Chapter 07

Teleportation

You can't copy an unknown qubit, and measuring destroys it — yet its exact state can be moved across the world using one entangled pair and two classical bits. Here is the whole trick, step by step and in symbols.

↩ before you start · keep these handy
·From Ch. 6: Alice and Bob each hold one half of a shared Bell pair (|00⟩+|11⟩)/√2.
·From Ch. 5 & Ch. 3: CNOT, H, X, Z are gates; a measurement returns a bit and collapses the state.
·From No-Cloning: an unknown qubit cannot be copied — which is why the original must die.
🔑 symbol decoder · every new mark, in plain words
|ψ⟩"psi" — the unknown qubit being teleported, α|0⟩ + β|1⟩. |Φ⁺⟩the shared Bell pair — the entangled "fuel" the protocol burns. the tensor product — writing several qubits side by side as one joint state. a, bAlice's two measurement outcomes — the two classical bits she phones to Bob. I, X, ZBob's possible corrections: do nothing, bit-flip, phase-flip (or both). ebitone shared entangled pair — the unit of entanglement teleportation consumes.
§1

Walk the protocol

📠 everyday picture

Teleportation is a fax machine that shreds the original. You feed in a page; the machine reads it — destroying it in the act — sends a short code down the phone line, and an identical page prints at the far end. No paper ever crossed the wire, only instructions. And because the original was shredded, there's never two copies in the world at once.

▸ stepteleport(ψ)
ALICE
ψ
⊗ ½ Bell
{{ bitsLabel }}
BOB
ψ
½ Bell ⊗
step {{ tpStepNum }} / 4
{{ tpCaption }}
bits → {{ tpBits }}Bob fix → {{ tpFix }}
recapOne unknown qubit moves using a shared Bell pair plus two classical bits — the original is destroyed en route.
§2

The same protocol, as a circuit

Every quantum algorithm is drawn the same way: horizontal wires are qubits, time flows left to right, and boxes are gates. Here is teleportation in that language — three wires, four gates, two measurements, and a classically-controlled fix on Bob's qubit.

▸ schematicteleport.circuit
|ψ⟩ q₂ q₃ Bell ALICE BOB H 2 classical bits → choose the fix Xᵃ Zᵇ |ψ⟩

The dashed lines are classical wires — ordinary bits, not qubits. They carry Alice's two measurement results to Bob, where they pick which of four corrections to apply. No quantum information ever travels along them.

recapDashed wires are classical bits — they pick Bob's correction and carry no quantum information.
§3 · the mathematics

The algebra, line by line

It looks like magic until you watch the algebra rearrange itself. Start with the unknown qubit beside the shared Bell pair (Alice's two qubits written first, Bob's last):

|ψ⟩ ⊗ |Φ⁺⟩ = (α|0⟩ + β|1⟩) ⊗ (|00⟩ + |11⟩)/√2

Apply Alice's CNOT then her Hadamard, and the whole thing regroups into four equal pieces — each tagged by the two bits she is about to read:

= ½ · [
|00⟩(α|0⟩ + β|1⟩)
+ |01⟩(α|1⟩ + β|0⟩)
+ |10⟩(α|0⟩ − β|1⟩)
+ |11⟩(α|1⟩ − β|0⟩)
]

Look at the trailing factor in each line — that is Bob's qubit. The instant Alice measures and gets some ab, Bob's qubit is left in exactly one of those four expressions — each a simple gate away from the original α|0⟩ + β|1⟩. That is why the corrections below are all he needs.

recapAfter Alice's CNOT + H the state splits into four equal branches, each tagging Bob's qubit with one known Pauli error.
§4 · the mathematics

Why two bits are enough

Alice's joint measurement randomly lands in one of four equally-likely outcomes. Each leaves Bob's qubit holding ψ up to a known Pauli error — and the two bits tell him exactly which one to undo:

00
apply I
01
apply X
10
apply Z
11
apply Z·X

Two facts keep the universe honest. No cloning: ψ is destroyed at Alice the instant she measures, so only one copy ever exists. No faster-than-light: until the classical bits arrive, Bob's qubit is an even mixture — useless. The state moved, but it rode on an ordinary phone call.

✎ worked example · Alice reads the bits 01
1.from the |01⟩ branch above, Bob's qubit is left as (α|1⟩ + β|0⟩)
2.the |0⟩ and |1⟩ amplitudes are swapped compared with α|0⟩ + β|1⟩
3.swapping the two slots is exactly what X does — the table says 01 → X
4.Bob applies X → α|0⟩ + β|1⟩ = ψ, recovered exactly. ✓
recapTwo bits name one of four Pauli fixes; no-cloning and the classical channel keep it honest.
⚠ what teleportation is not

Not a transporter. No matter or energy moves — only a state, a list of amplitudes. Bob already had a qubit; teleportation just reconfigures it to match ψ.

Not a copy. The original is destroyed at Alice's measurement, so there is never a second ψ in existence — the no-cloning theorem stays intact.

Not faster than light. Without Alice's two classical bits, Bob's qubit is indistinguishable from noise. The protocol runs no faster than that phone call.

✓ you can now
list the resources teleportation spends: one ebit + two classical bits
read Alice's two measurement bits and pick Bob's matching correction
explain why teleportation neither clones a qubit nor beats light speed
← 7b Superdense Coding next · 7c Key Distribution