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The Schroedinger-Robertson bound in anticommutator form and the Robertson bound on product domains

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space with inner product linear in the first argument (Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space), let $A, B$ be symmetric densely defined operators on $H$ (Symmetric, self-adjoint and essentially self-adjoint operators, Adjoint of a densely defined operator), and let $u \in D (A) \cap D (B)$ with $∥ u ∥ = 1$ satisfy in addition $u \in D (A B) \cap D (B A)$ , that is $B u \in D (A)$ and $A u \in D (B)$ . Put $a : = ⟨ A u, u ⟩, b : = ⟨ B u, u ⟩, x : = (A - a) u, y : = (B - b) u, Δ A : = ∥ x ∥, Δ B : = ∥ y ∥,$ and write ${A, B} u : = A B u + B A u$ and $[A, B] u : = A B u - B A u$ for the pairings at $u$ . Then:

(a) $a, b \in R$ , $x, y \in H$ , and $Re ⟨ x, y ⟩ = \frac{1}{2} ⟨ {(A - a), (B - b)} u, u ⟩ = \frac{1}{2} ⟨ {A, B} u, u ⟩ - a b, ⟨ [A, B] u, u ⟩ = - 2 i Im ⟨ x, y ⟩;$

(b) (Robertson bound) $Δ A Δ B \geq \frac{1}{2} ∣ ⟨ [A, B] u, u ⟩ ∣$ ;

(c) (Schroedinger-Robertson bound) $(Δ A)^{2} (Δ B)^{2} \geq (\frac{1}{2} ⟨ {(A - a), (B - b)} u, u ⟩)^{2} + \frac{1}{4} ∣ ⟨ [A, B] u, u ⟩ ∣^{2},$ that is, with the expanded anticommutator pairing $(Δ A)^{2} (Δ B)^{2} \geq (\frac{1}{2} ⟨ {A, B} u, u ⟩ - a b)^{2} + \frac{1}{4} ∣ ⟨ [A, B] u, u ⟩ ∣^{2}$ ;

(d) equality holds in (c) if and only if $x$ and $y$ are linearly dependent; in particular the Robertson bound (b) follows from (c) because the first square is nonnegative;

(e) if $A$ and $B$ are bounded everywhere-defined self-adjoint operators, then the product-domain hypothesis $u \in D (A B) \cap D (B A)$ holds for every unit $u \in H$ and clauses (a)-(d) apply to every unit $u$ .

No energy-time inequality and no apparatus statement is made.

Facts & Assumptions

Given: The Axiom of Countable Choice; a complex Hilbert space $H$ ; symmetric densely defined $A, B$ ; a unit $u \in D (A) \cap D (B)$ with $u \in D (A B) \cap D (B A)$ ; the numbers $a, b$ and vectors $x, y$ .

[F1]

Covariance-refined uncertainty. For symmetric densely defined $A, B$ and a unit $u \in D (A) \cap D (B)$ , with $a, b, x, y$ as above: $a, b$ are real and $x, y \in H$ ; ${Cov}_{u} (A, B) : = Re ⟨ x, y ⟩$ is real and symmetric; the refined bound $(Δ A)^{2} (Δ B)^{2} \geq [Re ⟨ x, y ⟩]^{2} + [Im ⟨ x, y ⟩]^{2}$ holds, with equality if and only if $x, y$ are linearly dependent; and if $u \in D (A B) \cap D (B A)$ then $⟨ [A, B] u, u ⟩ = - 2 i Im ⟨ x, y ⟩$ and $(Δ A)^{2} (Δ B)^{2} \geq [Re ⟨ x, y ⟩]^{2} + \frac{1}{4} ∣ ⟨ [A, B] u, u ⟩ ∣^{2}$ (Covariance-refined preparation uncertainty without operator products clauses (a)-(d)).

[F2]

Commutator pairing. If $u \in D (A B) \cap D (B A)$ then $([A, B] u, u) = ⟨ A B u, u ⟩ - ⟨ B A u, u ⟩ = - 2 i Im ⟨ A u, B u ⟩$ , and $⟨ x, y ⟩ = ⟨ A u, B u ⟩ - a b$ with real $a, b$ (Weak-form Robertson uncertainty on a common domain clauses (a) and (c)).

[F3]

Inner products and Cauchy-Schwarz. For $z \in C$ one has $∣ z ∣^{2} = (Re z)^{2} + (Im z)^{2}$ , and $∥ v ∥ ∥ w ∥ \geq ∣ ⟨ v, w ⟩ ∣$ with equality if and only if $v, w$ are linearly dependent (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F4]

Bounded operators. A bounded everywhere-defined self-adjoint operator on $H$ has domain $H$ , so for bounded self-adjoint $A, B$ the hypotheses $u \in D (A) \cap D (B)$ and $u \in D (A B) \cap D (B A)$ hold for every $u \in H$ (A bounded linear operator between normed spaces, Symmetric, self-adjoint and essentially self-adjoint operators).

Proof

technique · expand the two pairings of the centered vectors in terms of the anticommutator and commutator pairings using symmetry of the operators, then read the Schroedinger-Robertson inequality off the refined Cauchy-Schwarz bound

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

1.1givenF1

[Means and centered vectors] By [F1] the numbers $a$ and $b$ are real and $x = (A - a) u$ , $y = (B - b) u$ lie in $H$ , so $Δ A = ∥ x ∥$ and $Δ B = ∥ y ∥$ are the norms of the centered vectors.

2.1step 1.1F2algebra

[The real part is the anticommutator pairing] Since $B u \in D (A)$ and $A u \in D (B)$ , symmetry of $A$ gives $⟨ A B u, u ⟩ = ⟨ B u, A u ⟩ = \overline{⟨ A u, B u ⟩}$ and symmetry of $B$ gives $⟨ B A u, u ⟩ = ⟨ A u, B u ⟩$ . Adding, $⟨ {A, B} u, u ⟩ = 2 Re ⟨ A u, B u ⟩$ . By [F2], $⟨ x, y ⟩ = ⟨ A u, B u ⟩ - a b$ with $a b \in R$ , hence $Re ⟨ x, y ⟩ = Re ⟨ A u, B u ⟩ - a b = \frac{1}{2} ⟨ {A, B} u, u ⟩ - a b$ . Expanding the anticommutator of the centered operators, using $a, b \in R$ and $⟨ A u, u ⟩ = a$ , $⟨ B u, u ⟩ = b$ , gives $⟨ {(A - a), (B - b)} u, u ⟩ = ⟨ {A, B} u, u ⟩ - 2 a b$ , so the two displays for the real part agree.

3.1step 2.1F2algebra

[The imaginary part is the commutator pairing] By [F2], $⟨ [A, B] u, u ⟩ = - 2 i Im ⟨ A u, B u ⟩$ ; by step 2.1 and [F2], $Im ⟨ x, y ⟩ = Im ⟨ A u, B u ⟩$ , so $⟨ [A, B] u, u ⟩ = - 2 i Im ⟨ x, y ⟩$ . This completes (a).

4.1step 2.1step 3.1F1F3algebra

[The two bounds] By [F1] the refined bound $(Δ A)^{2} (Δ B)^{2} \geq [Re ⟨ x, y ⟩]^{2} + [Im ⟨ x, y ⟩]^{2}$ holds. Substituting step 2.1 and $∣ Im ⟨ x, y ⟩ ∣ = \frac{1}{2} ∣ ⟨ [A, B] u, u ⟩ ∣$ from step 3.1 gives (c); dropping the nonnegative first square gives (b).

5.1step 4.1F1

[Equality] By [F1] equality in the refined bound holds if and only if $x, y$ are linearly dependent; this is (d).

6.1givenF4∎

[Bounded case] If $A$ and $B$ are bounded everywhere-defined self-adjoint operators then $D (A) = D (B) = H$ and all products are everywhere defined, so the product-domain hypothesis of the statement holds for every unit $u \in H$ by [F4]; the displays of (a)-(d) are then unconditional in $u$ . This is (e).

The Schroedinger-Robertson bound in anticommutator form and the Robertson bound on product domains

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources