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A fixed Cartesian polarization is not a global transverse helicity frame

Statement refuted

Let $e_{1}, e_{2}, e_{3}$ be the standard coordinate basis of $R^{3}$ . The constant Cartesian circular vectors $e_{σ}^{0} = (e_{1} + i σ e_{2}) / \sqrt{2}$ , for $σ = \pm 1$ , form a transverse helicity frame for every nonzero momentum direction; consequently this fixed pair supplies a momentum-independent polarization factor on the transverse one-photon space.

Statement references

Transverse one-photon space and helicity projectors

Facts & Assumptions

Given: Countable Choice and the transverse helicity setting below; the additional full Choice premise [A1] is used only for the rotation-generator conclusion in step 3.1.

[A1]

For [F3] and step 3.1, assume the Axiom of Choice: every family of nonempty sets has a function selecting a member of each set. The fixed-pair refutation in steps 1.1–2.1 uses only the Countable Choice setting of [F4]–[F5]. The Axiom of Choice

[F1]

Each helicity fiber $L_{σ, n}$ lies in the transverse plane $n^{⊥}$ . Transverse one-photon space and helicity projectors

[F2]

The standard first Chern number of $L_{σ}$ is $- 2 σ \neq 0$ , and no global smooth unit helicity section exists. Helicity line bundles have nonzero Chern number

[F3]

Under [A1], fix $ℏ > 0$ and $a \in R^{3}$ with $∣ a ∣ = 1$ . On $D = {f \in C_{c}^{\infty} (M; C^{3}) : k \cdot f (k) = 0 for all k \in M}$ , put $L_{a} f = - i ℏ (a \times k) \cdot \nabla_{k} f$ and $S_{a}^{o p} f = i ℏ a \times f$ , with the cross product extended complex-linearly. The self-adjoint Stone generator of the rotation action $U (R) f (k) = R f (R^{- 1} k)$ , with convention $U (R_{a} (t)) = e^{- i t J_{a} / ℏ}$ and $R_{a} (t) = \exp (t [a]_{\times})$ , $[a]_{\times} v = a \times v$ , restricts on $D$ to $J_{a} = L_{a} + S_{a}^{o p}$ . Its orbital and spin terms generally do not separately preserve transversality. The full Choice premise is required by this supplier's Stone-theorem application. Transverse rotations preserve helicity and generate total angular momentum

[F4]

The momentum domain is $M = R^{3} ∖ {0}$ with $d ν = d^{3} k / (2 ∣ k ∣)$ , and $H_{T}^{ν}$ consists of $L^{2} (M, ν; C^{3})$ equivalence classes satisfying $k \cdot f (k) = 0$ for $ν$ -almost every $k$ . At each $k \in M$ , the transverse fiber is $T_{k} = {v \in C^{3} : k \cdot v = 0}$ , using the complex-bilinear extension of the real dot product. Transverse one-photon space and helicity projectors

[F5]

Under Countable Choice, every Borel subset of $R^{3}$ is Lebesgue measurable; closed coordinate boxes have measure equal to the product of their side lengths; $λ_{3}$ is a measure on the Lebesgue sigma-algebra; and measures are monotone on measurable inclusions. Countable Choice is already part of the transverse-space setting in [F4]. Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume Measures are monotone

Counterexample

Given: Countable Choice, a nonzero momentum $k \in M$ , and the constant vectors in the Statement; additionally [A1] for step 3.1.

To make the fixed-pair factorization claim precise, let $A = L^{2} (M, ν; C^{2})$ , with norm $∥ a ∥^{2} = \int_{M} (∣ a_{+} (k) ∣^{2} + ∣ a_{-} (k) ∣^{2}) d ν (k)$ . For each $k \in M$ define $Φ_{k} : C^{2} \to C^{3}$ by $Φ_{k} (z_{+}, z_{-}) = z_{+} e_{+}^{0} + z_{-} e_{-}^{0}$ , and define $Φ : A \to L^{2} (M, ν; C^{3})$ by $(Φ a) (k) = Φ_{k} (a_{+} (k), a_{-} (k))$ . The two fixed vectors are Hermitian-orthonormal, so $Φ$ is an isometry into the ambient vector-valued $L^{2}$ space. The fixed-pair factorization claim tested below is that this map takes values in $H_{T}^{ν}$ and is unitary onto it; pointwise, it claims that the same two vectors supply each transverse fiber $T_{k}$ .

1.1F1givenalgebra

Take $k = κ e_{1}$ with $κ > 0$ . Then $k \cdot e_{σ}^{0} = κ / \sqrt{2} \neq 0$ for either $σ$ , so neither fixed vector lies in the transverse plane $k^{⊥}$ . By [F1], neither can belong to the corresponding helicity line at this momentum.

1.2F2given

Independently, [F2] excludes any global smooth unit helicity eigenvector, even if the polarization vector is allowed to depend on momentum. Thus a smooth global helicity gauge cannot repair the fixed-frame claim.

1.3F4F5givenconstructalgebra

Let $A = B_{1 / 4} (e_{1})$ , $Q = [7 / 8, 9 / 8] \times [- 1 / 8, 1 / 8]^{2}$ , and $Q^{'} = [3 / 4, 5 / 4] \times [- 1 / 4, 1 / 4]^{2}$ . For $k \in Q$ , $∣ k - e_{1} ∣^{2} \leq 3 / 64 < 1 / 16$ , so $Q \subset A$ . For $k \in A$ every coordinate displacement from $e_{1}$ has absolute value less than $1 / 4$ , so $A \subset Q^{'}$ . The ball is open: if $k \in A$ , then the ball of radius $1 / 4 - ∣ k - e_{1} ∣ > 0$ about $k$ lies in $A$ by the triangle inequality. It is therefore Borel and Lebesgue measurable by [F5]. The box-volume formula and measure monotonicity in [F5] give $1 / 64 = (1 / 4)^{3} = λ_{3} (Q) \leq λ_{3} (A) \leq λ_{3} (Q^{'}) = (1 / 2)^{3} = 1 / 8$ . The triangle inequality gives $3 / 4 < ∣ k ∣ < 5 / 4$ on $A$ , so $2 / 5 \leq 1 / (2 ∣ k ∣) \leq 2 / 3$ there. Integrating these inequalities yields $0 < (2 / 5) λ_{3} (A) \leq ν (A) \leq (2 / 3) λ_{3} (A) < \infty$ . Thus $a = (1_{A}, 0)$ belongs to $A$ . For every $k \in A$ , $(Φ a) (k) = e_{+}^{0}$ and $k \cdot e_{+}^{0} = (k_{1} + i k_{2}) / \sqrt{2} \neq 0$ , since $k_{1} > 3 / 4$ . Thus $Φ a$ fails the transverse condition on the positive-measure set $A$ ; in particular, $Φ$ does not define a map from $A$ into $H_{T}^{ν}$ , and cannot be the proposed unitary factorization.

2.1step 1.1step 1.2step 1.3discharge-contradiction

The displayed fixed pair therefore is not a transverse helicity frame on all momentum directions, and the explicitly defined fixed-pair map fails to factor through the transverse one-photon space. The Chern obstruction separately excludes a global smooth helicity frame even when vectors are allowed to depend smoothly on momentum. This refutes the statement.

3.1A1F3given∎

Under the additional full Choice premise [A1], the rotation theorem [F3] also identifies the precise limitation of separating orbital and spin terms: their displayed differential expressions need not individually send the transverse domain into itself, whereas their sum does. Hence those separate restrictions do not give independent operators from $D$ to $H_{T}^{ν}$ on the common transverse domain. This refutes a decomposition by those particular restrictions; it does not rule out arbitrary Hilbert-space tensor-product isomorphisms.

A fixed Cartesian polarization is not a global transverse helicity frame

Statement refuted

Statement references

Facts & Assumptions

Counterexample

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