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.

Downward dipole coefficient normalization contract

Statement

Let $J \geq 1$ be an integer. In the ordered coupling $V_{1} \otimes V_{J}$ , let $c_{m, q} = C_{1, q; J, m}^{J - 1, m + q}$ for $m = - J, \dots, J$ and $q = - 1, 0, 1$ , with $c_{m, q} = 0$ when $∣ m + q ∣ > J - 1$ . The coefficients are those of the normalized coupled basis, with any allowed common phase choice for the spin- $(J - 1)$ summand. Then $\sum_{q = - 1}^{1} ∣ c_{m, q} ∣^{2} = (2 J - 1) / (2 J + 1)$ for every $m$ . Equivalently, defining $T_{q} : V_{J} \to V_{J - 1}$ by $T_{q} e_{J, m} = c_{m, q} e_{J - 1, m + q}$ , with an out-of-range output interpreted as zero, gives $\sum_{q = - 1}^{1} T_{q}^{†} T_{q} = ((2 J - 1) / (2 J + 1)) I_{V_{J}}$ . The $J = 0$ block is excluded.

Facts & Assumptions

Given: An integer $J \geq 1$ , the ordered coupling $V_{1} \otimes V_{J}$ , and the coefficients and maps in the Statement.

Goal: Prove the coefficient sum and the equivalent operator identity.

[F1]

The spin blocks have orthonormal magnetic bases, with $J_{z}^{(J)} e_{J, m} = m e_{J, m}$ and $J_{+}^{(J)} e_{J, m} = \sqrt{(J - m) (J + m + 1)} e_{J, m + 1}$ . The coupled coefficients are the product-basis coordinates of the normalized coupled vectors; their coupling order and common phase are specified in the Statement. Finite spin blocks, spherical tensors and Clebsch–Gordan phases

[F2]

The ordered product $V_{1} \otimes V_{J}$ is an orthogonal direct sum of invariant spin blocks. Its spin- $(J - 1)$ block has dimension $2 J - 1$ and an orthonormal basis $v_{M}$ , $M = - (J - 1), \dots, J - 1$ , of total $J_{z}$ eigenvectors of eigenvalue $M$ . The total spin operators are $J_{a}^{(1)} \otimes I + I \otimes J_{a}^{(J)}$ . Tensor spin composition

[F3]

All matrix entries and products are in the specified orthonormal bases. For a map $T : H \to K$ between these finite complex coordinate spaces, $T^{†} : K \to H$ is its adjoint, satisfying $⟨ T x, y ⟩_{K} = ⟨ x, T^{†} y ⟩_{H}$ for the inner products linear in the first argument; its matrix is the conjugate transpose. The finite-coordinate construction in the supplier verifies this identity and supplies existence. Finite operator and ket conventions

[F4]

If $W$ is a subspace of a finite-dimensional inner-product space with orthogonal decomposition $W \oplus W^{⊥}$ , its orthogonal projection takes $w + z$ to $w$ for $w \in W$ and $z \in W^{⊥}$ . We regard this as an endomorphism of the ambient space by the inclusion of $W$ . The orthogonal projection P_Wv is the W-component in V=W⊕ W^⊥

[F5]

The trace of a square matrix is the sum of its diagonal entries. Every trace below is taken in the specified magnetic or product basis. The trace tr(A) as the sum of the diagonal entries

Proof

1.1F1F2F3F4algebra

Let $W$ be the spin- $(J - 1)$ block and let $P$ be its orthogonal projection, viewed as an endomorphism of $V_{1} \otimes V_{J}$ . Thus $P (w + z) = w$ for $w \in W$ and $z \in W^{⊥}$ . The block and its orthogonal complement are invariant under each self-adjoint total operator, so $P$ commutes with every total operator. Indeed, if $S$ is such an operator, both $P S (w + z)$ and $S P (w + z)$ equal $S w$ . Since $(v_{M})_{M}$ is an orthonormal basis of $W$ , taking inner products with each $v_{M}$ gives $P x = \sum_{M} ⟨ x, v_{M} ⟩ v_{M}$ . Define $A$ on $V_{J}$ by $A_{m n} = \sum_{q = - 1}^{1} P_{(q, m), (q, n)}$ in the product basis. These finite matrix entries specify a linear endomorphism.

2.1step 1.1F1F2F3algebra

Fix $a \in {x, y, z}$ , and write $B = J_{a}^{(1)}$ and $D = J_{a}^{(J)}$ . Expanding the first-factor contribution gives $\sum_{q} [P, B \otimes I]_{(q, m), (q, n)} = \sum_{q, r} (P_{(q, m), (r, n)} B_{r q} - B_{q r} P_{(r, m), (q, n)}) = 0$ , because exchanging the finite indices $q, r$ in the second sum makes it identical to the first. The other contribution is $\sum_{q} [P, I \otimes D]_{(q, m), (q, n)} = \sum_{k} (A_{m k} D_{k n} - D_{m k} A_{k n}) = [A, D]_{m n}$ . Step 1.1 gives $[P, B \otimes I + I \otimes D] = 0$ , so $[A, J_{a}^{(J)}] = 0$ for each $a$ .

3.1step 1.1step 2.1F1F2F3F5algebra

Since $A$ commutes with $J_{z}^{(J)}$ , its entries between distinct magnetic labels vanish. Write $A e_{J, m} = a_{m} e_{J, m}$ . Commutation with $J_{+}^{(J)} = J_{x}^{(J)} + i J_{y}^{(J)}$ and its nonzero ladder coefficient for $- J \leq m < J$ give $a_{m + 1} = a_{m}$ . Thus $A = a I_{V_{J}}$ . Put $d_{M; q, m} = ⟨ v_{M}, e_{1, q} \otimes e_{J, m} ⟩$ , the product-basis coordinate of $v_{M}$ . The projection formula from step 1.1 gives $P_{(q, m), (q, m)} = \sum_{M} ∣ d_{M; q, m} ∣^{2}$ . Consequently the diagonal-sum definition of trace gives $tr A = \sum_{m, q} P_{(q, m), (q, m)} = tr P = \sum_{M} \sum_{q, m} ∣ d_{M; q, m} ∣^{2} = \sum_{M} 1 = 2 J - 1$ , using normalization of each $v_{M}$ . Also $tr A = (2 J + 1) a$ , hence $a = (2 J - 1) / (2 J + 1)$ . No change of basis for the trace is required.

4.1step 3.1F1F2F3algebra∎

Since $v_{M}$ has total magnetic label $M$ , its product-basis coordinate $d_{M; q, m}$ vanishes unless $M = m + q$ . Thus $P_{(q, m), (q, m)}$ is $∣ C_{1, q; J, m}^{J - 1, m + q} ∣^{2}$ when $∣ m + q ∣ \leq J - 1$ and is zero otherwise. Therefore $\sum_{q} ∣ c_{m, q} ∣^{2} = A_{m m} = (2 J - 1) / (2 J + 1)$ . By the conjugate-transpose formula, $(T_{q}^{†} T_{q})_{m n} = \sum_{ℓ = - (J - 1)}^{J - 1} \overline{(T_{q})_{ℓ m}} (T_{q})_{ℓ n}$ . For fixed $q$ , distinct $m$ give distinct output labels $m + q$ , so this matrix is diagonal with entry $∣ c_{m, q} ∣^{2}$ at $m$ ; summing proves the operator identity. A common phase change of the coupled basis does not affect either identity.

Depends on

Used by · 4

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.