I recently read "Atomic Adventures" by James Mahaffey. In the book (Chapter 10) he mentioned a theoretical way of faster-than-light communication. My brain instantly searched for flaws in the proposed system but I couldn't find any, I'm sure there is a reason but I just haven't thought of it yet. It would be very helpful to me if one of you could help me especially because I couldn't find any real information on the proposed system. (And I'm only in High School so there could just be a gap in my knowledge that would explain why this isn't possible very easily)

Here a description (edited for clarity): A source of entangled photons is placed half-way (slightly closer to the Sender) between the two points you want to communicate. One of the entangled photons is sent to the sender and the other one to the receiver. Before the Photons leave the source they're passed through a Polarization filter (Vertical) but only if they're going to the sender. The Sender has a Polarization filter on his end that can be at a 0° (Vertical)(Not changing anything) or be at a 45° angle relative to the vertical polarization and erase the Vertical/Horizontal polarization. The receiver has a 90° (Horizontal) Polarization filter before a detector. Since the photons are entangled if the 45° polarisation filter is in place ca. 25% of photons should go through the filter and if it isn't 50% should. You can now assign the values a digital 1 and 0 and you have FTL communications.

Small addition: You wouldn't send a single photon but packets of (for example) 100. If the 45° filter is in place 25% of photons will make it through, if the 45° filter isn't in place 50% of the photons will make it through.

2nd addition: So I found a diagram of the System published with the audiobook on audible. However, I have concluded that the System as shown wouldn't work since the entanglement is destroyed at the first polarization filter. If anyone disagrees please answer below, otherwise, I will close my question in 24h.

The FTL Communication System

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    $\begingroup$ The fact that 50% or 25% of photons are absorbed by filters on one path has no bearing on the photons on the other path. There is no experiment the "receiver" can do which will tell him anything about individual photons going to the "sender". $\endgroup$ – Charles Francis Aug 20 '20 at 19:28
  • $\begingroup$ @CharlesFrancis but they are entangled. If you're familiar with the three Polarization filter experiment you will know that 50% of Photons will go through one filter. If there're two filters (One Vertical & one Horizontal) 0% of Photons will make it trough and if there is a third one at a 45° angle to Vertical 25% of the light will go through. If you read up on it I believe that that is, in fact, true (if you have entangled photons) regardless of whether or not its all done by one photon or one of the filters is passed by the other entangled photon. $\endgroup$ – Dragongames DEV Aug 20 '20 at 19:52
  • $\begingroup$ Nonetheless, there is no experiment the "receiver" can do which will tell him anything about what happens to individual photons going to the "sender" $\endgroup$ – Charles Francis Aug 20 '20 at 19:55
  • $\begingroup$ @DragongamesDEV You are confused about the 50%. If you are talking about unpolarized light going through a polarizer then yes that will be reduced by 50%. That is different than pairs of correlated photons going through a separate 45° polarizer. $\endgroup$ – Bill Alsept Aug 20 '20 at 21:33

Let’s say the initial entangled state is $\big(|HV\rangle+|VH\rangle\big)/\sqrt2$ where H means horizontally polarised and V means vertically, and the place value notation is implicit to first entry being sender and second entry being receiver.

When the sender’s side photon passes through the initial vertical polariser the state is no longer in superposition and is updated to $|VH\rangle$.

As you can see, regardless of the later polarisation setting of the sender, the receiver always receives a horizontally polarised photon.

  • $\begingroup$ But doesn't the 45° Polarization filter effectively destroy the |VH⟩ position? Doesn't it basically reset to the HV + VH superposition because the 45° polarization mode destroys the Vertical / Horizontal polarization? $\endgroup$ – Dragongames DEV Aug 20 '20 at 19:45
  • $\begingroup$ It does so only for the sender’s photon. Remember, after passing through the first polariser, the state is no longer an entangled state. $\endgroup$ – Superfast Jellyfish Aug 20 '20 at 19:52
  • $\begingroup$ Sure for the ones that got absorbed not. But the ones that passed trough are still entangled with the ones that didn't go through. And the other ones don't matter, do they? But when the Superposition is reset by the 45° filter 50% of the entangled ones will make it trough and since all none entangled ones are still Vertically polarized they're going to get absorbed. So if 50% of 50% make it trough that is 25%. If the 45% filter isn't in place they stay the Vertically polarized ones are going to stay that way and 50% of the photons will pass trough. $\endgroup$ – Dragongames DEV Aug 20 '20 at 20:11
  • $\begingroup$ Superposition isn’t reset by the second polariser. Once a photon passes through the V-polariser, the sender’s photon is fixed to be V and the receiver’s is fixed to be H. The entanglement is broken. Anything further done to the sender’s photon is totally uncorrelated to the receiver’s. So it matters not what the second polariser setting is (of the sender), the outcome for the receiver is always H $\endgroup$ – Superfast Jellyfish Aug 20 '20 at 20:30
  • $\begingroup$ I believe that to be consistent with the OP's intent, your initial state won't work. He assumes (for at least some pairs) that both of the photons make it through the first set of vertical polarizers. $\endgroup$ – WillO Aug 20 '20 at 21:42

Before the Photons leave the source they're passed through a Polarization filter (Vertical).

Now the pair is in the state VV, which is not entangled. If the Sender polarizes vertically, they remain in the state VV. If he polarizes at 45 degrees, they collapse to one of the two states (V+H)V or (V-H)V. Either way, nothing makes it through the horizontal polarizer.


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