# Entanglement to speak faster than light

1. Oct 18, 2013

### friend

Suppose you construct two beams of entangled particles that travel in opposite directions at or near light speed. At two very distant places where each beam is aimed, you have a double slit experiment at the end of each beam. Measuring just one of the entangled particles will collapse the wavefunction for both so that knowing the properties of one particle tells you the property of the other particle.

So if covering one of the slits at the end of one beam will collapse the wavefunction at both ends, then will the double slit measurement at the other end show a fringe patern or not even though both slits are open at the other end? If so, then wouldn't this be a means of communicating instantaneously by chosing to cover one of the slits or not at one end?

2. Oct 18, 2013

### craigi

No pattern.

Entangled particles are only coherent, beyond a time horizon where their opposing particle has its path information erased, sufficiently to prevent any such pattern forming.

That's the spookiest thing I've seen in QM.

Last edited: Oct 18, 2013
3. Oct 18, 2013

### glappkaeft

I don't know if your particular method of measuring entanglement works works but there are ways to do it. Say your method works. How do you use it to send a message?

Last edited: Oct 18, 2013
4. Oct 18, 2013

### meBigGuy

I don't understand why you think covering the slit at one end causes decoherence at the other end. At one end the particle goes through 1 slit with no interference pattern. At the other end the partner particle still goes through two slits and interferes. Whether or not there is interference at one end has no effect on the other particle.

5. Oct 18, 2013

### craigi

Presumably the arrangement is such that the momenta are correlated so that if a particle is detected at one slit, it must go through the opposing slit in the other arm of the experiment.

As I've already mentioned, the reason that it doesn't work is that if the path information is available in one particle, the opposing particle isn't subject to self interference.

The path information can be erased by focussing the paths of one particle then interference can emerge in the other.

My guess would be that if you sufficienctly defocus an entangled particle then it can self interfere too. Which would give rise to the situation that you describe. At this point knowing which slit the opposing particle goes through becomes irrelevant to the interference pattern.

Last edited: Oct 18, 2013
6. Oct 19, 2013

### meBigGuy

I'm confused (which is nothing new, BTW, it's one of my hobbies). Maybe we are talking past each other. Let's start with a simple spin entangled system. In a simple entangled two particle system if I measure one particle, it is no longer entangled with the other particle and the other particle simply behaves as a particle with the properly correlated state (keeping it simple and excluding entanglement swapping scenarios). The measurement has no effect on the other particle's ability to self-interfere. The first particle could have gone through 1 slit, 2 slits (with self interference) or no slits, before measurement, it makes no difference. The second particle will then behave as an unentangled particle and can self interfere in a 2 slit experiment.

Maybe that is in no way related to the problem at hand.

You seem to be talking about producing particles that are somehow "path correlated". I've never heard of such a thing (but I'll be happy to learn).

7. Oct 19, 2013

### USeptim

This experiment has been test with the spin, that can only have some discrete values. But saying that if you know the path the particle has taken, you will know its momenta and therefore the others particle momenta it's another thing. The particle trayectory may suffer many fluctuations during the travel that make that its possition is not the same that the corresponding one for its momenta.

8. Oct 19, 2013

### craigi

How we resolve this new scenario really depends upon the cause of these fluctuations.

If these fluctuations are caused by things like reflection and lensing effects then the answer is the same.

Alternatively, if they are sufficient to be considered measurements then the way we resolve this depends upon where the fluctuation takes place. If it is before the slits then they become disentangled, then if there is sufficient uncertaintly on momenta, the 2 particles can go on to produce uncorrelated interference patterns. If instead it occurs after the slits then no self interferece is observed.

Last edited: Oct 19, 2013
9. Oct 19, 2013

### craigi

Due to conservation of momentum when an event produces an entangled pair their paths would be correlated. There may be uncertainty on their momenta, which is what allows their wave functions to pass through both slits, but that is not to say their momenta aren't correlated.

10. Oct 19, 2013

### friend

Actually, I'm trying to understand wavefunctions, and wavefunction collapse, when the wavefunction involves two particles that are entangled. It's my understanding that entangled particles involve only one wavefunction and that measuring just one of those particles collapses the entire wavefunction. Can a wavefunction that has collapsed self-interfere?

11. Oct 19, 2013

### craigi

Collapse of the wavefunction, means that the particle has been measured. If the observable is position, then the particle is found at a specific position and is no longer in a superposition of states. After that, the wavefunction spreads out again, spatially, due to the inherent uncertainty in the particle's momentum due to the measurement.

For self-interference to take place in the double slit experiement you require superpostion of positional states.

12. Oct 19, 2013

### Staff: Mentor

The wave function collapses to an eigenfunction of whatever observable you've measured; that will usually still be a superposition in some other basis.

Whether the post-collapse wave function spreads out again depends on whether it's an eigenfunction of the Hamiltonian, which is equivalent to asking whether the operator corresponding to the measured quantity commutes with the Hamiltonian. Position doesn't, so the wave function in the position basis does spread out after a measurement.

Free-particle position and momentum can be especially confusing in collapse interpretations because the eigenvalue spectrum is continuous so that measurement doesn't collapse the wave function down to a single eigenstate - instead we get a more or less tight range of values between $x$ and $x+\Delta{x}$.

13. Oct 19, 2013

### DrChinese

Entangled particles do not create an interference pattern. For them to do so, they must first be locally manipulated in a manner that will cause them to be no longer entangled.

14. Oct 19, 2013

### EskWIRED

So then do they form patterns indicative of particles? If one were to stream them at a double slit, would they form two "particle-like" patterns side by side?

15. Oct 19, 2013

### DrChinese

16. Oct 19, 2013

### craigi

My description of the reduction of positional states upon measurement, was an over-simplification, but the effect on emergence of the interference pattern is the same.

Is there a derivation of the Uncertainty Principle in terms of the eigenfunction of the Hamiltonian? We are talking about the same effect here, right?

Last edited: Oct 19, 2013