Alice holds a state she does not know
q0 carries |ψ⟩ = cos(θ/2)|0⟩ + e^{iφ} sin(θ/2)|1⟩. Choose it; Alice never learns θ or φ, and no measurement could tell her.
A Bell pair is shared
H then CNOT makes (|00⟩ + |11⟩)/√2 between q1 (Alice) and q2 (Bob). In the real link this pair is made at a relay and its halves travel as photons; it carries no information about |ψ⟩.
Alice rotates into the Bell basis
CNOT from q0 to q1, then H on q0. Look at the bars: all eight amplitudes are now non-zero, and every one of Bob's four possible states already contains |ψ⟩, up to a Pauli.
Alice measures: two classical bits
Each outcome has probability exactly ¼ whatever |ψ⟩ is, which is why the bits reveal nothing about it. Pick the outcome:
The bits travel at the speed of light
Until they arrive, Bob's qubit, averaged over Alice's four equally likely outcomes, is the maximally mixed state: its Bloch vector is zero. That is the no-signalling theorem in one line, and it is why nothing can go faster than the classical channel.
Bob corrects and holds |ψ⟩
X if m1 = 1, Z if m0 = 1. Fidelity with Alice's original state: 1. The original on q0 was destroyed by the measurement, so nothing was cloned. With a noisy pair of fraction f, the average fidelity would be (2f+1)/3 (qll/circuits/teleportation.py).