If it’s possible to use these phenomena to trasmit information (instantly or not) then couldn’t someone else use the same phenomena to trasmit different, wrong, info to the same place?
No, the protocol requires that Alice and Bob begin with the entangled pair B-C. Somebody else (call her Eve) can’t break in unless she can somehow entangle one of her particles (E, say) with C instead, but this requires access to either particle B or particle C.
No! Not at all! Your standard DSL modem, for one thing, doesn’t destroy anything on the sender’s end, so it can’t be “teleportation”.
Okay, maybe I oversimplified, but unless you want me to explain the entanglement of EPR states with the new qubit, some simplification is in order.
Okay, back from a weekend without internet access and I think someone needs to write up exactly what’s going on.
First you need a kind of particles with a property that takes two states. Spin of an electron, helicity of a photon, whatever. Call the states |1> and |0>.
Next you need an EPR pair, which is a pair of these particles in an entangled state, which is written 2[sup]-1/2[/sup](|10> - |01>). |10> means particle 1 is in state 1 and particle 2 is in state 0, and similarly for |01>. You take both of these states and put the particles in a superposition of them. Now neither particle has an independant value for the property, but is entangled with the other. Fly particle 2 to the other side of the galaxy.
Next introduce a third particle to be transfered. This is in the state (a|0> + b|1>), since any free particle is in a state of that form with |a|[sup]2[/sup] + |b|[sup]2[/sup] = 1. This is the value of the qubit to be teleported. When you add it to the EPR state (by interacting with particle 1) you get a total state for all three entangled particles of
(a|0> + b|1>)*2[sup]-1/2[/sup](|10> - |01>) = 2[sup]-1/2[/sup]a|010> - 2[sup]-1/2[/sup]a|001> + 2[sup]-1/2[/sup]b|110> - 2[sup]-1/2[/sup]b|101>
Now Alice (holding particle 1 and the original qubit) measures them. The outcome is random and gives one of four outcomes. The measurement also collapses the quantum state into one of four possible states. In each of these states, Alice’s particles are in the state she measured, while Bob’s is in some rotation of the original particle, but he can’t tell which. When Alice sends her two bits of information (which result came up) to Bob, they tell him what rotation to perform to recover the original state.
Unfortunately, I can’t recall the equations of this last part no matter how many times I read Sam’s book, which I didn’t think to bring with me to this side of the world. I’ll try to find the details and post later with them.
And immediately I find it.
The trick is that there are four “Bell pair states” such that the entangled state can be written as
1/2(|B[sub]0[/sub]>*(a|0>+b|1>)
- |B[sub]1[/sub]>*(-ia|0>+ib|1>)
- |B[sub]2[/sub]>*(b|0>-a|1>)
- |B[sub]3[/sub]>*(ib|0>+ia|1>))
So, Alice measures to see which B[sub]i[/sub] comes up, collapsing the state into one of the lines or the other. It should be clear that once Bob knows which Bell state came up, he can alter his particle’s state to recover (a|0> + b|1>).
Mathochist, you might be interested in learning about quantum stabilizer codes (if you don’t already know about them). The teleportation protocol can be described in stabilizer language more succinctly than by writing out the state vector explicitly, maybe making it look less like a coincidence of algebra. (I can describe it here briefly if you’d like. Good explanations of stabilizer codes are in most quantum-computing books (I like Nielsen&Chuang), or read Daniel Gottesman’s PhD thesis.)
Omphaloskeptic - thanks for the replies to my questions. Although the topic is going well beyond my knowledge, I’ll still keep an eye on this thread (although my guess is it’s gonna die a pretty quick death) 
Oh, I’ve seen it, but I figured that the basic algebra explanation would be more easily swallowed by the average reader. I like Lomonaco partly because I was at that conference and Sam’s one of my dissertation readers. 