This whole protocol is haunted by the no-cloning theorem: there is no operation that copies an arbitrary unknown quantum state. I worked through the proof and its consequences in a recent post, The No-Cloning Theorem in Quantum Computing: Why You Can’t Copy a Qubit, so I will not re-derive it here.

The single fact we need going forward: you cannot deterministically duplicate an unknown qubit. The qualifiers matter, because teleportation lives in the gaps between them. The theorem forbids copying unknown states (a known state you can re-prepare at will), it forbids deterministic, perfect copies, and it is a statement about unitary operations — and measurement, which teleportation leans on at the decisive moment, is not unitary.

Here is the distinction in one sentence: cloning would leave |ψ⟩ in two places; teleportation leaves it in exactly one.

Partway through the protocol, Alice’s message qubit is measured. Measurement collapses it into a classical basis state — a plain |0⟩ or |1⟩ carrying none of the original amplitudes. By the time Bob’s qubit holds |ψ⟩, Alice’s qubit demonstrably does not. The state was relocated, and the accounting is exact: one copy in, one copy out. No moment ever exists where two qubits both carry |ψ⟩, so there is nothing for the no-cloning theorem to object to.

Here is the whole protocol in a single circuit. Read it left to right; the amber dashed dividers mark the three stages. The message qubit q0 starts in the unknown state |ψ⟩, while q1 and q2 both start in |0⟩.

Stage 1 — Bell pair. Alice and Bob share an entangled pair: an H on q1 followed by a CNOT from q1 onto q2 prepares (|00⟩ + |11⟩)/√2 across the two halves.

Stage 2 — Alice measures. Alice folds her message into the pair (a CNOT from q0 onto q1, then an H on q0) and measures both of her qubits, collapsing them to two classical bits.

Stage 3 — Bob corrects. Depending on those two bits, Bob applies an X and/or a Z to q2 — the X controlled by Alice’s q1 bit, the Z by her q0 bit — and q2 emerges as |ψ⟩. The next section is the algebra that fixes exactly which correction each outcome needs.

Now the algebra behind that circuit. Three qubits: q0 carries |ψ⟩ = α|0⟩ + β|1⟩ (Alice), q1 is Alice’s half of the Bell pair, and q2 is Bob’s half. They pre-share |Φ+⟩ = (|00⟩ + |11⟩)/√2 on q1 q2.

Step 1 — the starting state. Tensor the message against the Bell pair:

Step 2 — Alice’s CNOT (control q0, target q1):

Step 3 — Alice’s Hadamard on q0. Substitute |0⟩ → (|0⟩+|1⟩)/√2 and |1⟩ → (|0⟩−|1⟩)/√2, then collect by the value of (q0 q1):

Each bracketed q2 state is the original |ψ⟩ acted on by a known Pauli. Alice measures, gets one of four outcomes (each with probability 1/4), sends the two bits to Bob, and Bob undoes the Pauli. The Case column ties each row back to the matching line in Step 3:

Bob applies Zm0 Xm1 — an X if m1 = 1, then a Z if m0 = 1 — and every branch lands back on α|0⟩ + β|1⟩ = |ψ⟩.

This version measures mid-circuit and uses real classical conditioning via if_test, the honest picture of the protocol. The verification trick: rather than read out Bob’s state, apply the inverse of the preparation to q2. If teleportation worked, that must collapse q2 to |0⟩ on every shot.

Before Bob learns (m0, m1), his qubit is an equal mixture of the four branch states. Averaging the four projectors gives the Pauli twirl:

for any |ψ⟩. Bob’s local state is identical no matter what Alice sent, so no information has reached him yet. The classical bits are not a formality — they are the only thing that carries the state across, and they travel no faster than light.