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Self-adjointness and the flow of a self-adjoint operator perturbed by a bounded operator commuting with its flow

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 $H$ be a nonzero complex Hilbert space (Hilbert space), let $A$ be a self-adjoint operator on $H$ with domain $D (A)$ (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators), and let $B$ be a bounded self-adjoint operator on $H$ (A bounded linear operator between normed spaces, Self-adjoint, positive, unitary and normal operators) with $B e^{i s A} = e^{i s A} B (s \in R),$ where $e^{i s A} = \int_{R} e^{i s λ} d E_{A} (λ)$ is the unitary group of $A$ in the spectral calculus (A self-adjoint operator generates a strongly continuous unitary group, Bounded borel pvm integral, Projection valued measure). Put $A + B$ on its natural domain $D (A + B) : = D (A) \cap D (B) = D (A)$ . Then:

(a) $A + B$ is self-adjoint;

(b) $e^{i s B} e^{i t A} = e^{i t A} e^{i s B}$ for all $s, t \in R$ ; the flows of $A$ and of the bounded $B$ commute;

(c) the family $s \mapsto P (s) : = e^{i s A} e^{i s B}$ is a strongly continuous one-parameter unitary group (Strongly continuous one-parameter unitary group), and its infinitesimal generator $G$ (Infinitesimal generator of a unitary group) is $G = i (A + B)$ : the quotient $\frac{1}{s} (P (s) x - x)$ converges as $s \to 0$ exactly for $x \in D (A)$ , with limit $i (A + B) x$ . Consequently $P (s) = e^{i s (A + B)}$ for every $s \in R$ .

Facts & Assumptions

Given: the Axiom of Choice and the Axiom of Countable Choice; a nonzero complex Hilbert space $H$ ; a self-adjoint operator $A$ ; a bounded self-adjoint operator $B$ commuting with every $e^{i s A}$ ; real $s, t, λ$ ; a unit vector or vector $x$ .

[F1]

Flow calculus. For a self-adjoint $T$ with spectral projection-valued measure $E_{T}$ and $U_{T} (s) : = e^{i s T}$ : $U_{T}$ is a strongly continuous one-parameter unitary group with $U_{T} (s)^{*} = U_{T} (- s)$ , $U_{T} (s) D (T) = D (T)$ and $T U_{T} (s) = U_{T} (s) T$ on $D (T)$ ; the quotient $\frac{1}{s} (U_{T} (s) x - x)$ converges as $s \to 0$ exactly for $x \in D (T)$ , with limit $i T x$ ; and for every bounded Borel $f : R \to C$ one has $∥ f (T) ∥ \leq \sup ∣ f ∣$ and $f (T) U_{T} (s) = U_{T} (s) f (T)$ , where $f (T) = \int f d E_{T}$ . The norm bound follows from the bounded PVM integral; the commutation follows because both factors are integrals against $E_{T}$ and the integral map is multiplicative, so $Φ_{E_{T}} (f) Φ_{E_{T}} (λ \mapsto e^{i s λ}) = Φ_{E_{T}} (f (λ) e^{i s λ}) = Φ_{E_{T}} (λ \mapsto e^{i s λ}) Φ_{E_{T}} (f)$ (Bounded borel pvm integral, Pvm integral is a star homomorphism). (A self-adjoint operator generates a strongly continuous unitary group, Bounded borel pvm integral).

[F2]

Resolvents. For $λ \neq 0$ put $r_{λ} (μ) : = (μ + i λ)^{- 1}$ and $R_{λ} : = Φ_{E_{A}} (r_{λ})$ . The bounded PVM integral gives $∥ R_{λ} ∥ \leq 1 / ∣ λ ∣$ . The spectral domain formula gives $\int μ^{2} ∣ r_{λ} (μ) ∣^{2} d E_{A, x} (μ) \leq ∥ x ∥^{2}$ , so $R_{λ} x \in D (A)$ for every $x \in H$ ; the unbounded Borel calculus product rule then gives $(A + i λ) R_{λ} = Φ_{E_{A}} (μ r_{λ} + i λ r_{λ}) = I$ on $H$ . For $y \in D (A)$ the same product rule gives $R_{λ} (A + i λ) y = Φ_{E_{A}} (r_{λ} (μ + i λ)) y = y$ . Thus $R_{λ} = (A + i λ)^{- 1}$ and $A + i λ : D (A) \to H$ is bijective. These claims use the spectral domain formula and bounded PVM norm bound (Spectral theorem for unbounded self-adjoint operators (PVM form), Bounded borel pvm integral) and the unbounded Borel calculus (Unbounded Borel functional calculus: domains, products, spectral mapping clause 2).

[F3]

Neumann series. If $R \in B (H)$ with $∥ R ∥ < 1$ , then $I - R$ is invertible and $(I - R)^{- 1} = \sum_{n \geq 0} R^{n}$ in operator norm (Neumann series and small perturbations of bounded inverses clause 1, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

Stone uniqueness. Two strongly continuous one-parameter unitary groups with the same infinitesimal generator agree at every time; equivalently $s \mapsto e^{i s T}$ is determined by the self-adjoint $T$ (Stone's theorem: unitary groups and self-adjoint generators, Infinitesimal generator of a unitary group).

[F5]

For bounded operators $S, T$ between compatible normed spaces, their composition satisfies $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ (Composition satisfies |ST|\le|S|,|T|).

Proof

technique · direct; prove self-adjointness of $A+B$ by the Neumann series for the resolvent of $A$, differentiate the conjugation curve to obtain the commutation of the two flows, and identify the generator of the product group by comparing difference quotients, concluding with Stone's theorem

Given: The data of the statement and [F1]-[F5].

1.1givenalgebra

[Symmetry and domain of $A + B$ ] The domain $D (A + B) = D (A)$ is a dense linear subspace of $H$ because $D (A)$ is dense and $B$ is everywhere defined and linear; for $x, y \in D (A)$ , using that $A$ is symmetric and $B$ is self-adjoint and bounded, $⟨ (A + B) x, y ⟩ = ⟨ A x, y ⟩ + ⟨ B x, y ⟩ = ⟨ x, A y ⟩ + ⟨ x, B y ⟩ = ⟨ x, (A + B) y ⟩$ , so $A + B$ is densely defined and symmetric.

1.2F2F3F5algebra

[Surjectivity of $(A + B) \pm i λ$ ] Fix $λ > ∥ B ∥$ (if $B = 0$ take $λ = 1$ ); by [F2] the map $(A + i λ) : D (A) \to H$ is a bijection with inverse of norm at most $1 / λ$ , so [F5] gives $∥ (A + i λ)^{- 1} B ∥ \leq ∥ B ∥ / λ < 1$ ; [F3] applied to $- R$ makes $I + R$ invertible on $H$ , where $R : = (A + i λ)^{- 1} B$ . Since $R (H) \subseteq D (A)$ , for any $y \in D (A)$ the vector $x : = (I + R)^{- 1} y$ satisfies $x = y - R x \in D (A)$ , so $I + R$ maps $D (A)$ onto $D (A)$ . For $x \in D (A)$ , $R x \in D (A)$ and $(A + B + i λ) x = (A + i λ) (I + R) x$ ; therefore $(A + B) + i λ$ maps $D (A)$ onto $H$ , and the same argument with $- λ$ shows that $(A + B) - i λ$ maps $D (A)$ onto $H$ .

1.3F1givenalgebra

[Adjoint inclusion] Let $y \in D ((A + B)^{*})$ and put $z : = (A + B)^{*} y$ ; for every $x \in D (A)$ one has $⟨ A x, y ⟩ + ⟨ B x, y ⟩ = ⟨ x, z ⟩$ , that is $⟨ A x, y ⟩ = ⟨ x, z - B y ⟩$ because $B^{*} = B$ and $B y \in H$ , and since $A$ is self-adjoint the existence of the vector $z - B y$ representing $x \mapsto ⟨ A x, y ⟩$ forces $y \in D (A)$ with $A y = z - B y$ , so $D ((A + B)^{*}) \subseteq D (A) = D (A + B)$ with $(A + B)^{*} y = (A + B) y$ there.

1.4F1givenalgebra

[Commutation of the flows] Fix $t$ and $x \in H$ and define $g (s) : = e^{i s B} e^{i t A} e^{- i s B} x$ ; by [F1] applied to the self-adjoint operator $B$ (whose domain is all of $H$ ) the difference quotients $\frac{1}{h} (e^{i (s + h) B} y - e^{i s B} y) \to i B e^{i s B} y$ and $\frac{1}{h} (e^{- i (s + h) B} y - e^{- i s B} y) \to - i e^{- i s B} B y$ for every $y$ , so the product rule gives $g^{'} (s) = e^{i s B} (i B e^{i t A} - i e^{i t A} B) e^{- i s B} x$ , and the hypothesis $B e^{i t A} = e^{i t A} B$ makes this derivative vanish for every $s$ ; hence $g$ is constant with value $g (0) = e^{i t A} x$ , that is $e^{i s B} e^{i t A} = e^{i t A} e^{i s B}$ for all $s, t$ , which is the first identity of (b).

1.5F1algebra

[Quotient of the product group on $D (A)$ ] Put $P (s) : = e^{i s A} e^{i s B}$ and let $x \in D (A)$ ; the algebraic identity $\frac{1}{s} (P (s) x - x) = e^{i s A} \frac{1}{s} (e^{i s B} x - x) + \frac{1}{s} (e^{i s A} x - x)$ holds for every $s \neq 0$ , and as $s \to 0$ the first term converges to $i B x$ because $\frac{1}{s} (e^{i s B} x - x) \to i B x$ in norm by [F1] and $e^{i s A} \to I$ strongly, while the second converges to $i A x$ by [F1]; hence the quotient of $P$ at $x$ converges, with limit $i (A + B) x$ .

2.1step 1.1step 1.3

[Self-adjointness] Step 1.1 gives $A + B \subseteq (A + B)^{*}$ and step 1.3 gives $(A + B)^{*} \subseteq A + B$ including domains, so $A + B = (A + B)^{*}$ , which is (a).

2.2step 1.4F1algebra

[Group property and strong continuity] By step 1.4, $P (s) P (t) = e^{i s A} e^{i s B} e^{i t A} e^{i t B} = e^{i s A} e^{i t A} e^{i s B} e^{i t B} = P (s + t)$ and $P (s)^{*} = e^{- i s B} e^{- i s A} = e^{- i s A} e^{- i s B} = P (- s)$ ; each factor is strongly continuous, and the estimate $∥ P (s) x - P (s_{0}) x ∥ \leq ∥ e^{i s A} (e^{i s B} x - e^{i s_{0} B} x) ∥ + ∥ (e^{i s A} - e^{i s_{0} A}) e^{i s_{0} B} x ∥$ shows that $P$ is strongly continuous, hence a strongly continuous one-parameter unitary group.

2.3step 1.4F1algebra

[No larger domain] Let $x \in H$ and suppose that $\frac{1}{s} (P (s) x - x)$ converges, to $y \in H$ ; by step 1.4, $P (s) = e^{i s B} e^{i s A}$ , so $e^{i s A} x = e^{- i s B} P (s) x$ and the identity $\frac{1}{s} (e^{i s A} x - x) = e^{- i s B} \frac{1}{s} (P (s) x - x) + \frac{1}{s} (e^{- i s B} x - x)$ holds; the first term converges to $y$ because $e^{- i s B} \to I$ strongly and $\frac{1}{s} (P (s) x - x) \to y$ , and the second converges to $- i B x$ by [F1] applied to the self-adjoint operator $- B$ , so $\frac{1}{s} (e^{i s A} x - x)$ converges and [F1] gives $x \in D (A)$ with $y = i (A + B) x$ .

3.1step 1.5step 2.2step 2.3

[Generator identification] By step 2.2 the family $P$ is a strongly continuous one-parameter unitary group with infinitesimal generator $G$ ; steps 1.5 and 2.3 show that the quotient of $P$ at $x$ converges exactly for $x \in D (A)$ , with limit $i (A + B) x$ ; hence $D (G) = D (A)$ and $G = i (A + B)$ , which is the generator statement of (c).

4.1step 1.4step 3.1F1F4∎

[Identification of the groups] By [F1] the family $s \mapsto e^{i s (A + B)}$ is a strongly continuous one-parameter unitary group whose quotient at $x$ converges exactly for $x \in D (A + B) = D (A)$ , to $i (A + B) x$ ; by step 3.1 it has the same infinitesimal generator $i (A + B)$ as $P$ , so Stone's theorem [F4] yields $P (s) = e^{i s (A + B)}$ for every $s$ , completing (c); (b) is step 1.4.

Remarks

Why the hypothesis is a commutation with the flow. The hypothesis $B e^{i s A} = e^{i s A} B$ for all $s$ is a commutation of the bounded operator with every element of the unitary group of $A$ ; it implies, but is not assumed as, a formal commutator identity on a small domain, and it is the exact form in which the model's internal matrices commute with the free flow.
Bounded perturbation. Clause (a) is the standard self-adjointness of a self-adjoint operator perturbed by a bounded self-adjoint operator; the flow hypothesis is not needed for it, only for the product formula of (b) and the generator statement of (c).

Self-adjointness and the flow of a self-adjoint operator perturbed by a bounded operator commuting with its flow

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources