# Delayed Choice Quantum Eraser [duplicate]

All delayed choice quantum eraser experiments I've seen record the signal photons reaching detector D0 and then use the data of the idler photons recorded at detectors D1, D2, D3, D4 to "filter out" the photons with which-path information at detector D0 to be able to see the interference pattern.

Now, what would one see at detector D0 if the experiment was performed every 1 second and changed to send the entangled idler photons to the moon? It takes more than a second for the idler photons to reach the moon, which is enough time for the photon stream to form a COMPLETE interference pattern (or not) at detector D0. At the moon, instead of using 4 detectors where 2 do detect which-path information and 2 don't, a device with only 2 detectors would alternate randomly every second between recording and not-recording the which-path information of the incoming entangled idler photons.

Recording the which-path information at the moon should produce an interference pattern on earth. So would the observer looking at detector D0 see the pattern randomly change from interference to non-interference every second? If yes, can he thus predict 1 second in the future what the device at the moon will do?

• The answer is always no, no matter how smart your perpetual motion machine concept is trying to be. – CuriousOne Dec 19 '14 at 10:33
• Entanglements are said to "live outside space and time". Even if the relativistic interval between the experiment on the Earth, and that on the Moon, is timelike, the entanglement behavior is independent of which particle was measured first. What matters is which pair of measurements is done on the particles, i.e. what is measured on B in combination with the measurement on A. This is the amazing fact with the entanglements. They behave as if A and B exchange information about what is measured on them, and decide together what to respond. Even if one is measured later. – Sofia Dec 19 '14 at 12:55
• @Björn M.P. : by the way, you make a mistake. You ask, "would the observer looking at detector D0 see the pattern randomly change from interference to non-interference every second?". How can an observer see a pattern built with only one particle? Every second he gets a particle, not a pattern. – Sofia Dec 19 '14 at 13:32
• – Hypnosifl Dec 19 '14 at 14:58
• @Sofia: During that 1 second period you would emit not just one photon, but a stream of photons. The point is that during that whole second those photons would haver their entangled part measured (or not) for which-path information on the moon, causing an interference pattern on D0 (or not) during that second of photon emission. – Björn M.P. Dec 19 '14 at 15:06

There is never a double-slit interference pattern in the total pattern of signal photons at D0, such a pattern at D0 can only be found by taking a coincidence count with some detector that detected the idler photons. See my answer here for details--I think your question is basically a duplicate of that one.

Incorporating some discussion in the comments into this answer, I'll add that complementarity between which-path information and interference (defined formally in a 1979 paper by Wootters and Zurek, the paper itself is not online but see this pdf for a discussion) only applies to the coincidence count between the signal photon and idlers that have been detected at a particular location (like detectors D3 or D4 in the diagram from my answer above). There's never double-slit interference in the total probabilities for the signal photons to land at different points on the screen (i.e. the probabilities you'd calculate if you don't know what happened to the idler, see this answer for an elaboration if it's not clear) if they have been entangled with idlers that could even potentially have been measured in a way that would determine the which-path information of the signal photon, regardless of what actually happens to the idlers. See for example the last page of "Induced Coherence and Indistinguishability in Optical Interference" by Zou et al., available as a pdf here, which says:

The disappearance of the interference pattern here is ... a consequence of the fact that the two possible photon paths ... have become distinguishable ... Whether or not this auxiliary measurement ... is actually made ... appears to make no difference. It is sufficient that it could be made, and that the photon path would then be identifiable, in principle, for the interference to be wiped out.

Likewise, page S275 of the paper "Quantum effects in one-photon and two photon-interference" by L. Mandel, online here, says:

If the different possible photon paths from source to detector are indistinguishable, then we have to add the corresponding probability amplitudes before squaring to obtain the probability. This results in interference terms as in Eq.(3). On the other hand, if there is some way, even in principle, of distinguishing between the possible photon paths, then the corresponding probabilities have to be added and there is no interference.

And page S279 discusses a particular example, and says (emphasis mine):

However, when $i_1$ is blocked, $D_i$ provides information about the source of the signal photon detected by $D_s$.For example, if the detection of a signal photon by $D_s$ is accompanied by the simultaneous detection of an idler photon by $D_i$, a glance at Fig. 6 shows immediately that the signal photon (and the idler) must have come from NL2. On the other hand, if the detection of a signal photon by $D_s$ is not accompanied by the simultaneous detection of an idler by $D_i$, then the signal photon cannot have come from NL2 and must have originated in NL1.

...

Needless to say, it is not necessary actually to carry out the auxiliary measurement with $D_i$; the mere possibility, in principle, that such a measurement could determine the source of the signal photon is sufficient to kill the interference of $s_1$ and $s_2$.

• I'll make a simpler question first then. During a 1 second period, a beam of photons is sent trough a double-slit and then trough a Glan-Thompson prism to generate an entangled pair of photons. The signal part of these photons is sent to detector D0 to form an interference (or not). The idler part of these photons is sent to a random destination far away. If no measurement is ever done on the idler photons, will there be an interference pattern on D0? – Björn M.P. Dec 19 '14 at 15:27
• No, if the total probabilities for the signal photons to be detected at various positions on the screen could change depending on what was done to the idlers, this would violate the no-communication theorem. I talked about this a little more in my answer here. – Hypnosifl Dec 19 '14 at 15:41
• What would be done to the signal photons is, they will simply hit a light-sensitive screen revealing their collapsed location. Just like the result of a normal double-slit experiment with visible light, but this being done to the signal part only. The idler part would hit the soil of the moon one second later. I would expect an interference pattern at D0 as there is no which-path information known (no detectors at the moon). – Björn M.P. Dec 19 '14 at 15:54
• No, with entangled systems, complementarity between which-path information and interference (defined formally in a 1979 paper, see this pdf) only applies to the coincidence count, there's never double-slit interference in the total probabilities for the signal photons if they have been entangled with idlers that could even potentially have been used to determine the which-path information. See for example the last page of this pdf, which says: – Hypnosifl Dec 19 '14 at 16:03
• "The disappearance of the interference pattern here is ... a consequence of the fact that the two possible photon paths ... have become distinguishable ... Whether or not this auxiliary measurement ... is actually made ... appears to make no difference. It is sufficient that it could be made, and that the photon path would then be identifiable, in principle, for the interference to be wiped out." – Hypnosifl Dec 19 '14 at 16:04