# Momentum of a photon when considered as a particle

According to dual nature of light, it is said to have both particle as well as wave nature. When we think of it as a wave, its momentum can be found out from De Broglie's equation i.e λ = h/mv, provided we know its wavelength. But how do we calculate the momentum of a photon when we think of it as a particle?

• The problem is that one should not think of photon as a particle with a certain location and speed. Photons are needed to describe the experimental facts that interactions with light (the e.m. field) require certain minimum chunks of energy proportional to the frequency of the light. – Jan Bos Dec 5 '17 at 12:37
• Please read this answer of mine here physics.stackexchange.com/questions/379866/… . You will see that the photon mathematically has the h*ν in its wavefunction, when considered a quantum particle, and it is the same frequency as the frequency of the classical light that will be built up by the zillions of photons. It is no coincidence because maxwell's equations are used for both quantum as operators, and classical as just differential equations for Eand B – anna v Jan 15 '18 at 20:13
• It doesn't make sense to talk about switching back and forth between thinking of a photon as a wave and thinking of it as a particle. It's both. When we think of it as a wave, its momentum can be found out from De Broglie's equation The de Broglie relation $\lambda=h/p$ is both a wave equation (left-hand side) and a particle equation (right-hand side, p being the momentum per particle). – Ben Crowell Jan 15 '18 at 20:48

$$E^2 = (pc)^2 + (mc^2)^2$$
We can see that when $m\rightarrow0$, $E = pc$. The energy of the photon can be found using $E=hf$ and you can solve for $p$.