Question

Does light have momentum?

Answer 1

We know that light is quantized in the form of photons. They are massless elementary particles that are essentially discrete packets of energy. The energy that they carry is nothing but quantized electromagnetic radiation. In such a case, determine the energy of a photon through a non-classical approach that does not consider the massiveness of the particle and ultimately arrive at an expression for the photon momentum. To this end, tentatively conclude any causes of photon momentum.
Formula Used:
Momentum $p = \dfrac{h\nu}{c}$

Complete answer:
We know that light is composed of massless particles called photons. Since they are not massive particles, it is easy to assume that they will not possess any momentum since momentum is usually defined as the product of mass times velocity. But this is not true since we cannot use a classical approach to a quantum notion since photons are essentially quantized energy packets.
Relativistically, the energy E of a free moving particle consists of two parts: rest-mass energy due to its mass $m_0$ when it is stationary, and kinetic energy due to its momentum $\;p$ while in motion. It is given as:
$E^2 = (m_0c^2)^2 + (pc)^2$, where c is the speed of light.
We know that the rest mass of a photon is zero, i.e., $m_0 = 0$, therefore the above equation becomes:
$E^2 = (pc)^2 \Rightarrow E=pc$
But we know from quantization of photons that each photon has an energy associated with its frequency $\nu$ given as:
$E = h\nu$, where h is the Planck’s constant indicating energy quantization.
Equating the two energy equations we get:
$pc = h\nu \Rightarrow p = \dfrac{h\nu}{c}$
But we know that wavelength $\lambda = \dfrac{c}{\nu}$. Plugging this in to the above equation we get:
$p = \dfrac{h}{\lambda}$
Thus, we see that a photon gets its momentum (inversely) proportional to the (wavelength) frequency of light that is quantized, irrespective of being a massless particle. We can therefore conclude that light indeed has momentum associated with it and is given as:
$p =\dfrac{h\nu}{c} = \dfrac{h}{\lambda}$

Note:
The above deduction exploited the dual nature of light, where we treated it as both a wave of specific frequency or wavelength and as a particle possessing momentum. It may be of importance to know that the disparity between the wave and particle nature of light was bridged by Young’s Double Slit Experiment, which involved the interference of two coherent light “waves” resulting in the formation of a distinct interference pattern producing bright and dark bands on a screen. This band pattern would not be expected if light consisted of classical entities or waves, since the light seemed to be differentially absorbed, hitting only specific regions of the screen at discrete points as individual “particles”, thus exhibiting a wave-particle duality.