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The propagation of the electromagnetic wave from a dipole-antenna is well described. The changing electric field creates a changing magnetic field, which creates a magnetic field and so on... the electromagnetic wave travels is all directions from the antenna (although with not the same intensity). We imagine the light the same electromagnetic wave, going in every direction from the source.

How could one single photon have only one direction? I mean at every point of its propagation the changing electric field creates a magnetic field in every direction (not just in the direction of the propagation). This would create an (elementary) wave source and it will propagate in every direction.

Emilio Pisanty
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redbaron
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    Photons are describing properties of quantum fields, they are not objects themselves. When a mechanical system changes 1J of potential energy into kinetic energy, then you don't say that a 1J "object of energy" has "gone some place". You accept that that 1J of energy is a change of a property of the system. This is no different. When we detect a visible photon, what we mean is that approx. $10^{-19}J$ of energy and $1\hbar$ of angular momentum were exchanged between the electromagnetic field and the detector. The wave function of the field tells us the likelihood of these exchanges. – CuriousOne Aug 03 '16 at 08:55
  • Now, if you want to get a semi-classical model of how quanta "move", then you have to look at the path integral formulation of quantum mechanics. Feynman has written a beautiful book about it called "QED: The Strange Theory of Light and Matter". Path integrals are, if you wish, the correct "interpretation" of quantum mechanics in terms of classical motion. Whether you will want to stick to the idea of any kind of motion after that is up to you. I certainly have given up on trying to interpret quantum objects that way after I understood what the path integral means. – CuriousOne Aug 03 '16 at 09:03
  • Thanks for your reply - I will look for that book - but the question still remains; a light beam definitely has a direction - how can you 'point', 'direct' a wave. – redbaron Aug 03 '16 at 09:17
  • You are basically asking for a quantum field description of a macroscopic wave. Since a photon is a quantum of a massless field, the energy that gets transferred is necessarily linked to a momentum, i.e. there has to be a direction of propagation. This is guaranteed by special relativity which combines energy and momentum into the four-momentum: https://en.wikipedia.org/wiki/Four-momentum. As a spin 1 boson the photon also carries a helicity, which is the equivalent of spin/angular momentum. A "lightwave" is a complex linear superposition of many such states with similar momentum. – CuriousOne Aug 03 '16 at 10:02
  • @redbaron it's better if you go with the photon particle approach. The only way to explain a light wave is with billions of coherent photons. – Bill Alsept Aug 05 '16 at 00:49
  • What are photons, electromagnetic radiation and radio waves – HolgerFiedler Aug 07 '16 at 19:54

2 Answers2

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How could one single photon have only one direction? I mean at every point of its propagation the changing electric field creates a magnetic field in every direction (not just in the direction of the propagation). This would create an (elementary) wave source and it will propagate in every direction.

The photon is an elementary particle in the standard model of particle physics. As such it is described by a quantum mechanical wavefunction , in complex numbers, whose complex concugate square gives the probability density of finding the photon at (x,y,z,t). It is only characterized by its energy=h*nu, and its spin which is + or -1h towards its direction of motion. Thus the image you have of it as a source of radially propagating fields is wrong. The confluence of photons builds up the electric and magnetic fields of the emergent classical electromagnetic wave, but the classical wave format cannot be cut down to describe a photon.

Quoting from this answer by Motl (to a different question)

the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of E and B vectors at each point. The probability density that a photon is found at a particular point is proportional to the energy density (E2+B2)/2 at this point. But again, the interpretation of B,E for a single photon has to be changed.

Have a look at this answer of mine on a similar question.

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anna v
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  • 'but the classical wave format cannot be cut down to describe a photon' - but one photon can interfere with itself (according to the single photon double slit experiment) - so it should be some kind of wave... – redbaron Aug 04 '16 at 06:01
  • @redbaron The photon does not interfere with itself. The probability of a photon hitting two slits has an interference pattern, due to the complex wavefunction of the solution of the boundary condition qm problem "photon scattering off two slits" see this single photon experiment http://www.sps.ch/en/articles/progresses/wave-particle-duality-of-light-for-the-classroom-13/ – anna v Aug 04 '16 at 06:17
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    The article you quoted says that the 'photon interferes with itself' - I've looked up the original article http://depts.washington.edu/jrphys/ph331/share/mach1.pdf and it says so, too. Can you explain the difference? From this experiment I would conclude that the photon (or electron in other experiments) take both ways as a wave would do, but while detecting, capturing it will become one single spot... – redbaron Aug 04 '16 at 08:15
  • @redbaron The wavefunction of the photon interferes with itself. That is the difference . In water waves it is the functional form of the energy transfer/amplitude that carries phases, i.e. sinusoidal behavior, which generate interference. In quantum mechanics it is the wavefunction that has sinusoidal behavior, and the interference appears in the probabilities, not in the mass or energy amplitudes, which are fixed. – anna v Aug 04 '16 at 13:39
  • @redbaron one photon cannot interfere with itself – Bill Alsept Aug 05 '16 at 00:55
  • I should qualify that the wavefunction of the photon obeys the boundary conditions of the problem, in the case of interferences seen as in the double slit single photon, the boundary conditions come from "photon scattering over two slits". A single photon's wavefunction with no boundary conditions will have no possibility to show interferences in its probability for localization, as @BillAlsept states – anna v Aug 05 '16 at 05:23
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(The previous version of this answer was so utterly wrong I tried to delete it after recovering my senses. Alas, I learned I cannot delete an accepted answer, so I've rewritten it.)

A single photon is some complicated disturbance of electric or magnetic fields, but photon is not defined neither as a point disturbance of these fields or a naive wave that would propagate from such a point disturbance.

The secret behind all types of waves is that they "propagate in every direction" if you restrict them first to a very narrow slit. A wave from a small drop of water has no single direction, but on the sea you have wide wave fronts going in a direction (and not created from many point disturbances).

The narrow slit and the wide front rule in quantum mechanics is officially named Heisenberg uncertainty principle. If you arrange to give a photon a precise direction experimentally, you can do that only by spreading the photon's position in space.

kubanczyk
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  • Thanks - I try not to see the photon as a billiard ball :-) let's take the analogy of the sea. Without having a look at the rest of the sea one can tell from just observing one certain point moving up and down what the wavelength is (knowing the frequency and the speed of propagation). So one single point has got a wavelength, right? I mean this does not mean that the 'photon' is 1 km long. But it could be 1km 'wide' (there is a probability to detect him 1 km apart from the direction of the beam) - is this what you meant? – redbaron Aug 03 '16 at 10:29
  • @redbaron: If all I give you is one point of a wave motion, you won't be able to tell what the wavelength is. You won't even be able to tell if there is a well defined wavelength (the spectrum could be that of white noise). With quanta it's even worse. If you measure the momentum precisely, you have to give up spatial information and if you want a measurement at or better around a certain point, then there will be a large momentum uncertainty. – CuriousOne Aug 03 '16 at 10:48
  • @CuriousOne I thought the lambda=c/f was applicable by the ocean waves - it does not apply to the light? – redbaron Aug 03 '16 at 12:32
  • @redbaron: How would you know the spectrum of either from a single measurement? You are confusing the problem by assuming monochromatic waves, but that information can not be gathered with just one measurement. It would, indeed, take an infinite number of measurements in either case. – CuriousOne Aug 03 '16 at 12:34
  • @CuriousOne - now everything is uncertain :-) so you say it's impossible to tell the wavelength of an ocean wave just by looking at its certain point? or you say it's not applicable for EM waves? – redbaron Aug 03 '16 at 12:54
  • @redbaron: A wavelength measurement requires that there are some identifiable periodic features in a field. One can not identify those features with a single measurement, or two or even three. We need to understand either the spectrum of the field or have a sufficient number of spatially diverse measurements of it. Wavelength is a macroscopic quantity and there is no one to one correspondence to a single photon energy-momentum. You can say the same thing about classical waves and a single spatial sample. – CuriousOne Aug 03 '16 at 19:22