If photon wavefunctions are omnidirectional and don't have a definite size, how comes those from the Sun don't all collapse on Mercury and some do actually reach Earth ?

I understand that wavefunctions describe a cloud of potential parameters, and that an actual particle can materialize at any point of the cloud, which has as many dimensions as needed to represent all characteristics of the particle to be.

But, does the wave function also represent the whole length and direction of the particle's trajectory (like : from the Sun to the Earth) ?

  • $\begingroup$ If you are thinking about wavefunctions in terms of the Schrodinger equation, that equation doesn’t apply to photons. (It’s non-relativistic; they’re as relativistic as you can get.) So you need to define what you think a “photon wavefunction” is, because there is no agreement among physicists that such a thing exists. $\endgroup$ – G. Smith Sep 18 at 19:21
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    $\begingroup$ You don't need to go to quantum mechanics to find problems that can't be solved with ray optics. For example, a diffraction grating can't be explained by ray optics. But it can be explained by classical wave optics without invoking any QM. $\endgroup$ – The Photon Sep 18 at 19:27
  • $\begingroup$ @G.Smith Do you mean psi(s,t) "Position-space wave functions" do not apply to photons ? en.wikipedia.org/wiki/… $\endgroup$ – Michelange Baudoux Sep 19 at 0:49
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    $\begingroup$ @MichelangeBaudoux, I mean that you don't need quantum mechanics to explain it. You can explain it quite well with classical physics. QM becomes important mainly in some situations where EM waves interact with matter. $\endgroup$ – The Photon Sep 19 at 0:56
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    $\begingroup$ @MichelangeBaudoux Yes, that’s what I mean. Photons are described by quantum electrodynamics (QED), a particular quantum field theory. Quantum field theories don’t have wave functions, or at least they didn’t when I learned quantum field theory. $\endgroup$ – G. Smith Sep 19 at 1:35

In quantum mechanics, which concept caters for light rays?

None, analogous to the fact that there is no concept for temperature in quantum mechanics either. Light rays are emergent from quantum mechanics the way thermodynamics emerges from statistical mechanics.

The photon is an elementary point particle in the standard model of particle physics. It has zero mass, spin 1 and energy = hν , where ν is the frequency the ensemble of such photons will show building the classical electromagnetic wave, h is Planck's constant.

Photons obey the quantized Maxwell's equation of quantum mechanics, with the corresponding wave function. The comlex conjugate square of the wavefunction gives the probability of finding the given photon at (x,y,z,t).

But, does the wave function also represent the whole length and direction of the particle's trajectory (like : from the Sun to the Earth) ?

The wavefunction of elementary particles leads to the probability of finding the particle. That is all. As to be called particles they have macroscopically to have a limited probability to be found outside a classical definition of a particle's footprint , no, the probability to be anywhere except close to a particle track is so small, it is zero.

This double slit experiment one photon at a time might help you acquire an intuition.


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 photon footprint at the left screen looks random, but the probability distribution accumulated at the right shows the classical interference pattern of a classical light wave of that frequency.

One could draw a direction, from the two slits to the individual photon footprint. What happens is that the individual wave function of the photon give this path as most probable, but because of the quantum mathematics, the boundary conditions, imposed by the width of the slits and the distance between them on the wavefunction (the same describes all these photons), gives the interference pattern on the right. The superposition is already in the wavefunction, induced by the boundary conditions.

You ask in a comment:

how do you get straight optical rays

It can be shown mathematically that the superposition of photons generates the classical electromagnetic field, and all its accoutrements. See here for example. The classical electromagnetic tools are so successful in describing light that not much stress is placed on the quantum mechanical wavefunction of the photon. Rays emerge from classical light analysis, are emergent.

  • $\begingroup$ "analogous to the fact that there is no concept for temperature in quantum mechanics either" - can you clarify what you mean by this statement, and how it squares with the healthy and long-lived literature on quantum thermodynamics? $\endgroup$ – Emilio Pisanty Sep 19 at 9:37
  • $\begingroup$ @anna v : Thank you so much ! Do I understand correctly that there is no direct causation from quantum to optical ray, because before that, a superposition of numerous photons, at a much larger scale, has to generate an electromagnetic field, wherefrom rays will emerge. $\endgroup$ – Michelange Baudoux Sep 19 at 10:48
  • $\begingroup$ @anna v : I found this nice representation of a photon, from an exepriment : cosmosmagazine.com/physics/what-shape-is-a-photon. Do we have a similar graphical representation of how a light ray emerges from the electromagnetic field ? As I am not litterate enough to make sense of the formulas showing on the link you mention, I also woul enjoy a more graphical representation of how an electromagnetic field emerges from photon superposition. $\endgroup$ – Michelange Baudoux Sep 19 at 10:51
  • $\begingroup$ @anna v : as for the double slit experiments, i'm bugged by the scale of those : sometimes, the slits are in the range of nanometers, but I often bump into comments saying you can try this at home with lasers and polarized glass. I'll post another question with that right away. $\endgroup$ – Michelange Baudoux Sep 19 at 10:58
  • $\begingroup$ @anna v : physics.stackexchange.com/q/503609/242752 $\endgroup$ – Michelange Baudoux Sep 19 at 11:37

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