advanced27 min studyLesson 427 in Physics

Quantum teleportation resource accounting

Track Bell-measurement branches, Pauli corrections, and the separate entanglement and classical communication resources of teleportation.

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What you’ll understand

  • Derive the four teleportation branches with a fixed factor order.
  • Apply the inverse Pauli correction to a concrete input.
  • Account for the consumed ebit and two communicated classical bits.

Teleportation transfers a quantum state between physical systems using previously shared entanglement and a later classical message. Alice need not know the amplitudes of the input, and no copy of that unknown state remains at her laboratory. To see why there is no faster-than-light signal or cloning paradox, keep three named factors throughout: input C, Alice's resource qubit A, and Bob's resource qubit B. An ideal complete Bell measurement and reliable classical feed-forward are explicit assumptions.

Re-express three qubits in a Bell basis

Alice and Bob initially share the Bell state with positive zero-zero and one-one amplitudes. Let the independent input on C have amplitudes alpha and beta. Expand the three-qubit product in the Bell basis of C tensor A. Collecting the B amplitudes gives four possible conditional vectors, related to the input by identity, Z, X, and XZ. Each branch has amplitude factor one half and probability one quarter, independent of the unknown normalized input.

Use the definitions ∣Φ±⟩=(∣00⟩±∣11⟩)/2|\Phi^\pm\rangle=(|00\rangle\pm|11\rangle)/\sqrt2 and ∣Ψ±⟩=(∣01⟩±∣10⟩)/2|\Psi^\pm\rangle=(|01\rangle\pm|10\rangle)/\sqrt2 on the stated factor order. In particular, the minus Psi branch gives XZ times the input, rather than ZX under this convention. The two differ only by a global sign, but preserving one convention makes corrections checkable and prevents relative-sign mistakes when the protocol is embedded in a larger circuit.

∣ψ⟩C∣Φ+⟩AB=12(∣Φ+⟩CA∣ψ⟩B+∣Φ−⟩CAZ∣ψ⟩B+∣Ψ+⟩CAX∣ψ⟩B+∣Ψ−⟩CAXZ∣ψ⟩B)|\psi\rangle_C|\Phi^+\rangle_{AB}=\frac12\left(|\Phi^+\rangle_{CA}|\psi\rangle_B+|\Phi^-\rangle_{CA}Z|\psi\rangle_B+|\Psi^+\rangle_{CA}X|\psi\rangle_B+|\Psi^-\rangle_{CA}XZ|\psi\rangle_B\right)
The factors are C,A,B; X and Z act only on B. This identity assumes an independent input and a perfect shared Bell pair. Each conditional B vector is normalized.↔ Scroll wide equations horizontally.

The classical message selects an inverse

Alice sends a two-bit label for her four-valued Bell result. Bob applies the corresponding inverse: identity, Z, X, or ZX. Operator products act right to left, so the last correction means first X and then Z. The corrected output equals the input for every alpha and beta. Linearity also preserves arbitrary correlations with an untouched reference, which is a stronger requirement than reproducing the probabilities of one chosen input basis.

Before Bob receives the label, his state is the average over the four equally probable Pauli branches. This average is the maximally mixed qubit, independent of the input. Alice cannot choose her Bell result to encode a desired instantaneous signal. The classical message does not describe the continuous amplitudes either; it merely identifies which of four known corrections is needed. Information becomes usable only after that message reaches Bob through an ordinary causal communication channel.

14(ρ+ZρZ+XρX+(XZ)ρ(XZ)†)=IB2,1 ebit+2 cbits ⟹ 1 transmitted qubit\frac14\left(\rho+Z\rho Z+X\rho X+(XZ)\rho(XZ)^\dagger\right)=\frac{I_B}{2},\qquad 1\ {\rm ebit}+2\ {\rm cbits}\ \Longrightarrow\ 1\ {\rm transmitted\ qubit}
The average holds for every normalized qubit density operator. The resource statement is for ideal deterministic teleportation, consuming one pre-shared Bell pair and two forward classical bits.↔ Scroll wide equations horizontally.

Count what the protocol consumes

The resource pair is consumed: after Alice's Bell measurement, A is part of the measured pair C:A rather than still sharing the original entanglement with B. Establishing that pair beforehand can itself require a quantum channel, purification, storage, and synchronization. Treating those costs as free is a resource convention, not a demonstration that classical communication alone transmits an arbitrary unknown state. Bob still needs a physical qubit and controlled local operations.

The input's quantum information is transferred, not its matter or energy. Alice's measured qubits do not retain a second independent copy. A postselected experiment that reports only one Bell outcome implements a conditional protocol with an associated success probability; it is not automatically the deterministic four-outcome scheme. Noisy entanglement, incomplete Bell measurement, miscalibrated Pauli operations, and mismatched reference frames change the realized channel. Their effect must be quantified separately from this exact ideal identity.

Correct the minus-Psi branch

Teleport ∣ψ⟩=(3∣0⟩+4∣1⟩)/5|\psi\rangle=(3|0\rangle+4|1\rangle)/5 using the stated Bell convention. Alice reports the minus-Psi outcome. Find Bob's conditional state and verify its correction.

  1. Normalization is 9/25+16/25=19/25+16/25=1. The minus-Psi branch applies XZ: Z first changes the one amplitude to −4/5, then X swaps amplitudes, giving (−4∣0⟩+3∣1⟩)/5(-4|0\rangle+3|1\rangle)/5.

  2. The branch probability is one quarter. Bob must invert XZ, whose adjoint is ZX, so he first applies X to obtain (3∣0⟩−4∣1⟩)/5(3|0\rangle-4|1\rangle)/5.

  3. Applying Z now restores the positive one amplitude, yielding (3∣0⟩+4∣1⟩)/5(3|0\rangle+4|1\rangle)/5. The correction uses the two-bit outcome label, with no estimate of either unknown amplitude required.

The result

Bob recovers the exact input after ZX. The branch occurs with probability 1/4 and is included in a deterministic protocol with feed-forward.

Explore the model

Interactive lab

Interactive lab · quantum

Inspect a phase-flip correction geometrically

For each fresh comparison, Reset and set both controls. At time zero, Reset and set both controls to angle 60 degrees and phase 90 degrees. Predict Bloch y=√3/2 and z=1/2. In a separate fresh run set angle 60 and phase −90 degrees; y changes sign while z and basis probabilities stay fixed, as for a Z phase flip. These are independently selected pure states, not a continuous control operation or a teleportation simulation; the lab has no Bell pair, measurement outcomes, or classical communication.

Inspect a phase-flip correction geometricallyA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA known Pauli correction changes state geometry without needing to reveal its unknown amplitudes.

Check your understanding

Before receiving Alice’s Bell-result message, what state does Bob have in ideal qubit teleportation when he averages over that unknown result?

Try it yourself

Take input ∣+⟩=(∣0⟩+∣1⟩)/2|+\rangle=(|0\rangle+|1\rangle)/\sqrt2. If Alice's Bell outcome is Phi-minus, find the conditional output before correction. Then average all four uncorrected branches and evaluate squared fidelity with the original input.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Two Pauli branches produce plus projectors and two produce minus projectors; their global phases do not affect density operators.

Compare with the solution

The Phi-minus branch is Z∣+⟩=∣−⟩Z|+\rangle=|-\rangle, so its squared fidelity with plus is zero. Identity and X give plus up to phase; Z and XZ give minus up to phase. Averaging gives (∣+⟩⟨+∣+∣−⟩⟨−∣)/2=I/2(|+\rangle\langle+|+|-\rangle\langle-|)/2=I/2. Its squared fidelity with the pure plus target is ⟨+∣I/2∣+⟩=1/2\langle+|I/2|+\rangle=1/2. Applying the outcome-dependent corrections restores fidelity one in every ideal branch.

Take it with you

  • Teleportation is a channel identity implemented with entanglement and classical feed-forward.
  • Two classical bits specify a correction, not a complete classical description of an unknown qubit.
  • The original shared ebit is consumed, and no usable signal reaches Bob before the message.

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