Is quantum entanglement fragile?

[kos]

What can cause disentanglement of the system. Could it happen even random?

[ajb]

Disentanglement occurs via interactions of the entangled system. This could be interactions with some background environment. In this sense it could be random.

[kos]

If you have composition from entangled particles and you can put some control for the first part of them does it mean that you auomaticly can control the behavior of the second part?

[ajb]

I am worried that control would break the entanglement. I am not an expert in quantum information theory and similar, so I am not really sure what one could exactly mean by control etc.

[Klaynos]

No. You can not control anything via entanglement.

[swansont]

Klaynos is right. Any sort of manipulation that determines the state of the particle will destroy the entanglement. There are evolutions that don't — you could do a spin-flip or polarization rotation to your particle(s) (a local unitary transformation). That would change the nature of the entanglement (e.g. from aligned spins to anti-aligned spins) but not the entanglement itself. The key being that at no point in this do you ever determine the spin state.

[Strange]

So Einstein only scores 1 out of 3 for his "spooky action at a distance" (not spooky and not an action, but it is at a distance).

[swansont]

So Einstein only scores 1 out of 3 for his "spooky action at a distance" (not spooky and not an action, but it is at a distance).

 

 

Still spooky, since whatever's going on is nonlocal.

[kos]

So first I gonna show you this article http://www.gizmag.com/quantum-entanglement-nuclei-university-chicago-argonne/40884/ . The natural question arising in my mind is the imaginable situation

With two objects. Both with the same number of composing particles and the same structure. Both in perfect entanglement between their corresponding internal structures. So we move them appart we exegerate control over one and the second object behaves the same way. It coppy the movement of fist object wich movevemnt we control . I imagine some sort of puppet-playing . Comment it ,please !

[Klaynos]

That's not how entanglement works. Read the replies above. You cannot exert control all you can do is collapse the wavefunction, the knowledge of doing this can only be transferred at the speed of light. Causality is not broken.

[kos]

Let's say we have two identical pawns . The coresponding composing particles of both are in perfect entanglement. We play chess on the board and move the first on F2 and on the other side of the world the second one pawn wich is part from another game moves for its own as the player dont know the connection between them and he is shocked !

[Strange]

That is not how entanglement works.

[kos]

The light speed is not the key point in my question. Let it be within the time needed for light speed reach the other pawn .

[Strange]

It still isn't how entanglement works. All it means is that there is a correlation between the properties you measure on the two objects. Doing something to one of them doesn't have any effect on the other.

[Klaynos]

It is a very common misconception. But it is a misconception.

[kos]

So if I move the first pawn that will not affect the second pawn since they are entangled

If I determine the spin of first groups s of particles it will determine automaticly the second part wich will corespond to some causal chain reaction wich will unlock some action that the object will do

So if I move the first pawn that will not affect the second pawn since they are entangledIf I determine the spin of first groups s of particles it will determine automaticly the second part wich will corespond to some causal chain reaction wich will unlock some action that the object will do

wich of two assumptions is more correct

[Strange]

Someone measuring the second set of particles will not (by themselves) see any difference.

 

Imagine you have several particles and each one is entangled with a particle that your friend on the other side of the world has. Now each of you are going to measure the spins of the particles. Before you do this, neither of you know what the spin of each particle is. When you measure the first particle, you find it is spin up, for example. Now you know that when your friend makes her measurement, she will get spin down. But apart from that, nothing else changes. She will see no change at her end (because she doesn't know what the sin is yet). She won't know which particle you measured or what result you got. But as soon as she makes a measurement she will now what values you will get (or have already got).

 

That's about it.

 

Afterwards, you can compare notes and find that the time between your measurements was less than the time it takes light to travel between you.

[Delta1212]

Let's say you have your two particles A and B.

 

If you mwah see the spin of A, the spin of B is also collapsed and is determinable from your measurement of spin A. There is no way to know that the wave function has collapsed from B's end, however. You can measure B, but that will immedoatelt collapse the wave function of both A and B, anyway. So if A has been measured, you can measure B and see that it has a definite spin (because A was measured and the wave function collapsed) or you can measure B and see that it has a definite spin (because you measured it).

 

There's no way, from B's end, to detect the moment that the wave function collapses because someone measured A. You could send a signal to B after measuring A letting the system there know "Hey, the wave function collapsed because I measured A, move the pawn" but that that point you could just set up a transmitter to say "Hey, move the pawn" without bothering with entanglement.

[swansont]

So if I move the first pawn that will not affect the second pawn since they are entangled

 

 

 

if their position is entangled, then you don't know where the pawns are. You can't talk about moving them at all, since you don't know where they are to begin with. Once you determine where the first pawn is you know where the second one is, but at that point they are no longer entangled.

[kos]

So if we do some change on B via A it will be just once because this change will destroy entenglement for next event

[Strange]

So if we do some change on B via A it will be just once because this change will destroy entenglement for next event

 

Yes. Also, there is no way of knowing that the change has occurred.

[kos]

If something happen due to entanglement it wkukd be only once

[Strange]

If something happen due to entanglement it wkukd be only once

 

Yes. Also, there is no way of knowing that something happened.

[Klaynos]

The person looking at b would not know the observation of a had been made.

[kos]

Is my lookung enought to dis terb the entanglement?