Edited question to face complaints about ambiguity
Following experimental setup.
Bob uses a nonlinear crystal to create two entangled photons A and B.
Consider a standard pair of EPR-entangled particles (correlated position and momentum values) with a joint wavefunction.
He sets up two double slits close to each other. I will refer to them as right and left double slit. (note that there are TWO double slits. Hence 4 slits total)
Photon A goes through the right double slit undisturbed.
Photon B runs through a long, coiled up, fiber optic cable, roughly 10 times the length of the distance to the moon, hence B will take about 10 seconds to come out at the other end.
Bob decides to either disturb the photon by measuring some of it's quantum states or not to disturb it, depending on a coin flip he does at 8 seconds into the experiment.
(we ignore that we would require repeaters for lengths higher than 200km to keep it simple)
Instead of Bob creating just 1 entangled photon pair, he creates a million entangled pairs and goes on to check two cases.
case 1) Timeline: Bob creates 1 million photon pairs. Bob checks the pattern on the screen behind the right slit, the 1 million non-delayed photons (An) created. At about 8 seconds, Bob flips a coin. The coin turns out to be HEADS.
Bob allows photon B to go through the left double slit undisturbed after coming out of the fiber optics cable at around 10 seconds.
–
case 2) Timeline: Bob creates 1 million entangled photon pairs. Bob checks the pattern on the screen behind the right slit, the 1 million non-delayed photons (An) created.
At about 8 seconds, Bob flips a coin which turns out to be TAILS.Bob disturbs the photon B by doing some measurement(setting up a unit-efficiency quantum detector directly behind one of the two slits of the left double-slit) such that it would allow the "which path" to be known when going through the double slit.
The question is. Will the pattern the 1 million entangled photons (An) created on the right screen in case 1), be discernible from the pattern in case 2), in the sense of being able to state with high probability which of the 2 cases applies, by analyzing the distribution of the photons (An), which hit the screen after passing through the right slit?
edit: Or otherwise stated. Would we be able to make a statement, at 0-1 seconds into the experiment, by observing the pattern behind the right slit the non-delayed photons have caused, in the form of "the experiment has a higher/lower probability to take the course of case 1 rather than case 2", beforehand?
If the answer to that question is yes, then doesn't Bob already know (with high probability) at the beginning of his experiment (0-1 seconds), what the result of his coin flip at 8 seconds will turn out to be?
Before answering the question, i would like you to describe the pattern we would see (behind the right slit created by the non-delayed photons) in case 1) vs case 2), if we were to actually perform this experiment.
This is not the same as the delayed choice quantum eraser. We are using a million photon pairs, which is a different scenario, as the objection of many was that we cannot tell the interference/no-interference pattern by just observing a single photon.
It's also different in the sense that we are just using two simple double slits, allowing us to not get lost in the details of different measuring devices.
edit: So i found a similar experiment proposed by John Hatter labeled "Superluminal information transmission using a double-double-slit apparatus"
In this experiment, the particles are entangled as follows: quote "Consider a standard pair of EPR-entangled particles (correlated position and momentum values) with a joint wavefunction"
He then uses a unit-efficiency quantum detector immediately behind A's left slit(B’s
left slit in my case) to extract the which path information and therefore destroy the interference pattern.
quoting part of the document
Now, of course, left to themselves, both detection screens should exhibit the
classic interference effect. Now however, take one of the double-slits (lets take
A) and put a small, unit-efficiency quantum detector immediately behind A’s
left slit. Let us be clear that the detector is directly behind A’s left slit and
therefore well in front of the screen. In fact, let the detector be so close to
the slit that if we ever get a hit, we can say with a high degree of certainty
which path the particle took. This should (of course) destroy the interference
pattern on the detection screen behind double-slit A since we now have whichpath
information for A. But consider our results derived above … since the
scattering potentials of A & B double-slits are conjugates, when the particles
encounter their respective double-slits, their paths will be correlated . In other
words if one particle uses the left-slit path of A, the other must use the left-slit
path of B…
(in my experimental setup B is the equivalent of A in the setup John Hatter proposed)
Conclusion:
It does not take much work to see how this could be used to perform superluminal
signalling. By allowing EPR-entangled particles to propagate in an
environment similar to that described above, and by sequentially destroying
and preserving the interference effect, information could be transmitted. Since
the interference effect is destroyed as soon as we gain knowledge of which-path
I hope that satisfies some as in defining the experimental setup more precisely.
Best Answer
Your question is muddled and unclear (in my professional opinion - you may think otherwise and that's OK), but I think I can make one thing clearer.
Note the keywords independent and local. If you look for correlations between the two experiments, then the situation can change. Thus, if you post-select the results from $A$ to include only the runs when $B$ performed experiment $B_1$ and obtained outcome $B_1(+)$, then the post-selected results from $A$ will depend, in general, on both the measurement and the outcome.
Having said this, I will continue to hammer on on what's been said before: your question is essentially undecidable unless you specify exactly what type of entanglement you imagine both photons to share. You seem to have a magical view of entanglement (which is alarmingly common) in which touching system $B$ will immediately and irrevocably change everything about system $A$, which is very, very far from the truth.
Allow me to elaborate by presenting two scenarios, which I believe are consistent with how you phrased your question:
Scenario 1
The photons are entangled in polarization, sharing the state $|HH\rangle+|VV\rangle$, but are otherwise indistinguishable.
In this case, nothing you do on $B$ will have any effect on the interference pattern (or lack thereof) shown in $A$, because the entanglement simply does not couple to the spatial modes.
Scenario 2
The photons are spatially entangled, in such a way that when photon $A$ hits its left slit then photon $B$ hits its right slit, and vice versa, i.e. they share the state $|LR\rangle+|RL\rangle$.
In this case, your actions on $B$, both your choice of measurement and the corresponding experimental outcome, can affect the results of $A$ if you postselect on the appropriate results on $B$, at least in principle. Other choices of measurement in $B$ will not have any effect on $A$. In particular:
If you perform a path measurement on $B$, where you detect which-way information for it, then (in this scenario) you also make available which-way information on $A$, which precludes it from showing an interference pattern. This is independent of the temporal sequence and spacing between the two measurements, and on whether the measurement in $B$ was randomized or not.
Suppose, on the other hand, that you do nothing with the photon in $B$ and allow it to run through the slits undisturbed. In this case (for this scenario), you will not observe any interference pattern on either photon. The reason for this is that each photon contains (in principle) which-way information on the other, so that the other one cannot interfere. The reduced state for both is $$\rho=\frac12\left(|L\rangle\langle L|+|R\rangle\langle R|\right),$$ and no coherent dynamics can be extracted from it. This is independent of the temporal sequence and spacing between the two measurements, and on whether the measurement in $B$ was randomized or not.
On the other hand, within this scenario there is indeed one measurement choice on $B$ which can restore the interference pattern on $A$, and it is of course the quantum eraser scheme. If you measure in the basis $\{|+\rangle,|-\rangle\}=\{|L\rangle+|R\rangle,|L\rangle-|R\rangle\}$, then the detector on $A$ will not turn up any interference pattern, but if you post-select and separate the counts on $A$ which coincided with $|+\rangle$ and $|-\rangle$ detections on $B$, then the blob will separate into two complementary interference patterns.
This behaviour is independent of the temporal sequence between the two measurements, their temporal separation, and whether the measurement in $B$ was randomized or not. If the measurement in $A$ is performed before the randomized decision to measure in $B$, then you have the delayed-choice quantum eraser experiment, which has been analyzed in detail elsewhere and which I will not attempt to clarify.
Note also that in the latter case, there is no measurement at $A$ that will enable you to predict the outcome in $B$. This is, again, exactly the same as with the delayed-choice quantum eraser, regardless of however many photons you send.
These two scenarios demonstrate that your question is ill-defined - the precise answer depends on information not contained in the question.
Some additional notes:
The measurement outcome is independent of however many photons you send through. Even in case 3 of scenario 2, if you try to send a million photons and observe the screen before you measure (or decide the measurement basis, or toss your coin, or whatever) on $B$, what you'll see in the screen in $A$ is... nothing, just an interferenceless blob. This is because each photon is an independent run of the experiment, and will have an independent result on the $|\pm\rangle$ basis, so if you add them up you are already performing a decision of what you'll do with the results from $B$.
The need for many photons to detect an interference pattern is a red herring; the "objection of many" is only raised by people with a hazy understanding of the quantum eraser experiment.
The goal is to detect coherence between $|L\rangle$ and $|R\rangle$ on the $A$ side, and this can be done without the need for a large number of photons, by being more clever about it. More specifically, you should collect the light behind each slit with an optical fiber and then shine them onto a beam splitter with detectors $M$ and $N$ on the output ports, calibrated so that $|L\rangle+|R\rangle$ will go exclusively into $M$ and $|L\rangle-|R\rangle$ will go exclusively into $N$. Thus a single count in $M$ rules out $|-\rangle$ and a single count in $N$ rules out $|+\rangle$.
This scheme is (provably) optimal. It will, of course, not give you enough information to completely determine the state from a single run of the experiment; this, however, is a fundamental limitation and the only way to get around it is via quantum state tomography.
This answer represents my best attempt in good faith to make sense of your question, which is a very tall proposition. If you feel I have not understood any aspect of the question, I would encourage you to think long and hard about how you are phrasing your post, instead of chucking more rep at it. Science depends crucially on its communication, and this includes making sure that one's audience can understand the material. If the entire audience says it is too obscure, blaming the audience will not benefit you at all; instead, try to clarify the material, and politely ask your audience what else you could do to make the point more evident. (If you want to, of course. Blaming the audience is lots of fun too.)