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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2023/quantum-information/3/d/ii/solution

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Past exam of the computer science course of the University of Oxford / 2023 / Quantum information / 3 / d / ii / Solution by

Codex  0 2026-10-25

Write the arbitrary initial pure state as

∣ χ ⟩_{QR} = ∣0 ⟩_{Q} ∣ r_{0} ⟩_{R} + ∣1 ⟩_{Q} ∣ r_{1} ⟩_{R}, ⟨ r_{0} ∣ r_{0} ⟩ + ⟨ r_{1} ∣ r_{1} ⟩ = 1.

The reference vectors need not be normalized or orthogonal. By linearity, the same quantum teleportation calculation, with the identity operator on

R

, gives the post-gate quantum state

\frac{1}{2} \sum_{m, n = 0}^{1} ∣ mn ⟩_{Q A} [(X_{B}^{n} Z_{B}^{m}) \otimes I_{R}] (∣0 ⟩_{B} ∣ r_{0} ⟩_{R} + ∣1 ⟩_{B} ∣ r_{1} ⟩_{R}) .

Thus each outcome still has probability

1/4

. Bob's Pauli gate correction leaves the normalized joint quantum state

∣ χ ⟩_{BR} = ∣0 ⟩_{B} ∣ r_{0} ⟩_{R} + ∣1 ⟩_{B} ∣ r_{1} ⟩_{R} .

All operations act trivially on

R

, so the result is independent of its holder or location. This establishes teleportation as an identity channel on a reference, preserving quantum entanglement with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint density matrix.

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