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Heisenberg flow of a bounded operator: conjugates, covariance and the equation of motion

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H \neq {0}$ be a complex Hilbert space (Hilbert space), let $T$ be self-adjoint (Symmetric, self-adjoint and essentially self-adjoint operators) and $U (t) = e^{i t T}$ as in Flow calculus of a self-adjoint generator: commutation, invariance and differentiability; for $A \in B (H)$ (A bounded linear operator between normed spaces) put $A (t) : = U (t)^{*} A U (t) = U (- t) A U (t) (t \in R) .$ Then:

(a) $A (t) \in B (H)$ with $A (0) = A$ , $∥ A (t) ∥ = ∥ A ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum) and $A (t)^{*} = A^{*} (t)$ ; in particular $A$ self-adjoint implies every $A (t)$ self-adjoint (Self-adjoint, positive, unitary and normal operators). Moreover $A (t + s) = U (s)^{*} A (t) U (s)$ for all $s, t$ .

(b) If $A D (T) \subseteq D (T)$ , then $A (t) D (T) \subseteq D (T)$ for every $t$ , and for every $x \in D (T)$ the map $t \mapsto A (t) x$ is differentiable at every $t$ with $\frac{d}{d t} A (t) x = - i [T, A (t)] x = - i (T A (t) - A (t) T) x .$

(c) For every $x \in H$ the map $t \mapsto A (t) x$ is continuous, and the uniform operator norm bound $∥ A (t) ∥ \leq ∥ A ∥$ holds.

(d) If $A$ is self-adjoint with spectral projection-valued measure $E_{A}$ (Spectral theorem for unbounded self-adjoint operators (PVM form)), then every $A (t)$ is self-adjoint with spectral projection-valued measure $E_{A (t)} (B) = U (- t) E_{A} (B) U (t) (B \subseteq R Borel),$ and the spectra agree, $σ (A (t)) = σ (A)$ .

Facts & Assumptions

Given: The Axiom of Choice and the Axiom of Countable Choice; a nonzero complex Hilbert space $H$ ; a self-adjoint operator $T$ with flow $U (t) = e^{i t T}$ ; a bounded operator $A \in B (H)$ ; vectors $x \in H$ ; real $s, t$ .

[F1]

Flow calculus. $U$ is a strongly continuous one-parameter unitary group with $U (t)^{*} = U (- t)$ , $U (t) D (T) = D (T)$ , $T U (t) = U (t) T$ on $D (T)$ , and for bounded Borel $h$ one has $h (T) U (t) = U (t) h (T)$ ; for $x \in D (T)$ the map $t \mapsto U (t) x$ is differentiable with derivative $i U (t) T x$ (Flow calculus of a self-adjoint generator: commutation, invariance and differentiability clause (a)-(c)).

[F2]

Adjoints and norms. For $B, C, D \in B (H)$ one has $(B C D)^{*} = D^{*} C^{*} B^{*}$ , $∥ B C D ∥ \leq ∥ B ∥ ∥ C ∥ ∥ D ∥$ and $∥ B^{*} ∥ = ∥ B ∥$ ; each $U (t)$ is unitary, so $∥ U (- t) B U (t) ∥ = ∥ B ∥$ and $U (t)^{*} = U (- t)$ (Hilbert-adjoint identities, The Hilbert-space adjoint of a bounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Self-adjoint, positive, unitary and normal operators).

[F3]

Bounded calculus under unitary conjugation. Let $E$ be a PVM on the Borel sets of $R$ of a self-adjoint operator and $W$ unitary. Then $B \mapsto W E (B) W^{*}$ is a regular PVM whose scalar measure in $x$ is $E_{W^{*} x}$ , and for bounded Borel $f$ one has $Φ_{W E (\cdot) W^{*}} (f) = W Φ_{E} (f) W^{*}$ (Projection valued measure, Bounded borel pvm integral, Pvm integral is a star homomorphism).

[F4]

Truncation definition and uniqueness. For a PVM $E$ and measurable $f$ one has $f (E) x = \lim_{n} Φ_{E} (f 1_{{∣ f ∣ \leq n}}) x$ for $x \in D (f (E))$ ; for bounded self-adjoint operators the representing regular PVM is unique (Integral of a measurable function against a projection-valued measure, Unbounded Borel functional calculus: domains, products, spectral mapping, Spectral theorem for unbounded self-adjoint operators (PVM form)).

Proof

technique · direct; conjugate the flow properties of [Flow calculus of a self-adjoint generator: commutation, invariance and differentiability](/physics/item/lem-nrqm-autonomous-dynamics-flow-calculus) by bounded operators, with the product-domain hypothesis used only where the differential equation is asserted

Given: The data of the statement, and [F1]-[F4].

1.1givenF1F2algebra

[Boundedness, norm and adjoint] $A (t) = U (- t) A U (t)$ is a product of bounded operators, so bounded, and $A (0) = A$ ; by [F2], $∥ A (t) ∥ = ∥ A ∥$ and $A (t)^{*} = U (t)^{*} A^{*} U (- t)^{*} = U (- t) A^{*} U (t) = A^{*} (t)$ ; hence $A$ self-adjoint implies $A (t)$ self-adjoint. For the covariance, $A (t + s) = U (- t - s) A U (t + s) = U (s)^{*} U (- t) A U (t) U (s) = U (s)^{*} A (t) U (s)$ by the group law. This is (a).

1.2givenF1

[Time-invariance of the product-domain hypothesis] Assume $A D (T) \subseteq D (T)$ . For $y \in D (T)$ one has $U (t) y \in D (T)$ by [F1], then $A U (t) y \in D (T)$ by hypothesis, and finally $U (- t) A U (t) y \in D (T)$ by [F1] again; thus $A (t) y = U (- t) A U (t) y \in D (T)$ .

1.3givenF1F3F4

[Conjugation of the spectral measure] Assume $A$ self-adjoint with spectral PVM $E_{A}$ . Fix $t$ and put $E^{'} (B) : = U (- t) E_{A} (B) U (t)$ . Then $E^{'}$ is a PVM: each $E^{'} (B)$ is an orthogonal projection because $U (- t) = U (t)^{*}$ is unitary and $E_{A} (B)$ is an orthogonal projection; $E^{'} (\emptyset) = 0$ , $E^{'} (R) = I$ ; and for pairwise disjoint Borel $B_{n}$ with union $B$ , the strong countable additivity of $E_{A}$ gives $\sum_{n} E^{'} (B_{n}) x = U (- t) (\sum_{n} E_{A} (B_{n}) U (t) x) \to U (- t) E_{A} (B) U (t) x = E^{'} (B) x$ because $U (- t)$ is bounded. It is regular because its scalar measure in $x$ is $E_{x}^{'} (B) = ∥ E^{'} (B) x ∥^{2} = ⟨ U (- t) E_{A} (B) U (t) x, x ⟩ = ⟨ E_{A} (B) U (t) x, U (t) x ⟩ = E_{A, U (t) x} (B)$ , a regular Borel measure.

2.1givenstep 1.1F1F2algebra

[Strong continuity] For $x \in H$ , $A (t) x - A (s) x = U (- t) A (U (t) - U (s)) x + (U (- t) - U (- s)) A U (s) x$ , and both terms tend to $0$ as $t \to s$ because $U$ is strongly continuous and $A$ is bounded; the norm bound is step 1.1. This is (c).

2.2givenstep 1.2F1algebra

[Equation of motion] Assume $A D (T) \subseteq D (T)$ and let $x \in D (T)$ . Decompose $A (t + s) x - A (t) x = U (- t - s) A (U (t + s) x - U (t) x) + (U (- t - s) - U (- t)) A U (t) x$ ; the first term divided by $s$ equals $U (- t - s) A \frac{1}{s} (U (t + s) x - U (t) x)$ , and $\frac{1}{s} (U (t + s) x - U (t) x) \to i U (t) T x$ by [F1], so it tends to $i U (- t) A U (t) T x = i A (t) T x$ . For the second term, $y : = A U (t) x$ lies in $D (T)$ by step 1.2, and $\frac{1}{s} (U (- s) y - y) \to - i T y$ by [F1] applied at $0$ to the group $s \mapsto U (- s)$ ; hence the second term divided by $s$ equals $U (- t) \frac{1}{s} (U (- s) y - y) \to - i U (- t) T y = - i T U (- t) y = - i T A (t) x$ , using the commutation $T U (- t) = U (- t) T$ on $D (T)$ from [F1]. Adding the two limits gives $\frac{d}{d t} A (t) x = i A (t) T x - i T A (t) x = - i (T A (t) - A (t) T) x$ . This is (b).

2.3givenstep 1.3F3F4algebra

[Identification as the spectral measure of $A (t)$ ] With $E^{'}$ as in step 1.3, [F3] gives $Φ_{E^{'}} (g) = U (- t) Φ_{E_{A}} (g) U (t)$ for bounded Borel $g$ : the identity holds for indicators by definition of $E^{'}$ , extends to simple functions by linearity, and passes to uniform limits by the norm estimate of Bounded borel pvm integral clause 1. Applying the truncation definition [F4] with $g_{n} (λ) = λ 1_{{∣ λ ∣ \leq n}}$ and $x \in H$ (the domain is all of $H$ because $A$ is bounded) gives $\int λ d E^{'} (λ) x = \lim_{n} Φ_{E^{'}} (g_{n}) x = \lim_{n} U (- t) Φ_{E_{A}} (g_{n}) U (t) x = U (- t) (\int λ d E_{A} (λ)) U (t) x = U (- t) A U (t) x = A (t) x$ , where the middle limit is interchanged with the bounded operator $U (- t)$ and $\int λ d E_{A} (λ) = A$ is the spectral representation of the bounded self-adjoint $A$ . Hence $E^{'}$ represents the bounded self-adjoint $A (t)$ , and by uniqueness [F4] it is the spectral PVM of $A (t)$ .

3.1givenstep 2.3F4

[Spectra] For a self-adjoint operator with spectral PVM $F$ , the spectrum consists of the points $λ$ with $F ((λ - ε, λ + ε)) \neq 0$ for every $ε > 0$ (Unbounded Borel functional calculus: domains, products, spectral mapping clause 4 with $f$ the identity). By step 2.3, $E_{A (t)} ((λ - ε, λ + ε)) = U (- t) E_{A} ((λ - ε, λ + ε)) U (t)$ is nonzero exactly when $E_{A} ((λ - ε, λ + ε))$ is; hence $σ (A (t)) = σ (A)$ .

4.1step 1.1step 2.1step 1.2step 2.2step 1.3step 2.3step 3.1∎

Steps 1.1, 2.1, 1.2, 2.2, 1.3, 2.3 and 3.1 prove (a)-(d).

Remarks

Where the hypothesis is needed. Clauses (a) and (c) require nothing beyond boundedness of $A$ ; clause (b), and with it the Heisenberg equation, uses the product-domain hypothesis $A D (T) \subseteq D (T)$ , which is exactly the condition under which the differentiated products are defined. A bounded observable whose spectral PVM commutes with that of $T$ satisfies the hypothesis trivially, because then $A$ maps every $D (h (T))$ into itself.
Transported measurable functional calculus. Step 2.3 shows more than the statement of (d): for every bounded Borel $g$ the operator $g (A (t))$ equals $U (- t) g (A) U (t)$ ; this is the operator form of the statement that the unitary flow is an automorphism of the algebra of bounded Borel functions of the observable.

Heisenberg flow of a bounded operator: conjugates, covariance and the equation of motion

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources