1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Jan Bos
    Commented Dec 5, 2017 at 12:37
  • $\begingroup$ 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 $\endgroup$
    – anna v
    Commented Jan 15, 2018 at 20:13
  • $\begingroup$ 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). $\endgroup$
    – user4552
    Commented Jan 15, 2018 at 20:48

1 Answer 1

1
$\begingroup$

From relativity, we know that the energy-momentum relation is given by:

$$ E^2 = (pc)^2 + (mc^2)^2 $$

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 $p$.

$\endgroup$
8
  • $\begingroup$ But when we find energy using the relation E=hf, we take into account the wave property of light. Considering the particle nature, it cannot have a frequency $\endgroup$ Commented Dec 5, 2017 at 5:54
  • 2
    $\begingroup$ A photon carries energy. If you divide the energy by Planck's constant then you get something with units of 1/time, so we call it "frequency." You can refuse to utter that word if you wish, but you'll find yourself in the minority - especially since the "frequency" of a particle is a useful quantum mechanical concept. $\endgroup$
    – J. Murray
    Commented Dec 5, 2017 at 6:04
  • $\begingroup$ @V.KiranBharadwaj Even if you don't want to use the Planck relationship for the frequency to assign the energy this is the relationship between the energy and momentum. Of course, blackbody spectrum and photoelectric effect don't leave you very much wiggle room in assigning the energy so you're going to end up either using the Planck relationship or going off into territory that contradicts reams of experimental evidence. $\endgroup$ Commented Dec 5, 2017 at 7:17
  • 1
    $\begingroup$ You could if you knew the photons energy through any other means. You have to remember that all matter exhibits both particle and wave properties, but the true nature of matter isn't either one of these things exclusively. So the ability of a particle to have momentum or frequency isn't necessarily just from it's wavelike behavior - it's a fundamental property of matter to have these. $\endgroup$ Commented Dec 5, 2017 at 7:57
  • 1
    $\begingroup$ @V.KiranBharadwaj yes wavelength is a convenient term. Actually 500nm would be the distance the photon travels during one full oscillation. $\endgroup$ Commented Dec 5, 2017 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.