About the closure: This Question is about why manipulating 1 particle of an entangled pair unitarily can’t cause measurable consequences on the other and how this manifests in the concretely described setup. The question this one got claimed as duplicate of and its answers not only don't contain any information about the setup described here, they also on a more abstract level do not even contain any mention of unitary time evolution and seem to generally apply to a completely different idea, namely measuring the state of 1 particle of an entangled pair and transmitting information to the other by collapsing the many particle state. I do not understand how this could be seen as a duplicate of the linked question.
Edit for clarity2: This proposal is very different from the classical EPR experiment in the aspect that measuring 1 particle to change the other is not part of this setup. Its all about unitary changes on 1 of the entangled particles, which lead to removal of the entanglement without measurement and thus measurable changes in the probability density of the other particle. I therefore changed the title, which previously contained “EPR”.
Edit for clarity: I am not asking whether faster then light communication is possible or not. From my information it is well established that it is not. However from my understanding the described setup would lead to instantaneous information transfer, so I am asking what exactly is wrong with my understanding of this setup as described in this question.
None of the other questions about FTL-information-transfer on this side, I have looked at, seem to address a sufficiently similar setup to answer this question. This includes the one suggested in the comments. Most of them just describe how doing a measurement on one of the entangled particles does not lead to observable consequences on the other, which is unrelated to my question as I am not trying to measure anything on the sender side.
So since this is not a duplicate I would kindly ask the people down voting this question to explain, what is wrong with it, so I can improve it, as right now I do not understand why you down vote.
To the actual question: For some reason, every time I think about the Einstein-Rosen-Podolsky-experiment my brain tries to construct ways for instantaneous information transfer. Usually it does not take me too long to see why they won't work, but in this case I am not sure.
So the basic idea is to instead of measuring one side to manipulate the other, we change one side unitarily to create a measurable effect on an interference pattern observed on the other side.
Lets first take a look at the interference measurement, that later is to be conducted with 1 of the EPR particles. For this I suggest to use a setup similar to a Mach-Zender-interferrometer, expect that the first beam splitter should be a polarizing one. We then add some additional components manipulating the phase and polarization on 1 path to ensure that for an incoming ray in a superposition of the 2 orthogonal polarization states we get the same behavior as in the Mach-Zender-experiment with one ray destructively annihilating.
Now my understanding would be that if we removed the polarization shifter P, the interference effect would vanish because then we have orthogonal Spin-states i.e.
$$||0\rangle|r_{D_1}\rangle – |1\rangle|r_{D_1}\rangle|^2 = ||0\rangle|r_{D_2}\rangle + |1\rangle|r_{D_2}\rangle|^2 = \langle0|0\rangle \langle r_{D_{1\setminus2}}|r_{D_{1\setminus2}}\rangle + \langle1|1\rangle \langle r_{D_{1\setminus2}}|r_{D_{1\setminus2}}\rangle \pm 2 \langle0|1\rangle \langle r_{D_{1\setminus2}}|r_{D_{1\setminus2}}\rangle = \langle0|0\rangle \langle r_{D_{1\setminus2}}|r_{D_{1\setminus2}}\rangle + \langle1|1\rangle \langle r_{D_{1\setminus2}}|r_{D_{1\setminus2}}\rangle := I $$
so equal intensity at both beam splitters, while with the polarization shifter we get
$$||0\rangle|r_{D_1}\rangle – |0\rangle|r_{D_1}\rangle|^2 = \langle0|0\rangle \langle r_{D_{1}}|r_{D_{1}}\rangle + \langle0|0\rangle \langle r_{D_{1}}|r_{D_{1}}\rangle – 2 \langle0|0\rangle \langle r_{D_{1}}|r_{D_{1}}\rangle = 0$$
at Detector 1 and
$$||0\rangle|r_{D_2}\rangle + |0\rangle|r_{D_2}\rangle|^2 = \langle0|0\rangle \langle r_{D_{2}}|r_{D_{1}}\rangle + \langle0|0\rangle \langle r_{D_{2}}|r_{D_{2}}\rangle + 2 \langle0|0\rangle \langle r_{D_{2}}|r_{D_{2}}\rangle = 2I$$
at Detector 2.
The idea is now to instead of the usage of the spin of the same particle to enable/disable the interference we use the state of a far away entangled particle to enable/disable the interference pattern. To do this we simply send 1 particle of our EPR pair into the above installation with the polarization shifter enabled, we will index this particle with $\uparrow$ and the other EPR-particle with $\downarrow$. So even though we shift the polarization for the upper EPR particle to have the same polarization in both paths of the upper apparatus, the state of the entangled other EPR particle will still be in different orthogonal states depending on the chosen way at the first polarizing beam splitter. Now if we do not manipulate the other particle I do expect the entanglement to destroy the interference pattern because:
$$||0\rangle_{\uparrow}|r_{D_1}\rangle_{\uparrow}|0\rangle_{\downarrow} – |0\rangle_{\uparrow}|r_{D_1}\rangle_{\uparrow}|1\rangle_{\downarrow}|^2 = \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle0|0\rangle_{\downarrow} + \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle1|1\rangle_{\downarrow} – 2 \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle1|0\rangle_{\downarrow} = \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle0|0\rangle_{\downarrow} + \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle1|1\rangle_{\downarrow} = I \neq 0$$
However if we manipulate the lower EPR particle in a way that independently of its initial polarization it ends up in the same state I would expect to see interference again.
$$||0\rangle_{\uparrow}|r_{D_1}\rangle_{\uparrow}|0\rangle_{\downarrow} – |0\rangle_{\uparrow}|r_{D_1}\rangle_{\uparrow}|0\rangle_{\downarrow}|^2 = \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle0|0\rangle_{\downarrow} + \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle0|0\rangle_{\downarrow} – 2 \langle0|0\rangle_{\uparrow} \langle r_{D_{1}}|r_{D_{1}}\rangle_{\uparrow} \langle0|0\rangle_{\downarrow} = 0$$
Now if this predicted behavior would be correct, it should be trivial to see how this could lead to faster then light information transfer. For each bit communicated one would of course have to make measurements on a whole bunch of EPR pairs to check if anything reaches detector 1 or not but that shouldn't be a problem, one can just exchange information about the planned procedure together with the EPR pairs, before separating them.
So the question is very simple: At what point exactly would the system behave differently then how I described it?
Best Answer
Main answer
Rephrasing the core idea
You would like to start with the state $$ |\Psi\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle - |11\rangle\right) $$ Then apply a unitary $U$ transformation to particle 1 such that \begin{eqnarray} U |0 x \rangle &=& |1 x\rangle \\ U |1 x \rangle &=& |1 x\rangle \end{eqnarray} where $x$ (describing the state of particle 2) can be either $0$ or $1$.
If you could find such a unitary, then by performing this operation on particle 1, you would find the state transformed to $$ U \Psi = \frac{1}{\sqrt{2}} |0\rangle \left(|0 \rangle - |1\rangle\right) $$ and then an observer looking at particle 2 would see an interference pattern. This would enable faster than light communication since the unitary could be applied at particle 1 at a spacelike distance from particle 2.
The problem
However, $U$ is not a unitary operator, since it is not one-to-one and thus cannot be inverted. So this is not a valid transformation of the quantum state (in the sense that it cannot be realized physically).
Another way to see this, is that if we form the density matrix $\rho=|\Psi\rangle\langle\Psi|$, and take the trace over particle 2 to get a reduced density matrix for particle 1, we get a mixed state (this is shown in detail under "Original answer"). A unitary matrix cannot transform a mixed state to a pure state.
More generally, given two state vectors $|A\rangle$ and $|B\rangle$, a unitary matrix cannot change the value of the inner product between them, $x = \langle A | B \rangle$. This follows from the definition of a unitary operator $U^\dagger U = 1$: $$ \langle A | B \rangle = \langle A | 1 | B \rangle = \langle A | U^\dagger U | B \rangle = \langle A' | B' \rangle $$ where $$ |A'\rangle = U |A \rangle, \ \ |B'\rangle = U | B\rangle $$ are the transformed state vectors.
Original answer
This is my original answer, which was based on misunderstanding the OP's idea, but I'm keeping it since it shows another interpretation of the general idea and a different way that would break down.
My attempt at rephrasing your idea
I believe the core of your idea is this:
If I could rephrase your idea in a simplified way (hopefully not overly simplified), then I believe this is essentially what you are saying.
Imagine a state like
$$ |\Psi\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle - |11\rangle\right) $$ Then, if we naively ignore the first qubit (corresponding to the "control" particle), the state of the second particle is a superposition $$ |\psi_2 \rangle = \frac{1}{\sqrt{2}} \left(|0\rangle - |1\rangle \right) \ \ [{\rm INCORRECT}] $$ Then if we send the second particle through an interferometer, we will see an interference effect due to the relative phases of $|0\rangle$ and $|1 \rangle$.
However, if we measure particle 1, and find it in state $|0\rangle$, then the full state collapses to $$ |\Psi\rangle = |00\rangle $$ Then the state of particle 2 is simply $|\psi_2\rangle=|0\rangle$, and there is no interference.
Therefore, you propose that by measuring particle 1, we can remove the interference pattern observed in an interferometer measuring particle 2, so we can know that particle 1 was measured "faster than light."
Where that breaks down
If I've understood you correctly, then the issue is that you haven't correctly understood what an observer will see for particle 2.
To understand the state of particle 2, it is best to use the formalism of a density matrix. The full density matrix for the whole system with particles 1 and 2 is $$ \rho_{12} = |\Psi\rangle \langle \Psi| = \frac{1}{2} \left(|00\rangle \langle 00| - |00\rangle \langle 11| - | 11 \rangle \langle 00 | + | 11 \rangle \langle 11 | \right) $$ Now the density matrix for an observer looking at particle 2 is obtained by tracing over particle 1 $$ \rho_2 = {\rm tr}_1 \rho_{12} = \frac{1}{2} \left(|0\rangle \langle 0 | + |1 \rangle \langle 1 |\right) $$ This is not a pure state -- it does not describe a superposition state where particle 2 is in a superposition of $0$ and $1$. Rather, it describes uncertainty in the state of particle 2. There is a 50% change particle 2 is in the $|0\rangle$ state, and a 50% chance it is in the $|1\rangle$ state. Crucially, neither the state $|0\rangle$ nor the state $|1\rangle$ will exhibit interference in an interferometer designed to look at the phase difference between the coefficients in a superposition of $|0\rangle$ and $|1\rangle$. So observer 2 does not observe an interference pattern. They simply will see some particles taking one path in the interferometer, and other particles taking the other path.
A measurement of particle 1 will not change this picture for observer 2. They will still see particle 2 either in state $0$ or state $1$. Only if the observers later compare the measurements of particles 1 and 2 can they see the outcomes of their measurements were correlated, but this comparison can only take place within the bounds of the speed of light.