Entanglement: Is it possible to prepare and reset probabilities to send information? [duplicate]

Question

I'm pretty certain that the answer to the question in the title is a no, but I don't understand why. I have some basic misunderstanding of quantum processes that I’d like clarified in the form of asking questions about a hypothetical scenario.

So, let’s say we have a pair of entangled photons…

1) Is it possible to prepare an entangled pair such that it can be known that, for example, the photon going down a certain path will be measured 25% of the time to be horizontally polarized and 75% likely to be vertically polarized? I guess this would be some sort of quantum decoherence? Partial collapse?

2) If that is possible, then would it also be possible to measure that photon’s diagonal polarization in order to "reset" the horizontal-vertical probability back to 50-50? I've seen a trick done where you take some polarizing filters and stack them horizontal on top of vertical, so that no light gets through. But when you slide a diagonally oriented filter between them, some light does get through, because the horizontal/vertical polarization information gets "reset" when measured against the diagonal filter.

3) If these two things are possible, then wouldn’t it be possible to send information along the entangled pair? If you sent bursts of 1000 photons and were able to measure them all, and found that the first burst had ~750 vertically polarized photons but the second burst had ~500 vertically polarized photons, then wouldn’t you know that the second burst had their diagonal measurements taken? Or would that not work because the extra measurements "break" the entanglement..?

4) Along with that, how many times can the photon pairs be measured before there are no longer entangled? Just once? Or maybe just once per state/qubit/aspect (such as polarization angle)? I understand that measuring the photon would necessarily change it (like with the polarization filter trick), but then how do scientists measure the polarization angles of two photons like in experiments of Bell’s inequality?

I know that somewhere along the line, what I've described cannot be possible. If it were, then you could easily send information faster than light. E.g., say you had a photon-entanglement-generator-and-probability-fixer station in between two planets, 1 light year from the planet Foo and .99 light years from the planet Bar. After a year's time, after the photon stream has propagated to both planets, scientists on Bar could measure diagonally polarization of 1000 photons to encode a 1 or not measure to encode a 0. Then scientists on the planet Foo could measure vertically polarity and decode the 1s and 0s based on the statistical readings. The Bar scientists could encode that such a message and the Foo scientists would "instantly" receive the message. Except it would actually be from the past, since Foo is 1.99 years outside the Bar light cone. So since this is impossible, what am I misunderstanding?

Thanks for answering, and thanks for putting up with yet another clueless but curious layman.

Answer 1

• 1. Without remarks as to how you would best prepare this state experimentally with photons, a pure entangled state of the form $$ |\phi\rangle \;=\; \tfrac{1}{2} \Bigl( |H\rangle_L |V\rangle_R \;-\; \sqrt3\, |V\rangle_L |H\rangle_R \Bigr) $$ describes the polarization situation you describe. (To consider this as an abstract two-qubit state, replace "H" and "V" with "0" and "1".) This is not a "maximally" entangled state, but as it doesn't factor into independent states of the two photons, it is nevertheless entangled.

• 2 & 3. Yes, filtering the photon in one of the beams (e.g the left-hand beam) gives rise to a photon is a uniform superposition of horizontal and vertical. However, there are no useful correlations between this "reset" photon, and the photon in the right-hand beam. Let's consider a filter for the polarization along the angle +45°, which is a projective measurement onto the basis $|\mathbin\nearrow\rangle, |\mathbin\searrow\rangle$. We have $$ |H\rangle = \tfrac{1}{\sqrt 2} \Bigl(|\mathbin\nearrow\rangle + |\mathbin\searrow\rangle\Bigr), \qquad |V\rangle = \tfrac{1}{\sqrt 2} \Bigl(|\mathbin\nearrow\rangle - |\mathbin\searrow\rangle\Bigr),$$ so that projecting the left-hand beam of the state $|\phi\rangle$ onto the state $|\mathbin\nearrow\rangle$ yields the (un-normalised) state $$\begin{align*} |\psi\rangle \;&=\; \tfrac{1}{2} \Bigl( \tfrac{1}{\sqrt 2}|\mathbin\nearrow\rangle_L |V\rangle_R \;-\; \tfrac{\sqrt3}{\sqrt 2}\, |\mathbin\nearrow\rangle_L |H\rangle_R \Bigr) \\&=\;|\mathbin\nearrow\rangle_L\;\otimes\;\tfrac{1}{2\sqrt 2}\Bigr(|V\rangle_R - \sqrt 3 |H\rangle_R\Bigr) \end{align*}$$ so that the left and right beams now share no entanglement whatsoever. Whatever further operations and observations you perform on the left, there will be no correlation that is observable on the right. The photons in the left and right beams have no enduring sort of connection or non-local interaction: their initial states were merely correlated — and as soon as you performed the measurement, you destroyed even that correlation which there was.
  
  Note that if you had instead filtered for the other polarization, you would instead obtain a state $$\begin{align*} |\psi'\rangle \;&=\; \tfrac{1}{2} \Bigl( \tfrac{1}{\sqrt 2}|\mathbin\searrow\rangle_L |V\rangle_R \;+\; \tfrac{\sqrt3}{\sqrt 2}\, |\mathbin\searrow\rangle_L |H\rangle_R \Bigr) \\&=\;|\mathbin\searrow\rangle_L\;\otimes\;\tfrac{1}{2\sqrt 2}\Bigr(|V\rangle_R + \sqrt 3 |H\rangle_R\Bigr) ; \end{align*}$$ notice that the state of the photon in the right-hand beam is different than if we selected $|\mathbin\nearrow\rangle$. That's because the entangled state $|\phi\rangle$ shows correlations not just in the H/V basis, but in every basis of measurement. While the person measuring the left-hand beam gets no information what someone on the right might measure in the H/V basis — they know that it will be biased towards H, but their measurement outcome of $|\mathbin\nearrow\rangle$ or $\mathbin\searrow\rangle$ will give them no further information — they do know what state the photon in the right-hand beam has collapsed to. That is in fact the defining characteristic of entanglement, as opposed to classical randomness: the correlation extends to multiple bases of measurement.

• 4. Each independent degree of freedom can sustain correlations with other degrees of freedom, on the same particle or other particles, in principle. In this case, all of the entanglement that the system had, aside from correlated uncertainties in their momenta (which you implicitly perform a weak measurement on, just by virtue of making a polarization measurement at a particular location after it was emitted), are in the polarizations. As the polarization of each beam is entirely involved in what entanglement there is (in a multi-particle scenario, you can consider entanglement spread across the system in a way that some particles are more involved in the entangled state than others), any single measurement will destroy all entanglement present.

Answer 2

Niel's answer shows in particular that your scheme does not work. But in fact you can show that as long as you cannot control the outcome of the measurement on Alice's side, then Bob cannot learn anything about Alice's choice of measurement. The only thing he can learn is what outcome Alice has obtained or will obtain, but that is random and outside her control.

To see that what I am saying is correct suppose we have some set of measurement operators $M_k \otimes 1$, where the first part of the Hilbert space is on Alice's side and the second part is on Bob's (and 1 is the identity operator on Bob's system). The measurement operators tell you what is the state of the system after doing the measurement. Now provided measurement outcomes cannot be controlled, then the average state after Alice has conducted her measurement is $$ \sum_k (M_k \otimes 1) \rho (M_k \otimes 1)^\dagger. $$ Bob's part of the state then is $$ \sum_k Tr_A\left[ (M_k^\dagger M_k \otimes 1) \rho\right] $$ due to the cyclic property of the trace and since $\sum_k M_k^\dagger M_k = 1$ (probabilities sum to one), we have $$ \sum_k Tr_A\left[ (M_k^\dagger M_k \otimes 1) \rho\right] = Tr_A\left[\rho\right] $$ So as far as Bob is concerned, the fact that Alice has conducted the measurement changed nothing at all about his measurement statistics.

The situation would be different if Alice could control the outcome, as then the sum would not need to be taken and the operators would not sum to 1. Quantum mechanics predicts this is not possible and the current experimental evidence seems to support this view. But it could be that one day experimental evidence will falsify quantum mechanics and FTL communication will be possible. Until then, entanglement is not useful for FTL communication.