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!
 A: 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. 
