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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?

Blacklight MG
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    waves are coherent states of photons. – Prahar Feb 02 '24 at 18:41
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    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 –  Feb 02 '24 at 18:43
  • @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. – Blacklight MG Feb 02 '24 at 18:46
  • @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? – Blacklight MG Feb 02 '24 at 18:52
  • my answer here shows the observation of how photons build up the electromagnetic wave https://physics.stackexchange.com/questions/113574/are-photons-electromagnetic-waves-quantum-waves-or-both/113588#113588 – anna v Feb 02 '24 at 20:53
  • @annav I'd understand this if it was the electron double slit experiment, due to it's quantum wierdness.. but why do photons appear as dots? I was expecting them to have a weak full wave by themselves. Am I mistaken if I think this is because, if I consider a a wave like a circle, the region that hits the sensor first has more intensity than the regions that hit it later, because they are comparatively a bit farther from the source? If we place a higher intensity light and a lower intensity light side by side, we may observe the lower intensity light as a dark region.. – Blacklight MG Feb 02 '24 at 21:03
  • @annav This video shows what I'm referring to: https://www.youtube.com/watch?v=p-OCfiglZRQ&t=52s – Blacklight MG Feb 02 '24 at 21:11
  • @BlacklightMG " but why do photons appear as dots?" that is why the mainstream standard model has photons as point quantum particles, their behavior described by a wavefunction that gives the probability of the photons location. This means there is no prediction that can be made for a single photon, an ensemble's location is needed. in the pictures they look random, but finally their locations end up consistent with classical light calculatins. – anna v Feb 03 '24 at 07:39
  • sorry but I cannot follow the video – anna v Feb 03 '24 at 07:42
  • Let us continue this discussion in chat. – Blacklight MG Feb 03 '24 at 10:10
  • What are photons, EM radiation and EM waves – HolgerFiedler Feb 04 '24 at 04:54

3 Answers3

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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.

Andrew Steane
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  • 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). – Blacklight MG Feb 02 '24 at 19:14
  • 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.

     – Blacklight MG Feb 02 '24 at 19:14
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    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. – Andrew Steane Feb 02 '24 at 19:17
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    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. – Andrew Steane Feb 02 '24 at 19:18
  • 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?

     – Blacklight MG Feb 02 '24 at 19:36
  • This makes sense numerically, but I don't get intuitively why the number of photons is square of the amplitude multiplied by volume.

    Also I'm considering all the cycles generated by a given energy OR all the cycles for a given duration generated at a constant power. I understand number of cycles is not related to the number of photons, but given the variations in amplitudes we should be able to calculate that..

     – Blacklight MG Feb 02 '24 at 19:38
  • I got it after thinking for a while. If I'm moving a charge back and forth to get a wave, the frequency is dependent on how fast I move the charges (same as doing it in water) - because the speed of light sets the wavelength. So, if I consider a single charge, E=hf also implies that a single cycle has an energy E=h (although it doesn't seem to be what's considered as a photon). That is, imagine moving a charge 1 unit up and 1 unit down, that will be 1 cycle. How fast we do it will determine the power, but the energy is all just in the motion. – Blacklight MG Feb 02 '24 at 20:46
  • So for frequency f, if we consider f cycles of the same amplitude, and we consider 1 cycle to have an energy E, then f cycles would have E=hf, and such a block would be what we call a photon. But I guess details on that is best asked as a different question. For now, this I'll mark this as the answer. – Blacklight MG Feb 02 '24 at 20:48
  • In other words, the part I missed was that if I moved a charge a a given power, the frequency would be the same. The only way I would be able to trade amplitude for frequency would be if I distributed the same power across a different number of charges. – Blacklight MG Feb 02 '24 at 20:53
  • Be careful that classical intution based on thinking about accelerating point charges definitely will not work at some point when it comes to understanding photons. If we could explain everything in terms of classical point charges and field we would never have to think about photons. – Luke Pritchett Feb 03 '24 at 20:30
  • @BlacklightMG I get the impression you are trying to learn from a very limited starting point and then trying to intuit the rest. It would be better to learn first some more about classical electromagnetism, and then some quantum mechanics, but I don't know if that is what you want to do. If a pulse of light has total energy $E$ and a reasonably well-defined frequency $f$ then the number of photons in that pulse is $E/(h f)$. – Andrew Steane Feb 04 '24 at 12:51
  • @LukePritchett but certainly it generates eelctromagnetic waves, and this answer tells that the classical wave is composed of photons. I was trying to understand how the energy of a photon is dependent on the frequency when frequency is something at the macroscopic scale. – Blacklight MG Feb 04 '24 at 13:09
  • @AndrewSteane I understood n=E/hf. I just want to understand what f has to do with the energy of a single photon, when frequency is at the macroscopic scale. Especially in an example case of AC radio transmission. – Blacklight MG Feb 04 '24 at 13:11
  • @BlacklightMG To understand that, you need to understand the quantum mechanics of simple harmonic motion. If a potential well has the form $V(x) = (1/2) m \omega^2 x^2$ then the energy levels form a ladder with spacing $\hbar \omega = h f$. For electromagnetic fields you can roughly equate the $x$ variable here to the electric field amplitude, but the full picture is a lot more subtle. – Andrew Steane Feb 04 '24 at 17:06
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Photons are quantized excitations in spatial modes of the electromagnetic field.

Jagerber48
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  • 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. – Blacklight MG Feb 02 '24 at 19:19
  • When charges accelerate they produce electromagnetic radiation. This true in the classical and quantum theories of electrodynamics. In quantum field theory the amplitude of the radiation is quantized. Each additional quantum of excitation in a certain mode is what people might call photons. I'm not sure what quantum "particles" are or why they have a place in quantum "field" theory. – Jagerber48 Feb 02 '24 at 19:45
  • If we are considering the amplitude of a wavepacket, how does the energy relate to frequency (E=hf)? If one wavepacket only has a set amplitude, and the total amplitude contains several photons (E=nhf), then a single wavepacket should have some set amplitude which is also connected to its frequency.. This is the confusing part for me. – Blacklight MG Feb 02 '24 at 20:31
  • @BlacklightMG there is a lot of history and many scientists use what is convenient for them. The "standard model" says everything is either a field or an excitation of the field (particle) ... so they apply this to the massless photon, but it's not a good fit. The photon acts in a local fashion, i.e. it is absorbed by a single molecule/atom, like a pixel in a camera a photocell in your eye ... hence the particle nature. Bear in mind that a field (like the EM field) is a perfect medium and we can never see what is going on inside it, no energy or cancellation/loss ever occurs in the medium. – PhysicsDave Feb 02 '24 at 22:46
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    @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. – PhysicsDave Feb 02 '24 at 22:52
  • We can observe the lake directly but not the EM field directly. Water is elastic .... so is the EM field but there is no way to study this phenomenon .... so nobody or scientist has ever investigated it. – PhysicsDave Feb 02 '24 at 22:53
  • @BlacklightMG “photons” are energetic excitation of the EM field mode. Each quanta has $\hbar \omega$ of energy. Therefore if you have a field with a larger amplitude it will have more quanta. The field amplitude is related to both $\hbar \omega$ (the energy per quanta) and the total number of quanta in the mode. – Jagerber48 Feb 03 '24 at 00:09
  • @Jagerber48 No, I was asking how a single photon will have an energy of hf. Forget multiple photons. I guess a single photon will always have the same amplitude? Your statement says that is not the case, as the amplitude is dependent on the frequency. I think this is better suited for a separate question, but why would amplitude be dependent on frequency? Is it because if electrons vibrate at a constant power, the distance travelled is greater for higher amplitudes, and hence the frequency will be lower and vice versa? – Blacklight MG Feb 03 '24 at 00:23
  • @Jagerber48 Also, if we take a wave of frequency f, is a photon not complete without a set number of cycles? As in, if the energy of a single cycle is constant, which happens the electron vibrates at a constant amplitude, wouldn't E=hf imply that a photon is comprised of f cycles (or at least a proportion of it)? Does an electron vibrate at constant amplitude or can it vary? – Blacklight MG Feb 03 '24 at 00:27
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    @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. – PhysicsDave Feb 03 '24 at 00:48
  • @PhysicsDave In radio, the electrons seem to be moving a distance of "drift velocity"x"time period" of the AC supply. This is different from changing orbitals. Also, how do free electrons generate electrons? Are they restricted on how far they can oscillate like the electrons bound to orbitals in atoms? – Blacklight MG Feb 03 '24 at 01:10
  • @BlacklightMG Once you fix (1) the frequency of the EM wave, (2) the number of photons, and (3) the spatial mode under consideration, then the amplitude of the EM wave if fixed. Note the field energy is linear in the number of photons but the field amplitude is related to the square root of the number of photons (so a two-photon field doesn't have twice the amplitude of a one-photon field). – Jagerber48 Feb 03 '24 at 01:48
  • @BlacklightMG photons are not based on "set numbers of cycles". Photons are quantum excitations of spatial modes the quantum EM field. The spatial mode is fixed in time and it is delocalized over space. This is why thinking about photons as particles is just weird. I recommend against it. Particles are localized, photons are delocalized. Trying to think about fields as being localized leads many a physics student to have the misconception that EM radiation consist of "trains" of photons which are composed of a few cycle of the sinusoidal field. This is just wrong. – Jagerber48 Feb 03 '24 at 01:55
  • photons are delocalized excitations of the delocalized spatial modes. – Jagerber48 Feb 03 '24 at 01:55
  • @PhysicsDave any charges can emit EM radiation if they accelerate. Any charges can also absorb EM radiation. EM radiation generation/absorption is not restricted to atoms. – Jagerber48 Feb 03 '24 at 01:56
  • Yes but 99.9999...etc are photons from excited atoms/molecules where the electron drops down to a lower orbital. Yes radio waves are possible and nuclear reactions too. – PhysicsDave Feb 03 '24 at 03:29
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    @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. – PhysicsDave Feb 03 '24 at 03:40
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    @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. – Jagerber48 Feb 03 '24 at 03:54
  • @Jagerber48 no source ... that's why it's a comment .... here on Earth most of the light comes from the sun ..... maybe we should stretch that number to 99.999999999999999999 etc etc – PhysicsDave Feb 03 '24 at 11:53
  • Let us continue this discussion in chat. – Jagerber48 Feb 03 '24 at 13:17
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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.

JEB
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