Timeline for How far back can you trace a photon?
Current License: CC BY-SA 3.0
9 events
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| Sep 5, 2023 at 19:53 | comment | added | Bill Alsept | @Marty Green, there is no proof that a photon can travel through two slits at the same time and of course the silver atom flips into either the spin-up or the spin-down states at the moment it passed through the magnetic field. That's what magnetic fields do. | |
| Dec 7, 2014 at 16:54 | comment | added | anna v | I should clarify this. there have been "which way" experiments since then that show the importance of the boundary conditions of the slits. It is the total solution photon+slits as boundary conditions. With minimum disturbance the interference does not get destroyed when the slit is known . Anyway I am sure that Feynman is talking of wavefunctions i.e. probabilities, and not of mass and energies in space. | |
| Dec 7, 2014 at 16:36 | comment | added | Marty Green | Then your argument is with Richard Feynmann, not me. I more or less quoted what he calls "Proposition A" (Vol. 3, chapter 1-5 in the Feynmann Lectures)...the very proposition which he concludes is false. According to Feynmann, it is simply not true that the photon must have passed through either one slit or the other. | |
| Dec 7, 2014 at 16:25 | comment | added | anna v | Yes for the double slit. Look at these cosmic gamma rays mpa-garching.mpg.de/lectures/ADSEM/WS0203_Obergaulinger.pdf . gamma rays are photons too. | |
| Dec 7, 2014 at 15:45 | comment | added | Marty Green | If there is no energy spread out, then the energy must be concentrated. And in your "standard quantum mechanics", we have local conservation of energy, which means the energy must be somewhere. So you're saying the photon existed from the moment it left the potassium atom. I suppose you're saying if it encountered a double slit along the way, it must have passed through either one or the other of the slits. | |
| Dec 7, 2014 at 15:36 | comment | added | anna v | I am using the standard quantum mechanical interpretation, which assigns probabilities of emission and detection and certainly no energy is spread out all over the universe in any decay. | |
| Dec 7, 2014 at 15:28 | comment | added | Marty Green | It seems like I'm saying it didn't become a photon until it entered the detector, and you're saying it was a photon from the moment it left the potassium atom. | |
| Dec 7, 2014 at 15:19 | comment | added | anna v | This is wrong: " The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years,". The quantum mechanical solution of the transition amplitude gives a PROBABILITY of finding the photon at the (x,y,z) at time t of the detector. The energy is not spread out. It comes in a quantum of energy that will excite a potassium atom to the analogous level it left. | |
| Dec 7, 2014 at 14:57 | history | answered | Marty Green | CC BY-SA 3.0 |