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Clebsch–Gordan total-rate identity

PDF p. 14: Eq. (S4.2), qualification on p. 15

Statement

Let $J \in {1, \frac{3}{2}, 2, \dots}$ and $m = - J, - J + 1, \dots, J$ . In the ordered coupling $V_{1} \otimes V_{J}$ , take $C_{1, δ; J, m}^{J - 1, m + δ}$ to be the coefficient of $∣ 1, δ ⟩ \otimes ∣ J, m ⟩$ in the normalized spin- $(J - 1)$ coupled vector of magnetic label $m + δ$ , and take it to be zero when that label is outside the spin- $(J - 1)$ range. Then $\sum_{δ \in {0, \pm 1}} ∣ C_{1, δ; J, m}^{J - 1, m + δ} ∣^{2} = (2 J - 1) / (2 J + 1)$ . The $J = 0$ ground manifold is excluded from the article's downward-transition sums.

Facts & Assumptions

Given: A spin label $J \geq 1$ , an admissible magnetic label $m$ , and the coefficient convention in the Statement.

Goal: Establish the displayed sum for each such $J$ and $m$ .

[F1]

The angular basis uses the ordered coupling and zero convention stated above. Its coefficients are the product-basis coordinates of normalized coupled vectors, with a common highest-weight phase for the spin- $(J - 1)$ summand. Angular-momentum basis and tensor labels Finite spin blocks, spherical tensors and Clebsch–Gordan phases

[F2]

The product decomposes orthogonally into invariant spin summands. The spin- $(J - 1)$ summand has dimension $2 J - 1$ and an orthonormal basis $v_{M}$ of total magnetic weights $M = - (J - 1), \dots, J - 1$ . The total spin components are $J_{a}^{(1)} \otimes I + I \otimes J_{a}^{(J)}$ . Tensor spin composition

[F3]

In $V_{J}$ , the magnetic basis diagonalizes $J_{z}^{(J)}$ , and $J_{+}^{(J)} ∣ J, m ⟩ = \sqrt{(J - m) (J + m + 1)} ∣ J, m + 1 ⟩$ with nonzero coefficient whenever $m < J$ . Finite spin blocks, spherical tensors and Clebsch–Gordan phases

[F4]

For integer $J \geq 1$ , the stated coefficient sum is the downward dipole coefficient normalization contract. Downward dipole coefficient normalization contract

[F5]

For a subspace $W$ of a finite-dimensional inner-product space with orthonormal basis $(v_{M})_{M}$ , its orthogonal projection is the linear map $P_{W} x = \sum_{M} ⟨ x, v_{M} ⟩ v_{M}$ , with image $W$ , kernel $W^{⊥}$ , and $P_{W}^{2} = P_{W}$ . Orthogonal projection is linear, and an orthonormal basis (eᵢ) of W gives P_Wv=∑ᵢ⟨ v,eᵢ⟩ eᵢ

[F6]

For a finite square matrix $B = (B_{i j})$ over $C$ , $tr B = \sum_{i} B_{i i}$ . Every trace below is taken in the specified magnetic or product basis. The trace tr(A) as the sum of the diagonal entries

[F7]

If $B \in M_{r \times s} (C)$ and $D \in M_{s \times r} (C)$ , then $tr (B D) = tr (D B)$ ; in particular this applies to square matrices of size three. For A∈ M_m× n(F) and B∈ M_n× m(F), tr(AB)=tr(BA)

Proof

1.1F1F2F4F5F6F7algebra

If $J$ is an integer, the result follows from [F4] with $q = δ$ and the coefficient convention of [F1]. It remains to consider half-integer $J \geq 1$ . Let $P$ be the orthogonal projection onto the spin- $(J - 1)$ summand, and define $A$ on $V_{J}$ by $A_{m n} = \sum_{δ = - 1}^{1} P_{(δ, m), (δ, n)}$ in the ordered product basis. The summand and its orthogonal complement are invariant under the self-adjoint total spin components, so $P$ commutes with them. Summing the diagonal entries in the first tensor factor of each such commutator gives $[A, J_{a}^{(J)}] = 0$ : the terms acting on the first factor cancel by finite matrix trace cyclicity. Indeed, for fixed $m, n$ , put $Q_{δ η} = P_{(δ, m), (η, n)}$ and $B = J_{a}^{(1)}$ . This contribution is $tr (Q B - B Q) = 0$ by [F6] and [F7], since both matrices have size three.

2.1step 1.1F2F3F5F6algebra

By commutation with $J_{z}^{(J)}$ , $A$ is diagonal in the magnetic basis. Commutation with $J_{+}^{(J)}$ and the nonzero ladder coefficients show that all its diagonal entries are equal, so $A = a I_{V_{J}}$ . Taking traces gives $(2 J + 1) a = tr A = tr P = 2 J - 1$ , hence $a = (2 J - 1) / (2 J + 1)$ . Here [F6] and the definition of $A$ give $tr A = \sum_{m, δ} P_{(δ, m), (δ, m)} = tr P$ . Writing $(v_{M})_{δ, m}$ for the product-basis coordinates, [F5] gives $tr P = \sum_{M} \sum_{δ, m} ∣ (v_{M})_{δ, m} ∣^{2} = \sum_{M} 1 = 2 J - 1$ by the normalization in [F2].

3.1step 2.1F1F2F5algebra∎

By [F5], expand $P$ in the orthonormal basis $v_{M}$ . A product basis vector with labels $(δ, m)$ has total magnetic weight $m + δ$ , so its diagonal entry of $P$ is the squared modulus of the coefficient in the Statement when that weight is admissible, and is zero otherwise. Consequently the sum of these three squared moduli is $A_{m m} = a$ , as required. A common phase change of the coupled vectors leaves the squared moduli unchanged.

Depends on

Used by · 3

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 14, Eq. (S4.2), and p. 15, paragraph following it