Forget the transporter from science fiction — nothing here dematerialises a body and beams it across space. Quantum teleportation is subtler and, in its own way, stranger: it moves the exact quantum state of one qubit onto another, distant qubit, without any qubit travelling between them. Alice holds a qubit in some unknown state |\psi\rangle = \alpha|0\rangle + \beta|1\rangle; when the protocol ends, Bob's far-away qubit is in precisely that state, and Alice's copy is gone. What crosses the gap is not the qubit but a pre-shared Bell pair plus two ordinary classical bits. That is the whole magic trick, and every piece of it will make sense by the end of this page.

What you start with

The protocol needs three qubits and one classical channel:

• The message qubit |\psi\rangle = \alpha|0\rangle + \beta|1\rangle, held by Alice. Its amplitudes \alpha, \beta are unknown — not even Alice knows them, and she is not allowed to measure them (a single measurement would collapse the state and destroy the information).
• A shared Bell pair |\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle), made ahead of time. Alice keeps one half, Bob takes the other half far away. This entanglement is the "quantum wire" the state will slide along.
• A classical channel — a phone line, an email, anything — that can carry two bits from Alice to Bob.

Label the three qubits q_0 (Alice's message), q_1 (Alice's half of the Bell pair) and q_2 (Bob's half). The full starting state is the product |\psi\rangle \otimes |\Phi^+\rangle.

The three moves

The whole protocol is just three steps:

1. Alice entangles and measures. She applies a CNOT from her message qubit q_0 (control) onto her Bell half q_1 (target), then a Hadamard on q_0. She measures both of her qubits in the computational basis, getting two classical bits m_0 m_1 — one of 00, 01, 10, 11. (This CNOT-then-H-then-measure is exactly a Bell-basis measurement of her two qubits.)
2. Alice phones Bob. She sends those two bits down the classical channel. That is all she sends — two bits, nothing quantum.
3. Bob corrects. Reading the two bits, Bob applies a matching Pauli gate to his qubit: I for 00, X for 01, Z for 10, or ZX for 11. His qubit is now exactly |\psi\rangle.

The circuit

Here is the standard picture. Alice's two wires meet a CNOT and a Hadamard, then two meters; the classical results (drawn as double lines) drop down to control an X and a Z on Bob's wire. Step through it and watch the state travel from the top wire to the bottom one.

Notice the two double lines — those are classical bits, not qubits. They are the only thing that crosses from Alice's side to Bob's, and without them Bob's gates have no idea which correction to make.

The algebra: one expansion does it all

Everything follows from expanding the starting state and regrouping. Begin with |\psi\rangle \otimes |\Phi^+\rangle on qubits (q_0, q_1, q_2):

Now apply Alice's CNOT (q_0 controls q_1), which flips the middle bit whenever the first is 1, then a Hadamard on q_0 (|0\rangle \to \tfrac{1}{\sqrt2}(|0\rangle+|1\rangle), |1\rangle \to \tfrac{1}{\sqrt2}(|0\rangle-|1\rangle)). Collecting the terms by Alice's two qubits gives the key identity:

Read this carefully. Each of Alice's four possible measurement outcomes (the first two kets) leaves Bob's qubit (the bracketed part) in a known Pauli image of |\psi\rangle — and all four have equal probability \big|\tfrac12\big|^2 \cdot 2 = \tfrac14. That is the whole trick in one line: no matter what Alice gets, Bob is one Pauli gate away from |\psi\rangle.

The correction table

The bracketed states above are exactly |\psi\rangle, X|\psi\rangle, Z|\psi\rangle and ZX|\psi\rangle. Since every Pauli is its own inverse, Bob just re-applies the same gate to undo it. Reveal the rows one at a time.

In the circuit this reads off cleanly: the bit from q_1 switches Bob's X on or off, and the bit from q_0 switches his Z. So the correction is Z^{m_0} X^{m_1}.

Worked example 1: Alice measures 00

Suppose Alice's meters read m_0 m_1 = 00. Look at the |00\rangle branch of the expansion: Bob's qubit is

Bob is already done — his qubit is the message. The correction is the identity I: he does nothing. Yet even here he still had to wait for Alice's phone call to know that no correction was needed. Until the bits arrive, his qubit, looked at alone, is a featureless 50/50 mix that carries no usable information about \alpha, \beta.

Worked example 2: Alice measures 11

Now suppose the meters read m_0 m_1 = 11. The |11\rangle branch leaves Bob holding

The amplitudes are scrambled and sign-flipped — definitely not |\psi\rangle. Because m_1 = 1 Bob first applies X (swapping |0\rangle \leftrightarrow |1\rangle):

Then, because m_0 = 1, he applies Z (which flips the sign of the |1\rangle part):

The combined correction Z^{1}X^{1} = ZX recovers |\psi\rangle exactly. (A harmless overall minus sign that could appear is a global phase, which never affects any measurement.)

Three things it does not do

Teleportation is famous for what it is careful not to violate:

• It does not clone. Alice's original |\psi\rangle is destroyed the instant she measures — her qubits collapse to plain classical bits. The state moves; it is never duplicated, so the no-cloning theorem is perfectly safe.
• It does not beat light. The entanglement alone tells Bob nothing; he must wait for the two classical bits, which travel no faster than light. No classical bits, no reconstruction — so teleportation cannot signal faster than light.
• Alice never learns the state. Her measurement yields two random bits, not \alpha or \beta. She ships an unknown state she could not have written down.