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Re:You can’t get entangled without a wormhole by jblow on Saturday December 07, 2013 @08:07PM (#6868511)
You are misunderstanding the physics. Suppose A and B are two quantum entities that result from a collision. They are entangled and moving away from each other. The math very clearly says that a modification of A's state will alter B's state after the time of collision. In fact, it must do so, since A and B do not have separate states. Their states are the same one little chunk of math.

This even works if the states are known! This is how quantum cryptography works. You put two particles into Bell state with each other, give one to someone else, and then decide later what message you want to send.

I recommend learning the math. It is pretty simple if you already have linear algebra.

Re:You can’t get entangled without a wormhole by yk on Saturday December 07, 2013 @08:34PM (#6868592)
Well, what I am saying is, that the state of two entangled photons (with polarization +,-) is

    s_e= a |+-> + b |-+> 
and not

    s_n= a |+-> + b |-+> + c |--> + d |++>
as is the case for two arbitrary photons. ( See also my post script to the original post.)
Re:You can’t get entangled without a wormhole by jblow on Saturday December 07, 2013 @10:01PM (#6868793)
I don't know why you think that?

a|+-> + b|-+> is a very special-case entanglement. Yes, you'll see that one mentioned in Wikipedia (and quantum cryptography and etc) because it's very simple.

The general form for the states of two photons is exactly as you have listed for s_n. For some coefficient values of a, b, c, d, the photons are "entangled", for some, they are not "entangled". How do you know which is which? It is whether you can factor the polynomial into two separate expressions of the form (x|+> + y|->). If you can do this, the photons are not entangled; if you cannot do it, they are entangled.

Since most sets of (a, b, c, d) represent unfactorable expressions, you would expect almost any expression chosen at random to represent an entangled pair. Clean un-entanglement is the rare exception.

Re:You can’t get entangled without a wormhole by oisadiusahd on Saturday December 07, 2013 @11:46PM (#6868986)
Clean un-entanglement is the rare exception.

This is true only for pure-state entanglement (which itself is a rare exception ;-).