Electromagnetic waves are generated by accelerating electric charges.

Photons on the other hand, tend to describe something different, specifically the particle nature of electromagnetic waves as detected experimentally.

Is it something like EM waves are like the ripples in water, but photons are like the individual molecules of water in a wavefront? Or is it like the amplitude of the wavefront?

According to this answer, phonons can be thought of as molecules of the medium, but photons don't move in a medium. If we were talking about current, we would have electrons acting as a waveguide, but EM waves are different.

(Is it too far off to think of the electromagnetic field as a medium? e.g. like a 2d matrix full of 0s, and if I place a +5 value in a corner, it spreads out to the three adjacent cells in the next step, and so on.)

Also, if there is a difference between the two, why are they always mentioned as synonymous? Is it a common mistake?

  • 1
    $\begingroup$ waves are coherent states of photons. $\endgroup$
    – Prahar
    Commented Feb 2 at 18:41
  • 1
    $\begingroup$ The reason electromagnetic waves and photons are often mentioned synonymously is because they are two complementary aspects of the same phenomenon, known as wave-particle duality $\endgroup$
    – user391340
    Commented Feb 2 at 18:43
  • $\begingroup$ @Ragnrok but I think there's something more to photons than just observed field fluctuations, according to some other answers, which is what prompted this question. $\endgroup$ Commented Feb 2 at 18:46
  • $\begingroup$ @Prahar coherent states would mean a bunch of photons travelling together? Say, a crest would be +5 and a trough would be -5, and both sides of the wave are made of a bunch of photons? $\endgroup$ Commented Feb 2 at 18:52
  • $\begingroup$ my answer here shows the observation of how photons build up the electromagnetic wave physics.stackexchange.com/questions/113574/… $\endgroup$
    – anna v
    Commented Feb 2 at 20:53

3 Answers 3


There are three main concepts here, not two:

  1. the photon,
  2. the quantum field mode (somewhat like a quantum wavefunction)
  3. the classical electromagnetic wave.

The relationship is as follows.

A quantum field mode is a continuous function of position and time, quite like a wave. It is also quite like a quantum wavefunction though not exactly the same. We are usually interested in quantum field modes having a well-defined frequency.

A photon is a way of talking about how much energy is in some quantum field mode. The more energy, the more photons.

A classical wave is an approximation to what happens when you have a large number of photons in a group of quantum field modes all of similar frequency and spatial shape.

All this can be made precise via the mathematics. Then main thing to note is that the electromagnetic wave, which can be thought of as oscillating electric and magnetic field, corresponds to a stream of a large number of photons, and the quantum field modes that are involved have a shape in space and time similar to the shape of the resulting electromagnetic wave.

  • $\begingroup$ I was thinking of how the photons would be distributed. If I take a given amount of energy and try to create waves out of it, if I consider all the waves generated for the time period (or for a given time period if I generate it at constant power), I'll either have to create high frequency waves of lower amplitude, or low frequency waves of higher amplitude. But we model photons of the high frequency wave as those with higher energy (hF), so there being a lower number of them would mean the same energy (nhF), vs more low energy photons in the other set of waves (Nhf). $\endgroup$ Commented Feb 2 at 19:14
  • $\begingroup$ But for a given time period, the high frequency wave will have more cycles (because frequency is just cycles per time period) and the low frequency wave will have fewer cycles. So does this mean the high frequency wave has less number of photons even though it has more cycles for the duration? Because otherwise the energy won't be conserved. I'm thinking of a cycle as being comprised of several photons, as that is what I inferred from your answer. $\endgroup$ Commented Feb 2 at 19:14
  • 1
    $\begingroup$ For a given energy, yes, a higher frequency wave has fewer photons. The number of cycles in a given time period is NOT related in any simple general way to the number of photons. Rather, the amount of energy passing through a given plane is proportional to the number of photons passing through that plane. $\endgroup$ Commented Feb 2 at 19:17
  • 1
    $\begingroup$ In terms of electric and magnetic fields, it is the square of the electric field amplitude (and magnetic field amplitude) multiplied by the volume which gives the energy (up to a constant factor) and this corresponds to the number of photons in that volume. $\endgroup$ Commented Feb 2 at 19:18
  • $\begingroup$ The part that confuses me is how the energy is not only proportional to the number of photons, but also to the frequency (E = nhf). What does frequency has to do with the number of photons? Since a photon is the energy in a quantum field mode, and each cycle is a stream of photons, shouldn't each photon in such a cycle have the same energy? As I've mentioned, higher frequency generates more cycles at a lower amplitude. But then that would mean the volume is lower (due to lower wavelength). So as per your comment, the total number of photons would be a^2v, compared to A^2V. Am I correct? $\endgroup$ Commented Feb 2 at 19:36

Photons are quantized excitations in spatial modes of the electromagnetic field.

  • $\begingroup$ I was thinking that when a charge moves back and forth, the resulting moving change in the EM field is what we call a photon. But anna v (a prominent user here, who's an experimental physicist) said that I'm confusing photons with electromagnetic light, which prompted me to make this question. She prompted me to this answer by her in a comment to my now deleted answer. By her comment, it seems like photons are observable particles, not the fluctuations in the electromagnetic field. $\endgroup$ Commented Feb 2 at 19:19
  • 1
    $\begingroup$ @BlacklightMG All photons are created by an atom/electron and all photons we observe are/will absorbed by a photon/electron. If we think of water waves in a lake, the created waves will continue forever, where waves overlap/superimpose/cancel the energy is stored in the elasticity of the water ... the energy is only released (like photons observed) when the wave crashes on the shore. The other aspect of the particle nature is because we can isolate single photons/clicks on a detector if we reduce intensity far enough. $\endgroup$ Commented Feb 2 at 22:52
  • 1
    $\begingroup$ @BlacklightMG A photon is released when the electron falls/accelerates from one orbital to another .... the photon is like 1 cycle .... if the electron gets excited again then it can repeat and release another different photon. In a radio antenna we can apply ac voltage and get all the electrons to accelerate in unison but each cycle separates the photons. $\endgroup$ Commented Feb 3 at 0:48
  • 1
    $\begingroup$ @BlacklightMG the antenna is an interesting situation ...in DC current we have drift velocity .... I am guessing that the AC signal has a decent amount of the electrons accelerating enough to generate photons. The electrons in antenna are not free when under the effect of the voltage/atoms. There are many better answers on this website you can search about the truly free electron case. $\endgroup$ Commented Feb 3 at 3:40
  • 1
    $\begingroup$ @PhysicsDave Do you have a citation for that 99.9999% number? Charges accelerate in plasmas, The Aurora Borealis, Compton scattering, Antennas, Plasmons Brehmstrallung etc.. In certain fields of physics we focus on atoms but we shouldn't forget there are other sources of EM radiation. $\endgroup$
    – Jagerber48
    Commented Feb 3 at 3:54

I recommend reading about the Glauber State (https://en.wikipedia.org/wiki/Coherent_state). It's a quantum state version of a coherent, monochromatic, EM plane wave--$Ae^{i(\vec k\cdot \vec x - \omega t)}$

The key takeaways are that it is not an eigenstate of the photon number operator, rather it's an eigenstate of the annihilation operator, the variance of the number of photons is equal to the mean number of photons, and as the mean number of photons becomes large, it approaches a classical plane wave.

Also: it is a sum of Fock states (eigenstates of the number operator), and is not the squiggly photon we see in Feynman diagrams.


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.