LemmaStatement: AI-generatedProof: AI-generated

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Nonvanishing downward coefficients for spin one and two

Statement

In the ordered coupling $V_{1} \otimes V_{J}$ , for $J = 1, 2$ , $m = - J, \dots, J$ and $q = - 1, 0, 1$ , let $c_{m, q} = C_{1, q; J, m}^{J - 1, m + q}$ when $∣ m + q ∣ \leq J - 1$ and $c_{m, q} = 0$ otherwise, and put $b_{J, m, q} = \sqrt{(2 J + 1) / (2 J - 1)} c_{m, q}$ . The coefficient $b_{J, m, q}$ is nonzero exactly when $∣ m + q ∣ \leq J - 1$ . For $J = 1$ the three allowed squared magnitudes are one. For $J = 2$ , for $q = - 1$ and $m = 0, 1, 2$ the squared magnitudes are $1 / 6, 1 / 2, 1$ ; for $q = 0$ and $m = - 1, 0, 1$ they are $1 / 2, 2 / 3, 1 / 2$ ; the $q = 1$ values are the reflected $q = - 1$ values. In each block $\sum_{q = - 1}^{1} B_{q}^{†} B_{q} = I_{V_{J}}$ , where $B_{q} : V_{J} \to V_{J - 1}$ is the linear map defined by $B_{q} e_{J, m} = b_{J, m, q} e_{J - 1, m + q}$ when $∣ m + q ∣ \leq J - 1$ and $B_{q} e_{J, m} = 0$ otherwise; adjoints are taken in the specified orthonormal bases.

Statement references

Finite operator and ket conventions

Finite spin blocks, spherical tensors and Clebsch–Gordan phases

Downward dipole coefficient normalization contract

Facts & Assumptions

[F1]

The coupled vectors are normalized highest-weight vectors followed by normalized total lowering, with the displayed spin ladder matrices. Finite spin blocks, spherical tensors and Clebsch–Gordan phases

[F2]

The sum of squared downward coefficients at fixed $m$ is $(2 J - 1) / (2 J + 1)$ . Downward dipole coefficient normalization contract

[F3]

In the specified orthonormal bases, the adjoint of a linear map between finite complex coordinate spaces is represented by its conjugate transpose, and operator products use matrix multiplication. Finite operator and ket conventions

Proof

Given: The finite data and hypotheses in the Statement.

Goal: Establish the asserted dimension or support identity.

1.1F1algebra

For $J = 1$ the normalized total-spin-zero vector is $(e_{1, - 1} \otimes e_{1, 1} - e_{1, 0} \otimes e_{1, 0} + e_{1, 1} \otimes e_{1, - 1}) / \sqrt{3}$ : total raising and lowering annihilate it by the ladder formulas. Multiplication by $\sqrt{3}$ proves the three stated magnitudes. A permitted common phase has no effect.

2.1F1F2F3step 1.1algebra∎

For $J = 2$ the total-spin-one highest vector, in ordered coordinates $(q, m) = (- 1, 2), (0, 1), (1, 0)$ , has coefficients $(\sqrt{3 / 5}, - \sqrt{3 / 10}, \sqrt{1 / 10})$ . Its norm is one and total raising kills it: the two resulting coordinates have coefficients $\sqrt{2} \sqrt{3 / 5} - 2 \sqrt{3 / 10} = 0$ and $- \sqrt{2} \sqrt{3 / 10} + \sqrt{6} \sqrt{1 / 10} = 0$ . Total lowering divided by $\sqrt{2}$ gives at total magnetic label zero the coefficients $(\sqrt{3 / 10}, - \sqrt{2 / 5}, \sqrt{3 / 10})$ in coordinates $(- 1, 1), (0, 0), (1, - 1)$ . A second normalized lowering gives the reflected highest-vector coefficients at total label minus one. Squaring and multiplying by $5 / 3$ proves the table and nonvanishing. Terms outside these three total labels vanish. For fixed $q$ , nonzero columns of $B_{q}$ have distinct output basis vectors; by [F3], $B_{q}^{†} B_{q}$ is diagonal with $m$ th entry $∣ b_{J, m, q} ∣^{2}$ , including zero for forbidden outputs. Scaling [F2] by $(2 J + 1) / (2 J - 1)$ gives their sum as the identity.

Depends on

Used by · 2

Dependency tree · 2 levels

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