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One maybe interesting question please!

In quantum point of view, the electromagnetic waves (EMWs) consist of photons. However, if there are only very very few photons, can they form a wave-like macro EM field?

OR

If a spherical monochromatic EMW (frequency is $\nu$) propagates and decays into very low level of energy flux density, e.g., for every square meter, the energy flux is far less than 1*$h\nu$ per second, then, does the EM fields still exist there?

OR

If the EMW is extremely weak (by value of corresponding energy flux density), can the electric field and magnetic field still exist in the spacetime and still propagate in the shape of waves? Or, in this case, is the form of wave only meaning the quantum wave function to indicate the probability of where the photons appear?

Thank you very much!

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  • $\begingroup$ You can have a wave of marching soldiers but no wave if there is only a few. $\endgroup$ – Bill Alsept Dec 18 '18 at 2:37
  • $\begingroup$ But a single packet still shows wave properties, ex diffraction. $\endgroup$ – PhysicsDave Dec 18 '18 at 3:27
  • $\begingroup$ You may be able to answer your own question by researching the single photon double slit experiement. It's a pretty spectacular fringe case that's right along the line of questioning you're going down. $\endgroup$ – Cort Ammon - Reinstate Monica Dec 18 '18 at 3:55
  • $\begingroup$ I think when charge particle accelerates the changing electric field generates magnetic field and vice versa and this pattern goes on in a straight line at speed of light turns out to be a photon. $\endgroup$ – user6760 Dec 18 '18 at 4:04
  • $\begingroup$ interesting, marching soldiers, packet, double slit,....your comments are very helpful, thank you all! $\endgroup$ – Wein Dec 18 '18 at 19:13
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The smallest EM wave is generated by single electrons in atoms and has discrete energy levels, which we can call a photon. This small EM wave tends to propagate in one direction where its E and M fields are strongest, the solution to Maxwell's equation says the E and M fields are well confined to sinusoids in a certain direction. However the wave function is a different function for the photon and it describes a probablilty nature so that the photon has a small chance of being anywhere but the greatest chance of going in a straight direction. So for Q1 a few photons do form a field but localized, but for Q2 and Q3 I am not an expert but would say that the QM photon description is not the same as the Maxwell EM field, so the EM field is not measurable everywhere but QM says it possible.

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  • $\begingroup$ actually I feel the edge between the pictures of EMWs and photons is not very clear to me....., thank you! $\endgroup$ – Wein Dec 18 '18 at 19:15
  • $\begingroup$ The EM field can be very calm like a pond with no waves but can be excited by a pebble, with certain barriers the water wave could be directed to look like a photon wave. The smallest EM wave is the photon from a single atom. $\endgroup$ – PhysicsDave Dec 18 '18 at 20:08
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This has been studied experimentally, how one photon at a time builds up the interference pattern that classical electromagnetic wave models perfectly.

singphot

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The little dots clear on the left are the footprints of the photons as they interact with the camera's recording surface. The photons is posited as an elementary particle in the standard model of particle physics, and are called particles because of the point interaction with matter , like a classical particle.

They are in reality quantum mechanical entities building up interference patterns in space and time, not seen in classical particles. As the number of photons increases towards the right, the interference of the classical beam becomes clear. The same plots have been measured for electrons .

However, if there are only very very few photons, can they form a wave-like macro EM field?

No, photons are localized, (within the Heisenberg uncertainty principle) as seen above. They do not have an electromagnetic field in real spacetime, though their complex wavefunction carries the information.

As you see , single photons are points, not spread over space-time

If a spherical monochromatic EMW (frequency is ν) propagates and decays into very low level of energy flux density, e.g., for every square meter, the energy flux is far less than 1*hν per second, then, does the EM fields still exist there?

No, there are only single photons as seen above

is the form of wave only meaning the quantum wave function to indicate the probability of where the photons appear?

The wave nature of the photon is in the wavefunction description, as modeled , for example,here

photwav

so it is just probabilities.

How in quantum field theory single photons bbuild up by superposition of their wavefucntions the classical wave is explored here , but a mathematics background in quantum field theory is necessary to understand it.

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    $\begingroup$ I have to tug on one thing: I think part of the reason it’s difficult for laypersons to stop thinking about particles as classical objects is statements like “photons are points.” The best we can say about any particle is that it INTERACTS at a specific location x,y,z, but the probability of it interacting with your measuring device at that specific x,y,z is given by the wavefunction squared. The fact that we see localized interactions doesn’t necessarily mean the photon (or any particle) is a localized bullet in flight before we detect it, but saying “it’s a point” implies that. $\endgroup$ – JPattarini Dec 18 '18 at 5:16
  • $\begingroup$ @JPattarini it is a point within the HUP limits , and certainly it is axipomatically posited as a point in the standard model of physics, which describes all particle data very well with this axiom. Maybe it will be a string in the future, but it is a point now. $\endgroup$ – anna v Dec 18 '18 at 5:19
  • $\begingroup$ What do u mean when you say the photon do not have an em field in real spacetime? What is it that's oscillating, photon as a single point or its probability wavefunction? (I'm ready to throw away maxwell eq if need be.) $\endgroup$ – user6760 Dec 18 '18 at 5:32
  • $\begingroup$ @user6760 The probability oschilates. If you read the link the photon wavefunction is a solution of the quantized maxwells equation, that is why E and B are there. Other formalisms have the A potential there, but it is the wavefunction that superposed with the boundary conditions of the two slits , oscillates in the above example. $\endgroup$ – anna v Dec 18 '18 at 5:39
  • $\begingroup$ great thanks for so much details you provide, however, I am confused that if the wavefunction (in quantum language to decide the probability of behavior of photons) and the waveform of the EMWs (in classical point of view), and they are both ``waves'',,,,,,but with totally different physical meaning....so, are they connected each other please? $\endgroup$ – Wein Dec 18 '18 at 19:18
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Here is a point which may make things clearer for you.

The practical answer to your question depends strongly on the particular wavelength of the electromagnetic wave you are interested in. This is because (as you mention) an electromagnetic disturbance can be modeled either as the emission of photons or waves. The wave model is most convenient for dealing with wavelengths longer than infrared and the photon picture is most convenient for dealing with wavelengths shorter than that.

Knowing that the energy of a photon E is equal to hc/lambda, big lambda means little E. For example, a radio frequency source at a wavelength of 3 meters (100mHz) produces photons of ~4 x 10^-7 eV, which is tiny on a per-photon basis. Note that a typical 100mHz FM radio station will be broadcasting anywhere from 50,000 watts to 250,000 watts of RF power; in this wavelength regime, then, it is customary to deal with the wave picture.

When the energy per photon is of order ~1eV, the photon picture becomes more convenient.

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  • $\begingroup$ very helpful answer, thank you! You mentioned the wavelength, and it is really a key issue. $\endgroup$ – Wein Dec 18 '18 at 19:14

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