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The generator of a unitary group is closed and skew-adjoint

Statement

Assume Countable Choice and Dependent Choice. The generator $G$ of a strongly continuous one-parameter unitary group $U$ is densely defined, closed, and skew-adjoint: $G^{*} = - G$ . Consequently $T = - i G$ is self-adjoint with $D (T) = D (G)$ .

Facts & Assumptions

[A1]

Both $Q_{\pm} (λ)$ have range $D (G)$ and satisfy $Q_{\pm} (λ) y = \int_{0}^{\infty} e^{- λ t} U (\pm t) y d t$ with $∥ Q_{\pm} (λ) ∥ \leq 1 / λ$ ; also $(λ \mp G) Q_{\pm} (λ) = I$ and $Q_{\pm} (λ) (λ \mp G) = I$ on $D (G)$ (Laplace resolvents of a unitary group).

[A2]

$⟨ U (t) x, U (t) y ⟩ = ⟨ x, y ⟩$ and $t \mapsto ⟨ U (t) x, U (t) y ⟩$ is differentiable at $0$ with derivative $⟨ G x, y ⟩ + ⟨ x, G y ⟩$ when $x, y \in D (G)$ , by the definition of $G$ and sesquilinearity and continuity of the inner product (Infinitesimal generator of a unitary group, Strongly continuous one-parameter unitary group, Hilbert space, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A3]

A densely defined symmetric operator $S$ with $ran (S \pm i) = H$ is self-adjoint (Range criterion for self-adjointness, Symmetric, self-adjoint and essentially self-adjoint operators).

[A4]

A bounded everywhere-defined inverse to $1 - G$ puts $1$ in the resolvent of $G$ and forces its graph closed, with convention $R_{G} (1) = (I - G)^{- 1}$ (Resolvent and spectrum of an unbounded operator, Densely defined, closed and closable operators, and cores). The exact limit argument is also given below.

[A5]

The Bochner integral is linear by passage from simple integral approximations, and its norm is bounded by the integral of the norm (Bochner-integrable function, Bochner integral norm inequality). Compact Newton--Leibniz, the Countable Choice Riemann/Lebesgue bridge and monotone convergence compute $\int_{a}^{\infty} λ e^{- λ t} d t = e^{- λ a}$ for $a \geq 0$ , $λ > 0$ , by the primitive $- e^{- λ t}$ on $[a, N]$ followed by $N \to \infty$ (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, Monotone convergence for the integral).

[A6]

The adjoint domain consists of vectors making $x \mapsto ⟨ S x, y ⟩$ bounded on $D (S)$ , with $⟨ S x, y ⟩ = ⟨ x, S^{*} y ⟩$ in the first-variable-linear convention (Adjoint of a densely defined operator).

Proof

technique · direct

Given: Countable Choice, Dependent Choice, and a strongly continuous unitary group $U$ with generator $G$ .

1.1

$D (G)$ is dense: for $y \in H$ and $λ > 0$ , [A1] gives $λ Q_{+} (λ) y \in D (G)$ and $λ Q_{+} (λ) y - y = \int_{0}^{\infty} λ e^{- λ t} (U (t) y - y) d t .$ Given $ε > 0$ , strong continuity supplies $δ > 0$ such that $∥ U (t) y - y ∥ < ε$ for $0 \leq t \leq δ$ ; the integral norm is then at most $ε + 2 ∥ y ∥ e^{- λ δ}$ . Letting $λ \to \infty$ and then $ε ↓ 0$ proves $λ Q_{+} (λ) y \to y$ . Taking positive integer $λ \to \infty$ supplies an approximating sequence in $D (G)$ for every $y$ . Thus $D (G)$ is dense.

A1A5given

1.2

$G$ is skew-symmetric: the function $t \mapsto ⟨ U (t) x, U (t) y ⟩$ is constant with value $⟨ x, y ⟩$ for $x, y \in D (G)$ , so its derivative at $0$ vanishes, that is $⟨ G x, y ⟩ + ⟨ x, G y ⟩ = 0$ ; equivalently $⟨ G x, y ⟩ = - ⟨ x, G y ⟩$ .

1.3

$G$ is closed: with $R = Q_{+} (1)$ , [A1] and [A4] already imply closedness. Explicitly, if $x_{n} \in D (G)$ , $x_{n} \to x$ and $G x_{n} \to y$ , then $x_{n} = R (x_{n} - G x_{n}) \to R (x - y)$ since $R$ is bounded. Uniqueness of limits gives $x = R (x - y) \in D (G)$ and $(I - G) x = (I - G) R (x - y) = x - y$ , hence $G x = y$ . The sequential graph criterion is valid in the norm metric under the assumed Countable Choice.

A1A4

2.1

On $D (T) = D (G)$ set $T = - i G$ . For $x, y$ in this domain, step 1.2 gives $⟨ T x, y ⟩ = - i ⟨ G x, y ⟩ = i ⟨ x, G y ⟩ = ⟨ x, T y ⟩$ , so $T$ is symmetric. It is densely defined by step 1.1. It is closed: convergence of $x_{n}$ and $T x_{n}$ implies convergence of $G x_{n} = i T x_{n}$ , so step 1.3 applies. The correct signed formulas are $T + i I = i (I - G)$ and $T - i I = - i (I + G)$ . Both ranges equal $H$ by the two signs of [A1] at $λ = 1$ ; multiplication by a nonzero scalar preserves surjectivity. Hence [A3] makes $T$ self-adjoint.

A1A3step 1.1step 1.2step 1.3

3.1

The adjoint domain of $T = - i G$ equals that of $G$ : multiplication of the scalar functional in [A6] by $- i$ preserves boundedness in both directions. For $y$ in this domain, $⟨ T x, y ⟩ = - i ⟨ G x, y ⟩ = - i ⟨ x, G^{*} y ⟩ = ⟨ x, i G^{*} y ⟩$ ; uniqueness of the representing vector gives $T^{*} = i G^{*}$ . Step 2.1 gives $T^{*} = T = - i G$ including domains, hence $D (G^{*}) = D (G)$ and $G^{*} = - G$ .

A6step 2.1

4.1

The conclusions are density, closedness and skew-adjointness from steps 1.1, 1.3 and 3.1, and self-adjointness of $- i G$ from step 2.1. The zero space and constant identity group satisfy the same identities directly. The declared Countable Choice and Dependent Choice match the Laplace supplier; Countable Choice also licenses the range/adjoint and integration interfaces. Only strictly positive Laplace parameters are used.

A1A3A5A6step 1.1step 1.3step 2.1step 3.1∎

Depends on

Used by

Stone's theorem: unitary groups and self-adjoint generators Theorem

Dependency tree · two levels

78 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Roland Schnaubelt, Evolution Equations (lecture notes) (standard reference, not scraped)
Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)