# Is superluminal communication possible?

1. Sep 2, 2011

### bryanvann

I know the answer to this is most likely "No", and that there have been other threads on this topic before. But recently someone explained a technique for superluminal communication using entanglement and watching for wave form collapse to transmit data. I've always know this not to be possible, but I didn't know how to refute what he was describing. Any insight you folks could give on this particular case would be much appreciated. Note: I'm not a physicist, just a computer guy so please go easy on the math/notation. :)

Quick summary:
Use the act of measuring or not measuring one entangled particle to cause wave form collapse (or not) in the other entangled particle.

Proposed technique:
Alice wants to send data to Bob. They already have someone in the middle sending them bursts of entangled particles every second. They are also setup with a double-slit experiment so if neither of them observes which slit the electrons go through they should each see an interference pattern. When Alice starts measuring which slit the electrons are going through it causes waveform collapse and both stop seeing an interference pattern. Alice then turns the measurement apparatus on/off to transmit data to Bob superluminally.

I'm pretty sure there's something wrong with this approach but not sure what it is. It does rest on the premise that measuring one entangled particle causes a collapse in the other so that they either both see interference or not. Is this correct? If so, what else could be the problem?

For reference I've read this paper, which touched on some of the ideas refuting superluminal communication, but didn't specifically talk about using the technique of observation to cause collapse in the other particle:
http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

I also looked at the non-communication theorem, which didn't seem to apply directly either since it seemed more concerned with changing something about the particle rather than simply wave form collapse. In this case it seems pretty clear that Bob either sees the interference pattern or not.
http://en.wikipedia.org/wiki/No-communication_theorem

If you have any other references to point me to that would be great. Thanks!

2. Sep 2, 2011

### Ryan_m_b

Staff Emeritus
I'm no expert in this but as I understand it the problem is Bob has no way of knowing if his particles have collapsed until he measures them. At that point he has no way of knowing if he has caused their collapse or if Alice has. Take a look at this analogy;

- Alice and Bob both have black boxes
- In each box is a spinning coin
- There is a button on the top of each box, when pressed it opens the lid one second later
- Less than a second after the button is pressed both coins fall (one heads, the other tails).

So if Bob pushes his button and sees tails he has noway of knowing if it was spinning until he pressed the button or if Alice previously pressed her button.

P.S, Welcome to Physics Forums

3. Sep 2, 2011

### bryanvann

Hi Ryan,

Thanks for your reply. Isn't the difference though that in this experiment Bob's only measurement is by having the beam hit a card on the other side of the slit. He'd either see interference or not. Is it not the case that the act of Alice measuring which slit the electron went through causes Bob to not see an interference pattern?

The other way I was thinking about it is that Alice's measurement is before it goes through the slit, while Bob's measurement is after. So assuming Alice and Bob are equally far from the emitter, Alice's measurement would happen before Bob's and also before the electrons go through the slits on his side (i.e. by the time it goes through Bob's slit its already collapsed into a particle).

I know there's got to be something wrong with this reasoning, but can't put my finger on it.

4. Sep 2, 2011

### Ryan_m_b

Staff Emeritus
Annnnnnnnnnd now you're beyond my knowledge in the field :tongue2: hopefully someone can come along and enlighten us both.

5. Sep 2, 2011

### Cthugha

Sources that produce entangled photons are always emitting incoherent light. That means you will not see interference when you shine this light on a double slit. If you increase the coherence of that light, you automatically break entanglement. You will notice that in all experiments where you see interference patterns using entangled light, they are present in the coincidence counts only. This two-photon interference is a trademark of entangled states, while single-photon coherence as seen in a double slit is a trademark of coherent light. These two properties can even be shown to be complementary.

See A. F. Abouraddy et al., "Demonstration of the complementarity of one- and two-photon interference", Phys. Rev. A 63, 063803 (2001).

So in a nutshell the interference pattern will not disappear when you perform a measurement on the other photon because the pattern was never there.

6. Sep 2, 2011

### bryanvann

Ah, ok thanks for the explanation! That makes sense.