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Intermediate

Quantum Teleportation

A qubit's state crosses the room without the qubit — and it still needs two ordinary bits, sent the slow way, to get there.

Quantum Teleportation visualization

Three qubits, all in |0⟩. Alice holds the first two, Bob holds the third — and Bob may be anywhere.

What you are looking at

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 the right-hand sphere.

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.

The two bits are the whole protocol

Turn off Apply Bob’s correction and step through all four outcomes. One of them — 00 — arrives correct by luck. The other three arrive as a full-length arrow pointing somewhere else: rotated by X, by Z, or by both. 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.

The four Bell outcomes, their probabilities, the correction each one calls for, and whether Bob’s state matches Alice’s.
c₁c₀ProbabilityBob appliesResult
0025.0%identity — nothing to doexact match
0125.0%Zexact match
1025.0%Xexact match
1125.0%X then Zexact match

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

Why this is not faster-than-light communication

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.

Nothing was copied

Look at the left sphere after the measurement: Alice’s two qubits are left in a definite computational state, ±z, carrying no trace of θ or φ. 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.

Build it yourself

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.