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9.5

Quantum teleportation

One Bell pair plus one two-bit phone call delivers an unknown quantum state to the far side intact: four measurement outcomes, four correction gates, all worked by hand — with the no-cloning theorem proved and the not-faster-than-light point cleared up along the way.

Recommended first

After this section you should be able to

  • Work through the teleportation protocol in full: expand the three-qubit state in the Bell basis and list all four outcomes with their corrections
  • Prove the no-cloning theorem and show how it blocks the naive "quantum fax machine"
  • Explain why the classical channel is indispensable and why the protocol does not violate relativity
  • Balance the resource ledger: 1 ebit + 2 classical bits = 1 qubit delivered

Last section’s conclusion felt a little frustrating: entanglement’s correlations are strong enough to violate the Bell inequality, yet cannot carry a single bit of messaging. This section shows what entanglement is really for — not sending messages with it, but sending states.

The problem at the door: how do you mail an unknown quantum state?

Alice holds a qubit in a state that even she does not know:

∣χ⟩=α∣0⟩+β∣1⟩(9.5.1)\ket{\chi}=\alpha\ket{0}+\beta\ket{1}\tag{9.5.1}

She wants a faraway Bob to end up owning this state. Box up the particle and courier it? Decoherence en route would destroy the state (chapter 10’s subject). Then how about faxing: measure it, phone the readings to Bob, and let him rebuild a copy?

Two walls facing the classical approach

Wall one: you cannot measure it all. Section 9.1 told us a single measurement yields only 1 bit, and the state collapses on the spot — the rest of the information in α,β\alpha,\beta goes up in smoke. With a single specimen, (α,β)(\alpha,\beta) simply cannot be measured.

Wall two: you cannot copy it. Fine — photocopy a few first and measure at leisure? Quantum mechanics welds that road shut too:

The no-cloning theorembasic~4 min

Claim: there is no unitary UU that achieves U∣χ⟩∣e⟩=∣χ⟩∣χ⟩U\ket{\chi}\ket{e}=\ket{\chi}\ket{\chi} for an arbitrary unknown state ∣χ⟩\ket{\chi} (∣e⟩\ket{e} being a fixed blank page).

Proof (one clean blow from linearity). Suppose it works on the two basis states:

U∣0⟩∣e⟩=∣0⟩∣0⟩,U∣1⟩∣e⟩=∣1⟩∣1⟩(9.5.2)U\ket{0}\ket{e}=\ket{0}\ket{0},\qquad U\ket{1}\ket{e}=\ket{1}\ket{1}\tag{9.5.2}

For the superposition ∣+⟩=∣0⟩+∣1⟩2\ket{+}=\frac{\ket{0}+\ket{1}}{\sqrt2}, linearity forces

U∣+⟩∣e⟩=∣00⟩+∣11⟩2(9.5.3)U\ket{+}\ket{e}=\frac{\ket{00}+\ket{11}}{\sqrt2}\tag{9.5.3}

But “successful cloning” would require

∣+⟩∣+⟩=∣00⟩+∣01⟩+∣10⟩+∣11⟩2(9.5.4)\ket{+}\ket{+}=\frac{\ket{00}+\ket{01}+\ket{10}+\ket{11}}{2}\tag{9.5.4}

The two are not equal (section 9.2 computed the inner product: 1/21/\sqrt2). Contradiction — no cloning machine exists. ■\blacksquare

Linear evolution can only turn “copying basis states” into “creating entanglement” — CNOT’s failed photocopy was not poor craftsmanship but a law of nature. Incidentally: if cloning were possible, section 9.4’s entanglement correlations would become a faster-than-light telegraph (Bob could clone a thousand copies and measure each to tell which axis Alice had measured along) — so no-cloning and no-signalling are tied to the same rope.

Cannot measure it all, cannot copy it — a dead end, it seems. The way out, found by Bennett and five co-authors in 1993, is exquisite: don’t read it — let entanglement carry it for you.

The full protocol, by hand

Setup in advance: Alice and Bob each hold half of the Bell state ∣Φ+⟩23=∣00⟩+∣11⟩2\ket{\Phi^+}_{23}=\frac{\ket{00}+\ket{11}}{\sqrt2} (particle 2 to Alice, particle 3 to Bob). Alice also holds particle 1, in the state ∣χ⟩\ket{\chi} to be sent.

particle 1 |χ⟩  ──●──[H]──[measure]──┐m₁ ─────────────┐
                  │                  │m₂ ──┐          │
particle 2 ───────⊕───────[measure]──┘     │(phone)   │(phone)
                                           ↓          ↓
particle 3 ───────────────────────────────[Xᵐ²]─────[Zᵐ¹]── |χ⟩
          Alice's side                 Bob's side
Expanding the three-qubit state in the Bell basis: all four outcomes, none missedadvanced~9 min

Step 0: the initial state.

∣Ψ0⟩=∣χ⟩1⊗∣Φ+⟩23=12(α∣000⟩+α∣011⟩+β∣100⟩+β∣111⟩)(9.5.5)\ket{\Psi_0}=\ket{\chi}_1\otimes\ket{\Phi^+}_{23} =\frac{1}{\sqrt2}\Big(\alpha\ket{000}+\alpha\ket{011}+\beta\ket{100}+\beta\ket{111}\Big)\tag{9.5.5}

(The three subscripts are particles 1, 2, 3 in order.)

Step 1: put on different glasses for particles 1 and 2. Rather than measuring the two particles separately, Alice performs a Bell-basis measurement — asking “which of the four Bell states are particles 1 and 2 in?” First invert the computational basis into the Bell basis:

∣00⟩=∣Φ+⟩+∣Φ−⟩2,∣11⟩=∣Φ+⟩−∣Φ−⟩2,∣01⟩=∣Ψ+⟩+∣Ψ−⟩2,∣10⟩=∣Ψ+⟩−∣Ψ−⟩2(9.5.6)\ket{00}=\frac{\ket{\Phi^+}+\ket{\Phi^-}}{\sqrt2},\quad \ket{11}=\frac{\ket{\Phi^+}-\ket{\Phi^-}}{\sqrt2},\quad \ket{01}=\frac{\ket{\Psi^+}+\ket{\Psi^-}}{\sqrt2},\quad \ket{10}=\frac{\ket{\Psi^+}-\ket{\Psi^-}}{\sqrt2}\tag{9.5.6}

Step 2: substitute and sort by Bell state. Replace the particle-1,2 part of ∣Ψ0⟩\ket{\Psi_0} term by term and patiently collect like terms (write out every one):

∣Ψ0⟩=12[∣Φ+⟩12⊗(α∣0⟩+β∣1⟩)3+∣Φ−⟩12⊗(α∣0⟩−β∣1⟩)3+∣Ψ+⟩12⊗(α∣1⟩+β∣0⟩)3+∣Ψ−⟩12⊗(α∣1⟩−β∣0⟩)3]\begin{aligned} \ket{\Psi_0}=\frac12\Big[ &\ket{\Phi^+}_{12}\otimes(\alpha\ket{0}+\beta\ket{1})_3\\ +&\ket{\Phi^-}_{12}\otimes(\alpha\ket{0}-\beta\ket{1})_3\\ +&\ket{\Psi^+}_{12}\otimes(\alpha\ket{1}+\beta\ket{0})_3\\ +&\ket{\Psi^-}_{12}\otimes(\alpha\ket{1}-\beta\ket{0})_3 \Big] \end{aligned}

This is a pure identity — nobody has done anything yet, and Bob’s particle 3 already has four states “pre-hung” on it, each just one Pauli gate away from ∣χ⟩\ket{\chi}.

Step 3: Alice measures. Each Bell state carries amplitude 12\tfrac12, so the four outcomes have probability 14\tfrac14 each (independent of α,β\alpha,\beta — Alice’s readout contains no information whatsoever about ∣χ⟩\ket{\chi}; we will need this later). After collapse, four cases:

Alice obtainsBob’s particle 3 becomesBob’s correction
∣Φ+⟩\ket{\Phi^+}α∣0⟩+β∣1⟩\alpha\ket{0}+\beta\ket{1}Do nothing (II)
∣Φ−⟩\ket{\Phi^-}α∣0⟩−β∣1⟩\alpha\ket{0}-\beta\ket{1}ZZ (remove the relative minus sign)
∣Ψ+⟩\ket{\Psi^+}α∣1⟩+β∣0⟩\alpha\ket{1}+\beta\ket{0}XX (swap back up and down)
∣Ψ−⟩\ket{\Psi^-}α∣1⟩−β∣0⟩\alpha\ket{1}-\beta\ket{0}ZXZX (first XX, then ZZ)

Step 4: make the phone call. Alice encodes her result into two classical bits (m1,m2)(m_1,m_2) and sends them to Bob, who applies Zm1Xm2Z^{m_1}X^{m_2} according to the table. Verify the trickiest row: when she obtains ∣Ψ−⟩\ket{\Psi^-},

ZX(α∣1⟩−β∣0⟩)=Z(α∣0⟩−β∣1⟩)=α∣0⟩+β∣1⟩=∣χ⟩(9.5.7)ZX(\alpha\ket{1}-\beta\ket{0})=Z(\alpha\ket{0}-\beta\ket{1})=\alpha\ket{0}+\beta\ket{1}=\ket{\chi}\tag{9.5.7}

All four cases converge: Bob’s particle is exactly in ∣χ⟩\ket{\chi}, while Alice’s particles 1 and 2 have collapsed onto a Bell state — ∣χ⟩\ket{\chi} has vanished from her side without a trace.

Implementation detail: the Bell measurement itself can be done with section 9.2’s circuit — CNOT (1 controlling 2) followed by HH on 1 maps the four Bell states onto the four computational basis states ∣m1m2⟩\ket{m_1 m_2}; then just measure both (see the circuit diagram above).

The picture

What just happened? No particle flew from Alice to Bob. What was transported is the state — the entire structure described by the pair of complex numbers (α,β)(\alpha,\beta) — which changed hosts through the “pre-laid pipeline” of entanglement. The marvel: (α,β)(\alpha,\beta) are continuous parameters (writing them down takes infinitely many decimal places), yet only a two-bit phone call was made. Where did the rest of the information come from? It hid in the Bell pair — the Bell experiments told us entangled correlations outclass every classical mechanism, and teleportation is the first time that super-strength was “cashed in” as a working protocol.

The mathematics

The resource ledger (neither side of the equation can be shorted):

1 ebit+2 cbit ⟶ 1 qubit(9.5.8)1\ \text{ebit}+2\ \text{cbit}\ \longrightarrow\ 1\ \text{qubit}\tag{9.5.8}

Consumable: the used Bell pair has collapsed; sending a second state takes a fresh pair. The two classical bits cannot be skipped: the correction is a four-way choice, and one bit short leaves Bob guessing. Next section shows the “dual” of this ledger: 11 ebit ++ 11 qubit →2\to 2 cbit.

Key formulas

  • No-cloning theorem

    ∄ U: U∣χ⟩∣e⟩=∣χ⟩∣χ⟩  ∀∣χ⟩\nexists\,U:\ U\ket{\chi}\ket{e}=\ket{\chi}\ket{\chi}\ \ \forall\ket{\chi}

    Linearity ⇒ copying basis states only creates entanglement; two faces of the no-signalling coin

  • Bell-basis expansion

    ∣χ⟩1∣Φ+⟩23=12∑k∣Bk⟩12⊗(σk∣χ⟩)3\ket{\chi}_1\ket{\Phi^+}_{23}=\frac12\sum_{k}\ket{B_k}_{12}\otimes(\sigma_k\ket{\chi})_3

    A pure identity: each Bell state pairs with one Pauli correction σ_k ∈ {I, Z, X, ZX}

  • Correction rule

    ∣χ⟩=Zm1Xm2 ∣ψ3(m1m2)⟩\ket{\chi}=Z^{m_1}X^{m_2}\,\ket{\psi_3^{(m_1m_2)}}

    Two classical bits encode exactly the four-way choice

  • Resource equation

    1 ebit+2 cbit→1 qubit1\ \text{ebit}+2\ \text{cbit}\to 1\ \text{qubit}

    The Bell pair is destroyed on use; the classical channel sets the speed limit

Self-check4 questions

  1. 1.

    Alice performs the Bell measurement. What is the probability of each of the four outcomes? (Independent of the state ∣χ⟩ being sent.)

    0% relative tolerance
  2. 2.

    Alice obtains ∣Ψ⁺⟩ and Bob’s particle is in α∣1⟩ + β∣0⟩. Which gate should he apply?

  3. 3.

    Why can teleportation not transmit information faster than light?

  4. 4.

    Which statements about the protocol’s resource costs are correct? (Select all that apply.)

    Select all that apply

What comes next

Teleportation trades “1 ebit + 2 classical bits” for delivering 1 qubit. What happens if you run the trade backwards — first hand Bob half an entangled pair, then have Alice mail him 1 qubit: how many classical bits does that buy? The answer is 2: a single qubit can supposedly yield only 1 bit (section 9.1), yet entanglement lets it carry an “overload”. Next section: superdense coding.

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