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Momentum Fourier Jacobian: the unitary momentum representation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), let $d \geq 1$ and $ℏ > 0$ . Write $F_{2}$ for the unitary Plancherel transform extending $\hat{f} (ξ) = \int_{R^{d}} f (x) e^{- 2 π i x \cdot ξ} d x$ (Fourier transform on complex L1 classes, Plancherel theorem); the $L^{2}$ spaces are Hilbert spaces by Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ and Hilbert space. For $ψ \in S (R^{d})$ , denote by $\hat{ψ}$ the continuous representative of $F_{2} ψ$ given by that absolutely convergent integral.

(a) For $ψ \in S (R^{d})$ the momentum wavefunction

$ψ_{p} (p) : = (2 π ℏ)^{- d / 2} \int_{R^{d}} e^{- i p \cdot x / ℏ} ψ (x) d x$

is well defined, bounded and continuous, and $ψ_{p} (p) = (2 π ℏ)^{- d / 2} \hat{ψ} (p / (2 π ℏ))$ for every $p$ ; here $\hat{ψ}$ is the stated continuous representative, not a point evaluation of an $L^{2}$ equivalence class.

(b) The dilation operator $D_{ℏ} g (p) : = (2 π ℏ)^{- d / 2} g (p / (2 π ℏ))$ is a surjective linear isometry of $L^{2} (R^{d}; C)$ onto itself, with inverse $D_{ℏ}^{- 1} g (ξ) = (2 π ℏ)^{d / 2} g (2 π ℏ ξ)$ .

(c) The formula $U_{0} ψ : = ψ_{p}$ defines a linear isometry $S (R^{d}) \to L^{2} (R^{d})$ (for the Lebesgue spaces $d x$ and $d p$ ) that extends uniquely to a surjective linear isometry $U : L^{2} (R^{d}, d x) \to L^{2} (R^{d}, d p)$ ; on Schwartz functions $U = D_{ℏ} F_{2}$ , and for every $ψ \in L^{2} (R^{d})$ ,

$∥ U ψ ∥_{L^{2} (d p)} = ∥ ψ ∥_{L^{2} (d x)} .$

The normalization statement is: the substitution $ξ = p / (2 π ℏ)$ carries $d ξ = (2 π ℏ)^{- d} d p$ , and the factor $(2 π ℏ)^{- d / 2}$ in $U_{0}$ is exactly what makes $U$ an isometry with respect to $d p$ . No physical identification of $U ψ$ with a measured momentum distribution is made.

Facts & Assumptions

Given: $d \geq 1$ , $ℏ > 0$ ; a Schwartz function $ψ \in S (R^{d})$ ; the Plancherel transform $F_{2}$ and the $L^{1}$ transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ .

[F1]

$F_{2}$ is a surjective complex-linear isometry $L^{2} (R^{d}) \to L^{2} (R^{d})$ that extends the $L^{1}$ Fourier transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ (Plancherel theorem, Fourier transform on complex L1 classes).

[F2]

If $f_{n} \to f$ pointwise and $∣ f_{n} ∣ \leq g$ for a fixed integrable $g$ , then $\int f_{n} \to \int f$ (Dominated convergence).

[F3]

An invertible linear map $T$ of $R^{d}$ scales Lebesgue measure by $∣ \det T ∣$ on measurable sets (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); for nonnegative measurable $h$ the integral identity $\int h (T x) d x = ∣ \det T ∣^{- 1} \int h (y) d y$ follows by approximation with simple functions and monotone convergence (Monotone convergence for the integral).

[F4]

$S (R^{d})$ contains $C_{c}^{\infty} (R^{d})$ , which is dense in $L^{2} (R^{d})$ (Smooth compactly supported functions of an open set are dense in $L^{2}$ , Schwartz space and its seminorms); in particular every Schwartz function lies in $L^{1} \cap L^{2}$ . The $L^{2}$ space is complete by Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ .

[F5]

A bounded linear map from a dense subspace of a normed space into a Banach space extends uniquely with the same norm (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm).

Proof

1.1givenF2F4algebra

[Absolute convergence and continuity] For $ψ \in S (R^{d})$ , $∣ e^{- i p \cdot x / ℏ} ψ (x) ∣ = ∣ ψ (x) ∣$ is integrable by [F4], so the defining integral converges absolutely for every $p$ and $∣ ψ_{p} (p) ∣ \leq (2 π ℏ)^{- d / 2} ∥ ψ ∥_{1}$ . Continuity in $p$ follows from [F2] applied to $e^{- i p_{n} \cdot x / ℏ} ψ \to e^{- i p \cdot x / ℏ} ψ$ with the integrable dominating function $∣ ψ ∣$ .

2.1step 1.1F1algebra

[Identification with the Plancherel transform] For $ψ \in S$ , put $ξ = p / (2 π ℏ)$ , so $2 π x \cdot ξ = p \cdot x / ℏ$ ; comparing the defining integral of $ψ_{p}$ with the $L^{1}$ Fourier transform [F1] gives $ψ_{p} (p) = (2 π ℏ)^{- d / 2} \hat{ψ} (ξ) = (2 π ℏ)^{- d / 2} F_{2} ψ (p / (2 π ℏ))$ , the last equality because $ψ \in L^{1}$ and $F_{2}$ extends $\hat{\cdot}$ . Hence $U_{0} = D_{ℏ} \circ F_{2}$ on $S (R^{d})$ .

2.2step 1.1F3algebra

[The dilation is a unitary] For $g \in L^{2}$ the substitution $p = (2 π ℏ) ξ$ is an invertible linear map with determinant $(2 π ℏ)^{d}$ , so by [F3] applied to $h = ∣ g ∣^{2}$ , $\int ∣ D_{ℏ} g (p) ∣^{2} d p = (2 π ℏ)^{- d} \int ∣ g (p / (2 π ℏ)) ∣^{2} d p = (2 π ℏ)^{- d} (2 π ℏ)^{d} \int ∣ g (ξ) ∣^{2} d ξ = ∥ g ∥^{2}$ . Thus $D_{ℏ}$ is a linear isometry; its inverse $D_{ℏ}^{- 1} g (ξ) = (2 π ℏ)^{d / 2} g (2 π ℏ ξ)$ is likewise an isometry and the two compositions are the identity, so $D_{ℏ}$ is a surjective isometry.

3.1step 2.1step 2.2F1algebra

[Norm preservation on Schwartz functions] By steps 2.1 and 2.2 and unitarity of $F_{2}$ [F1], $∥ ψ_{p} ∥_{2} = ∥ D_{ℏ} F_{2} ψ ∥_{2} = ∥ F_{2} ψ ∥_{2} = ∥ ψ ∥_{2}$ for $ψ \in S (R^{d})$ ; moreover $U_{0}$ is linear on $S (R^{d})$ because the integral and the dilation are linear.

4.1step 2.2step 3.1F1F4F5algebra∎

[Unique unitary extension] The subspace $S (R^{d})$ is dense in $L^{2}$ because it contains the dense subspace $C_{c}^{\infty} (R^{d})$ [F4]; by [F5] the isometry $U_{0}$ on $S (R^{d})$ has a unique bounded linear extension $U : L^{2} (R^{d}, d x) \to L^{2} (R^{d}, d p)$ with the same norm, so $∥ U ψ ∥ = ∥ ψ ∥$ for every $ψ \in L^{2}$ . Surjectivity: for arbitrary $h \in L^{2} (R^{d}, d p)$ , the vector $ψ : = F_{2}^{*} D_{ℏ}^{- 1} h$ lies in $L^{2}$ , and since $U$ agrees with $D_{ℏ} F_{2}$ on the dense set $S$ while both sides are bounded, $U ψ = h$ ; hence $U$ is onto. The norm identity of clause (c) is exactly $∥ U ψ ∥_{2} = ∥ ψ ∥_{2}$ , and the normalization discussion is the determinant computation of step 2.2 read at $g = F_{2} ψ$ .

Momentum Fourier Jacobian: the unitary momentum representation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources