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In electromagnetics you learn that an electromagnetic wave is a perturbation of the electric and magnetic fields that propagate in space according to the wave equation.

This makes sense when you are dealing with radio transmission: an electric current perturbates the electromagnetic field and this change propagates in space; when this perturbation meets a conductor, it will induce a current in the conductor and thus energy is transmitted. So in a vacuum you can easily identify the wave as the region of space where the $E$ and $B$ fields have a different intensity from the "background" (assuming an empty space)

However, when you are dealing with single photons as if they were particles, where are the located in space?

Let's say that a photon hits an electron and raises its energy level. Where would the photon "hit" the electron? Is it when the perturbance enters in a certain area of the electron orbital? Or when it exits? And most importantly, if another photon comes and hits the electron again, how can you differentiate between the two? And how can a photon be a "quantum" of energy if the wave is a continuous perturbation?

So, the mother question is: where is a photon located compared to the electromagnetic wave of which it is an expression?

Qmechanic
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Mauro F.
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  • I don't have a full answer, but I'd like to point out that in states with a fixed number of photons the expectation value of the electric field actually vanishes so it's going to be hard to do E&M with that. You might be more interested in coherent states. – Wihtedeka Sep 13 '21 at 21:17
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    this may help https://physics.stackexchange.com/q/90646/ – hyportnex Sep 13 '21 at 22:17
  • What is the characteristic of a photon – HolgerFiedler Sep 15 '21 at 04:07
  • Also https://physics.stackexchange.com/q/462565/123208 "a photon doesn't have a strict position observable. A photon can't be strictly localized in any finite region of space. It can be approximately localized, so that it might as well be restricted to a finite region for all practical purposes" – PM 2Ring Sep 15 '21 at 05:13

2 Answers2

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The particle physics answer:

A photon is a point elementary particle of zero charge, zero mass, spin one and energy =$hν$ where $h$ is the Planck constant and $ν$ is the frequency of classical electromagnetic radiation that a large ensemble of same energy photons builds up.

where is a photon located compared to the electromagnetic wave of which it is an expression?

The location of quantum entities depends on the solution of the quantum mechanical equation for the specific boundary conditions , and can only be given as "probabiity of measuring the photon of a monochromatic beam at (x,w,z) at time t".

This basic experiment in wave particle duality with one photon at a time can give an intuition about photons and how they build up light:

singlephoton

1.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.

It is an (x,y,z) measurement on the screen, that can give the location of each photon footprint, a $Δθ$ $Δφ$ with respect to the beam direction . The accumulation in the last frames shows the expected interference pattern of the classical wave of which they are a component. One sees the probability that can be calculated from the quantum mechanical solution for the boundary conditions"photon scattering on two specific slits" in the last frame.

That is why one needs quantum mechanics theory, because it can mathematically model the behavior of particles and the wave particle duality and predict new set ups. How quantum fields build up the classical electromagnetic wave is outlined here , ( it needs a background in quantum field theory ).

BTW,

Let's say that a photon hits an electron and raises its energy level. Where would the photon "hit" the electron?

Photons interact with whole atoms when scattering, not individual electrons. The energy is absorbed or scattered by the whole atom, and the photon has to be within a $(Δx, Δy, Δz)$ of the atom to have a quantum mechanical probability to interact with it.

anna v
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  • Most of the above sayings are referred to the interference of a "wave" properties. What is the quantum-mechanics equation of a photon? – ytlu Sep 15 '21 at 16:53
  • @ytlu one way of calculating it, there are others (using the maxwell potentials) https://arxiv.org/abs/quant-ph/0604169 – anna v Sep 15 '21 at 17:48
  • Really nice answer, I think this should be the accepted one. Can you please tell me what do you think when the EM wave is traveling, do the photons build up the wave the same way you can see the waves on the screen (interference pattern)? – Árpád Szendrei Sep 15 '21 at 22:08
  • @ÁrpádSzendrei I do not think so. At a specific time t (traveling you ask)for the individual wave functions the phases should not change each photon's to each other photon , so that images could form with incoherent beams (light) when scattering from appropriate surfaces, with a mirror for example. – anna v Sep 16 '21 at 04:18
  • @annav What is Maxwell potential? – ytlu Sep 16 '21 at 16:23
  • https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Potential_field_approach 'this alternate way of the classical equations can also be used as a quantum mechanical equation givnit a wavefunction – anna v Sep 16 '21 at 18:54
  • @annav It is Maxwell equation. How the quantization $h\nu$ comes out of the equation? – ytlu Sep 18 '21 at 18:22
  • The phenomena which you mentioned in the post can all be explained well enough by wave properties of the radiation. I simply cannot get your idea about the wave function (in term of spatial location) about a photon. – ytlu Sep 18 '21 at 18:26
  • It is not my idea, it is the standard model of particle physics that treats photons on par with the other particles in the table. https://en.wikipedia.org/wiki/Standard_Model – anna v Sep 18 '21 at 18:34
  • In standard model, did it mention about the spatial extension of a photon? – ytlu Sep 19 '21 at 19:18
  • In this answer you said Photons interact with whole atoms when scattering. This is ambiguous and misleading. The interaction hamiltonian has only one-electron and photon interaction. The atom potential provides the electron the initial and final states. Atomic charge or potential were not participated in the interaction hamiltonian. – ytlu Sep 19 '21 at 19:33
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A photon is a monochromatic electric magnetic wave (EMW). The extension of this wave must be considerable very large compared to its wave length to neglect the short wave train effect, which must be made of many waves of variant frequencies. The total photon-electron scattering is a rather low efficient process. Multiple photos scattering between photon-electron interaction did occur but in a very low probability.

Then what is the meaning of a photon?

A photon is a measure of the energy unit for a monochromatic EMW, $h \nu$. In the interaction of electron and EMW, the energy transfer between them must be an integer number of this unit, $n h \nu$.

It make no sense at all??

Yes. But it is life. The energy of a EMW has been quantized after 1905 as Plank use the concept of photon to derive the Plank distribution of the average energy of EMW under thermal equilibrium, and then the Einstein explained the photo-electron effect using the concept of photon.

Therefore, accept the concept of photon as a base unit of energy for the electric magnetic wave, don't think it as an particle in real space. Investigating it spatial extension (as a real particle) definitely will drive you crazy.

ytlu
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