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The spatial rotation action on the component-valued wave-function space

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice), the choice hypotheses of the cited spectral suppliers. Fix an integer $r \geq 1$ and $ℏ > 0$ , let $H^{(r)} = L^{2} (R^{3}; C^{r})$ with inner product linear in the first argument (Complex Lp classes and Euclidean test-function conventions, The space $L^{p} (μ)$ as the quotient by null functions, Hilbert space), let $Q_{j}, P_{k}$ be the componentwise position and momentum operators (Componentwise position and momentum operators on the component-valued wave-function space clause (a)), and let $S^{(r)} = S (R^{3}; C^{r})$ and $W^{m, 2} (R^{3}; C^{r})$ be the componentwise Schwartz and Sobolev classes (Schwartz space and its seminorms, Integer-order Sobolev spaces and their norms). Let $S O (3)$ be the special orthogonal group with its Euclidean topology (The orthogonal group $O (3)$ and the rotation group $S O (3)$ , $S O (3)$ is a subgroup of the general linear group, with $R^{- 1} = R^{T}$ ) and for $R \in S O (3)$ define, componentwise, $(U (R) f)_{l} (x) : = f_{l} (R^{- 1} x) (x \in R^{3}, l = 1, \dots, r) .$ Let $U^{(r)}$ be the unitary momentum representation of The momentum representation of the componentwise kinematics and $M_{R}$ the bounded operator on $L^{2} (R^{3}, d p; C^{r})$ given by $(M_{R} g) (p) : = g (R^{- 1} p)$ . Then:

(a) $U (R)$ is a well-defined unitary operator for every $R$ , with $U (I) = i d$ , $U (R S) = U (R) U (S)$ and $U (R)^{- 1} = U (R^{- 1})$ for all $R, S \in S O (3)$ ; and for every $f \in H^{(r)}$ the orbit map $R \mapsto U (R) f$ is continuous on $S O (3)$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form). Moreover $U (R) S^{(r)} = S^{(r)}$ and $U (R) W^{m, 2} (R^{3}; C^{r}) = W^{m, 2} (R^{3}; C^{r})$ for $m = 1, 2$ ;

(b) $U^{(r)} U (R) U^{(r) - 1} = M_{R}$ . For $a \in R^{3}$ define the maximal directional operators $Q (a) : = M_{a \cdot x}, P (a) : = U^{(r) - 1} M_{a \cdot p} U^{(r)},$ where each multiplier has its maximal domain, namely the classes for which its product is in $L^{2}$ . Thus $D (Q (a)) = {f : (a \cdot x) f \in L^{2}}, D (P (a)) = {f : (a \cdot p) U^{(r)} f \in L^{2}} .$ They are self-adjoint (for $a = 0$ they are the zero operator on all of $H^{(r)}$ ). On the common coordinate domains they agree with $\sum_{l} a_{l} Q_{l}$ and $\sum_{l} a_{l} P_{l}$ , respectively. The full conjugation identities, including domains, are $U (R) Q_{j} U (R)^{- 1} = Q (R e_{j}), U (R) P_{k} U (R)^{- 1} = P (R e_{k}),$ $U (R) D (Q_{j}) = D (Q (R e_{j})), U (R) D (P_{k}) = D (P (R e_{k})) .$ In particular, an individual coordinate domain is preserved when $R e_{j} = \pm e_{j}$ (respectively $R e_{k} = \pm e_{k}$ ); no such preservation is asserted for an arbitrary rotation. The expressions $\sum_{l} R_{l j} Q_{l}$ and $\sum_{l} R_{l k} P_{l}$ in full operator identities mean these maximal directional realizations.

(c) with the free operator $H_{0} : = \frac{1}{2 m_{0}} \sum_{k = 1}^{3} P_{k}^{2}$ for a parameter $m_{0} > 0$ , whose domain is $W^{2, 2} (R^{3}; C^{r})$ , one has $U (R) H_{0} U (R)^{- 1} = H_{0}$ ; and for a real bounded measurable $V : R^{3} \to R$ with multiplier $M_{V}$ (multiplication by $V$ ) and $V_{R} (x) : = V (R^{- 1} x)$ one has $U (R) M_{V} U (R)^{- 1} = M_{V_{R}}$ , so that $U (R) H_{V} U (R)^{- 1} = H_{V_{R}}$ for the free-plus- $V$ operator $H_{V}$ with domain $W^{2, 2} (R^{3}; C^{r})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and the Axiom of Choice; $r \geq 1$ ; $ℏ > 0$ ; the space $H^{(r)}$ ; the componentwise operators $Q_{j}, P_{k}$ ; $R, S \in S O (3)$ ; $f \in H^{(r)}$ ; $ψ \in S^{(r)}$ ; the momentum unitary $U^{(r)}$ and the multipliers $M_{R}$ ; a real bounded measurable $V$ .

[F1]

Componentwise kinematics. $Q_{j}, P_{k}$ are self-adjoint with domains $D (Q_{j}) = {f : x_{j} f_{l} \in L^{2} \forall l}$ , $D (P_{k}) = {f : D_{k} f_{l} \in L^{2} \forall l}$ , and on $S^{(r)}$ one has $Q_{j} f = x_{j} f$ , $P_{k} f = - i ℏ D_{k} f$ componentwise; $S^{(r)} \subseteq D (Q_{j}) \cap D (P_{k})$ and both operators map $S^{(r)}$ into itself (Componentwise position and momentum operators on the component-valued wave-function space clauses (a),(c), Schwartz space and its seminorms).

[F2]

Momentum representation. $U^{(r)}$ is unitary and satisfies $U^{(r)} D (P_{k}) = D (M_{p_{k}})$ and $U^{(r)} P_{k} = M_{p_{k}} U^{(r)}$ ; on Schwartz classes it acts by the componentwise Fourier transform with normalization $(2 π ℏ)^{- 3 / 2}$ , and it maps $S^{(r)}$ onto itself (The momentum representation of the componentwise kinematics clauses (a),(b), Componentwise position and momentum operators on the component-valued wave-function space clause (b), Schwartz space and its seminorms).

[F3]

Substitution and components. For an orthogonal $R$ and $f \in L^{1} (R^{3}; C)$ one has $\int f (R^{- 1} x) d x = \int f (y) d y$ , because the linear substitution scales Lebesgue measure by $∣ \det R ∣ = 1$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, $S O (3)$ is a subgroup of the general linear group, with $R^{- 1} = R^{T}$ ). A class $f = (f_{l})$ of $H^{(r)}$ is determined by its components, and componentwise operations preserve the $L^{2}$ norm: $∥ f ∥^{2} = \sum_{l} ∥ f_{l} ∥^{2}$ (Unitary identifications of the component-valued wave-function space clause (a), The induced length is a norm).

[F4]

Schwartz calculus. $ψ \in S^{(r)}$ means each component $ψ_{l}$ is smooth with $\sup_{x} ∣ x^{α} \partial^{β} ψ_{l} (x) ∣ < \infty$ for all multi-indices $α, β$ ; hence each component and each of its derivatives is bounded, and $S^{(r)} \subseteq W^{2, 2} (R^{3}; C^{r})$ (Schwartz space and its seminorms, Integer-order Sobolev spaces and their norms). For an invertible linear substitution $A$ with $∣ A x ∣ \leq ∥ A ∥ ∣ x ∣$ , the chain rule gives $\partial^{β} (ψ_{l} \circ A) (x) = \sum_{∣ γ ∣ = ∣ β ∣} c_{γ} \partial^{γ} ψ_{l} (A x)$ with constants depending only on $A, β$ , so the seminorms of $ψ_{l} \circ A$ are bounded by finitely many seminorms of $ψ_{l}$ ; in particular $S^{(r)}$ is invariant under composition with an invertible linear map.

[F5]

Sobolev and Fourier characterization. For a scalar position-space class $f$ , let $g = U^{(1)} f$ be its momentum representative in the normalization of [F2], with $p = 2 π ℏ ξ$ . Then $f \in W^{1, 2}$ exactly when $p_{k} g \in L^{2}$ for all $k$ , and $f \in W^{2, 2}$ exactly when $∣ p ∣^{2} g \in L^{2}$ . Moreover $∥ f ∥_{W^{2, 2}}$ is equivalent to $∥ (1 + ∣ p ∣^{2}) g ∥_{2}$ for fixed $ℏ > 0$ . These statements apply componentwise (Integer-order W^{k,2} and H^k agree with equivalent norms, Plancherel theorem, Integer-order Sobolev spaces and their norms).

[F6]

Density and extension. $S^{(r)}$ is dense in $H^{(r)}$ (Schwartz space is dense in L2); a bounded linear operator on a dense subspace extends uniquely to the whole space with the same norm (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm); if a sequence $g_{n} \to g$ pointwise and $∣ g_{n} ∣ \leq h$ with $h$ integrable, then $\int g_{n} \to \int g$ (Dominated convergence); and for $u, v \in R^{3}$ one has $∣ u \cdot v ∣ \leq ∣ u ∣ ∣ v ∣$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

Proof

technique · direct; define the rotation action componentwise, prove unitarity and the representation property by orthogonal substitution, transport the action to the momentum representation for the operator conjugations, and obtain strong continuity from dominated convergence and density

Given: The data of the statement and [F1]-[F6].

1.1F3givenalgebra

[Well-definedness and unitarity] Let $R \in S O (3)$ . For a representative $f$ of a class in $H^{(r)}$ the function $x \mapsto f (R^{- 1} x)$ is measurable and its components satisfy $\int ∣ f_{l} (R^{- 1} x) ∣^{2} d x = \int ∣ f_{l} (y) ∣^{2} d y$ by [F3], so the assignment is well defined on classes, linear, and isometric; it is surjective with inverse given by $R^{- 1}$ because composition with $R^{- 1}$ undoes it on representatives, so $U (R)$ is a unitary operator and $U (R)^{- 1} = U (R^{- 1})$ .

2.1step 1.1F3F4algebra

[Representation and Schwartz invariance] For $R, S \in S O (3)$ and $f$ one has $(U (R) U (S) f)_{l} (x) = (U (S) f)_{l} (R^{- 1} x) = f_{l} (S^{- 1} R^{- 1} x) = f_{l} ((R S)^{- 1} x) = (U (R S) f)_{l} (x)$ , since $(R S)^{- 1} = S^{- 1} R^{- 1}$ by [F3]; and $U (I) = i d$ . For $ψ \in S^{(r)}$ each component $ψ_{l} \circ R^{- 1}$ is smooth with all seminorms finite by [F4] applied to $A = R^{- 1}$ , and the same argument with $R$ in place of $R^{- 1}$ gives $U (R) S^{(r)} = S^{(r)}$ .

2.2step 1.1F1givenalgebra

[Conjugation of position] For $f \in H^{(r)}$ , the condition $U (R)^{- 1} f \in D (Q_{j})$ is, by orthogonal substitution, exactly $(\sum_{l} R_{l j} x_{l}) f \in L^{2}$ . On this domain the componentwise computation gives $(U (R) Q_{j} U (R)^{- 1} f) (x) = (R^{- 1} x)_{j} f (x) = (R e_{j} \cdot x) f (x)$ . Thus the conjugate is exactly the maximal multiplier $Q (R e_{j})$ , and $U (R) D (Q_{j}) = D (Q (R e_{j}))$ . More generally $Q (a)$ is self-adjoint: for $a \neq 0$ choose a rotation taking $e_{j}$ to $a / ∣ a ∣$ , and the same computation makes $Q (a)$ a real scalar multiple of a unitary conjugate of the self-adjoint $Q_{j}$ ; the case $a = 0$ is immediate. On the common coordinate domain multiplication distributes to give $Q (a) f = \sum_{l} a_{l} Q_{l} f$ .

3.1step 1.1step 2.1F4F6givenalgebra

[Strong continuity] Let $ψ \in S^{(r)}$ and $R_{0} \in S O (3)$ . For $R$ in a bounded neighbourhood of $R_{0}$ one has $∣ R^{- 1} x ∣ \geq \frac{1}{2} ∣ x ∣$ and $∣ R_{0}^{- 1} x ∣ \geq \frac{1}{2} ∣ x ∣$ , so with the radial majorant $h (x) : = \sup_{∣ y ∣ \geq ∣ x ∣ / 2} \max_{l} ∣ ψ_{l} (y) ∣$ one has $∣ (U (R) ψ - U (R_{0}) ψ)_{l} (x) ∣^{2} \leq 4 h (x)^{2}$ for every $x$ , where $h \in L^{2}$ because $ψ$ is Schwartz; moreover $(U (R) ψ)_{l} (x) \to (U (R_{0}) ψ)_{l} (x)$ for every $x$ as $R \to R_{0}$ , by continuity of $ψ_{l}$ and of $R \mapsto R^{- 1} x$ . Dominated convergence gives $∥ U (R) ψ - U (R_{0}) ψ ∥ \to 0$ , and for $ε / 3$ -close $ψ$ to $f$ in the sense of [F6] the triangle inequality and unitarity of step 1.1 give $∥ U (R) f - U (R_{0}) f ∥ \leq 2 ∥ f - ψ ∥ + ∥ U (R) ψ - U (R_{0}) ψ ∥ < ε$ for $R$ close to $R_{0}$ ; hence the orbit map is continuous.

3.2step 2.1F2F3F6givenalgebra

[Momentum-space form of the action] Let $ψ \in S^{(r)}$ and compute componentwise for $p \in R^{3}$ , using the substitution $y = R^{- 1} x$ , $x = R y$ of [F3] and $R^{T} = R^{- 1}$ : $(U^{(r)} U (R) ψ)_{l} (p) = (2 π ℏ)^{- 3 / 2} \int e^{- i p \cdot x / ℏ} ψ_{l} (R^{- 1} x) d x = (2 π ℏ)^{- 3 / 2} \int e^{- i p \cdot R y / ℏ} ψ_{l} (y) d y = (2 π ℏ)^{- 3 / 2} \int e^{- i R^{T} p \cdot y / ℏ} ψ_{l} (y) d y = (U^{(r)} ψ)_{l} (R^{- 1} p) = (M_{R} U^{(r)} ψ)_{l} (p)$ . Both $U^{(r)} U (R) U^{(r) - 1}$ and $M_{R}$ are bounded operators (products of unitaries), and they agree on the dense subspace $U^{(r)} S^{(r)} = S^{(r)}$ of [F2]; [F6] extends the identity to all of $L^{2} (R^{3}, d p; C^{r})$ .

4.1step 3.2F2F5givenalgebra

[Sobolev invariance] By [F2] and [F5], $W^{1, 2} (R^{3}; C^{r})$ and $W^{2, 2} (R^{3}; C^{r})$ correspond in momentum space to the conditions $p_{k} g \in L^{2}$ for every $k$ , respectively $∣ p ∣^{2} g \in L^{2}$ . By step 3.2 the rotated representative is $g_{R} (p) = g (R^{- 1} p)$ . Set $q = R^{- 1} p$ , so $d p = d q$ and $p_{k} = \sum_{l} R_{k l} q_{l}$ ; therefore each $p_{k} g_{R} (p)$ is the rotation-substituted finite linear combination $\sum_{l} R_{k l} q_{l} g (q)$ , which is in $L^{2}$ when every $q_{l} g (q)$ is, and the converse follows by applying the same argument to $R^{- 1}$ . For the second-order condition, $∣ p ∣^{2} g_{R} (p) = ∣ q ∣^{2} g (q)$ under the same measure-preserving substitution, and the converse again follows with $R^{- 1}$ . Thus $U (R) W^{m, 2} = W^{m, 2}$ for $m = 1, 2$ .

4.2step 3.2step 2.2F2givenalgebra

[Conjugation of momentum] By [F2] and step 3.2, the momentum representation of $U (R) P_{k} U (R)^{- 1}$ is $M_{R} M_{p_{k}} M_{R}^{- 1}$ . Orthogonal substitution shows that its domain is exactly ${g : (R e_{k} \cdot p) g \in L^{2}}$ , and its action there is multiplication by $(R^{- 1} p)_{k} = \sum_{l} R_{l k} p_{l}$ . Transporting by $U^{(r)}$ gives exactly $P (R e_{k})$ and $U (R) D (P_{k}) = D (P (R e_{k}))$ . For any nonzero $a$ , the same rotation-and-scaling argument as step 2.2 makes $M_{a \cdot p}$ and hence $P (a)$ self-adjoint; for $a = 0$ this is immediate. On the common momentum domain $P (a) f = \sum_{l} a_{l} P_{l} f$ by [F2] and distribution of the multiplier. This proves the full directional identities of (b).

5.1step 4.1step 4.2F1givenalgebra

[Free operator and multipliers] Since $U (R) P_{k} U (R)^{- 1} = \sum_{l} R_{l k} P_{l}$ on the maximal directional domains by step 4.2 and $U (R) W^{2, 2} = W^{2, 2}$ by step 4.1, also $U (R) P_{k}^{2} U (R)^{- 1} = \sum_{l, l^{'}} R_{l k} R_{l^{'} k} P_{l} P_{l^{'}}$ on $W^{2, 2}$ ; summing over $k$ and using $\sum_{k} R_{l k} R_{l^{'} k} = δ_{l l^{'}}$ for orthogonal $R$ gives $U (R) (\sum_{k} P_{k}^{2}) U (R)^{- 1} = \sum_{l} P_{l}^{2}$ , so $U (R) H_{0} U (R)^{- 1} = H_{0}$ . For $f \in H^{(r)}$ one has $(U (R) M_{V} U (R)^{- 1} f)_{l} (x) = V (R^{- 1} x) f_{l} (x)$ by step 1.1, so $U (R) M_{V} U (R)^{- 1} = M_{V_{R}}$ on all of $H^{(r)}$ ; adding the two identities on the common domain $W^{2, 2}$ gives $U (R) H_{V} U (R)^{- 1} = H_{V_{R}}$ , which is (c).

6.1step 1.1step 2.1step 3.1step 3.2step 4.1step 4.2step 2.2step 5.1∎

[Assembly] Steps 1.1, 2.1, 3.1 prove unitarity, the representation property, the Schwartz invariance, strong continuity and, with step 4.1, the Sobolev invariance of (a); steps 3.2, 4.2, 2.2 prove (b); step 5.1 proves (c).

Remarks

Convention. With $U (R) f = f \circ R^{- 1}$ the position operator transforms with the inverse matrix, $U (R) Q_{j} U (R)^{- 1} = \sum_{l} R_{l j} Q_{l}$ ; this is exactly the conjugation identity computed in Binney's (4.26) with $U^{†} (α) x U (α) = R (α) x$ . The invariant statement used by the physics items is the multiplier identity $U (R) M_{V} U (R)^{- 1} = M_{V \circ R^{- 1}}$ and the criterion $V \circ R^{- 1} = V$ .
Scope. The lemma concerns the spatial factor only. No rotation action on the internal factor $C^{r}$ and no SU(2) lift are introduced; those belong to the orbital-and-spin development.

The spatial rotation action on the component-valued wave-function space

Statement

Facts & Assumptions

Proof

Remarks

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