Quantum Teleportation

Pick a state with the two sliders — any point on the sphere, including the ones with a phase, which is the half of a qubit a classical bit cannot describe. Then walk the seven steps and watch both spheres. The left one is the payload qubit itself, not a label for what Alice chose: once it stops pointing where she prepared it, the prepared direction stays on only as a faint dashed ghost, so you can still compare it with Bob’s arrow.

The thing to watch for is the moment Bob’s arrow disappears. From the instant the pair is shared until the instant the classical bits land, his qubit has no direction at all: the drawing shows a cloud, not a short arrow, because a maximally mixed state genuinely has nowhere to point. Every measurement Bob can make on it comes out a coin flip, whatever Alice did.

Turn off Apply Bob’s correction and step through all four outcomes. For a generic state one of them — 00 — arrives correct by luck; the other three arrive rotated by X, by Z, or by both. (For special states like |0⟩ or |+⟩ a second branch happens to be exact too — the table shows which.) Bob’s qubit is in a perfectly good pure state; it is simply the wrong one, and nothing he can do to it alone will tell him which.

Alice’s two bits are the missing information. They cost two bits because there are exactly four possibilities, each with probability 0.25 no matter what state she sent — which is also why the bits are safe to shout across a room: on their own they say nothing about the state.

With the correction on, all four rows read exact match — the protocol works for every outcome, not on average.

The tempting story is that Alice’s measurement instantly changes Bob’s qubit, so something crossed the gap instantly. Something did change — but nothing Bob can detect.

Before the bits arrive, Bob does not know which of the four branches he is in, so the state he actually has is the average of all four. For this input that average has Bloch length 0.000 — zero, to within floating point. Slide θ and φ anywhere you like: it stays zero. His statistics are identical for every state Alice could have sent, and for the case where she sent nothing at all. There is no measurement that distinguishes them, so there is no message.

The information Bob is missing travels in the classical channel, at the speed a classical channel travels, and the protocol simply waits for it. That is the step labelled Send 2 bits, and it is the only step in which nothing quantum happens at all.

Watch the left sphere as you step. At Bell basis the payload is already entangled with Alice’s half of the pair, and its arrow shrinks to length |cos θ| along x: the phase φ — the half a classical bit could never carry — is the first thing to go, and for a state on the equator nothing is left but a cloud. After Measure it sits at ±z, a definite computational state carrying no trace of θ or φ. The dashed ghost is only a reminder of what was prepared; the qubit no longer points there. The state was not duplicated and then transmitted — it was destroyed at one end and rebuilt at the other.

That is not a design choice, it is forced. If teleportation left the original intact you would have a perfect cloning machine, and cloning an unknown quantum state is impossible. The protocol pays for the copy it makes by giving up the original, every time.

The circuit simulator can express this circuit for real: place a measurement, choose which classical bit it records into, then place an X and a Z bound to those bits. The preset Teleportation (real classical wires) loads the finished version — press Run and watch the Bell result change from shot to shot while qubit 2 comes out right every time.