I've been reading/watching a lot of stuff about the double slit experiments. I was watching a youtube video about the delayed quantum eraser experiment and it was really interesting and got me thinking.

What if the detector is placed, a photon is emitted, and before the photon reaches the slits or the detector itself(depending on which of these is closer to the photon source), the detector is removed far away so that no measurement can be made? This is repeated over time for each photon. What would be the output of this experiment?

Will it prevent the interference pattern from appearing? Has an experiment like this ever been done? Perhaps there is a link?

The reason I'm asking this is because I've read that a photon doesn't experience time since it is moving at light-speed. So, perhaps for the photon, it wouldn't matter at what time the detector existed in front of the screen and this would cause the wave function to collapse?

In this(https://youtu.be/8ORLN_KwAgs?t=5m41s at 5:41) video by PBS SpaceTime, he explains how the 'past is re-written' by the quantum eraser. Basically, the interference pattern disappears when the particles hit the 'delayed detector'. I was wondering what would happen if the detectors that track which slit the photon came through were to be removed after the pattern has formed on the screen, but before they reach the detectors and thus an actual detection doesn't take place. Wouldn't this be somewhat equivalent to placing the detector and taking it away before it can detect which slit the particle came through? Or is the video inaccurate? Perhaps he simplified the facts for the sake of the audience?

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    $\begingroup$ Yes, if there is no detector, then there is no measurement. Photons aren’t so spooky that they can sniff around for the trace of a detector that was there earlier. $\endgroup$ – knzhou Sep 9 '18 at 22:37
  • $\begingroup$ @knzhou Hmm, from what I understand from the 'delayed choice quantum eraser' experiment, the interference pattern vanished even though one of the entangled particles was measured only after the other one had reached the screen. Couldn't something similar happen in this case as well? $\endgroup$ – Aj_ Sep 9 '18 at 22:42
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    $\begingroup$ I've seen a lot of "photons don't experience" time stuff lately. I would avoid that type of reasoning, as photons in the lab (n=1.000273) experience plenty of time. $\endgroup$ – JEB Sep 9 '18 at 22:44

The delayed-choice quantum eraser experiment is really not as amazing as one might expect, and it highlights the importance of understanding correlations. QM allows for some really strange correlations that cannot be described with "local hidden variables" -- the principle that if I have only one ball and put it in one of two boxes and ship those two boxes to opposite sides of the world -- say China and Argentina -- someone who opens the box in Argentina receives instantaneous information about the presence or absence of a ball in the box in China because that information is carried "locally" with the box. So QM allows for some level of information transfer that cannot be described this way.

But let's say you do the delayed-choice quantum eraser experiment on a similarly massive scale. Someone in China observes some sequence of flashes on a grid of photomultipliers arranged along a $z$-axis. If they were to make a histogram, these flashes would look like two overlapping bell curves, some pattern $\frac12 |f_0(z)|^2 + \frac12 |f_1(z)|^2.$

Let me be a bit clearer about what that means. If they just had the two slits separately, $|f_{0,1}(z)|^2$ are these bell curves they'd normally observe with just one or the other slit open. If the rest of their apparatus were not there, they would record $\frac12 |f_0(z) + f_1(z)|^2$ with both slits open, which would show a wavy interference pattern. But there is a part of their apparatus, which rotates the polarization of light, so that this interference pattern gets "erased" because the incident photons have been rotated so their polarizations are now at a 90-degree angle to each other, and with these two opposite "quarter-wave plates" that each rotate by 45 degrees in opposite directions, the Chinese researchers already expect to observe $\frac12 |f_0(z)|^2 + \frac12 |f_1(z)|^2$ in their histogram, no matter how the photon was originally polarized.

Here's the trick, these measurements are not just numbers $z$ but also come with some sort of ID number or sequence number, so it's $(0, z_0), (1, z_1), (2, z_2), \dots$. Some team in Argentina also has a bunch of qubits $q_{0,1,2,\dots}$ that correspond to these sequence numbers, too.

The team in Argentina can of course choose not to measure these qubits, in which case the above two-bell-curve histogram is really the only pattern that the Chinese researchers see. If they can keep qubits alive for a long time they could hypothetically delay their measurement an arbitrary time -- though in practice qubits are unbelievably fragile.

But if they want, they can choose one of two measurements. Neither one affects what has already been measured directly, and they generate a stream of bits, 0 or 1.

If they then send these bits up to China, the Chinese researcher can correlate the two and use them to split up their signal into two new histograms.

If the Argentinians choose one measurement, then once this is done, the two histograms will show $\frac12 |f_0|^2$ and $\frac12 |f_1|^2$ and this is all somewhat boring, "We can (apparently) tell you which slit the photon went through."

But if the Argentinians choose another measurement, then once this is done, the two histograms will show $\frac14 |f_0(z) + f_1(z)|^2$ and $\frac14 |f_0(z) - f_1(z)|^2$. Surprisingly this also sums to the same overlapping bell curves, but they both look like wavy interference patterns. And filtering out the second one (which is the role of the "coincidence counter" that you will see in the experimental setups for this), an Argentinian might say "oh, remember how you set up your apparatus to always 'measure' what slit the photon went through by rotating the polarization? Well I just 'erased' that measurement with my own!"

But there's nothing terribly "spooky" here like some folks appear to sensationalize. Like if the Chinese were to actually measure polarization for the photons we're sending through their apparatus, then normally if you started with purely up-down polarization, you could discriminate between whether the photons went through slit 0 or 1, with a polaroid filter oriented 45 degrees one way or the other. But the photons we sent them, which were entangled with the qubits we sent to Argentina, were not purely polarized up-down: half of them were polarized left-right. Accordingly they can measure those polarizations but they no longer really tell them what slit the photons went through. And if they put a sheet of up-down polaroid filter in front of the two slits to make sure, "yes now all of these are polarized up-down and we can measure which slit the photon went through," our Argentinian colleagues would not be able to improve on their information anymore because the Chinese would be missing half of the data points, and it would turn out that everything "interesting" is only interesting if you have both halves of the data.

  • $\begingroup$ Thanks for the detailed explanation. I think I understand it now to some extent. I've learned the lesson to not take these pop science videos in the literal sense. $\endgroup$ – Aj_ Sep 18 '18 at 1:37

in a quantum point of view when a single photon emitted to the double slits, there is a wave function that its wavefront splits by two slits and these two combine to form a fringe pattern due to the path difference when reaching to the screen.

the key point here is coherency, it is important to have a coherency in Young's experiment. watching photon in each slit clears the fringes pattern because of the act of watching photon changes the phase and destroy coherency.

so in your case, before the photon reaches the two slits you want to watch it. in this case, we have fringes because any change in photon's phase before splitting does not matter. and photon's phase value divided into same copies of original photon and form a fringes pattern.

see 3 states that described in below: enter image description here

  • $\begingroup$ Sorry, I realized I wasn't clear. Basically, the detector is removed before the photon reaches the detector itself or the slit, whichever is nearer to the photon source. But only after the photon has been emitted. $\endgroup$ – Aj_ Sep 9 '18 at 23:04
  • $\begingroup$ watch this also: youtube.com/watch?v=Xmq_FJd1oUQ $\endgroup$ – Persian_Gulf Sep 9 '18 at 23:30
  • $\begingroup$ @Aj_ We can do the double slit experiment with ancient starlight. Is the photon really supposed to keep track of where everything with eyes on Earth has been for the past few billion years? Quantum is weird but not that weird. $\endgroup$ – knzhou Sep 9 '18 at 23:50
  • $\begingroup$ @knzhou: Thanks for the reply. I know my understanding of physics is limited and I am probably over-simplifying stuff. Anyway, I've added some more information in the question. Tell me what you think. $\endgroup$ – Aj_ Sep 10 '18 at 0:04
  • $\begingroup$ @Persian_Gulf: Thanks for the detailed explanation. I've added some more details to the question. Do tell me what you think. $\endgroup$ – Aj_ Sep 10 '18 at 0:05

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