Note that I'm looking for a layman explanations for these layman ideas.


Can someone help me understand why my idea wouldn't work?

Explain the the Delayed Choice Quantum Eraser?

First, I saw this interpretation of the experiment: Video I (at $4:30$)

Then, an idea of how to "cheat on lottery", by sending information to the past: Video II

Then, explained why it would fail here: Video III (at $4:00$)

But I have another idea of how to try something like that.

Theoretical Experiment - predict future

Leave the board on the earth, and the detectors A & B on the moon. Let's say you can send the first pairs to the earth board and their pairs to the moon detectors.

If you turn the machine, you see a scattered pattern, meaning that there are detectors on moon that will detect it few seconds in the future. Same case as the initial experiment, other than the distance to detectors is greater. So far, so good?

But what if there is the second person on the moon?

He then decides to replace the detectors with the eraser, before the photons reach them (You can synchronize watches to know when the machine was started, so he can replace them in that one second, assuming you can do that or just assume greater distance instead). Meaning if you start the machine and see a wave pattern, it means someone will replace the detectors with erasers in a couple of seconds. If you see a scattered pattern again, it means he will decide not to replace detectors with erasers, or fail to do so for some reason. You know the future?

But you can't change it since you can't send him a message to change his mind faster than the second pairs hit his detectors/erasers. (Faster than the speed of light)

Thus you can predict the future, if it involved a choice of whether to detect the path or to erase it?

Based on operating under premise that this is true, if you assume you have a theoretical way to trap light indefinitely, you could send messages to the past?

Example - alter future

Instead of using one such setup, why not use one for each bit of information? Each setup will end to be either $0$ or $1$, depending on if it has a wave pattern or just scattered dots on its board.

Lets say you play lottery and want to send results back.

Then, lets say you start the machine and trap the moon photons until you see the results. Then, you encode those results in binary and travel to the moon, and replace the detectors/erasers accordingly. Then you release the trapped light.

Now, look at when you first started the machine. The boards on earth will, after turning the machine on, yield wave patterns and you can read that as $0$s, or will be scattered and read as $1$s. After reading each bit of information, you got the results because you know you will send them back in the future, after the results really get announced.

But what if you decide not to send yourself lottery numbers after you get them ? That means you shouldn't have received them in the first place.

But what if you do not get them, and still decide to send them back. That means you should've got them in the first place.

This paradox makes me thing something is wrong.

Is the first idea valid?

What is wrong with the second idea?

Why wouldn't a similar idea ever work?

Help me understand of what the things I got wrong? Because I feel like I did not grasp a proper idea of these quantum properties.


closed as unclear what you're asking by WillO, Jon Custer, ZeroTheHero, Qmechanic Apr 5 '17 at 21:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Voted to close as unclear what you're asking, as I would for any question that begins by talking about "the board", "the detectors" and "the first pair" with no context, as if the reader is supposed to be able to read the mind of the poster. $\endgroup$ – WillO Apr 5 '17 at 19:06

It is not going to work. This is because regardless of what the person on the Moon does, the person on Earth will never see an interference pattern. He can only see an interference pattern if the person on the Moon sends him a message telling him: 1) At what time he used a "detector", and if he detected a signal or not. 2) At what time he used an "eraser", and if he detected a signal or not in that case.

Using that information, the person on Earth will see an interference pattern only if he plots the photons that were detected using the "eraser" or not detected if using the "detector".

  • $\begingroup$ So you need the information from the detectors/erasors to be able to read the board. This clears up the things. They should've stated that more directly in the video. Thanks! $\endgroup$ – Vepir Apr 6 '17 at 11:42

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