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--Diagram from Wikipedia of the experiment of Kim et al. (1999)

Original research paper: A Delayed Choice Quantum Eraser

I want to know whether the landing position of the signal photon (photon that is detected at $D_0$) on the screen at $D_0$ is affected by anything we do to the idler photon (photon that is detected at one of the other four detectors $D_1$, $D_2$, $D_3$, $D_4$) in the Delayed Choice Quantum Eraser experiment performed by Kim et al.


4 Answers 4


Since the idler and signal photons are entangled and not separable in your reference: they must be treated as a single quantum system.

It is the future measurement context of the pair (a biphoton) that determines the possible outcomes. That context will also consider the things that the individual component photons might or might not encounter along their way. Filters, lenses, beam splitters, etc. can all be factors, as you see in your reference. Some things are not likely to have an impact on the overall context, such as mirrors or wave plates.

I don't know if it is fair to say (i) "a change to the idler setup affects the outcome of the signal photon". It is fair to say that (ii) "a change to the idler setup affects the statistical outcomes of the signal and idler, considered together".

To a degree, the differences in the above 2 statements is semantics. Which in turn essentially means bringing "interpretations" into the picture. And that's a wormhole, as we all know. Unquestionably, a change in a quantum setup from configuration A to configuration B has the potential to cause a visible change in the statistical outcomes. But you can't really say that the idler photon itself is responsible for that change any more than you can say that the signal photon changes the idler.

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    – SuperCiocia
    Jun 1 at 2:20

I want to know whether anything we do to the idler photon affects the landing position of the signal photon on the screen at $D_0$ in the Delayed Choice Quantum Eraser experiment performed by Kim et al.

To put it simply, as long as the idler photon is separated from the signal photon, there is nothing you can do to the idler photon(s) to alter the probability distribution of landing positions for the signal photon(s). If it were possible, you would be able to take advantage of this and communicate to other parts of the universe faster than light.

Even a slight probability distribution change means you can get an edge and violate causality as understood in special relativity.

Saying that the probability distribution cannot be altered at a distance is the most we can say. We cannot make any stronger statement without getting lost in "interpretations."

Note, however, there is a difference between $$ P(\text{signal photon detected at some position }x) $$ versus the conditional probabilities $$ P(\text{signal photon detected at some position }x \,|\, \text{idler photon detected at }D_{1}) $$ and $$ P(\text{signal photon detected at some position }x \,|\, \text{idler photon detected at }D_{2}). $$ The latter two probabilities are different than the first probability, but that's only due to conditionality. In order to know the latter two, you need information about where the idler photon was detected. Again, we can't say anything stronger if we want to stick to QM proper.

  • $\begingroup$ Is it safe to assume idler and signal photons independent of each other as soon as they are released? I mean photons can take any path if you see how Feynmann diagrams are drawn. There are paths that even signal photon comeback and hit idler detectors or interact with idler temporarily. $\endgroup$ Jun 7 at 5:43
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    $\begingroup$ Independent? In terms of correlations or causation? If you are talking in terms of correlations, the fact that the conditional probabilities are different means they are not independent. (Note that conditional dependence isn't even a quantum phenomena.) Now in terms of causation? That is a question I genuinely cannot answer and people cannot decide on no matter how much debate goes on. Each interpretation of QM says something entirely different regarding the underlying causation. $\endgroup$ Jun 7 at 5:51
  • $\begingroup$ @DukeWilliam I found your series of questions you've been posting interesting, but it seems like you are very hung up on the DCQE when really you should be thinking about the nature of entanglement (at separated distances) itself. That's a much broader topic, and people in the debate of whether there's any "causation" going on just can't agree with each other no matter how hard they try. I apologize if I am misunderstanding you or missing your question. $\endgroup$ Jun 7 at 5:55
  • $\begingroup$ This a late response but No need to apologise. You are right I am obsessed with DCQE because of conflicting and misguided explanations even among well-recognised theoretical physicists about it. In the process, I learned that physics is so divergent that it is easier to find two physicists that differ on a subject than agree on it. There is something about DCQE that became apparent to me after going through dozens of research papers.. $\endgroup$ Jun 21 at 15:15
  • $\begingroup$ Actually, the reason for no directly visible interference patterns at any detector is the destruction of photon at BBO and the creation of the entangled pair. Its wavefunction collapses completely and a new wavefunction emerges. The superposition of two slits path becomes one slit path. From there afterwards, only one slit diffraction pattern is possible unless there is a new double slit or beam splitter-like device introduced later down the path. $\endgroup$ Jun 21 at 15:22

If the signal and idler photons were created where the source EM field was influenced by both slits ... then they have a common "interference" property. Once you have made your apparatus setup you can employ mid-flight electronic shutters/mirrors/polarizer objects in the optical paths however coincidence detection will become impossible. If the point of the experiment is to measure "interference" (which is due to the double slit effect on the EM field) any mid-flight changes result in new paths ... i.e. no interference.

Additionally as you attempt to make the idler path very long it’s probability of being travelled becomes less (fibres are usually employed).... but signals can still be measured.

In general many physics papers have discussed "faster than light" communication BUT presently most physicists have debunked these interpretations. There are certainly some believers .... but like in Star Trek, I don't see any transporters or time machines being developed any time soon.

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    – Buzz
    May 29 at 19:16

Based on the discussions in response to the answers provided, it seems to me that the issue to be resolved is a simple matter of looking at the unitary process represented in the specific case where interference is reproduced.

I am going to use a mathematical approach. Without math the discussion would just be too cumbersome and I think it is the lack of mathematics that causes the confusion.

We start just after the slit before the nonlinear crystal. There the state is (ignoring normalization) $$ |\psi\rangle = |s_1\rangle + |s_2\rangle , $$ where $s_1$ and $s_2$ represent the two slits. Because they are in superposition, they would produce the well-known interference associated with a double slit experiment. However, when we then send it through the nonlinear crystal, we get $$ |\psi\rangle = |H\rangle_1|V\rangle_1 + |H\rangle_2|V\rangle_2 , $$ where $H$ and $V$ represent horizontal and vertical polarization of the down-converted photons, and the subscripts indicate the slits. These two terms are now separated and then recombined in different ways for the different detections. The ones we are interested in are those that would be able to reproduce interference. Let's assume the interference observed at $D_0$ is observed in the horizontal polarization. So the vertical polarization state are manipulated to give the associated measurements.

For the interference require the sums and difference (still ignoring normalization): $$ |+\rangle = |V\rangle_1 + |V\rangle_2 , $$ $$ |-\rangle = |V\rangle_1 - |V\rangle_2 , $$ which means that $$ |V\rangle_1 = |+\rangle + |-\rangle , $$ $$ |V\rangle_2 = |+\rangle - |-\rangle . $$ Now I can replace the last two equations into the expression for the state. Then it becomes, with some rearrangement of terms $$ |\psi\rangle = (|H\rangle_1 + |H\rangle_2)|+\rangle + (|H\rangle_1 - |H\rangle_2)|-\rangle . $$ Here the states $|+\rangle$ and $|-\rangle$ are measured at $D_1$ and $D_2$, respectively, in coincidence with the measurement at $D_0$, which would then respectively measure $|H\rangle_1 + |H\rangle_2$ and $|H\rangle_1 - |H\rangle_2$. Clearly, they produce the interference that we are looking for. They process is simply because of unitary transformation applied to the one system in an entangled state which then becomes a unitary transformation on the other system.

Discussion: The nonlinear parametric down-conversion process which converts the incoming photon after the slit into a pair of entangled photons is a unitary process. What this means is that a pure state (which we have assumed for the photon after the slit) remains a pure state even though it changed. Therefore, it remains coherent. As a result it can maintain momentum and energy conservation that are required for such a parametric process. It is in fact these conservation properties that give rise to entanglement in the wave vectors and frequencies.

So when we select a single photon after the slit, which is then converted into two photons by the BBO, it contains the full information in the bi-photon wave functions that allows us, through the appropriate manipulation of one of the two photons, to produce the interference pattern as a single-photon wave function at $D_0$. However, since it is just one photon, we can only observe one dot. To see the complete interference pattern, we need to repeat the experiment to build up the statistics from which the interference pattern emerges.

That is the only reason for the ensemble. Every photon produces its own interference. (As Dirac phrased it: every photon only interferes with itself.) But to see this pattern we need lots of photons to build up the statistics.

  • $\begingroup$ Optically wave function collapses at the BBO. Superposition of two slit path become one slit path because photon now cannot take both paths through double slits as it is destroyed completely at BBO into an entangled pair. Now this is a non-reversible process. A single photon/ or single entangled pair is no longer in the superposition state of two slits. It is basically one slit diffraction pattern. As a result superposition states of two slits now can be only applied to the entire ensemble of photons through the double slit, but not for any single photon/entangled pair $\endgroup$ Jun 22 at 5:15
  • $\begingroup$ This is in stark contrast with a simple double slit experiment where we can say any individual photon, as well as the ensemble, is in a superposition of states through slit 1&slit 2. Although we need the ensemble to have an interference pattern. $\endgroup$ Jun 22 at 5:15
  • $\begingroup$ @DukeWilliam. You are incorrect that the wave collapses at the BBO crystal .... where did you hear that? If you make this incorrect assumption then you will always be unable to understand the nature of light especially "interference". $\endgroup$ Jun 22 at 14:13
  • $\begingroup$ @DukeWilliam: clearly the experimental results show that interference does happen. So, you must conclude that your reasoning must be flawed, right? The thing is that although the photon that passed through the slits is replaced by a pair of photons, the information is still contained in the state produced by the BBO. This can therefore be used to recover the interference. $\endgroup$ Jun 23 at 2:18
  • $\begingroup$ The thing is interference pattern we recovered is for the entire ensemble. It is a statistical and analytical pattern. It is a common misunderstanding that a single photon or pair retains its superposition states of two slit paths even after the SPDC at BBO. It is a very important fact that is lost under the unneeded retrocausal hype. Because most physicists focus on the wrong places in this specific experiment. $\endgroup$ Jun 23 at 2:55

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