LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Spin-two downward coefficients and detection-code output populations

Statement

In the ordered spin coupling $V_{1} \otimes V_{2} \to V_{1}$ , choose the common coupled phase for which $C_{1, - 1; 2, 2}^{1, 1}$ is positive. Define $B_{δ} e_{2, m} = \sqrt{5 / 3} C_{1, δ; 2, m}^{1, m + δ} e_{1, m + δ}$ , with out-of-range terms zero. Put $a = (e_{2, - 2} + e_{2, 2}) / \sqrt{2}$ and $b = e_{2, 0}$ . Let $∣ 0 ⟩ = (1, 0)^{T}$ and $∣ 1 ⟩ = (0, 1)^{T}$ be the ordered standard orthonormal basis of $C^{2}$ , and define the linear map $W : C^{2} \to V_{2}$ by $W ∣ 0 ⟩ = a$ , $W ∣ 1 ⟩ = b$ . Then $\sum_{δ} B_{δ}^{†} B_{δ} = I_{V_{2}}$ . The nonzero actions on the two codewords are $B_{- 1} a = e_{1, 1} / \sqrt{2}, B_{+ 1} a = e_{1, - 1} / \sqrt{2}, B_{- 1} b = e_{1, - 1} / \sqrt{6}, B_{+ 1} b = e_{1, 1} / \sqrt{6}, B_{0} b = - \sqrt{2 / 3} e_{1, 0} .$ In particular $B_{0} a = 0$ . The population vectors, in output order $m = - 1, 0, 1$ , of $\sum_{δ} B_{δ} ∣ a ⟩ ⟨ a ∣ B_{δ}^{†}$ and $\sum_{δ} B_{δ} ∣ b ⟩ ⟨ b ∣ B_{δ}^{†}$ are $(1 / 2, 0, 1 / 2)$ and $(1 / 6, 2 / 3, 1 / 6)$ . They remain these diagonal matrices if arbitrary phases are put on the two nonzero magnetic amplitudes of $a$ before applying the $B_{δ}$ .

Facts & Assumptions

[F1]

Use the specified spin matrices and the coupled highest-weight and lowering definition of each Clebsch–Gordan coefficient. Finite spin blocks, spherical tensors and Clebsch–Gordan phases

[F2]

For $J = 2$ the downward coefficient sum is $3 / 5$ , and multiplying all components by $\sqrt{5 / 3}$ makes their adjoint-product sum the identity. Downward dipole coefficient normalization contract

[F3]

Use the specified finite orthonormal bases, adjoints, outer products and ordered tensor coordinates. Finite operator and ket conventions

Proof

Given: The ordered coupling, phase and code vectors in the statement.

Goal: Compute the needed coefficients and normalized output populations.

1.1givenF1F3algebra

In the total magnetic-weight-one product basis $e_{1, - 1} \otimes e_{2, 2}$ , $e_{1, 0} \otimes e_{2, 1}$ , $e_{1, 1} \otimes e_{2, 0}$ , write the highest-weight vector as $(x, y, z)$ . The total raising equation gives $\sqrt{2} x + 2 y = 0$ and $\sqrt{2} y + \sqrt{6} z = 0$ . With positive $x$ and unit norm, these equations give $(x, y, z) = (\sqrt{3 / 5}, - \sqrt{3 / 10}, \sqrt{1 / 10})$ . Lowering this vector once and twice, with the spin-one factors $\sqrt{2}$ , gives $v_{1, 0} = \sqrt{3 / 10} (e_{1, - 1} \otimes e_{2, 1} + e_{1, 1} \otimes e_{2, - 1}) - \sqrt{2 / 5} e_{1, 0} \otimes e_{2, 0},$ and $v_{1, - 1} = \sqrt{1 / 10} e_{1, - 1} \otimes e_{2, 0} - \sqrt{3 / 10} e_{1, 0} \otimes e_{2, - 1} + \sqrt{3 / 5} e_{1, 1} \otimes e_{2, - 2}$ . These are the coupled vectors by the definition in [F1].

2.1F1F2F3step 1.1algebra∎

Reading their product-basis coefficients and multiplying by $\sqrt{5 / 3}$ gives exactly the five stated actions and $B_{0} a = 0$ . The identity $\sum B_{δ}^{†} B_{δ} = I$ follows from [F2]. For $a$ , its two magnetic components have different nonzero jump labels, so their output cross terms never occur within a single $B_{δ}$ conjugation. For $b$ , the three outputs have their stated squared amplitudes. Squaring and summing therefore gives the two diagonal population vectors, unchanged by phases on the two components of $a$ . A different common coupled phase multiplies all $B_{δ}$ by the same unit scalar and changes none of these conjugation or product claims.

Depends on

Used by · 3

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.