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When for example, a photon is emitted from an atom, does that photon propagate through spacetime in all directions away from the atom in the form of a sphere (Wave Function) and then at some point along its propagation it is absorbed, reflected or refracted (detected) and at that point do we then consider it a Point Particle?

Mike R.
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  • Yes, the photon leaves the emitter as a wave, not necessarily a sphere, but as what is described by its specific wave function. Eventually the photon is detected as a particle. Between these two events, it is possible that the wave is reflected or refracted without detection. See this for details: https://physics.stackexchange.com/questions/368333/when-light-reflects-off-a-mirror-does-the-wave-function-collapse/423528#423528 – safesphere Aug 21 '18 at 08:45
  • So then with regards to reflection and refraction of a wave function, are they consequnces which are analogous to certain energies that are required to make an electron jump states. Are reflections and refractions interactions that like energies that are not exact enough, will not produce a detection? – Mike R. Aug 21 '18 at 11:06
  • Reflection and refraction are not events where electrons absorb/release photons. @safesphere was saying (I am pretty sure) that between your events of emission and detection of the photon, the photon could also undergo reflection or refraction without counting as "detection" as you have stated in your question. – BioPhysicist Aug 21 '18 at 11:36
  • @AaronStevens "Elastic scattering " is just a label. The actual QFT process is described by the operator of annihilation of the initial photon pushing the electron off shell (making it virtual to mediate the process). Then the operator of creation of a new photon puts the electron back on shell. Massless particles are not little balls that can bounce. They must follow the hull geodesic between the emission and absorption. The correct explanation of why this is not "detection" is that the photon interacts not with one electron, but with a superposition of many. See the link in my comment above. – safesphere Aug 21 '18 at 17:27
  • @MikeR. An atom can absorb a photon due to the electron changing levels. However a free electron cannot absorb a photon, because both energy and momentum cannot be conserved by a massive particle absorbing a massless particle. When an electron absorbs a photon, energy is not conserved. This is allowed for a very short time by the uncertainty principle and makes the electron virtual (off mass shell). Then the electron emits a new photon that also violates energy conservation, but in the exactly opposite way thus restoring the energy balance and making the electron real (on mass shell) again. – safesphere Aug 21 '18 at 19:18
  • @safephere how is it then that a mirror which is ultimately an arrangement of atoms can reflect "light" (in the form of a wave function) but an atom that absorbs an electron would bring about a detection. What is the threshold that determines whether or not an event will lead to a detection or merely reflecting of a wave function? – Mike R. Aug 21 '18 at 19:28
  • A photon can reflect or refract in transparent media like glass. An example is glass that refracts visible photons ( actually of 93% of them) but reflects IR photons ( a high percentage). Probably glass can even absorb photons of some correct energy. You are correct in focusing on the allowable electron energy levels of the material, that has a big impact. Also the spacing of the atoms in the material , the density and other factors will influence absorption, refraction reflection, diffraction. – PhysicsDave Aug 21 '18 at 20:51
  • Scientists use Maxwell/Fresnell's equations and publish tables of "permittivity" for many materials. The permittivity values also change with the wavelength of photons used. – PhysicsDave Aug 21 '18 at 20:52
  • @PhysicsDave So when a photon that is emitted from a source, the wave function describes the charatneriscs of that photon, that photon interacts with atoms (on a mirror) for example. What is interacting with the mirror? it can't be the wave function, right because the wave function merely describe the photons characteristics. – Mike R. Aug 21 '18 at 22:40
  • In a vacuum there is nothing, no mass, empty, nothing. Yet light can travel thru it and still there is not mass added, so what's up? We (science) say that all space is capable of supporting or has a field, for the photon it's the EM field. Somehow this field can transmit and the EM energy ( photon) or transfer the energy to an electron. There are equations and theories to describe all of science's observations, some basic and some super complicated. For the photon of the EM field we say it has an E component and M component described well by a sinusoidal waveform, i.e. the wave function. – PhysicsDave Aug 21 '18 at 23:21
  • Sorry just now saw your comment. A wave means a collective behavior described as a superposition. In a double slit experiment a photon flies through both slits as a superposition of two separate photons. When a photon bouncess off the mirror, exactly as in the double slit experiment, the photon does not bounce off just one electron, but off all the electrons in the mirror. So the photon exhibits the collective wave behavior, because it acts as a superposition of multiple photons. If you detect at the mirror, off which electron the photon actually bounced, then the wave function would collapse. – safesphere Aug 22 '18 at 20:49

2 Answers2

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A photon is effectively a wave until it is detected. The classical two-slit interferometer, used with light so dim that only one photon is present in the apparatus at a time, shows that each photon's wave function passes through both slits - which is inconsistent with the photon being a point particle during its propagation. However, the wave function only has spherical symmetry if it is emitted from an infinitesimal point. In the case of the two-slit interferometer, the part of the wave downstream from each slit has approximately the form of a cylindrical wave.

S. McGrew
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A single photon gets emitted in a specific direction. Example, a pebble dropped in a pond but the pebble was dropped in the middle of 3 logs in a "U" shape, thus the wave is directed and this is a good analogy for the photon. But the wave is still spreading out and can be affected by many other objects in the pond. Why the photon/atom behaves this way is likely because of the interaction with the other electrons or nucleus of the atom when the photon is emitted.

A photon can interact with objects it passes close too, this is diffraction, it changes the direction but the photon wave function is the same.

A photon can be refracted or reflected, the wave function is destroyed but immediately replaced by a new one as the electrons do the work.

You can consider the photon a point particle in many experiments/explanations, ex why is one light brighter than the other, i.e more particles. The use of the particle explanation is widely used. You can consider the photon a wave when your are studying its propagation, diffraction patterns etc.

Like water you would typically use the particle explanation when its raining or the wave explanation to explain for example a tsunami.

PhysicsDave
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