38-3 Photons, Momentum, Compton Scattering, Light Interference

38-3 Photons, Momentum, Compton Scattering, Light Interference#

Prompts

  • A photon has momentum even though it is massless. Write \(p\) in terms of \(E\), \(f\), and \(\lambda\). (Hint: from relativity, \(E = pc\) for \(m=0\).)

  • Describe Compton scattering: x rays scatter from electrons. Why does the scattered wavelength increase? What is the Compton shift \(\Delta\lambda\) formula?

  • As the Compton scattering angle \(\phi\) increases from 0° to 180°, what happens to the scattered photon’s wavelength, energy, and momentum? What happens to the electron’s kinetic energy?

  • The Compton wavelength of the electron is \(h/(m_e c) \approx 2.43\) pm. For scattering at 90°, what is \(\Delta\lambda\)?

  • In the single-photon double-slit experiment, photons arrive one at a time yet an interference pattern builds up. How is this explained? What is a probability wave?

Lecture Notes#

Overview#

  • A photon has momentum \(p = E/c = hf/c = h/\lambda\) despite having no mass. Photon–matter interactions transfer both energy and momentum, like particle collisions.

  • Compton scattering (1923): x rays scatter from electrons; the scattered wavelength increases by \(\Delta\lambda = (h/m_e c)(1 - \cos\phi)\). This confirmed the photon model and demonstrated momentum transfer.

  • Light as probability wave: The probability of detecting a photon at a point is proportional to the light intensity (square of the field amplitude). Single-photon double-slit experiments show that each photon behaves as a wave that determines where it may be absorbed.

Photon momentum#

Although a photon is massless, it carries linear momentum. From the relativistic relation \(E^2 = (pc)^2 + (mc^2)^2\) with \(m = 0\):

(356)#\[ p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda} \]

When a photon is absorbed or scattered by matter, both energy and momentum are transferred—as in a collision between particles.

Compton scattering#

In 1923, Arthur Compton directed x rays onto a carbon target and measured the wavelength of x rays scattered at various angles. Classical puzzle: An electron driven by an oscillating wave should re-radiate at the same frequency. The scattered x rays should have the same wavelength as the incident beam—but they don’t. The scattered wavelength is longer.

Photon interpretation: Treat the interaction as a “collision” between a photon and an initially stationary electron. The photon transfers energy and momentum to the electron and recoils. Energy conservation: \(hf = hf' + K\), so the scattered photon has less energy and thus longer wavelength.

Compton shift:

(357)#\[ \Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\phi) \]

where \(\phi\) is the angle between the incident and scattered photon directions. The quantity \(h/(m_e c) \approx 2.43\) pm is the Compton wavelength of the electron.

  • \(\phi = 0°\): forward scattering; \(\Delta\lambda = 0\).

  • \(\phi = 90°\): \(\Delta\lambda = h/(m_e c) \approx 2.43\) pm.

  • \(\phi = 180°\): backscattering; \(\Delta\lambda = 2h/(m_e c)\) (maximum).

Unshifted peak: X rays scattered from tightly bound electrons (effectively from the whole atom) have negligible \(\Delta\lambda\) because the atom’s mass is much larger than \(m_e\). These produce the peak at the incident wavelength.

Example

X rays \(\lambda = 22\) pm scatter at \(\phi = 85°\). \(\Delta\lambda = (2.43\,\text{pm})(1 - \cos 85°) \approx 2.43 \times 0.91 \approx 2.2\) pm. So \(\lambda' \approx 24.2\) pm. The fractional energy loss \((E - E')/E = (\lambda' - \lambda)/\lambda'\) increases as \(\lambda\) decreases.

Light as a probability wave#

Standard double-slit: Light produces interference fringes. A small detector clicks when it absorbs a photon; the click rate varies with position, matching the fringe pattern. We cannot predict when a photon will be detected, but the probability of detection at a point is proportional to the light intensity there.

Single-photon version: With an extremely dim source (one photon at a time), interference fringes still build up over time. Each photon is detected at a random point, but the distribution of many detections matches the interference pattern. The photon travels as a probability wave—we interpret the wave amplitude as governing the probability of absorption at each point. We cannot say which slit a given photon “went through”; we only know it was emitted and later absorbed.

Particle vs wave: When light interacts with matter (absorption, emission, scattering), it behaves like particles (photons) transferring energy and momentum at a point. When light propagates, we describe it as a wave that determines detection probabilities. Both descriptions are needed.

Summary#

  • Photon momentum \(p = h/\lambda = E/c\); massless but carries momentum.

  • Compton scattering \(\Delta\lambda = (h/m_e c)(1 - \cos\phi)\); photon loses energy to electron; confirms photon model.

  • Compton wavelength \(h/(m_e c) \approx 2.43\) pm; \(\Delta\lambda\) independent of incident \(\lambda\).

  • Probability wave: detection probability \(\propto\) intensity; single-photon double-slit builds fringes from many random detections.