Wave-Particle Duality of Electromagnetic Radiation

The question this section answers

In its interaction with matter, light behaves as if it were made up of packets of energy called quanta. Later work confirmed that these quanta also carry momentum. That raised the question of whether a particle can properly be associated with electromagnetic radiation, and the answer came from Compton's scattering experiments.

Compton scattering

• In 1923 A. H. Compton made a monochromatic beam of X-rays of wavelength $\lambda$ incident on a graphite sheet and measured the intensity of the scattered rays in different directions as a function of wavelength.
• Although the incident beam had a single wavelength $\lambda$, the scattered intensity was maximum at two wavelengths: one equal to $\lambda$, and another $\lambda'$ larger by an amount $\Delta\lambda$. This $\Delta\lambda$ is the Compton shift, and it depends on the scattering angle.
• Compton explained this by treating the incidence of X-rays on graphite as a collision of X-ray photons with the electrons of graphite, like a collision of billiard balls. Energy and momentum are transferred in the collision, so the scattered photons have lower energy, hence lower frequency and longer wavelength.
• The Compton shift is

$$\Delta\lambda = \dfrac{h}{m_e c}\,(1 - \cos\theta)$$

where $\theta$ is the scattering angle.

• The shift depends only on the scattering angle and not on the incident wavelength. It cannot be explained by wave theory; and letting $h \to 0$ recovers the result the wave theory would give, which is the standard test of whether the new picture is right.
• Compton scattering was the third experiment establishing the particle nature of radiation, and the one that showed a photon possesses momentum. The other two are black body radiation (mostly infrared near room temperature) and the photoelectric effect (visible and ultraviolet).

Properties of a photon

• A photon has an associated momentum along with the energy it carries.
• All photons of electromagnetic radiation of a particular frequency have the same energy and the same momentum.
• Photons are electrically neutral and are not deflected by electric or magnetic fields.
• Photons can have particle-like collisions with other particles such as electrons.
• In a photon-particle collision, the energy and momentum of the system are conserved, but the number of photons is not conserved: photons can be absorbed, and new photons can be created.
• Photons can transfer their energy and momentum during collisions and disappear. They are created when we turn on a light.
• A photon always moves with the speed of light and is never at rest.
• The mass of a photon is not defined as it is for a Newtonian particle. Its rest mass is zero in all frames of reference.

Momentum of a photon

$$p = \dfrac{E}{c}$$

valid for a massless particle travelling with the speed of light, by Einstein's special theory of relativity. Using $E = h\nu$ and $\lambda\nu = c$,

$$p = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}$$

• This relation, $E = pc$, is valid for a photon but not for an electron or a proton, which have non-zero rest mass. That is the standard multiple-choice trap.

Where each picture works

• Wave effects — interference and diffraction — are seen when the slit width or the separation between two slits is smaller than or comparable to the wavelength of light. If the slits are wide or far apart, the patterns change and the wave nature is no longer obvious.
• Particle effects are seen in black body radiation, the photoelectric effect and Compton scattering.
• Some phenomena can be explained by either theory.

The conclusion

Some observations can be explained only by treating light as a wave, others only by treating electromagnetic radiation as a stream of photons with a definite quantum of energy and momentum. It is therefore essential to accept that both characters hold good: one dominates in some situations, the other in the rest. Both physical models must be kept to explain careful experimental observations. This dual character, which turned out to be needed not just for light but for the whole electromagnetic spectrum, is called the wave-particle duality of electromagnetic radiation.

Standard results collected

$$E = h\nu = \dfrac{hc}{\lambda} \qquad p = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda} \qquad \Delta\lambda = \dfrac{h}{m_e c}(1 - \cos\theta)$$

• Rest mass of a photon is zero; $E = pc$ holds for a photon only.
• For sources of equal power, the number of photons emitted per second is $N = P\lambda/hc$, so red light gives more photons per second than blue light of the same power.