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Transverse-polarization completeness and angular average

PDF p. 12: Eq. (S3.12) and following text

Statement

Assume the Axiom of Countable Choice. Let $S^{2} = {n \in R^{3} : ∣ n ∣ = 1}$ have polar surface measure $σ$ , and let $e_{1}, e_{2}, e_{3}$ be the standard real coordinate basis. For each $n \in S^{2}$ , let $ε_{1}, ε_{2}$ be an orthonormal basis of the transverse subspace $n^{⊥} = {v \in C^{3} : \sum_{r = 1}^{3} v_{r} n_{r} = 0}$ , with respect to $⟨ u, v ⟩ = \sum_{r = 1}^{3} u_{r} \overline{v_{r}}$ . Thus transverse means perpendicular to $n$ , and orthonormal means $⟨ ε_{a}, ε_{b} ⟩ = δ_{a b}$ . For $i, j \in {1, 2, 3}$ , $\sum_{a = 1}^{2} (e_{i} \cdot ε_{a}) \overline{(e_{j} \cdot ε_{a})} = δ_{i j} - n_{i} n_{j}, \int_{S^{2}} (δ_{i j} - n_{i} n_{j}) d σ (n) = \frac{8 π}{3} δ_{i j} .$ Here $e_{i} \cdot v = v_{i}$ and $δ_{i j}$ is one when $i = j$ and zero otherwise. In particular, the angular integral is $8 π / 3$ on the diagonal and zero off the diagonal.

Statement references

The Axiom of Countable Choice (AC_ω)

Real and complex inner product spaces, with the inner product linear in the first argument

Orthogonal vectors and subspaces, orthogonal and orthonormal sets, and orthonormal bases

The polar surface set function on the unit sphere

Transverse projector and isotropic angular contraction

Facts & Assumptions

Given: Countable Choice, the sphere, measure, transverse bases and indices as in the statement.

Goal: Prove the component completeness identity and the stated angular integral.

[F1]

Countable Choice says that every sequence of nonempty sets has a choice function. The Axiom of Countable Choice (AC_ω)

[F2]

The displayed inner product uses the linear-first convention, and $σ$ is the polar surface measure on the unit sphere. Real and complex inner product spaces, with the inner product linear in the first argument The polar surface set function on the unit sphere

[F3]

Under Countable Choice, for an orthonormal complex basis of $n^{⊥}$ , the transverse-angular-average lemma gives $\sum_{a} ε_{a, i} \overline{ε_{a, j}} = δ_{i j} - n_{i} n_{j}$ and $\int_{S^{2}} (δ_{i j} - n_{i} n_{j}) d σ (n) = 8 π δ_{i j} / 3$ . Transverse projector and isotropic angular contraction

Proof

1.1givenF1F2F3

For each fixed $n$ , the given basis satisfies the hypotheses of [F3]. Since $e_{i} \cdot ε_{a} = ε_{a, i}$ , its first identity is exactly the displayed completeness identity.

2.1F3step 1.1algebra∎

The second identity of [F3] gives the integral of the right-hand side of step 1.1. Substituting $δ_{i j} = 1$ for $i = j$ and $δ_{i j} = 0$ otherwise gives the two stated values. The integrand is the pointwise projector expression, so no measurable choice of transverse basis is needed.

Depends on

Used by · 1

Off-diagonal correlators vanish in an isotropic photon bath

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.

Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 12, Eq. (S3.12) and following text