"Quantum Entanglement Could Stretch Across Time"

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http://www.wired.com/wiredscience/2011/ ... anglement/

That is truly fascinating. Imagine the breakthroughs in Quantum Computing that we have the fortune of being alive to see

Even with entanglement FTL data transmission is not possible. All entanglement can do is change one random piece of crap into another random piece of crap instantly. In order to transmit data one must correlate both ends of the transmission to find out what the message really was.

ruveyn

I hope I live long enough to see machine with 128 q bits operate. I do not plan on holding my breath until then.

ruveyn

True, but even at one bit at a time, that it happens instantly regardless of distance means you have the means of some form of instant transmission over vast distances.

Better than current methods with equally huge limitations.

Can binary be transmitted using Quantum Entanglement?

No. Even with entanglement a FTL Morse telegraph lamp is not going to work. In order to determine what is sent both sides must be correlated to determine what the message was. This comparison of the two ends can only be done at less than the speed of light.

ruveyn

Forgive me if I'm completely wrong, but as far as I know the principle is based on interlinked Spin-Up, Spin-Down states. If both observers take "Spin-Up = 1, down = 0", and wonder off, could one observer then transmit information instantaneously?

In entanglement experiments all that each side can see one glob of random crap turn into another glob of random crap. All globs of random crap tent to look alike.


Have a look at this:
"The General Case

This is the section for people who know some quantum mechanics. This would probably be clearer in terms of the density matrix, but I didn't want to limit my audience further. Here I will consider a system consisting of two entangled subsystems, because I believe it will be clearer, but I think the explanation is only trivially different when dealing with a system comprised of N entangled subsystems. We will work in the interaction picture here, meaning effectively that we will ignore the free evolution of the system and only worry about additional evolution introduced by the hypothetical communication process.

We consider, then, a system comprised of two subsystems A and B. There are two observers, who we will again call Alice and Cooper. Alice can only interact with subsystem A and can only measure subsystem A. Likewise Cooper can only interact with and measure subsystem B. I will show that no matter what interactions with A or measurements of A Alice makes, the measurements of B by Cooper will be unaffected.

When we say Cooper can only measure B this means simply that if subsystems A and B are in a separable (unentangled) state, then the value of a measurement by Cooper will be independent of the state of A. This implies the operator representing the dynamical quantity being measured must have the form

O2 = 1A*O2B

possibly apart from some multiplicative factor. Thus, if the system is the separable state

|ψ> = |φ>A*|θ>B

then the expectation value of the measurement is

<O2>ψ = <ψ|O2|ψ> = <θ|O2B|θ>B

satisfying our requirement that it is independent of the state of A. This tells us that Cooper is measuring B only. If this criterion were not met then he would not even need entanglement to transmit information.

We also require that Alice may only introduce interactions with subsystem A, meaning that she may only introduce evolution of the system with an operator of the form

U1 = U1A*1B

This is the class of evolutions that only effect subsystem A in a separable state. Likewise, Cooper can introduce interactions on subsystem B.

Once we have defined what it means for each observer to interact with and measure only his own subsystem, now we can examine whether they may use entanglement to transmit information. We now assume that the system begins in an arbitrary state, which may have any sort of entanglement.

|ψ> = Σj,k ajk|φj>A*|θk>B

We represent the state in terms of an orthonormal basis of eigenstates of the observables that Alice and Cooper will measure. First we need to show that any measurement on A will not effect the expectation value of a measurement on B. Ok, well, first we find an expression for the expectation value of Cooper's measurement of B for the general state.

<O2>ψ = Σj',k' Σj,k conj(aj'k')ajk<φj'|φj>A<θk'|O2B|θk>B_ = Σj,k,k'conj(ajk')ajk<θk'|O2B|θk>B

If Alice measures the system A and finds it in the state |φl> then the total state of the system will be

|ψ>m(l) = 1/sqrt(P(l)) Σk alk|φl>A*|θk>B

where P(l) is the probability that the system will be found in the state |φl>. For this measured state, the expectation value of Cooper's measurement on B will be

<O2>ψm(l) = <ψ|O2|ψ>m(l) = 1/P(l) Σk,k' conj(alk')alk<φl|φl>A<θk'|O2B|θk>B = 1/P(l) Σk,k' conj(alk')alk<θk'|O2B|θk>B

Now this isn't the same as what we had for the general state. That doesn't look good, right? Except, remember we're talking about just one measurement. We found the expectation value for a measurement of B provided that A is in the state |ψl> every time, but that won't happen, because as we discussed, Alice can't control what result she gets. Cooper can't tell anything from just one measurement (other than that the result is possible, which doesn't depend on Alice's measurement). He has to perform many measurements and look at the distribution before he can tell if Alice has done anything. If Alice and Cooper perform this process many times, Alice will get a different result each time with probability P(l). So, then the actual expectation value Cooper would find would be

<O2>ψm = Σl P(l)<O2>ψm(l) = Σl,k,k' conj(alk')alk<θk'|O2B|θk>B

which is exactly what he would have found for the unmeasured state. Thus, Cooper cannot tell that Alice made any measurement.

Now you might get trickier and ask, "Well, what if Alice and Cooper also evolve their states as in quantum teleportation?" It turns out that doesn't help either. We can suppose that each observer performs a local evolution on his system before and after the measurement of A. The evolution could just be the identity, so we've also included the cases where one or both don't evolve their respective systems. Each performs a local evolution on his system at the outset, then the state of the total system becomes

|ψ'> = Σj,k ajkUA1|φj>A*UB1|θk>B

which gives an expectation value for the measurement of B

<O2>ψ' = Σj',k' Σj,k conj(aj'k')ajk<φj'|adj(UA1)UA1|φj>A<θk'|adj(UB1)O2BUB1|θk>B

but

<φj'|adj(UA1)UA1|φj> = <φj'|φj>=δjj'

so

= Σj,k',k conj(ajk')ajk<θk'|adj(UB1)O2BUB1|θk>B

For simplicity, we may now write the state after the first evolution in our general form

|ψ'> = Σj,k bjk|φj>A*|θk>B

so the expectation value for measurement of B before the any measurement of A is

<O2>ψ' = Σj,k,k'conj(bjk')bjk<θk'|O2B|θk>B

Now, we suppose Alice measures A and then both Alice and Cooper perform local evolutions on their systems. The measured and evolved state is then

U2|ψ'>m(l) = 1/sqrt(P(l)) Σk blkUA2|φl>A*UB2|θk>B

Again, we must take the expectation value for each possible measured state and add them all up with their probabilities. This yields

<O2>m = Σl P(l) <O2>m(l) = Σl,k',k conj(blk')blk<φl|adj(UA2)UA2|φl>A<θk'|adj(UB2)O2BUB2|θk>B
= Σl,k',k conj(blk')blk<θk'|adj(UB2)O2BUB2|θk>B

If we consider what the expectation value would have been if we started with |ψ'> and only Cooper applied his evolution operator, we would have

|ψ''> = Σj,k bjk|φj>A*UB2|θk>B

<O2>ψ'' = Σj,k',k conj(bjk')bjk<θk'|adj(UB2)O2BUB2|θk>B

identical to the value with the measurements and evolution on A. So there you have it. No matter what Alice and Cooper do here, they cannot use the entanglement alone to transmit information. Now if Alice sends Cooper the result of her measurement via a classical channel and then Cooper chooses his evolution based on that result, they could teleport a quantum state, but then they would be constrained by the speed of their classical signal, which should presumably be subluminal."

ruveyn

I won't pretend to understand even 1/4 of that, but if you had a pair of devices sharing an entangled pair (one-half in each device), could not two people go to vastly different and distant locations and use the entanglement to transmit information to one another even at a very limited bandwidth?

Certainly in such a device, the value of 1 and 0 would be pre-determined via the construction of the device, right?

I don't think anyone is going to argue with your logic Ruveyn!

... I've just read your post and... anyone know a cure for a brain haemorrhage?

No you can't. You need to of studied quantum mechanics to understand what Ruveyn wrote but basically, it was a mathematical derivation showing that this can't be done. Let's say that that we have two people Alice and Cooper (I'm using the same two names Ruveyn did). Alice has in her possession, a system of particles and Cooper has in his possession a system of particles, where Alice's system of particles is entangled to Cooper's system of particles in the sense that for every particle in Alice's system that spin-up, for example, the corresponding particle in Cooper's system is spin-down and vica-versa. Now if Alice decides to make a number of measurements to her system to determine which of her particles are spin-up and spin-down, she will know instantly which states Cooper's corresponding particles are in. However, Alice cannot control which states (spin-up or spin-down) any of her particles turn out to be in when she does her measurement. Therefore her system of particles would just be a random mix of spin-up and spin-down states and so would Cooper's. You can only notice the correlation when you compare the results of the measurements because both systems are random, thus no message is transmitted from one to the other. That's what Ruveyn meant when he said "it just changes one piece of random crap into another piece of random crap". That's basically what the first part of the derivation in Ruveyn's post says.

In the second part, it shows what would happen if Alice tries to send a message to Cooper by first measuring her system and changing it to form a message. The reason why this doesn't work is because any tampering Alice does with her system of particle would destroy the entanglement and thus doesn't effect the outcome of Cooper's measurement on his system. Either way, Alice cannot use the entanglement to send a superluminal message to Cooper whatsoever.

Got you. It's just random crap, and it's inverse, and any alteration breaks the entanglement. That's a pity

Quantum entanglement isn't useless. It's just not useful for transmitting superluminal messages. For example, electron tunneling microscopes wouldn't be possible without quantum entanglement. Other technologies use it as well.

Agreed. Quantum Entanglement is required for Quantum Teleportation (not quite what it sounds like), and for Quantum Key Distribution (necessary for the common sorts of Quantum Cryptography).