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The one-mode Schrödinger formulas and their phase

Statement

Assume Countable Choice and let $λ \neq 0$ . On $L^{2} (R)$ define $U_{q} f (x) = f (x - q)$ and $V_{p} f (x) = e^{i λ p x} f (x)$ . Then $U_{q} V_{p} = e^{- i λ q p} V_{p} U_{q}$ . The phase-corrected operators $W_{λ} (q, p) : = e^{i λ q p / 2} U_{q} V_{p}$ satisfy $W_{λ} (q, p) W_{λ} (q^{'}, p^{'}) = e^{- i λ (q p^{'} - p q^{'}) / 2} W_{λ} (q + q^{'}, p + p^{'})$ and are exactly the one-mode Schrödinger operators, acting by $(W_{λ} (q, p) f) (x) = e^{i λ p (x - q / 2)} f (x - q) .$

Facts & Assumptions

Given: $λ \neq 0$ , Lebesgue $L^{2} (R)$ , and the operators in the Statement.

[F1]

Under Countable Choice, $L^{2} (R)$ with its integral pairing is a complex Hilbert space; translation preserves Lebesgue measure, so $U_{q}$ is a well-defined isometry with inverse $U_{- q}$ ( $L^{2}$ with the integral pairing is a Hilbert space, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F2]

For every real $s$ , $∣ e^{i s} ∣ = 1$ , so multiplication by $e^{i λ p x}$ is unitary on complex $L^{2}$ and the scalar phase $e^{i λ q p / 2}$ has modulus one ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Complex Haar L^p spaces and compactly supported functions).

[F3]

The Weyl phase convention is $e^{- i λ ω (v, w) / 2}$ with $ω ((q, p), (q^{'}, p^{'})) = q p^{'} - p q^{'}$ (Regular Weyl systems and irreducibility).

[A1]

Countable Choice is the hypothesis used to make $L^{2} (R)$ a Hilbert space (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: The data in the Statement.

1.1A1F1F2givenalgebra

By [F1] and [F2], $U_{q}$ and $V_{p}$ are unitary. Direct evaluation gives $(U_{q} V_{p} f) (x) = e^{i λ p (x - q)} f (x - q)$ and $(V_{p} U_{q} f) (x) = e^{i λ p x} f (x - q)$ , so $U_{q} V_{p} = e^{- i λ q p} V_{p} U_{q}$ .

2.1givenstep 1.1algebra

The formula in the Statement yields $(W_{λ} (q, p) f) (x) = e^{i λ q p / 2} e^{i λ p (x - q)} f (x - q) = e^{i λ p (x - q / 2)} f (x - q)$ , proving the asserted Schrödinger formula directly.

3.1F3step 1.1step 2.1algebra∎

The commutation identity from step 1.1 gives $V_{p} U_{q^{'}} = e^{i λ p q^{'}} U_{q^{'}} V_{p}$ . Therefore $W_{λ} (q, p) W_{λ} (q^{'}, p^{'}) = e^{i λ ((q p + q^{'} p^{'}) / 2 + p q^{'})} U_{q + q^{'}} V_{p + p^{'}}$ . The exponent coefficient satisfies $((q p + q^{'} p^{'}) / 2 + p q^{'}) = ((q + q^{'}) (p + p^{'}) - (q p^{'} - p q^{'})) / 2$ , so this is $e^{- i λ (q p^{'} - p q^{'}) / 2} W_{λ} (q + q^{'}, p + p^{'})$ , as required by [F3].

Boundary cases

The displayed formula requires no regularity or irreducibility result; its algebraic product identity also holds at $λ = 0$ .
The two basic operator families commute up to the displayed scalar; omitting the factor $e^{i λ q p / 2}$ changes the Weyl convention.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Roland Speicher, Mathematical Aspects of Quantum Mechanics, Lecture Notes (standard reference, not scraped)