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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) ?

Michelange Baudoux
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  • 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. – G. Smith Sep 18 '19 at 19:21
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    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. – The Photon Sep 18 '19 at 19:27
  • @G.Smith Do you mean psi(s,t) "Position-space wave functions" do not apply to photons ? https://en.wikipedia.org/wiki/Wave_function#Position-space_wave_functions – Michelange Baudoux Sep 19 '19 at 00:49
  • @ThePhoton Do you mean that Quantum physics doesn't explain how light is propagating through space ? – Michelange Baudoux Sep 19 '19 at 00:52
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    @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. – The Photon Sep 19 '19 at 00:56
  • @ThePhoton Thanks for the precision. I vaguely read that QM can't be reconciled with cosmology, I also read that light waves have nothing in common with QM wavefunctions. So yes, you can explain rays with older models. But as QM states that light is also a particle, and as that statement is fundamental to the theory, and as movement is so essential to what light is, I would like to know if QM (in any of its flavor) would state anything about macroscopic movement of light through vacuum. – Michelange Baudoux Sep 19 '19 at 01:06
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    @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. – G. Smith Sep 19 '19 at 01:35
  • @G.Smith : Thank you so much. Then how do QED field operators explain the movement of photons through space ? – Michelange Baudoux Sep 19 '19 at 01:58
  • in quantum physics there are no trajectories for any kind of particle, because it is impossible to have a well-defined position and a well-defined momentum at the same time. Particles have a probability amplitude to get from interaction point A to interaction point B, but we can’t talk about the path in between. On fact, one formulation of quantum physics says that they must be treated as if they take take all possible paths. – G. Smith Sep 19 '19 at 02:04
  • See https://en.wikipedia.org/wiki/Path_integral_formulation “It replaces the classical notion of a single, unique classical trajectory for a system with a sum, or functional integral, over an infinity of quantum-mechanically possible trajectories to compute a quantum amplitude.” – G. Smith Sep 19 '19 at 02:07
  • @G.Smith : Thanks, this is exactly the object of my question : given that, at quantum level, the path of a photon is the sum of all possible paths, how do you get straight optical rays from that on cosmic scale, given that there is nothing to interfere with photons in the vacuum? – Michelange Baudoux Sep 19 '19 at 02:23
  • Each path contributes a complex probability amplitude. Most paths destructively interfere, but paths near the classical trajectory constructively interfere. In this way, a classical path “emerges” from a quantum description involving all possible paths. – G. Smith Sep 19 '19 at 02:29
  • @G.Smith "Quantum field theories don’t have wave functions", this is incorrect. All quantum field theories use the plain wave wavefunction of the corresponding particle as a representation of the field , on which creation and annihilation operators workd. These plane waves obey the corresponding equation, Dirac for fermions, Klein Gordon for bosons and a quantized maxwell for photons, although how to do that has various versions. ( I link to one) QFT is a quantum mechanical calculational tool , but is based on wavefunctions. – anna v Sep 19 '19 at 04:39
  • @annav I disagree with you, as does Steven Weinberg. Please see https://physics.stackexchange.com/questions/171057/is-maxwells-field-the-wave-function-of-the-photon where he is quoted as saying “Certainly the Maxwell field is not the wave function of the photon.” Some of the answers explain why. – G. Smith Sep 19 '19 at 05:01
  • @G.Smith can one disagree with mathematics? – anna v Sep 19 '19 at 05:03
  • As @dmckee put it in a comment to that question, “There is, of course, a relationship between the classical electromagnetic field and photons, but it is much more subtle than the field being the wave function of a photon which is nonsensical for several reasons starting with Ján's post below. I generally assume that people who write things like that are either lying to children (if I think they actually know better) or parroting something they heard (if I have no reason to believe they no better). I presume that Weinberg means exactly what he says.” – G. Smith Sep 19 '19 at 05:07
  • @G.Smith I do not presume to judge Weinberg out of context. In my answer I have given links that show how mathematically from photon fields one ends up with the classical electromagnetic wave. The fact that all quantum field theories are based, represent the field by plane wave wavefunctions of the corresponding equations is also mathematically indisputable. Quantum field theory is based on the postulates principles and laws of quantum mechanics. In QFT onshell particles have to be represented by wavepackets (but that is another story) – anna v Sep 19 '19 at 05:57
  • @G Smith, anna v : Anna intially stated that all particles, even the photon, have a wavefunction in QFT. G. Smith explained that the photon wavefunction is not the electromagnetic wave function. I think both propositions may stand together and are true. Please read answerfrom anna hereunder. – Michelange Baudoux Sep 19 '19 at 11:04

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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.

singlphdbslt

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.

anna v
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  • "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? – Emilio Pisanty Sep 19 '19 at 09:37
  • @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. – Michelange Baudoux Sep 19 '19 at 10:48
  • @anna v : I found this nice representation of a photon, from an exepriment : https://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. – Michelange Baudoux Sep 19 '19 at 10:51
  • @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. – Michelange Baudoux Sep 19 '19 at 10:58
  • @anna v : https://physics.stackexchange.com/q/503609/242752 – Michelange Baudoux Sep 19 '19 at 11:37
  • @MichelangeBaudoux quantum mechanics has no meaning without knowing (going into) mathematics.. It is not philosophy, there are no simple diagrams for complicated processes, (other than feynman diagrams of cours) – anna v Sep 19 '19 at 18:10
  • @anna v : allright, but for math illiterates that is a bit Bohring.

    At least, please, give me two numbers : a scale at which those emergences occur, so that I can represent myself where the three models (wave function, EM field, rays) are a relevant representation and transition to each other.

     – Michelange Baudoux Sep 19 '19 at 19:52
  • @anna v : not all diagrams are that simple : I do understand experimental schemas (double slit experiments, Aspect's intrication experiments) some principles of the standard model, and great images like this : https://phys.org/news/2015-03-particle.html – Michelange Baudoux Sep 19 '19 at 20:04
  • All correct mathematical models of light are consistent with each other in the hierarchy a)quantum level, classical level . The specific numbers where it is ok to ignore the quantum level depend on the specific problem and its boundary conditions, i.e. on mathematics. Hand waving about things just introduces irrelevant concepts. Particularly for photons the classical equations hold very well, and photons can be ignored unless in specific boundary conditions, as "one photon at a time" , interactiion with atoms/molecules/lattices (photoelectric, black body, spectra).. There are no two numbers. – anna v Sep 20 '19 at 03:27
  • The relevant number that will give for the specific problem the answer whether the quantum regime is evident are h (planck's constant). This used with the boundary conditions , if it can be assumed zero , shows the classical regime, but one has to put it in the relevant to the problem equations to see this. It is not a recipe. – anna v Sep 20 '19 at 03:30
  • @anna v : " are h (Planck's constant) and the boundary conditions" ? – Michelange Baudoux Sep 20 '19 at 13:50
  • @ anna v : "a)quantum level, b)classical level ?" – Michelange Baudoux Sep 20 '19 at 13:51
  • No, both are necessary checks to know if quantum frame mathematics is needed. or can be ignored and go to classical directly .mathematics is necessary – anna v Sep 20 '19 at 15:09
  • @ anna v : I thought there were two typos in the two PREVIOUS answers of yours. The quotations marks were trying to hint you at those typos. Now I sorrily can't figure out what your last answer applied to. I can make sense of "mathematics is necessary " though :-) – Michelange Baudoux Sep 21 '19 at 13:56