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I'm trying to wrap my head around the "Delayed Choice Quantum Eraser" experiment and how events in the future affect light in the past.

I'm sure I'm wrong but to me this seems to indicate that photons experience no time. Since time slows down when approaching the speed of light, a photon(which travels at the speed of light) experiences no time: the photon at the destination is the "same" photon at the source, meaning there can be no different state for the photon because inbetween source and destination time does not pass for it. The Photon would appear to be "locked" in time. So altering a photon in the future alters the photon in the past, all the way up to the source of the light.

Just for an analogy: if I was a photon and was "locked" in time, I would be like a statue not being able to change my state and if someone would draw a mustache on me, the mustache would be there yesterday and the day before that, all the way to eternity.

Now please tell me how I got it wrong :)

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  • $\begingroup$ $\gamma$ is only around 41 in air, which is a long way from infinity. $\endgroup$ – JEB Oct 9 at 14:44
  • $\begingroup$ It's meaningless to contemplate the time and space coordinates of a photon in its own frame. There is no frame in which the photon is at rest. $\endgroup$ – garyp Oct 9 at 14:44
  • $\begingroup$ yes garyp, my analogy is far fetched because I(the statue) would be at rest whereas a photon is in motion, but I was just explaining what I meant be "locked in time". For a more realistic analogy you can consider that I(the statue) am moving at the speed of light. $\endgroup$ – Cristi B Oct 9 at 14:47
  • $\begingroup$ What's your interaction with it going to be? What about it are you going to change? $\endgroup$ – simon at rcl Oct 9 at 14:54
  • $\begingroup$ Photons are not localized particles with a certain position at a certain time. They're non localized in time and space quanta of the e.m. quantum field. There is no easy analogy of them with an example of the world around us. A understanding of QFT is unavoidable to understand how they come about. $\endgroup$ – Jan Bos Oct 9 at 14:59
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Photons are elementary particles, part of the SM, they are traveling at speed c in vacuum, when measured locally.

The world line (or worldline) of an object is the path that object traces in 4-dimensional spacetime. It is an important concept in modern physics, and particularly theoretical physics.

https://en.wikipedia.org/wiki/World_line

Now photons do move along lightlike worldlines. You are correct, because on the lightlike worldline of a photon, the spacetime distance is 0. The photon does not have a reference frame, so it does not make sense to say " what would it look like from the photon's frame". But, since the spacetime distance for the photon is 0, you are correct, saying that the emission and absorption are casually connected for the photon and we could say the photon experiences both emission and absorption in one.

Now you are saying that altering the photon in the future (on the fly) would alter the photon back in time to the source.

In reality, you can use the double slit experiment with a detector, positioned after the slits, to show how there will be no interference pattern because the detector interacted with the photons after passing through the slits.

In this case the wavefunction of the photons is collapsed (after passing through the slits) and the phases of the photons are lost, and there will be no interference pattern.

When the electron suffers inelastic scattering, it is localized; this means that its wavefunction collapses and after the measurement act, it propagates roughly as a spherical wave from the region of interaction, with no phase relation at all with other elastically or inelastically scattered electrons,” Frabboni said. “The experimental results show electrons through two slits (so two bright lines in the image when elastic and inelastic scattered electrons are collected) with negligible interference effects in the one-slit Fraunhofer diffraction pattern formed with elastic electrons.

https://physics.stackexchange.com/a/452027/132371

The example is for electrons but works the same for photons. In the example there is a crystal positioned after the slits, and this is inelastic scattering which means the photons interact with the detector, thus there will be no interference pattern.

You could argue that this is like changing the photon back in time, but in reality this is because of QM and the wavefunction is set up so that interacting with the photons collapses the wavefunction and the phases are lost and there will be no interference pattern. You could say that photons do not experience time or that they experience the whole timescale between emission and absorption in one, but photons do not have a reference frame.

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  • $\begingroup$ From what I gather, the "Double slit" experiment shows collapse of wavefunction AFTER the detector, while the "Delayed Choice" experiments shows collapse of wavefunction BEFORE any action on the photon. It would be interesting to see if other properties of the photon are affected the same way, meaning future action affecting the past(I regard the collapse of wavefunction as a "state" or "property" of the photon which changes). Do you know of such experiments? $\endgroup$ – Cristi B Oct 9 at 16:33
  • $\begingroup$ It's confusing to talk about photons travelling. It's the field that are propagating. When they interact say with a charge it goes in chunks of "photons" that are not localized. This instantaneous collapse you mention illustrates this. $\endgroup$ – Jan Bos Oct 9 at 16:38
  • $\begingroup$ @JanBos you are correct. and it propagates in all directions outwards, when it interacts, it gets localized, before that, the photon is not localized. $\endgroup$ – Árpád Szendrei Oct 9 at 16:43
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    $\begingroup$ @CristiB "Crucially, physicists choose which measurement to make after the light pings off the satellite halfway through its 10-millisecond round-trip, they report 25 October in Science Advances. Again, the delayed decision seems to reach back in time, defining how the photon behaved after it left the first beam splitter." sciencemag.org/news/2017/10/… $\endgroup$ – Árpád Szendrei Oct 9 at 19:49
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Retrocausality is unnecessary to describe DCQE behavior. Nothing is working backwards in time (at least not anymore than in any other basic QM process, for fans of the Transactional interpretation).

In the DCQE, remember that the recovered interference patterns from the “no which-way info” arms of the experiment are complimentary - e.g. the bright bands of one arm would line up with the dark bands of the other arm.

This is very important - it means that when you first detect an entangled photon at a location (x,y) on your “idler” detection plate, you can say with certainty that at least one possible option is off the table for where the sister photon will later be found:

Using the notation from the Wiki article on the Kim et al setup, if your idler detector receives a photon at an (x,y) that corresponds to a trough in the interference pattern recovered from coincidence counting with the D1 detector, you can have high confidence that the sister photon will show up at either D3, D4, or D2 (since these 3 produce patterns still consistent with where your idler photon showed up on D0).

So far from requiring any spooky retrocausality, you can see that the photon arriving at D0 provides you enough information to update your expectations about where it’s entangled sister may be found in the future. That’s all.

The choice to measure either which-way or interference will then only further reduce the options for where the signal photon can be detected. Choose to measure which-way, and you’ve now got high confidence it can only be detected at either D3 or D4. Measure interference, and you can be confident that it’ll show up at D2 (because again, the location on D0 you already know corresponds to a dark band for the D1 recovered interference pattern).

The choice of whether to measure interference or not has nothing to do with retrocausality or anything of the sort - it can be described purely in forward-working reductions in possibilities as you gain more info about the system - first based on the D0 reading, and next based on your choice of what to measure.

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    $\begingroup$ Wow, this is a great answer. Thank you. $\endgroup$ – Connor McCormick Oct 25 at 5:34

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