bright it is, and Thomson is its low-energy limit.
rₑ = e²/(4πε₀mₑc²) = 2.818 fm, the classical electron radius; total Thomson σT = 0.665 b
Give the quantum a momentum, not just an energy
Planck's E = hν prices energy exchange; the sharper claim, Einstein's of 1916, is that a quantum of a mode with wavevector k also carries directional momentum p = ħk, of magnitude h/λ. Package both as one null four-vector q = (ħω/c, ħk): its self-contraction q⋅q = (ħω/c)² − ħ²|k|² vanishes because ω = c|k|, which is the statement E = pc for a massless particle. In the field language of later units this is enforced by the formalism itself — the mode's creation operator a†ₖ adds exactly ħk to the field's momentum. The classical alternative is definite: a free electron driven by a wave oscillates at the driving frequency and reradiates at that same frequency in every direction, so Thomson scattering predicts no shift at all. The momenta at stake are small but not hopeless: a 71.1 pm X-ray photon carries p = h/λ = 9.32 × 10⁻²⁴ kg m s⁻¹, enough to kick a free electron to a few percent of c.
Contract once and the unmeasured electron drops out
Conservation is q + p = q′ + p′, four equations. A spectrometer sees only the scattered photon, so eliminate the recoil electron wholesale: write p′ = p + q − q′ and contract each side with itself. The left side is the mass shell, p′⋅p′ = mₑ²c²; on the right, q⋅q = q′⋅q′ = 0 and p⋅p = mₑ²c² cancel it, leaving one scalar equation, p⋅(q − q′) = q⋅q′. Now choose the frame: in the electron's initial rest frame p = (mₑc, 0), so the left side is mₑħ(ω − ω′), while q⋅q′ = (ħ²ωω′/c²)(1 − cos θ) in the metric (+,−,−,−). Divide through and convert frequencies to wavelengths: λ′ − λ = (h/mₑc)(1 − cos θ). The move to remember is the squaring: contracting the conservation law is how collision kinematics deletes whatever the detector does not measure, and it is the same trick that later defines the Mandelstam variables.
Read the law in energy, the way a detector does
A Bragg spectrometer reads wavelengths; a scintillator reads energies. Solve the shift for the scattered energy: E′ = E/[1 + ε(1 − cos θ)] with ε = E/mₑc² and mₑc² = 511 keV, and the electron takes K = E − E′. For the 662 keV line of caesium-137, ε = 1.295. At θ = 180° the denominator is 1 + 2ε = 3.591, so E′ = 184 keV — the backscatter peak, from photons that turned around in the shielding and re-entered the crystal — while K = 478 keV is the Compton edge, the endpoint of the recoil-electron continuum inside the detector. The formula also hides a ceiling: as E → ∞, E′(180°) → mₑc²/2 = 255.5 keV, so no backscattered photon, whatever its source, reaches 256 keV — which is why the backscatter peaks of ¹³⁷Cs, ⁶⁰Co and ²²Na all crowd into the 170–215 keV region however far apart their photopeaks sit.
Kinematics fixes the line; field theory fixes its brightness
Nothing in the four-momentum algebra says how many photons scatter through a given angle: conservation laws locate the line, they do not populate it. The rate needs a dynamical theory of the electron–photon coupling — the Klein–Nishina cross-section of 1929, one of the first results of quantum electrodynamics. Its low-energy limit is exactly Thomson's classical formula, dσ/dΩ = (rₑ²/2)(1 + cos²θ) with rₑ = 2.818 fm and total σT = 0.665 b, so the classical calculation survives as the ε → 0 limit of the quantum one, as correspondence demands. As ε grows the total cross-section falls — to about 0.26 b at 662 keV — and the angular distribution tilts forward; Compton scattering nevertheless stays the dominant photon interaction from a few hundred keV up to where pair production overtakes it, near 5 MeV in lead and only past 20 MeV in carbon. Keep the division of labour explicit: the shift tests kinematics with ħ in it; the intensity pattern tests the dynamics.
What the shift establishes — and what it merely permits
Be precise about the epistemics, because this experiment is often over-claimed. The measured shift is the single length h/mₑc = 2.426 pm: it fixes that ratio, not h itself, and not the existence of photons. A semiclassical calculation — radiation treated as a classical wave, Doppler-shifted once into and once out of the frame of a continuously recoiling electron — reproduces the same Δλ, and the Bohr–Kramers–Slater proposal of 1924 went further, keeping energy and momentum conserved only on statistical average. What killed those readings was not the shift but the correlations. Bothe and Geiger put a recoil-electron counter and a scattered-photon counter in time coincidence and found the pairs arriving together far above chance; Compton and Simon checked in a cloud chamber that each electron's angle matched its photon's through cot φ = (1 + ε) tan(θ/2), event by event. Momentum is conserved in each individual scattering, not merely on average — that is the quantisation this experiment establishes.
Bound electrons break the free-electron premise
Every step so far assumed a free electron at rest, and no target supplies one. Two corrections write themselves into the spectrum. First, quasi-free electrons are not at rest: their initial momenta Doppler-spread the shifted line, and the measured lineshape — the Compton profile J(pz) — is a one-dimensional projection of the momentum density |φ(p)|², which is why Compton spectroscopy is a working probe of momentum-space wavefunctions in solids. Second, an electron bound more tightly than the energy on offer cannot recoil alone: the whole atom takes the momentum, so mₑ in the shift is replaced by the atomic mass — about 21,900 mₑ for carbon — and that line moves by a negligible 10⁻⁴ pm. That is the unmodified line in Compton's own graphite data. The 17.4 keV Mo Kα photon hands roughly 0.6 keV to an electron at 90°: far above the ~11 eV binding of carbon's valence electrons, comparable to its 284 eV K shell. One target, two populations, two lines.
Change one variable at a time
Make the relationship visible.
Drive E from 100 to 1200 keV and watch the right-hand end of the solid curve flatten toward the fixed dashed line — a backscattered photon never clears mₑc²/2 = 256 keV. Then leave E at 662 keV and push θ to 180°: K reads 478 keV, the Compton edge of every caesium-137 spectrum.
SHIFT Δλ2.43 pm
SCATTERED E′288 keV
ELECTRON K374 keV
RECOIL ANGLE φ23.5 deg
Live interpretationSHIFT Δλ: 2.43 pm. SCATTERED E′: 288 keV. ELECTRON K: 374 keV. RECOIL ANGLE φ: 23.5 deg
Catch the common trap
Explain before calculating.
Mo Kα X-rays of λ = 71.1 pm scatter off graphite, and the spectrum at θ = 90° is recorded. What does the model of this lesson predict for it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCompton's own geometry: Mo Kα X-rays of wavelength 71.1 pm scatter off graphite. Find the wavelength observed at θ = 90°, the fractional shift, and the fractional shift the same collision would produce on 500 nm green light.
- Compute the electron's Compton wavelength once: λC = h/mₑc = 6.626 × 10⁻³⁴ / (9.109 × 10⁻³¹ × 2.998 × 10⁸) = 2.43 × 10⁻¹² m = 2.43 pm.
- At θ = 90°, 1 − cos θ = 1, so Δλ = λC = 2.43 pm and λ′ = 71.1 + 2.43 = 73.5 pm.
- Fractional shift on the X-ray: 2.43/71.1 = 3.4% — comfortably resolved by a Bragg spectrometer.
- On green light the absolute shift is identical, but 2.43 pm/500 nm = 4.9 × 10⁻⁶ — invisible inside any spectral line width, which is why the experiment had to be done with X-rays.
Answerλ′ = 73.5 pm. The fractional shift is 3.4% at 71.1 pm but only 4.9 × 10⁻⁶ at 500 nm — same Δλ, different visibility.
MediumA caesium-137 source sits beside a NaI scintillation detector, emitting 662 keV gamma rays. Predict the energies of the backscatter peak and the Compton edge — the two Compton features every ¹³⁷Cs spectrum shows — and check they are consistent.
- Work in units of the electron rest energy: ε = E/mₑc² = 662/511 = 1.295.
- The scattered energy is E′ = E/[1 + ε(1 − cos θ)]; both features live at θ = 180°, where 1 − cos θ = 2 and the denominator is 1 + 2ε = 3.591.
- Backscatter peak: E′(180°) = 662/3.591 = 184 keV — photons that turned around in the shielding and then photo-absorbed in the crystal.
- Compton edge: Kₘₐₓ = E − E′(180°) = 662 − 184 = 478 keV — the largest energy a single scatter can leave with an electron inside the crystal.
- Consistency: 184 + 478 = 662 keV, the photopeak; and 184 keV < mₑc²/2 = 256 keV, as every backscatter peak must be.
AnswerBackscatter peak at 184 keV, Compton edge at 478 keV; they sum to the 662 keV photopeak.
HardA 511 keV annihilation photon scatters through θ = 90° off a free electron at rest. Find E′ and the electron's kinetic energy, then verify four-momentum conservation componentwise: get the electron's momentum from its energy and from the two photon momenta, and its recoil angle both ways.
- With E = mₑc², ε = 1 and 1 − cos 90° = 1, so E′ = 511/(1 + 1) = 255.5 keV; energy conservation gives K = 511 − 255.5 = 255.5 keV — the photon hands over exactly half its energy.
- Mass-shell route: pₑc = √(K² + 2K mₑc²) = √(255.5² + 2 × 255.5 × 511) = √(65,280 + 261,121) = √326,401 = 571.3 keV.
- Vector route: the incoming photon carries (511, 0) keV/c, the outgoing (0, 255.5) keV/c, so the electron takes (511, −255.5) keV/c, of magnitude √(511² + 255.5²) = √326,401 = 571.3 keV/c. Identical — the contraction that produced E′ already enforced the mass shell.
- Recoil angle from components: tan φ = 255.5/511 = 0.500, so φ = 26.6° on the opposite side of the beam from the photon.
- Cross-check with the angle relation the Compton–Simon cloud chamber tested: cot φ = (1 + ε) tan(θ/2) = 2 × tan 45° = 2.000, so tan φ = 0.500 and φ = 26.6° again.
AnswerE′ = K = 255.5 keV, pₑc = 571.3 keV, φ = 26.6° — the mass-shell and component routes agree exactly.