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I have limited knowledge of QM but I know some about electromagnetism. I have read some about the photon description and I am confused.

How can EM waves be viewed as a continuous, changing field but at the same time be considered as consisting of a finite number of discrete particles?

For example, when considering EM radiation power, the inverse square law applies based upon surface area expansion as you move from the radiating source. How does this work with the photon description?

Urb
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FourierFlux
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2 Answers2

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The mathematical consistency can be shown, but it needs quantum field theory (QFT) to follow the mathematics. You can read how classical fields and particle emerge from QFT here.

In this image you can get an intuition of the complexity of the build up of classical E and B fields:

photspin

It is just displaying the spins for the photons, which are either +1 or -1, nevertheless the circular polarization is built up by the photons, the connection being the left and right polarization.

This is clear if one understands that it is the complex wavefunctions of the photons which are superposed into a $Ψ$ . The real measurements of E an B come from the $Ψ^*Ψ$ probability distribution, and thus it is not a simple build up.

The wavefunctions of the individual photons are solutions of a quantized Maxwell's equations, and they carry the E and B fields in the complex function:

photwave

that is why the emergent classical wave is consistently built up.

anna v
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  • Ok thanks, classically the wave equation can be solved for in free space, if you have a perfect spherical radiator you can view it as spheres moving outward from the radiator. In the photon description, are the photons uniformly spaced going out in all directions like the EM wave? Like when talking about a photon, does it have a single specific direction and velocity? Or can a photon go in all directions at once? – FourierFlux May 02 '18 at 16:58
  • Ooops, by wave equation I meant maxwells equations can be solved for producing a wave. – FourierFlux May 02 '18 at 17:08
  • The photon follows a probabilistic quantum mechanical path, such that in superposition with zillions of others of the same energy, they build up the classical wave ,their energy being = h*nu, where nu is the frequency of the light built up. The complex conjugate of the photon wave function multiplied by the wave function builds up the classical electric field. ( in the illustration the red arrow is the classical electric field, the most probable manifestation of the zillions of photons – anna v May 02 '18 at 18:09
  • Thanks, does a photon have a definite size/length? Basically I think I understand the probabilistic argument but what I don't understand is if a Photon is itself a discrete entity in space? I'm trying to make the bridge between discrete and continuous. For example when EM radiation hits something, does the radiation hit at a finite number of points consisting of the photons that makeup the light? Could maxwells equations be kind of thought of as a continuous ideal of a limit of discontinuous functions but given scale that continuous ideal works? – FourierFlux May 02 '18 at 18:24
  • The photon in the present standard model of particle physics is an elementary point particle with zero mass and spin 1. Classical light is a wavefront of zillions of photons , each a point particle . Look at this answer https://physics.stackexchange.com/questions/386905/does-a-photon-have-any-measurable-size-as-a-particle/388347#388347 – anna v May 02 '18 at 19:10
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Consider diffraction through a slit or a grating, where classically the intensity on the detector varies due to interference. Detection of light occurs by some electronic electric dipole transition. Such transitions are discrete events which occur with probability $|\langle \psi_i|\vec{E}| \psi_f \rangle|^2$, so $E^2$ is the probability distribution of detecting a photon. The classical field describes the average value of a statistical distribution of detections.

Andrew
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my2cts
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