Empirical/QM: Quantum teleportation (E5) #
Category: 3-Local (promotion-ready to 2-Framework on demand).
Quantum teleportation (Bennett et al. 1993): Alice transmits an unknown qubit
|ψ⟩ = α|0⟩ + β|1⟩ to Bob using one shared Bell pair |Φ⁺⟩ and two classical
bits. Alice performs a Bell-basis measurement on her input qubit together with
her half of the entangled pair; the two-bit outcome tells Bob which of the four
Pauli corrections {I, Z, X, ZX} to apply to his half to recover |ψ⟩.
This file delivers the algebraic core — the branch-conditional form
(§0.1 QM-validity layer, branch-conditional reading; see specs/qm-empirical-tests.md):
teleState_factorises— the three-qubit input state is genuinely|ψ⟩₁ ⊗ |Φ⁺⟩₂₃(amplitude factorisationΨ(a,b,c) = ψ(a)·Φ⁺(b,c)).teleportation_bell_expansion— the teleportation identity: re-expanding in Alice's Bell basis on qubits (1,2),Ψ = ½ Σₖ |Bellₖ⟩₁₂ ⊗ (Cₖ|ψ⟩)₃, withCₖ ∈ {I, Z, X, XZ}the Pauli image in branchk. So in each Bell-measurement branch Bob holds a Pauli image of|ψ⟩.teleportation_branch_recovers_input— applying the branch correction recovers|ψ⟩exactly in all four branches (branch-conditional; the measurement-collapse / ¼-branch-probability layer is out of scope, see the honesty note below).
The three qubits live in EuclideanSpace ℂ (Fin 2 × Fin 2 × Fin 2) (matching
Empirical/QM/Multipartite/GHZ.lean), with qubit 1 = Alice's input, qubit 2 =
Alice's half of the pair, qubit 3 = Bob's half. This is the dual of
superdense coding (Resources/SuperdenseCoding.lean): there one entangled qubit
carries two classical bits; here two classical bits + one entangled qubit carry
one quantum state.
Honesty note #
This is the QM-validity layer in branch-conditional form: it proves the
Bell-basis expansion and per-branch recovery as exact algebraic identities. The
measurement step proper — collapsing the four-branch superposition to a single
classical outcome with probability ¼ each — is the measurement-update notion the
corpus does not yet have (the same LF5 obligation that gates BB84). The
no-signalling content ("Bob's marginal is I/2 independent of the branch before
he learns the outcome") likewise needs the reduced-state / partial-trace
machinery absent from Mathlib; it is not asserted here.
Source #
Bennett, Brassard, Crépeau, Jozsa, Peres, Wootters 1993, Phys. Rev. Lett. 70, 1895. Experimental: Bouwmeester et al. 1997, Nature 390, 575.
States #
The Bell state |Φ⁺⟩ = (|00⟩ + |11⟩)/√2 on qubits (2,3).
Equations
- CSD.Empirical.QM.Teleportation.bellPhiPlus = (↑√2)⁻¹ • (EuclideanSpace.single (0, 0) 1 + EuclideanSpace.single (1, 1) 1)
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|Φ⁻⟩ = (|00⟩ − |11⟩)/√2.
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|Ψ⁺⟩ = (|01⟩ + |10⟩)/√2.
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- CSD.Empirical.QM.Teleportation.bellPsiPlus = (↑√2)⁻¹ • (EuclideanSpace.single (0, 1) 1 + EuclideanSpace.single (1, 0) 1)
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|Ψ⁻⟩ = (|01⟩ − |10⟩)/√2.
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The full three-qubit input |ψ⟩₁ ⊗ |Φ⁺⟩₂₃, written out in components:
(1/√2)(α|000⟩ + α|011⟩ + β|100⟩ + β|111⟩). (The tensor structure is verified
by teleState_factorises.)
Equations
- One or more equations did not get rendered due to their size.
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Bob's conditional vectors (the Pauli images of |ψ⟩) #
Branch |Φ⁺⟩: Bob holds I|ψ⟩ = |ψ⟩.
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Branch |Φ⁻⟩: Bob holds Z|ψ⟩ = α|0⟩ − β|1⟩.
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Branch |Ψ⁺⟩: Bob holds X|ψ⟩ = β|0⟩ + α|1⟩.
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Branch |Ψ⁻⟩: Bob holds XZ|ψ⟩ = −β|0⟩ + α|1⟩.
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Pauli correction matrices #
Pauli X = !![0, 1; 1, 0]. (Not in Gates/SingleQubit.lean, which has only
H, S, T, Z.)
Equations
- CSD.Empirical.QM.Teleportation.teleX = !![0, 1; 1, 0]
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Pauli Z = !![1, 0; 0, −1].
Equations
- CSD.Empirical.QM.Teleportation.teleZ = !![1, 0; 0, -1]
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The |Ψ⁻⟩-branch correction ZX = !![0, 1; −1, 0] (apply X then Z).
Equations
- CSD.Empirical.QM.Teleportation.teleZX = !![0, 1; -1, 0]
Instances For
Faithfulness: the input state is genuinely |ψ⟩ ⊗ |Φ⁺⟩ #
The teleportation identity (Bell-basis expansion) #
Teleportation identity. Re-expanding the input |ψ⟩₁ ⊗ |Φ⁺⟩₂₃ in Alice's
Bell basis on qubits (1,2):
Ψ(a,b,c) = ½ ( Φ⁺(a,b)·(I|ψ⟩)(c) + Φ⁻(a,b)·(Z|ψ⟩)(c) + Ψ⁺(a,b)·(X|ψ⟩)(c) + Ψ⁻(a,b)·(XZ|ψ⟩)(c) ).
So a Bell measurement on qubits (1,2) collapses Bob's qubit 3 to a Pauli image of
|ψ⟩, the image determined by the two-bit outcome.
Per-branch recovery #
Branch |Φ⁺⟩ recovery: I · (I|ψ⟩) = |ψ⟩.
Branch |Φ⁻⟩ recovery: Z · (Z|ψ⟩) = |ψ⟩.
Branch |Ψ⁺⟩ recovery: X · (X|ψ⟩) = |ψ⟩.
Branch |Ψ⁻⟩ recovery: ZX · (XZ|ψ⟩) = |ψ⟩.
Teleportation recovers the input in every branch. Bundles the four
per-branch corrections: with the two-bit Bell-measurement outcome, Bob's Pauli
correction returns |ψ⟩ exactly.