# Why won't this information sending work?

1. Nov 5, 2012

### bob900

A large number of particles are entangled and then sent out as a stream in opposite directions to two far away points. At one point, a sender attempts to transmit a binary string bit by bit : for a 1 bit he measures the position of N (fixed constant) particles, to a high precision. For a 0 bit, he does nothing for these N particles (i.e. does not measure anything).

At the other point, a receiver measures the momentum of every N particles in the stream coming towards him, and notes the standard deviation (uncertainty). If the deviation in momentum is large, he knows the sender measured the position of the partner particle to a high precision (which collapsed the state to an eigenstate with a wide momentum spread), and so notes a 1. If the momentum deviation is not large, then he notes a 0.

Why won't this work?

Thanks,
Bob

2. Nov 5, 2012

### Drakkith

Staff Emeritus
That is not how the uncertainty principle works. One can measure the position and momentum of a particle with any precision you want, it is that you cannot prepare identical particles to have identical values of momentum and position upon measurement. Also entanglement only works for certain properties, like spin, polarization, etc. Not for momentum.

3. Nov 6, 2012

### bob900

The position/momentum uncertainty principle says that for particles prepared in an identical state ψ, the standard deviation of position measurements (σ_x) will be related to the standard deviation of momentum measurements (σ_p), by the following : σ_x * σ_p >= h/2

In my example, measuring the position of an entangled particle to a high precision at one end, will collapse the state vector of the two particle system to a position eigenstate ψ_p (i.e. where the both particles have a sharply defined position). By the uncertainty principle, measurements of momentum on position eigenstates (where σ_x is close to 0) will have a very large σ_p. Observing such a large σ_p for momentum measurements on the second particle, should convey information that the position of the first has been measured.

What do you mean? It sure seems like entanglement works for momentum position duality - in fact it was used in the EPR paper. The wiki on EPR says :

Incidentally, Bell used spin as his example, but many types of physical quantities—referred to as "observables" in quantum mechanics—can be used. The EPR paper used momentum for the observable. Experimental realisations of the EPR scenario often use photon polarization, because polarized photons are easy to prepare and measure.

4. Nov 6, 2012

### Bill_K

He can measure the position of each particle to a high precision, but that does not mean that the N values he gets will be close together. They may still have a large deviation. This is not something he has any control over, it comes from the initial conditions when the particles were prepared. The momenta of the other particles may consequently have a small deviation. Although the first observer thinks he is sending a 1, the second observer would record a 0.

5. Nov 6, 2012

### DrChinese

If I measure Alice's stream for position, I place Bob's stream into complete uncertainty and so Bob will see the momentum standard deviation respecting the HUP. If I DON'T measure Alice's stream for position, Bob's stream will remain as a random series of momentum values and so Bob will again see standard deviation respecting the HUP. If the momentum values in this case were not randomly distributed, then they could not be entangled.

You see the same thing either way. So no signal can be sent.

Another way to envision the result is as follows: you have ANY stream of particles in unknown states. You measure the position on the stream, so now the momentum is completely uncertain. This is the same result as above, really the same example as measuring Alice is effectively like measuring entangled partner Bob.

To eliminate any question about the point: momentum and position are conjugate partners i.e. they do not commute. They can be entangled, as can be many other observables.

6. Nov 6, 2012

### Drakkith

Staff Emeritus
Interesting. I had thought otherwise, my mistake. This is why I usually stay out of the QM forum!