LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Four-branch open-graph pattern for CZ

Statement

Take vertices $1, 2$ as inputs and $3, 4$ as outputs, with edges ${1, 2}, {1, 3}, {2, 4}$ . Prepare $C Z_{12} C Z_{13} C Z_{24} (∣ ψ ⟩_{12} ∣ + ⟩_{3} ∣ + ⟩_{4})$ and measure vertices $1, 2$ in the $J (0) = H$ basis. For outcomes $b_{1}, b_{2} \in {0, 1}$ , the branch operator from the two input wires to the two output wires is

$K_{b_{1} b_{2}} = \frac{1}{2} (X^{b_{1}} H \otimes X^{b_{2}} H) C Z .$

Each of the four branches has probability $1 / 4$ for every normalized input, and the terminal correction $(H X^{b_{1}} \otimes H X^{b_{2}})$ yields the corrected branch $\frac{1}{2} C Z ∣ ψ ⟩$ .

Facts & Assumptions

Given: The ordered wire convention, the open-graph pattern definition, the one-bit branch identity, and the controlled-phase gate $C Z$ .

[F1]

One-bit teleportation has branch operator $2^{- 1 / 2} X^{b} J (α)$ for each outcome (One-bit J-gate teleportation branch identity).

[F2]

The graph pattern branch map applies the graph-edge unitary before the ordered measurement bras and preserves the stated input/output order (Finite open-graph adaptive measurement pattern).

[F3]

$J (0) = H$ by J-gate basis and exact phase identities, and $C Z = diag (1, 1, 1, - 1)$ by Controlled Phase Rotation.

Proof

1.1F1F2F3algebra

The vertical edge ${1, 2}$ applies $C Z$ to the arbitrary two-qubit input before the two horizontal teleportation branches; the horizontal edges ${1, 3}$ and ${2, 4}$ implement the one-bit identity independently on the two wires. Thus [F1]–[F3] give the branch operator $K_{b_{1} b_{2}} = \frac{1}{2} (X^{b_{1}} H \otimes X^{b_{2}} H) C Z$ .

2.1step 1.1algebra

In the declared wire order, the four branch operators are $K_{00} = \frac{1}{2} (H \otimes H) C Z$ , $K_{01} = \frac{1}{2} (H \otimes X H) C Z$ , $K_{10} = \frac{1}{2} (X H \otimes H) C Z$ , and $K_{11} = \frac{1}{2} (X H \otimes X H) C Z$ . Each factor after $1 / 2$ is unitary, so $K_{b_{1} b_{2}}^{†} K_{b_{1} b_{2}} = I / 4$ and every normalized input has probability $1 / 4$ .

3.1step 2.1F3algebra∎

Since $(X^{b} H)^{†} = H X^{b}$ , applying $H X^{b_{1}} \otimes H X^{b_{2}}$ to the branch cancels both local factors and leaves $\frac{1}{2} C Z ∣ ψ ⟩$ . All four branches have the same corrected normalized output $C Z ∣ ψ ⟩$ .

Depends on

Used by

One-edge graph state and two-wire CZ pattern Example

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415, §20.6, p. 180 (standard reference, not scraped)
Robert Raussendorf, Daniel E. Browne, and Hans J. Briegel, Measurement-based quantum computation on cluster states, CNOT correlation proof, pp. 20–21 (standard reference, not scraped)