Quantum Teleportation

No matter is transported, no faster-than-light signal is sent, and nothing about quantum teleportation breaks relativity — despite the name. What actually happens is more interesting: Alice can transmit a qubit's exact state to Bob using only a shared entangled pair, two classical bits sent over an ordinary channel, and zero direct quantum transmission of the state itself.

What you need

• An unknown state |φ⟩ = α|0⟩+β|1⟩ that Alice wants to send (she doesn't need to know α, β — that's the whole point).
• One Bell pair (|00⟩+|11⟩)/√2 shared in advance between Alice and Bob, one qubit each.
• A classical communication channel — a phone call is enough.

Why the correction depends on the outcome

All four of Alice's possible outcomes are equally likely (25% each), and each one leaves Bob's qubit in a related but different state — α|0⟩+β|1⟩ rotated by some combination of X and Z. Bob doesn't know which until Alice's two classical bits arrive; once they do, the correction is completely deterministic. This is exactly why the classical channel is required: without it, Bob is holding one of four possible states with no way to know which, which carries zero usable information about |φ⟩ on its own.

What was and wasn't transmitted

The original qubit C is destroyed by Alice's measurement — this is required, not incidental: the no-cloning theorem forbids a perfect copy from existing at both ends simultaneously, and teleportation respects it exactly by destroying the original as a side effect of extracting the classical bits. No information traveled faster than light: Bob's qubit is in one of four possible states immediately after Alice's measurement, and it takes the classical bits, limited to light speed, to tell him which correction turns it into |φ⟩. Until those bits arrive, Bob's qubit is — to him — indistinguishable from a fresh, uncorrelated qubit.