I wanted to ask about an idea I got about the Double Slit Experiment. As you may know, you can use the Double Slit Experiment to have particles form an interference pattern with themselves, however, observation of what slit they went through causes this pattern do disappear, even if that observation happened after the particle landed on the detector... What would happen if you did this same experiment, but for example 4 at the same time. Lets say you wanted to send back the information 1001, so you decide that an interference pattern is 1, and no interference pattern is 0, you send the particles through, but delay observation of what slit they went through until a time after you observed where they landed. Then, you would remove the detectors from the 1st and 4th experiment in the line to cause it to be unknown what slit they went through, in effect causing an interference pattern, but for the 2nd and 3rd you did not remove the detectors, causing no interference pattern. Would this allow you to send information back in time? If not, what would stop it from happening? What would the results of what patterns appeared be? Thank you for taking time to read this, and replying if you do so.

  • $\begingroup$ Despite how it is occasionally taught, a measurement in QM does not have to involve a conscious observer. If the detector took the measurement, it doesn't matter if you erase that information without personally looking at it, there will be no interference pattern. $\endgroup$ – Chris Nov 18 '17 at 5:17
  • $\begingroup$ Are you talking about "it doesn't need to be a human to observe what slit it went through"? I am talking about removing the detectors entirely between the time they hit the main detector to see it and the detector to gather the info of what slit, so there is no detection of what slit changed in the future $\endgroup$ – Herbert Duggan Nov 18 '17 at 6:52
  • $\begingroup$ By the time you see or don't see an interference pattern, a measurement has already been taken or not taken. There's no way around that. $\endgroup$ – Chris Nov 18 '17 at 7:19

I think you are a little unclear about the quantum eraser experiment. If you read the wikipedia page for quantum eraser, you will see that their is no interference pattern visible in either case. The interference only emerges when you trace back the photons from both the detectors individually and then compare them with the previous done experiment to obtain two mutually exclusive interference patterns which in effect has no interference.

  • $\begingroup$ Sorry, I do not think I was clear enough. If it does not reach any detector at all it should form a normal interference pattern (Am I correct on that? If not, thank you for clearing it up for me) If so, I was talking about keeping it from accessing any detector if you wanted to show a 1 (the pattern being there), and letting it run as normal to show a 0 (no pattern) $\endgroup$ – Herbert Duggan Nov 18 '17 at 4:28
  • $\begingroup$ Yeah, according to my understanding the entangled photons don't have an interference patterns without tracing them back. Look at the simulated recording at en.wikipedia.org/wiki/Delayed_choice_quantum_eraser $\endgroup$ – Rishabh Jain Nov 19 '17 at 4:56
  • $\begingroup$ Yea, I was talking about the photo here: upload.wikimedia.org/wikipedia/commons/thumb/2/21/… If you took that, had it so the light hits D0, then it has a delay before reaching PS, and if you blocked it from ever reaching PS or D1-D4, wouldn't it then just act like the double slit experiment done completely normally, creating a normal interference pattern? So you could block it before PS to create the 1, and not block it to have the blob of everything as a 0 $\endgroup$ – Herbert Duggan Nov 19 '17 at 6:56
  • $\begingroup$ No, you would not see the interference pattern because the pair of photons are entangled. For proper information, refer to algassert.com/quantum/2016/01/07/…. While searching for an appropriate link, I was surprised to find out the number of wrong misinterpretations of delayed quantum erasure experiments which are completely nonsense . Beware of those explanations. $\endgroup$ – Rishabh Jain Nov 20 '17 at 8:51

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