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The free flow as a unimodular multiplier in the momentum representation

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice) and the choice hypotheses of the cited spectral suppliers. Fix an integer $r \geq 1$ , a mass parameter $m > 0$ and $ℏ > 0$ , let $H^{(r)} = L^{2} (R^{3}; C^{r})$ with the componentwise operators $Q_{j}, P_{k}$ of Componentwise position and momentum operators on the component-valued wave-function space, let $U^{(r)}$ be the momentum unitary of The momentum representation of the componentwise kinematics, and let $H_{0} : = \frac{1}{2 m} \sum_{k = 1}^{3} P_{k}^{2}, D (H_{0}) = W^{2, 2} (R^{3}; C^{r}),$ which is self-adjoint with the displayed domain (Real multiplication operators and free plus bounded-potential Schrödinger operators clause (c) applied componentwise, Componentwise position and momentum operators on the component-valued wave-function space clause (a)). Write $M_{a}$ for multiplication by a measurable function $a$ on $L^{2} (R^{3}, d p; C^{r})$ and put $f_{t} (p) : = e^{- i t ∣ p ∣^{2} / (2 m ℏ)}$ (The complex exponential by its power series). Then:

(a) $U^{(r)} H_{0} U^{(r) - 1} = M_{∣ p ∣^{2} / (2 m)}$ as self-adjoint operators, with domain $D (M_{∣ p ∣^{2} / (2 m)}) = {g \in L^{2} : ∣ p ∣^{2} g \in L^{2}}$ ;

(b) $U^{(r)} e^{- i t H_{0} / ℏ} U^{(r) - 1} = M_{f_{t}}$ for every $t \in R$ ;

(c) for every $ψ \in H^{(r)}$ , every $t$ and almost every $p$ , $(U^{(r)} e^{- i t H_{0} / ℏ} ψ) (p) = f_{t} (p) (U^{(r)} ψ) (p), ∣ (U^{(r)} e^{- i t H_{0} / ℏ} ψ) (p) ∣ = ∣ (U^{(r)} ψ) (p) ∣;$ consequently the spectral outcome law of each $P_{k}$ is independent of $t$ ; and $e^{- i t H_{0} / ℏ} S^{(r)} = S^{(r)}$ for every $t$ , where $S^{(r)} = S (R^{3}; C^{r})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and the Axiom of Choice; integers $r \geq 1$ ; $m, ℏ > 0$ ; the space $H^{(r)}$ ; the componentwise operators $Q_{j}, P_{k}$ ; the momentum unitary $U^{(r)}$ ; the free operator $H_{0}$ ; a vector $ψ \in H^{(r)}$ ; a Schwartz class $u \in S^{(r)}$ ; real $t$ .

[F1]

Free operator. $H_{0} = \frac{1}{2 m} \sum_{k} P_{k}^{2}$ with domain $W^{2, 2} (R^{3}; C^{r})$ is self-adjoint; it is the componentwise assembly of the scalar free operators of Real multiplication operators and free plus bounded-potential Schrödinger operators clause (b), which are self-adjoint with domain $W^{2, 2} (R^{3}; C)$ per component and equal $- \frac{ℏ^{2}}{2 m} Δ$ on Schwartz classes (Real multiplication operators and free plus bounded-potential Schrödinger operators clauses (b),(c), Componentwise position and momentum operators on the component-valued wave-function space clause (a), Integer-order Sobolev spaces and their norms).

[F2]

Momentum representation. $U^{(r)}$ is unitary, $U^{(r)} D (P_{k}) = D (M_{p_{k}})$ with $U^{(r)} P_{k} = M_{p_{k}} U^{(r)}$ , and $U^{(r)}$ maps $S^{(r)}$ onto itself and acts there by the componentwise Fourier transform with normalization $(2 π ℏ)^{- 3 / 2}$ (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]

Domain in momentum form. With $p$ the momentum variable of [F2], $W^{2, 2} (R^{3}; C^{r}) = {ψ \in H^{(r)} : ∣ p ∣^{2} U^{(r)} ψ \in L^{2}}$ and $W^{1, 2} (R^{3}; C^{r}) = {ψ : p_{k} U^{(r)} ψ \in L^{2} \forall k}$ as sets, by the Fourier characterization of integer-order Sobolev spaces read in the normalization of [F2] (Integer-order W^{k,2} and H^k agree with equivalent norms, Plancherel theorem, Integer-order Sobolev spaces and their norms).

[F4]

Conjugation calculus. For a unitary $W$ and a self-adjoint $K$ with spectral PVM $E_{K}$ , the operator $W K W^{- 1}$ with domain $W D (K)$ is self-adjoint with PVM $W E_{K} (\cdot) W^{- 1}$ and flow $e^{- i t W K W^{- 1} / ℏ} = W e^{- i t K / ℏ} W^{- 1}$ (Unitary conjugation, generator transport and strong commutation of self-adjoint observables clause (a)), and if two self-adjoint operators have a common domain $D$ and agree on a dense linear subspace of $D$ , then they are equal: their difference is symmetric on $D$ and vanishes on a dense subspace, hence is zero (Symmetric, self-adjoint and essentially self-adjoint operators).

[F5]

Multiplier spectral data. For a real measurable finite-valued function $a$ the multiplication operator $M_{a}$ on $L^{2} (R^{3}, d p; C^{r})$ is self-adjoint with domain ${g : a g \in L^{2}}$ and spectral PVM $M_{1_{a^{- 1} (B)}}$ ; for bounded Borel $h$ the bounded calculus satisfies $Φ_{M_{a}} (h) = M_{h \circ a}$ and $∥ M_{g} ∥ = {ess sup}_{p} ∣ g (p) ∣$ for bounded measurable $g$ ; these are the componentwise versions of the scalar multiplier facts (Spectral measure of a real L2 multiplication operator, Spectral theorem for unbounded self-adjoint operators (PVM form), Bounded borel pvm integral, Unbounded Borel functional calculus: domains, products, spectral mapping, Integral of a measurable function against a projection-valued measure).

[F6]

Density. $S^{(r)}$ is dense in $H^{(r)}$ (Schwartz space is dense in L2).

Proof

technique · direct; identify the momentum-space form of the free operator as a multiplier, obtain the flow by the conjugation calculus and read the modulus invariance and Schwartz preservation off the multiplier formula

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

1.1F1F2F3F4F5F6givenalgebra

[Momentum form of the free operator] Put $M : = M_{∣ p ∣^{2} / (2 m)}$ , self-adjoint with domain $D (M) = U^{(r)} W^{2, 2}$ by [F3] and [F5], and put $A : = U^{(r) - 1} M U^{(r)} = U^{(r) *} M U^{(r)}$ ; by [F4] $A$ is self-adjoint with domain $D (A) = U^{(r) - 1} D (M) = W^{2, 2}$ . On a Schwartz class $u \in S^{(r)}$ one has $P_{k}^{2} u \in S^{(r)}$ by [F2] and hence $H_{0} u = \frac{1}{2 m} \sum_{k} P_{k}^{2} u$ by [F1], while $U^{(r)} H_{0} u = M_{∣ p ∣^{2} / (2 m)} U^{(r)} u$ by [F2]; thus $A u = H_{0} u$ for every $u \in S^{(r)}$ . Both operators are self-adjoint with the common domain $W^{2, 2}$ and agree on the dense subspace $S^{(r)}$ [F6], so $A = H_{0}$ by [F4], which is (a).

2.1step 1.1F4F5givenalgebra

[Flow] Applying the flow identity of [F4] to $A = U^{(r) - 1} M U^{(r)} = H_{0}$ gives $U^{(r) - 1} e^{- i t M / ℏ} U^{(r)} = e^{- i t H_{0} / ℏ}$ , hence by the functional calculus of [F5] $M_{f_{t}} = e^{- i t M / ℏ}$ , $f_{t} (p) = e^{- i t ∣ p ∣^{2} / (2 m ℏ)}$ , and therefore $U^{(r)} e^{- i t H_{0} / ℏ} U^{(r) - 1} = M_{f_{t}}$ , which is (b).

3.1step 2.1F2givenalgebra

[Pointwise phase and modulus] Writing $g : = U^{(r)} ψ$ , step 2.1 gives $U^{(r)} e^{- i t H_{0} / ℏ} ψ = M_{f_{t}} g$ with $∣ f_{t} (p) ∣ = 1$ for every $p$ , so the two displays of (c) hold for almost every $p$ .

3.2step 2.1F2F5givenalgebra

[Schwartz preservation] Let $u \in S^{(r)}$ ; by [F2], $g : = U^{(r)} u \in S^{(r)}$ . The function $p \mapsto f_{t} (p) g (p)$ is smooth, and every seminorm $\sup_{p} ∣ p^{α} \partial^{β} (f_{t} g) (p) ∣$ is finite because $f_{t}$ is a unimodular exponential of a polynomial and hence has derivatives of polynomial growth while $g$ is Schwartz; thus $f_{t} g \in S^{(r)}$ , and $U^{(r) - 1} (f_{t} g) \in S^{(r)}$ by [F2] again. Applying the same argument to $- t$ gives the reverse inclusion, so $e^{- i t H_{0} / ℏ} S^{(r)} = S^{(r)}$ .

4.1step 3.1F2F5givenalgebra

[Momentum laws] By [F2], $E_{P_{k}} (B) = U^{(r) - 1} M_{1_{B} (p_{k})} U^{(r)}$ , so for a unit $ψ$ the law is $μ_{ψ}^{P_{k}} (B) = ∥ M_{1_{B} (p_{k})} U^{(r)} ψ ∥^{2} = \int_{{p_{k} \in B}} ∣ (U^{(r)} ψ) (p) ∣^{2} d p$ ; by step 3.1 the modulus $∣ (U^{(r)} e^{- i t H_{0} / ℏ} ψ) (p) ∣$ equals $∣ (U^{(r)} ψ) (p) ∣$ almost everywhere, so the integral is independent of $t$ .

5.1step 1.1step 2.1step 3.1step 3.2step 4.1∎

[Assembly] Step 1.1 proves (a), step 2.1 proves (b), and steps 3.1, 3.2 and 4.1 prove (c): the pointwise phase and modulus, the Schwartz invariance and the momentum laws.

Remarks

What is not claimed. No free propagator integral kernel, no pointwise trajectory of the particle, and no assertion that the momentum representative's modulus is measurable-free (it is an $L^{2}$ class; the invariance is that of the class of the modulus) appear in this lemma.
Scope. The statement is purely mathematical, about the free operator on $L^{2} (R^{3}; C^{r})$ ; the physical reading of the momentum laws is that of the observable and probability postulate.

The free flow as a unimodular multiplier in the momentum representation

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources