University Physics V · Introduction to Quantum Information · 20.7

No-Cloning, Teleportation & the Classical Channel

Two consequences of the same linearity. An unknown qubit cannot be copied — and it can still be moved, if you spend a shared Bell pair, measure the original out of existence, and send two ordinary bits. Learn to do the algebra in the Bell basis, and to say what each of those resources pays for.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum mechanics is linear, and that one fact both closes a door and opens another. A copier would be a unitary U with U(|ψ⟩|s⟩) = |ψ⟩|ψ⟩ for every |ψ⟩; take the inner product of that equation for two inputs and unitarity gives ⟨ψ|φ⟩ = ⟨ψ|φ⟩², so only orthogonal states — classical bits — can be duplicated. Yet the same linearity lets an unknown state be moved.

Give Alice the state |ψ⟩_C and one half of a Bell pair |Φ+⟩_AB, the other half to Bob, and rewrite the product |ψ⟩_C|Φ+⟩_AB in the Bell basis of Alice's two qubits: it is ½Σₖ |βₖ⟩_CA ⊗ σₖ|ψ⟩_B with σₖ ∈ {I, Z, X, XZ}. Bob's qubit already differs from |ψ⟩ only by a Pauli operator, indexed by an outcome Alice has not yet obtained. She measures in the Bell basis — CNOT, Hadamard, two σ_z readouts — gets two bits m₁m₂ with probability ¼ each whatever |ψ⟩ is, sends them, and Bob applies Z^m₁X^m₂.

The costs are exact. The Bell pair is consumed; Alice's original ends as a σ_z eigenstate carrying nothing about α or β, which is how the protocol obeys no-cloning; and until the two bits arrive Bob holds ¼Σₖ σₖ|ψ⟩⟨ψ|σₖ† = I/2, which is how it obeys the speed of light. One ebit plus two classical bits move one qubit, and neither ingredient alone moves anything.

Simple definition: Teleportation transfers an unknown qubit |ψ⟩ from Alice to Bob by a Bell-basis measurement on |ψ⟩ and her half of a shared |Φ+⟩ pair, two classical bits, and Bob's correction Z^m₁X^m₂ — no copy is made, because linearity forbids one.
Example: For |ψ⟩ = |+⟩ = (|0⟩ + |1⟩)/√2 the |Φ−⟩ outcome, bits m₁m₂ = 10, leaves Bob holding Z|+⟩ = |−⟩ — orthogonal to the input, fidelity 0 — and that outcome arrives with probability ¼ like the other three; the bit m₁ = 1 tells him to apply Z, which restores |+⟩.

No-cloning from linearityU(|ψ⟩⊗|s⟩) = |ψ⟩⊗|ψ⟩ for all ψ ⇒ ⟨ψ|φ⟩ = ⟨ψ|φ⟩² ⇒ ⟨ψ|φ⟩ ∈ {0, 1}

Only orthogonal or identical states survive: a copier for the σ_z eigenbasis exists, a copier for all of C² does not.

U unitary on H⊗H, |s⟩ a fixed blank state; take the inner product of the copy equation for ψ with the one for φ

Bell basis of C²⊗C²|Φ±⟩ = (|00⟩ ± |11⟩)/√2, |Ψ±⟩ = (|01⟩ ± |10⟩)/√2

The Bell measurement projects onto these four; CNOT then H on the first qubit rotates them onto |m₁m₂⟩ for two σ_z readouts.

orthonormal; |Φ−⟩ = (Z⊗I)|Φ+⟩, |Ψ+⟩ = (X⊗I)|Φ+⟩, |Ψ−⟩ = −(XZ⊗I)|Φ+⟩

Teleportation identity|ψ⟩_C|Φ+⟩_AB = ½[ |Φ+⟩|ψ⟩ + |Φ−⟩Z|ψ⟩ + |Ψ+⟩X|ψ⟩ + |Ψ−⟩XZ|ψ⟩ ]

Each branch has probability ¼ whatever α and β are, and Bob's qubit is |ψ⟩ up to a Pauli that Alice knows and Bob does not — yet.

Bell states on CA, Paulis act on B; |ψ⟩ = α|0⟩ + β|1⟩ never appears in the CA factor

Bob before the bits: the Pauli twirlρ_B = ¼ Σₖ σₖ|ψ⟩⟨ψ|σₖ† = I/2, σₖ ∈ {I, Z, X, XZ}

Bob's statistics are independent of ψ until the bits arrive: no signalling, and no information sits in the pair alone.

Bloch vector r → ¼(r + Zr + Xr + XZr) = 0: each component is flipped by two of the four

Correction and resource countBob applies Z^m₁ X^m₂1 ebit + 2 cbits → 1 qubit

Four equiprobable corrections need log₂4 = 2 bits, sent at light speed or slower — the protocol is no faster than its channel.

m₁ from qubit C after H, m₂ from qubit A, both σ_z readouts; the pair and the original are both consumed

Imperfect pair: Werner singlet fraction Ff = ⟨ψ|ρₒᵤₜ|ψ⟩ = (2F + 1)/3, F = ⟨Φ+|ρ_AB|Φ+⟩

Beats the measure-and-prepare limit 2/3 exactly when F > ½, the same threshold at which the Werner pair becomes entangled.

ρ_AB = p|Φ+⟩⟨Φ+| + (1 − p)I/4 with p = (4F − 1)/3; f is the same for every pure input, dimensionless

Linearity forbids the copier

Suppose a machine copies an unknown qubit: a unitary U on C²⊗C² with U(|ψ⟩|s⟩) = |ψ⟩|ψ⟩ for every |ψ⟩, |s⟩ a fixed blank. Ask it only to copy |0⟩ and |1⟩, U|0⟩|s⟩ = |00⟩ and U|1⟩|s⟩ = |11⟩, and linearity fixes what it does to |+⟩: U|+⟩|s⟩ = (|00⟩ + |11⟩)/√2 = |Φ+⟩. The copy we wanted is |+⟩|+⟩ = (|00⟩ + |01⟩ + |10⟩ + |11⟩)/2, a product of Schmidt rank one; what linearity delivered is a Bell state of Schmidt rank two, and |⟨++|Φ+⟩|² = ½. The general form takes the inner product of the copy equation for |ψ⟩ with the one for |φ⟩: unitarity keeps the left side at ⟨ψ|φ⟩ while the right side is ⟨ψ|φ⟩², so ⟨ψ|φ⟩ is 0 or 1. Orthogonal states can be copied, which is all a classical bit ever asked for; a general unit vector in C² cannot. The theorem has teeth. A cloner run repeatedly would let some copies be read in σ_z and others in σₓ, estimating θ and φ to any precision from a single system, and would let Alice signal through a shared pair faster than light. What survives is approximate: the best universal 1→2 cloner reaches fidelity 5/6 = 0.833, and measuring then re-preparing manages 2/3.

Three qubits, and the identity that does all the work

Alice holds the unknown |ψ⟩_C = α|0⟩ + β|1⟩ and qubit A; Bob holds qubit B; A and B were prepared in |Φ+⟩_AB = (|00⟩ + |11⟩)/√2. The joint state in C²⊗C²⊗C², ordered C, A, B, is |ψ⟩_C|Φ+⟩_AB = (α|000⟩ + α|011⟩ + β|100⟩ + β|111⟩)/√2. Nothing has been sent; the move is algebra. Rewrite the CA pair in the Bell basis with |00⟩ = (|Φ+⟩ + |Φ−⟩)/√2, |11⟩ = (|Φ+⟩ − |Φ−⟩)/√2, |01⟩ = (|Ψ+⟩ + |Ψ−⟩)/√2, |10⟩ = (|Ψ+⟩ − |Ψ−⟩)/√2, then collect Bob's qubit in each branch: ½[|Φ+⟩_CA(α|0⟩ + β|1⟩) + |Φ−⟩_CA(α|0⟩ − β|1⟩) + |Ψ+⟩_CA(α|1⟩ + β|0⟩) + |Ψ−⟩_CA(α|1⟩ − β|0⟩)]. The four B factors are |ψ⟩, Z|ψ⟩, X|ψ⟩ and XZ|ψ⟩. Two things are already decided. Each branch has norm ½, so every Bell outcome has probability ¼ whatever α and β are, and Alice's record will carry no information about the state. And Bob's qubit in every branch is |ψ⟩ up to one of four unitaries he can undo — if he knows which. In NumPy, state = np.kron(ψ, bell) with bell = np.array([1, 0, 0, 1])/np.√(2) is the 8-vector; project it onto each |βₖ⟩⊗I and read the branches off.

Alice measures in the Bell basis, and the original is gone

A Bell measurement is projective, with Pₖ = |βₖ⟩⟨βₖ| ⊗ I_B; no detector reads it directly, so Alice rotates the Bell basis onto the computational one. CNOT with C as control and A as target sends |Φ±⟩ → |±⟩|0⟩ and |Ψ±⟩ → |±⟩|1⟩; a Hadamard on C then sends |±⟩ → |0⟩, |1⟩. Two σ_z readouts follow, m₁ from C and m₂ from A: |Φ+⟩ → 00, |Φ−⟩ → 10, |Ψ+⟩ → 01, |Ψ−⟩ → 11. Read the bits as the Pauli Bob must undo: m₁ = 1 means Bob's branch carries a Z, m₂ = 1 that it carries an X. By Lüders' rule the CA pair is left in the product |m₁⟩|m₂⟩, and the whole system is |m₁⟩|m₂⟩ ⊗ σₖ|ψ⟩. Look at what Alice still holds: two σ_z eigenstates containing nothing about α or β. That is not a flaw; it is the first section enforced. Were qubit C still |ψ⟩ when Bob finishes, the protocol would have made two copies of an unknown state from one, which linearity forbids. Teleportation moves a state precisely by consuming the original, and the pair goes with it: A and B, which began entangled, end in a product with everything else, so a second qubit needs a second pair.

Before the bits arrive, Bob holds I/2

Bob knows the protocol but not Alice's outcome, so his qubit is the mixture over the four branches: ρ_B = ¼(|ψ⟩⟨ψ| + Z|ψ⟩⟨ψ|Z + X|ψ⟩⟨ψ|X + XZ|ψ⟩⟨ψ|ZX). Do the sum on the Bloch vector. Conjugation by Z flips rₓ and r_y; by X flips r_y and r_z; by XZ = −iY flips rₓ and r_z. Each component is left alone by two of the four and flipped by the other two, so the average vector is zero and ρ_B = I/2, the same maximally mixed state Bob held as his half of |Φ+⟩ before Alice did anything. Take |ψ⟩ at Bloch angle 60° in the x–z plane, r = (0.866, 0, 0.5). The four images are (±0.866, 0, ±0.5), and Bob's fidelity |⟨ψ|σₖ|ψ⟩|² with the input is 1, 0.25, 0.75 and 0 for outcomes 00, 10, 01 and 11: average exactly ½, the fidelity of a random guess. This is no-signalling in one line, Σₖ Tr_CA[(Pₖ ⊗ I)ρ(Pₖ ⊗ I)] = Tr_CA ρ = ρ_B, and it is why the light-speed limit survives. Nothing Bob can measure changes when Alice measures; it changes when her two bits reach him.

Two bits, one Pauli, and the resource ledger

The classical channel carries m₁m₂ and Bob applies Z^m₁X^m₂, X first and then Z. On the |Ψ−⟩ branch he holds XZ|ψ⟩ and ZX(XZ)|ψ⟩ = |ψ⟩ exactly; on |Φ−⟩, Z(Z|ψ⟩) = |ψ⟩; on |Ψ+⟩, X(X|ψ⟩) = |ψ⟩. Fidelity 1 on every branch, with no phase left over. Two bits are the minimum. The four outcomes are equiprobable, so Alice's record has entropy 2 bits, and any shorter message leaves Bob uncertain between at least two Paulis, whose mixture has fidelity below 1 for a generic |ψ⟩. Two bits are also nowhere near enough to carry the state itself: θ and φ are continuous, and 2 bits pick one of four. The state is in neither the bits nor the pair; it is in the combination, and the ledger reads 1 ebit + 2 cbits → 1 qubit. Its mirror image is superdense coding, 1 ebit + 1 qubit → 2 cbits. On hardware the ledger has been paid over 143 km of free space between two Canary Islands and, with the Micius satellite, from the ground to orbit at up to 1400 km, at fidelities around 0.80 against the classical 2/3.

Run it in NumPy, then degrade the pair

Build the eight-component vector with np.kron(ψ, bell), ordered C, A, B. Alice's circuit is two 8×8 matrices, np.kron(CNOT, I2) and np.kron(np.kron(H, I2), I2); after both, state.reshape(2, 2, 2)[m1, m2, :] is Bob's unnormalised branch and its squared norm the probability — print it and read 0.25 four times, whatever psi you typed. Normalise, apply matrix_power(Z, m1) @ matrix_power(X, m2), and np.allclose against psi passes on all four branches. Now replace the pair by a Werner state ρ_AB = p|Φ+⟩⟨Φ+| + (1 − p)I/4, whose singlet fraction is F = ⟨Φ+|ρ_AB|Φ+⟩ = (3p + 1)/4. Teleportation is linear in ρ_AB: the |Φ+⟩ part delivers |ψ⟩, and the I/4 part delivers I/2 whatever Alice reads and whatever Bob applies, since σ(I/2)σ† = I/2. So ρₒᵤₜ = p|ψ⟩⟨ψ| + (1 − p)I/2, a depolarising channel with fidelity f = (1 + p)/2 = (2F + 1)/3. At F = 0.85 that is 0.90; at F = ½ it is exactly the classical 2/3, and F = ½ is also where the Werner state stops being entangled. The figure below draws both bars.

Change one variable at a time

Make the relationship visible.

Interactive model

Input Bloch angle θ (x–z plane)60 °

Bell outcome m₁m₂ read as 0–32

Shared-pair singlet fraction F1.00

Step the outcome slider 0 to 3 at F = 1: the dashed arrow visits the four reflections of the input and the before-bits bar swings from 1 to 0 while the corrected bar stays at 1. Then lower F to 0.50: the corrected bar sinks onto the classical 2/3 line, exactly where the pair stops being entangled.

SHRINK p = (4F − 1)/31.000

FIDELITY BEFORE BITS, THIS m₁m₂0.250

FIDELITY AFTER Z^m₁X^m₂1.000

MARGIN OVER CLASSICAL 2/30.333

Live interpretationSHRINK p = (4F − 1)/3: 1.000. FIDELITY BEFORE BITS, THIS m₁m₂: 0.250. FIDELITY AFTER Z^m₁X^m₂: 1.000. MARGIN OVER CLASSICAL 2/3: 0.333

Catch the common trap

Explain before calculating.

Misconception check

The error is to think Alice's measurement does the teleporting and the two bits are a formality: the instant she reads her Bell outcome, Bob's qubit collapses to |ψ⟩, so the state has crossed the gap faster than light. It has not. Conditioned on her outcome Bob holds σₖ|ψ⟩, and since the four outcomes are equiprobable and independent of |ψ⟩, everything he can measure is the mixture ¼Σₖ σₖ|ψ⟩⟨ψ|σₖ† = I/2 — exactly what he held before she began. The opposite error, that the two bits carry the state so teleportation is classical communication in disguise, fails on counting: 2 bits select one of four Paulis and cannot encode two continuous parameters, and a measure-and-prepare strategy through the same channel reaches average fidelity 2/3, never 1. The state arrives only when the entangled pair and the bits are combined; neither alone carries it.

Alice teleports |ψ⟩ = 0.6|0⟩ + 0.8|1⟩ through a shared |Φ+⟩ pair. Her CNOT–H Bell measurement returns m₁m₂ = 01, the |Ψ+⟩ outcome. Conditioned on that outcome, what is Bob's qubit before her bits reach him, and what must he apply?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA proposed copier is a unitary U on C²⊗C² satisfying U|0⟩|0⟩ = |0⟩|0⟩ and U|1⟩|0⟩ = |1⟩|1⟩. Use linearity to find what it does to |+⟩|0⟩, compare with the wanted |+⟩|+⟩, and compute the fidelity between the two.

Linearity: U(|+⟩|0⟩) = (U|0⟩|0⟩ + U|1⟩|0⟩)/√2 = (|00⟩ + |11⟩)/√2 = |Φ+⟩. The output is forced; U has no freedom left on this input.
The wanted copy is |+⟩|+⟩ = (|00⟩ + |01⟩ + |10⟩ + |11⟩)/2, a product state of Schmidt rank 1; |Φ+⟩ has Schmidt rank 2 and is entangled, so the two cannot be the same vector.
Overlap: ⟨++|Φ+⟩ = (1/2)(1/√2)(1 + 1) = 1/√2, so the fidelity is |⟨++|Φ+⟩|² = 1/2.
Check with the inner-product form: cloning both |0⟩ and |+⟩ would need ⟨0|+⟩ = ⟨0|+⟩², but ⟨0|+⟩ = 1/√2 = 0.707 while its square is 0.5. Only ⟨0|1⟩ = 0 passes, which is why the σ_z pair alone can be copied.

AnswerU|+⟩|0⟩ = |Φ+⟩ = (|00⟩ + |11⟩)/√2, not |+⟩|+⟩; fidelity |⟨++|Φ+⟩|² = 1/2. A copier for the σ_z basis fails on σₓ eigenstates.

α = √3/2 = 0.866, β = 0.5. From ½[|Φ+⟩|ψ⟩ + |Φ−⟩Z|ψ⟩ + |Ψ+⟩X|ψ⟩ + |Ψ−⟩XZ|ψ⟩], Bob's branches as column vectors are 00 → (0.866, 0.5); 10 → Z|ψ⟩ = (0.866, −0.5); 01 → X|ψ⟩ = (0.5, 0.866); 11 → XZ|ψ⟩ = (−0.5, 0.866). Each carries the factor ½, so each probability is ¼.
Corrections Z^m₁X^m₂: 00 → I; 10 → Z; 01 → X; 11 → ZX. Check the last: X(−0.5, 0.866) = (0.866, −0.5), then Z gives (0.866, 0.5) = |ψ⟩ with no leftover phase.
Bloch vector of |ψ⟩: r = (2αβ, 0, α² − β²) = (0.866, 0, 0.5), polar angle 60°. The four branches have rₖ = (0.866, 0, 0.5), (−0.866, 0, 0.5), (0.866, 0, −0.5), (−0.866, 0, −0.5); their mean is (0, 0, 0), so ρ_B = ¼Σₖ σₖ|ψ⟩⟨ψ|σₖ† = I/2.
Fidelity in each branch, (1 + r⋅rₖ)/2: 00 → 1; 10 → (1 − 0.75 + 0.25)/2 = 0.25; 01 → (1 + 0.75 − 0.25)/2 = 0.75; 11 → (1 − 1)/2 = 0. Weighted by ¼ each the average is 0.5, the fidelity of a random guess.
Direct check of two: ⟨ψ|Z|ψ⟩ = α² − β² = 0.75 − 0.25 = 0.5, squared 0.25; ⟨ψ|XZ|ψ⟩ = 0.866 × (−0.5) + 0.5 × 0.866 = 0, squared 0. Both agree with the Bloch form.

AnswerBranches |ψ⟩, Z|ψ⟩, X|ψ⟩, XZ|ψ⟩, each with probability ¼, undone by I, Z, X, ZX. Before the bits ρ_B = I/2; branch fidelities 1, 0.25, 0.75 and 0 average to ½.

HardAlice and Bob share a Werner pair ρ_AB = F|Φ+⟩⟨Φ+| + (1 − F)(I₄ − |Φ+⟩⟨Φ+|)/3 with F = 0.85. Show that teleportation through it is a depolarising channel, find the output fidelity, and give the F below which measure-and-prepare (fidelity 2/3) does as well and the F needed for fidelity 0.95.

Rewrite the pair as ρ_AB = p|Φ+⟩⟨Φ+| + (1 − p)I₄/4. Matching the |Φ+⟩ weight gives p + (1 − p)/4 = F, so p = (4F − 1)/3; at F = 0.85, p = 0.80. Check: 0.80 + 0.20/4 = 0.85.
The protocol is linear in ρ_AB. On the |Φ+⟩⟨Φ+| part Bob receives |ψ⟩⟨ψ| exactly. On the I₄/4 part A and B are uncorrelated, so Bob's qubit is I/2 whatever Alice reads and whatever Pauli he applies, since σ(I/2)σ† = I/2; Alice's outcome probabilities stay ¼ each, so no bias enters.
Hence ρₒᵤₜ = p|ψ⟩⟨ψ| + (1 − p)I/2, a depolarising channel with Bloch shrink p, and ⟨ψ|ρₒᵤₜ|ψ⟩ = p + (1 − p)/2 = (1 + p)/2 = (2F + 1)/3, the same for every pure input.
F = 0.85: f = (1.70 + 1)/3 = 0.90. Cross-check: (1 + 0.80)/2 = 0.90.
Classical threshold: (2F + 1)/3 = 2/3 gives F = 1/2, p = 1/3. A Werner state is entangled only for F > 1/2, so the pair beats measure-and-prepare exactly when it is entangled. For f = 0.95: 2F + 1 = 2.85, F = 0.925.

Answerρₒᵤₜ = 0.80|ψ⟩⟨ψ| + 0.20 I/2, fidelity 0.90. Teleportation beats the classical 2/3 only for F > 1/2, the entanglement threshold; fidelity 0.95 needs F = 0.925.