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Flow calculus of a self-adjoint generator: commutation, invariance and differentiability

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )), the choice hypotheses of the cited spectral and functional-calculus suppliers. Let $H \neq {0}$ be a complex Hilbert space (Hilbert space), let $T$ be self-adjoint on $H$ (Symmetric, self-adjoint and essentially self-adjoint operators) with spectral projection-valued measure $E$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Projection valued measure), and put $U (t) : = e^{i t T} = \int_{R} e^{i t λ} d E (λ) (t \in R)$ in the sense of the integral of a measurable function against a projection-valued measure (Integral of a measurable function against a projection-valued measure). Then:

(a) $U$ is a strongly continuous one-parameter unitary group (Strongly continuous one-parameter unitary group) with $U (0) = I$ , $U (t + s) = U (t) U (s)$ and $U (t)^{*} = U (t)^{- 1} = U (- t)$ for all $s, t$ ; moreover $U (t) D (T) = D (T)$ , $T U (t) = U (t) T$ on $D (T)$ , and the difference quotient $\frac{1}{t} (U (t) x - x)$ has a norm limit as $t \to 0$ exactly for $x \in D (T)$ , where that limit is $i T x$ (A self-adjoint operator generates a strongly continuous unitary group, Infinitesimal generator of a unitary group).

(b) For every bounded Borel function $h : R \to C$ and every $t \in R$ one has $h (T) U (t) = U (t) h (T)$ ; in particular $E (B) U (t) = U (t) E (B)$ for every Borel set $B \subseteq R$ .

(c) For $x \in D (T)$ the map $t \mapsto U (t) x$ is differentiable at every $t \in R$ , with $\frac{d}{d t} U (t) x = i U (t) T x = i T U (t) x;$ for $x \in H ∖ D (T)$ the map $t \mapsto U (t) x$ is differentiable at no point of $R$ .

Facts & Assumptions

Given: The Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )); a nonzero complex Hilbert space $H$ (Hilbert space); a self-adjoint operator $T$ on $H$ with spectral projection-valued measure $E$ (Symmetric, self-adjoint and essentially self-adjoint operators, Spectral theorem for unbounded self-adjoint operators (PVM form), Projection valued measure); the operator $U (t) = \int e^{i t λ} d E (λ)$ (Integral of a measurable function against a projection-valued measure); a bounded Borel function $h : R \to C$ ; vectors $x \in H$ and real $s, t$ .

[F1]

Generated group and generator. $U$ is a strongly continuous one-parameter unitary group (Strongly continuous one-parameter unitary group) with $U (t) D (T) = D (T)$ , $T U (t) = U (t) T$ on $D (T)$ , and the difference quotient $\frac{1}{t} (U (t) x - x)$ converges as $t \to 0$ exactly for $x \in D (T)$ , the limit then being $i T x$ (A self-adjoint operator generates a strongly continuous unitary group); by Infinitesimal generator of a unitary group this says that the generator of $U$ is $i T$ with domain $D (T)$ .

[F2]

Bounded Borel calculus. For bounded Borel $f, g : R \to C$ the operator $Φ_{E} (f) = \int f d E$ is the unique bounded operator with $⟨ Φ_{E} (f) x, y ⟩ = \int f d E_{x, y}$ for all $x, y \in H$ , obtained as the norm limit of integrals of simple functions along uniform approximations; and $Φ_{E} (f g) = Φ_{E} (f) Φ_{E} (g)$ , $Φ_{E} (\overline{f}) = Φ_{E} (f)^{*}$ , $Φ_{E} (1_{R}) = I$ (Bounded borel pvm integral, Unbounded Borel functional calculus: domains, products, spectral mapping, Pvm integral is a star homomorphism).

[F3]

The flow as a function of $T$ . $U (t) = Φ_{E} (f_{t})$ with $f_{t} (λ) = e^{i t λ}$ , and $E (B) = Φ_{E} (1_{B})$ for every Borel $B \subseteq R$ (Integral of a measurable function against a projection-valued measure, Spectral theorem for unbounded self-adjoint operators (PVM form)).

Proof

technique · direct; read the group properties and the domain characterization from the cited generation lemma, then prove the commutation with the bounded calculus and the everywhere-differentiability statements

Given: A self-adjoint $T$ with spectral PVM $E$ ; $U (t) = e^{i t T}$ ; a bounded Borel $h$ ; $x \in H$ ; $s, t \in R$ .

1.1givenF1

Clause (a) is exactly the statement of the cited generation lemma, read in the notation $U (t) = e^{i t T}$ : it gives the group law, unitarity, $U (t)^{*} = U (- t)$ , the invariance $U (t) D (T) = D (T)$ , the commutation $T U (t) = U (t) T$ on $D (T)$ , and the characterization of the difference quotient with limit $i T x$ (A self-adjoint operator generates a strongly continuous unitary group, Infinitesimal generator of a unitary group).

1.2givenF2F3algebra

For bounded Borel $h$ and fixed $t$ , [F2] and [F3] give $h (T) U (t) = Φ_{E} (h) Φ_{E} (f_{t}) = Φ_{E} (h f_{t})$ , and multiplication of complex functions is commutative, so $Φ_{E} (h f_{t}) = Φ_{E} (f_{t} h) = Φ_{E} (f_{t}) Φ_{E} (h) = U (t) h (T)$ ; this is the first assertion of (b).

1.3givenF1algebra

Let $x \in D (T)$ . At $t = 0$ the difference quotient converges to $i T x$ by [F1]. For fixed $t$ and $s \to 0$ the group law gives $U (t + s) x - U (t) x = U (t) (U (s) x - x)$ , hence $\frac{1}{s} (U (t + s) x - U (t) x) = U (t) \frac{1}{s} (U (s) x - x) \to U (t) i T x = i U (t) T x$ , because $U (t)$ is bounded and the limit in the last factor exists; the identity $i U (t) T x = i T U (t) x$ is the commutation in [F1]. The derivative $t \mapsto i U (t) T x$ is continuous because $U$ is strongly continuous and $U (t) T x$ depends continuously on $t$ ; hence $t \mapsto U (t) x$ is differentiable at every $t$ with the displayed derivative.

1.4givenF1algebra

Let $x \in H ∖ D (T)$ and suppose that $t \mapsto U (t) x$ were differentiable at some $t_{0}$ with derivative $y$ . Then for $s \to 0$ , using the group law and boundedness of $U (- t_{0})$ , $\frac{1}{s} (U (s) x - x) = U (- t_{0}) \frac{1}{s} (U (t_{0} + s) x - U (t_{0}) x) \to U (- t_{0}) y$ , so the difference quotient at $0$ would converge, contradicting [F1], which says that it converges exactly for $x \in D (T)$ . Hence $t \mapsto U (t) x$ is differentiable at no point of $R$ .

2.1step 1.2F3

Taking $h = 1_{B}$ for a Borel set $B$ in step 1.2 and using $E (B) = Φ_{E} (1_{B})$ from [F3] gives $E (B) U (t) = U (t) E (B)$ , the second assertion of (b).

3.1step 1.1step 1.2step 2.1step 1.3step 1.4∎

Steps 1.1, 1.2, 2.1, 1.3 and 1.4 establish clauses (a), (b) and (c).

Remarks

Sign and scaling conventions. The physical reading of this lemma uses $T = - K / ℏ$ for a self-adjoint energy operator $K$ and a positive constant $ℏ$ ; then $D (T) = D (K)$ and $U (t) = e^{- i t K / ℏ}$ in the spectral calculus, while the derivative statement of clause (c) becomes $i ℏ \frac{d}{d t} U (t) x = K U (t) x$ for $x \in D (K)$ . The identification of the group generated by $- i K / ℏ$ with the spectral-calculus exponential is the uniqueness statement of Stone's theorem; it is used, not reproved, on the physics pages.
Only the domain carries the equation. Clause (c) shows that the norm-differentiability of the flow is equivalent to membership in $D (T)$ ; the flow itself, clause (a), is defined on all of $H$ and is continuous there.

Flow calculus of a self-adjoint generator: commutation, invariance and differentiability

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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