Can scatter interference patterns from entangled pairs carry information?

I am having problems wrap around why you can't send information faster than light with a similar setup to the quantum eraser experiment. ( https://arxiv.org/abs/quant-ph/9903047v1 )

What I mean is, isn't the fact of whether I measured the photon or not, some sort of information? If it is not interfering, that means I've measured it, if it is, i didn't. Since this will happen instantaneously, I can send information by measuring or not within some fixed time-steps and then observing the pattern on the screen on the other side.

I am sure this is impossible and there are good reasons for it, but I don't know what they are. Thanks for explaining. Also, if that can be done without too much specialistic physics lingo, that would be great. I am an electronics engineer (hence the question) and the farthest I got in modern physics was special relativity. Thank you!

(Edit): I am not sure I asked this correctly, what I meant was, could the interference pattern itself or its absence be considered a hidden variable?

In my understanding, the paper focuses on proving that knowing which path is not a hidden variable (Bell's inequalities), but it doesn't say anything about the collapse of the wave function. Is a wave function collapse a hidden variable? Maybe paper doesn't explain this because it is a stupid question that makes no sense to people that know the experiment...

But if it does, then this is a time-machine!

• I can post the design for the simple "time machine" of sorts that you can do with this, if anyone is interested – user27221 May 9 '18 at 19:21
• The design of the "time machine" would be as such: many 2-slit diffraction patterns with entangled photons being measured for which path they take. On the measuring side, path is longer. The decision to measure or not is the message; with many experiments in parallel, the diffract pattern or not should appear by adding parallel results. The different in light paths size is how much in the future you measure. – user27221 Nov 13 '19 at 0:44
• Perhaps these answers answer your question a little? physics.stackexchange.com/questions/554586/… and a bit longer physics.stackexchange.com/questions/553309/… – HolgerFiedler May 31 at 4:38
• @HolgerFiedler I spent like 2 years thinking about this and I finally "understand" it. Actually it is more of "accept" it than anything. I get what everyone is saying but no one hit the nail in the head with this question. The answer to my question is NO. Scatter patterns only occur when you try to guess which is which and you need multiple measurements for that and then the faster than light travel thing is over. I found at some point some formulations of people describing how the math works out. Im glossing over a bunch of terminology I can't follow but I think I got the gist of it – user27221 Jul 26 at 17:29
• After a bit of research I also figured out (I think I did) exactly what would happen if I tried to do my experiment: I would see no interference pattern. As almost if my measurements changes nothing, like nature was saying "there is no need to send any information because nothing changes". I came to some exoteric conclusions, or interpretations, if you will, about what's happening. The first is "there is no need to send any information", which is equivalent to the pilot wave theory, I think. – user27221 Jul 26 at 17:37

Measuring the photon state does add information to your knowledge but it doesn't allow information to travel faster than the speed of light. For example, suppose you have an interstellar lottery with 2 ticket numbers only, 0 and 1. You found out the lottery is going to go to the ticket no. 0. Now, you want to convey this information to your friend, at say, vega star, that you need to buy the lottery with ticket number 0. No matter how you prepare any experiment with a way to communicate this 1 bit, you actually won't be able to. To see this more clearly, suppose you have an entangled pair of Bell state $$\frac{1}{\sqrt{2}}(\left|00\right> + \left|11\right>)$$ shared between you and your friend at vega. If you don't know about the Bell state, it just means that if you measured your qubit(or interference pattern) and it is 0, the wavefunction will collapse and your friend at vega will also get a 0 if he measures it. Similarly, the argument is the same for 1. Now, suppose you agree beforehand that whatever you measure, you should buy the ticket with that number. The problem here is that there is no way for you to force a measurement to be 0, which in turn would make the qubit at vega 0 instantaneously. All you can hope is a 50-50 chance of it being measured 0 or 1. So, you basically can't cheat in this interstellar lottery using quantum entanglement.