The question of retrocausality

[Andre_86]

TL;DR Summary

The question of retrocausality in the experiment of the quantum eraser with deferred choice. The well-known formulation of the experiment does not contain retrocausality. However, I came up with other experimental conditions in which retrocausality is present. I'd like to know where I am not right.

Hello all. There is the essence of the experiment in this link:
https://en.m.wikipedia.org/wiki/Delayed-choice_quantum_eraser
You can see the essence of the original experiment under this text.

Excerpt from the text:
By using a coincidence counte, the experimenters were able to isolate the entangled signal from photo-noise, recording only events where both signal and idler photons were detected (after compensating for the 8 ns delay).

• When the experimenters looked at the signal photons whose entangled idlers were detected at D1 or D2, they detected interference patterns.
• However, when they looked at the signal photons whose entangled idlers were detected at D3 or D4, they detected simple diffraction patterns with no interference.

Even if you remove the translucent mirrors BSa, BSB, the intervention at the dacha D0 will not be observed, because the two images of interference that will occur will have opposite highs and lows, which will give the usual diffraction. That is, it is impossible to obtain any information about which sensors will get photons, D1, D2 or D3, D4, until the moment of measurement and comparison of coincidence counters. There is no retrocausality.
But i don't know why there is next experiment retrocausality is not real. You can see the essence of the experiment under this:

The translucent mirror Bsc we can turn on and off by applying voltage. Let the mode of the on mirror 0 and the off mirror 1. After a time t0 after the start of the experiment, the first 10,000 photons reach the sensor D0. Let the mirror Bsc be in mode 0. This means that the tangled pairs of these photons are fed to the translucent mirror. The main difference between this experiment and the original is that the probability of passing the photon in different directions is different. 1. If the photon came from hole A, it will bounce off with a probability of 70%, and will get to D2.
2. If the photon came from hole B, it will pass through the mirror with a probability of 70%, and will get into D2.
Therefore, with a probability of 70% the photon will get into the sensor D2. However, information from which hole it is in nature is missing, so these photons will form an interference.
Therefore, approximately 70% of the photons out of 10,000, which is 7,000 photons, will contribute to the interference image from the D2 sensor. 30% of the photons will contribute to the interference from the sensor D1. In this case, the imposition of maxima and minima of interferences on the sensor D0 should not occur, because one interference has a higher intensity than the other.
If the mirror is in mode 1, then we will get a normal diffraction image on the sensor D0.
Does this not mean that the observation on the screen D0 of interference or diffraction is possible without a match counter? Because we can conclude that the presence of interference on D0 indicates that in the future the mirror will be in mode 0.

Thank you for your reading and where is my mistake?

 

[andrewkirk]

Does this not mean that the observation on the screen D0 of interference or diffraction is possible without a match counter?

We always observe an interference pattern in the full set of readings at D0 because they are just the results of a standard double-slit experiment. We don't even need QM to predict the interference pattern. We can predict it using Huygens' 17th century wave theory. The splitting prism PS and all equipment beyond that are irrelevant for that purpose, as they only affect the matching process.

In your set-up you will see interference patterns in all the of the four following subsets of photon-strike records at D0:

- the full set
- the subset that matches to readings at D1
- the subset that matches to readings at D2
- the combination of the previous two

There is no retro-causality in any of this.

 

[vanhees71]

There is of course no retro-causality but only delayed choice. The Wikipedia article is pretty ununderstandable, because it misses the information that type-II parametric down conversion was used and that the which-way information for photons detected at D3 or D4 is due to the polarization entanglement of the s-i-photon pairs. Taking this into account it's clear that D03 and D04 coincidences don't show a two-slit but only a single-slit interference pattern while the D01 and D02 coincidences show it.

 

[Andre_86]

In your set-up you will see interference patterns in all the of the four following subsets of photon-strike records at D0:

- the full set
- the subset that matches to readings at D1
- the subset that matches to readings at D2
- the combination of the previous two

There is no retro-causality in any of this.

Thank you! I meant the following manifestation of retrocausality: if the mirror is in the on mode, we will see the sum of interference patterns from sensors D1, D2, hence interference. If we make the mirror transparent, we will see the usual diffraction. However, we determine the on / off mode after fixing the photons at D0. So, we can know off or on before registering on D1,D2 and before choice. I mean it, when i said about retro-causality

 

[Cthugha]

We always observe an interference pattern in the full set of readings at D0 because they are just the results of a standard double-slit experiment. We don't even need QM to predict the interference pattern. We can predict it using Huygens' 17th century wave theory.

Sorry, but the first part is not really correct. There is never an interference pattern in the full set of readings at D0.

This is indeed just the result of a standard double-slit experiment - with light that lacks the necessary spatial coherence to show an interference pattern for the given slit geometry. But indeed: yes, the pattern one will see can be predicted using Huygens' theory.

The standard way of making incoherent light show an interference pattern is to filter it to increase its spatial coherence. A simple pinhole is sufficient to do so. The whole postselection process is some kind of Rube Goldberg machine that is the most sophisticated way to create filtering by a pinhole: You use entangled light, put a narrow detector that is narrower than the full beam and thus acts as an effective pinhole in the path of one of the photons and then use postselection, so the "partner" beam becomes effectively filtered as well.

 

[Andre_86]

Sorry, but the first part is not really correct. There is never an interference pattern in the full set of readings at D0.

I understand why we can not look interference, because full set of readings is combination of two interference pattern and we look diffraction. I don't understand why combination two interference which have different intensity is not interference. Why one interference pattern not domination by second interference pattern. Thank you!

 

[Cthugha]

First: you seem to want to create a beam splitter that has 70% reflection/30% transmission on one side and 70% transmission/30% reflection on the other side. This is impossible and violates conservation of energy unless that mirror is also lossy (as translucent mirrors are). You could just put an absorbing filter directly in front of D1 and get essentially the same scenario you suggest.

This directly makes it obvious why it will not work: You are just throwing away data to filter on. Nothing you can do on the side of D1 and D2 has any effect on the detection events at D0. The full pattern at D0 never changes, no matter what you do on the other side.

 

[Andre_86]

First: you seem to want to create a beam splitter that has 70% reflection/30% transmission on one side and 70% transmission/30% reflection on the other side. This is impossible and violates conservation of energy unless that mirror is also lossy (as translucent mirrors are). You could just put an absorbing filter directly in front of D1 and get essentially the same scenario you suggest.

This directly makes it obvious why it will not work: You are just throwing away data to filter on. Nothing you can do on the side of D1 and D2 has any effect on the detection events at D0. The full pattern at D0 never changes, no matter what you do on the other side.

Thank you for the explanation! I understand my mistake