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Graph-domain Ehrenfest identities and the conditional timescale bound

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space), let $K, A$ be self-adjoint operators on $H$ , let $I \subseteq R$ be a nonempty open interval, fix $ℏ > 0$ and let $ψ : I \to H$ satisfy, for every $t \in I$ :

$ψ (t) \in D (A)$ , and the curve is $C^{1}$ into $D (A)$ with the graph norm of $A$ (Unbounded linear operators: domain, graph and extension, Densely defined, closed and closable operators, and cores), meaning that both $t \mapsto ψ (t)$ and $t \mapsto A ψ (t)$ are $C^{1}$ into $H$ ;
$K ψ (t) \in D (A)$ , that is $ψ (t) \in D (A K)$ ; the additional condition $A ψ (t) \in D (K)$ (that is $ψ (t) \in D (K A)$ ) is imposed only in the clauses that state it;
$ψ^{'} (t) = - \frac{i}{ℏ} K ψ (t)$ in $H$ .

Then, writing $f (t) : = ⟨ A ψ (t), ψ (t) ⟩$ and $(K A - A K) ψ : = K (A ψ) - A (K ψ)$ for $ψ \in D (A K) \cap D (K A)$ :

(a) if hypotheses 1 and 3 and $K ψ (t) \in D (A)$ for all $t$ hold (that is, $ψ (t) \in D (A K)$ ), then $f$ is $C^{1}$ on $I$ and $f^{'} (t) = - \frac{2}{ℏ} Im ⟨ A ψ (t), K ψ (t) ⟩;$ if moreover $A ψ (t) \in D (K)$ for all $t$ (that is, $ψ (t) \in D (K A)$ ), then also $f^{'} (t) = \frac{i}{ℏ} ⟨ (K A - A K) ψ (t), ψ (t) ⟩$ ;

(b) if $(A_{t})_{t \in I}$ is a family of bounded self-adjoint operators that is $C^{1}$ in operator norm and $ψ$ satisfies 1 and 3 above (no product-domain hypothesis), then $g (t) : = ⟨ A_{t} ψ (t), ψ (t) ⟩$ is $C^{1}$ with $g^{'} (t) = ⟨ A_{t}^{'} ψ (t), ψ (t) ⟩ - \frac{2}{ℏ} Im ⟨ A_{t} ψ (t), K ψ (t) ⟩;$ if in addition $A_{t} ψ (t) \in D (K)$ and $K ψ (t) \in D (A_{t})$ for all $t$ , then this equals $⟨ A_{t}^{'} ψ (t), ψ (t) ⟩ + \frac{i}{ℏ} ⟨ [K, A_{t}] ψ (t), ψ (t) ⟩$ ;

(c) if moreover $∥ ψ (t) ∥ = 1$ for all $t$ and for a fixed self-adjoint $A$ as in (a) one has $f^{'} (t_{0}) \neq 0$ at some $t_{0} \in I$ , then with the preparation spreads $Δ K (t_{0}) : = ∥ (K - ⟨ K ψ (t_{0}), ψ (t_{0}) ⟩) ψ (t_{0}) ∥, Δ A (t_{0}) : = ∥ (A - f (t_{0})) ψ (t_{0}) ∥,$ and the timescale $τ_{A} (t_{0}) : = Δ A (t_{0}) / ∣ f^{'} (t_{0}) ∣$ , the bound $Δ K (t_{0}) τ_{A} (t_{0}) \geq \frac{ℏ}{2}$ holds;

(d) if $f^{'} (t_{0}) = 0$ then the definition of $τ_{A} (t_{0})$ supplies no timescale, and no universal time operator, apparatus error or energy-time observable is asserted by (a)-(c).

Facts & Assumptions

Given: self-adjoint $K, A$ on a complex Hilbert space $H$ , an open interval $I$ , a curve $ψ$ satisfying hypotheses 1-3 with $K ψ (t) \in D (A)$ (and $A ψ (t) \in D (K)$ where a clause states it), the function $f (t) = ⟨ A ψ (t), ψ (t) ⟩$ and the notation $(K A - A K) ψ = K (A ψ) - A (K ψ)$ .

[F1]

A self-adjoint operator is closed and symmetric; $T^{*} = T$ , $D (T) = D (T^{*})$ , and $⟨ T v, w ⟩ = ⟨ v, T w ⟩$ for $v, w \in D (T)$ (Symmetric, self-adjoint and essentially self-adjoint operators, Adjoint of a densely defined operator, Densely defined, closed and closable operators, and cores).

[F2]

Hypotheses 1 and 2: $t \mapsto ψ (t)$ and $t \mapsto A ψ (t)$ are $C^{1}$ into $H$ , and $K ψ (t) \in D (A)$ ; $A ψ (t) \in D (K)$ is assumed only for the commutator clauses. Because $A$ is closed and $t \mapsto ψ (t)$ is $C^{1}$ into $D (A)$ , the derivative of $t \mapsto A ψ (t)$ is $A ψ^{'} (t)$ : the graph-norm difference quotients converge to $ψ^{'} (t)$ , and applying the closed operator $A$ to the limit gives $(A ψ)^{'} (t) = A ψ^{'} (t)$ (Unbounded linear operators: domain, graph and extension, Densely defined, closed and closable operators, and cores).

[F3]

For symmetric $K, A$ and a unit $u \in D (K) \cap D (A)$ , clause (b) of Covariance-refined preparation uncertainty without operator products gives $Δ K Δ A \geq ∣ Im ⟨ (A - ⟨ A u, u ⟩) u, (K - ⟨ K u, u ⟩) u ⟩ ∣ = ∣ Im ⟨ A u, K u ⟩ ∣$ , since both means are real. No product-domain hypothesis is needed for this bound. The weak-form Robertson lemma of Weak-form Robertson uncertainty on a common domain: for symmetric $K, A$ , a unit $u \in D (K) \cap D (A)$ with $u \in D (K A) \cap D (A K)$ satisfies $Δ K Δ A \geq \frac{1}{2} ∣ ⟨ [K, A] u, u ⟩ ∣$ , where $[K, A] u = K A u - A K u$ (clause (c)); the covariance form of the same bound is Covariance-refined preparation uncertainty without operator products clause (d).

Proof

1.1givenF2algebra

[Differentiability of $f$ ] By [F2], $(A ψ)^{'} (t) = A ψ^{'} (t)$ and both factors $A ψ (t)$ and $ψ (t)$ are $C^{1}$ into $H$ ; the product rule for the inner product gives $f^{'} (t) = ⟨ (A ψ)^{'} (t), ψ (t) ⟩ + ⟨ A ψ (t), ψ^{'} (t) ⟩ = ⟨ A ψ^{'} (t), ψ (t) ⟩ + ⟨ A ψ (t), ψ^{'} (t) ⟩$ .

1.2givenF1algebra

[Bounded time-dependent observable] Let $A_{t}$ be a bounded self-adjoint family, $C^{1}$ in operator norm, and $g (t) = ⟨ A_{t} ψ (t), ψ (t) ⟩$ . By the product rule for the three $C^{1}$ factors, $g^{'} = ⟨ A_{t}^{'} ψ, ψ ⟩ + ⟨ A_{t} ψ^{'}, ψ ⟩ + ⟨ A_{t} ψ, ψ^{'} ⟩$ ; self-adjointness of $A_{t}$ combines the last two terms into $2 Re ⟨ A_{t} ψ^{'}, ψ ⟩ = 2 Re (- \frac{i}{ℏ} q)$ with $q : = ⟨ A_{t} K ψ, ψ ⟩$ , and for $q = x + i y$ one has $- \frac{i}{ℏ} q = \frac{y - i x}{ℏ}$ , so $2 Re (- \frac{i}{ℏ} q) = \frac{2 y}{ℏ} = \frac{2}{ℏ} Im q$ . Since $A_{t}$ is self-adjoint, $q = ⟨ A_{t} K ψ, ψ ⟩ = ⟨ K ψ, A_{t} ψ ⟩ = \overline{⟨ A_{t} ψ, K ψ ⟩}$ , hence $Im q = - Im ⟨ A_{t} ψ, K ψ ⟩$ and $g^{'} = ⟨ A_{t}^{'} ψ, ψ ⟩ - \frac{2}{ℏ} Im ⟨ A_{t} ψ, K ψ ⟩$ . If in addition $A_{t} ψ \in D (K)$ and $K ψ \in D (A_{t})$ , then $⟨ K A_{t} ψ, ψ ⟩ = ⟨ A_{t} ψ, K ψ ⟩$ and $⟨ A_{t} K ψ, ψ ⟩ = ⟨ K ψ, A_{t} ψ ⟩ = \overline{⟨ A_{t} ψ, K ψ ⟩}$ , so $⟨ [K, A_{t}] ψ, ψ ⟩ = 2 i Im ⟨ A_{t} ψ, K ψ ⟩$ and $- \frac{2}{ℏ} Im ⟨ A_{t} ψ, K ψ ⟩ = \frac{i}{ℏ} ⟨ [K, A_{t}] ψ, ψ ⟩$ , completing (b).

2.1givenstep 1.1F1algebra

[The commutator form] Since $ψ^{'} (t) \in D (A)$ by hypothesis 1 and $A$ is symmetric [F1], $⟨ A ψ^{'} (t), ψ (t) ⟩ = ⟨ ψ^{'} (t), A ψ (t) ⟩$ , so $f^{'} (t) = 2 Re ⟨ ψ^{'} (t), A ψ (t) ⟩$ . Substituting $ψ^{'} = - \frac{i}{ℏ} K ψ$ and writing $z : = ⟨ A ψ (t), K ψ (t) ⟩$ , we get $f^{'} = 2 Re (- \frac{i}{ℏ} ⟨ K ψ, A ψ ⟩) = 2 Re (- \frac{i}{ℏ} \overline{z})$ ; since $Re (- \frac{i}{ℏ} \overline{z}) = Re (- \frac{i}{ℏ} (Re z - i Im z)) = - \frac{1}{ℏ} Im z$ , this gives $f^{'} = - \frac{2}{ℏ} Im ⟨ A ψ, K ψ ⟩$ .

3.1givenstep 2.1F1algebra

[The commutator pairing] Assume in addition $A ψ (t) \in D (K)$ . Then the pairings $⟨ K A ψ, ψ ⟩ = ⟨ A ψ, K ψ ⟩$ (symmetry of $K$ with $A ψ (t) \in D (K)$ ) and $⟨ A K ψ, ψ ⟩ = ⟨ K ψ, A ψ ⟩ = \overline{⟨ A ψ, K ψ ⟩}$ (symmetry of $A$ with $K ψ (t) \in D (A)$ ) are defined, so $⟨ (K A - A K) ψ, ψ ⟩ = z - \overline{z} = 2 i Im z$ with $z$ as in step 2.1, and $\frac{i}{ℏ} (z - \overline{z}) = \frac{i}{ℏ} \cdot 2 i Im z = - \frac{2}{ℏ} Im z$ , which is the second displayed form of $f^{'}$ . This proves (a).

3.2givenstep 2.1F3algebra

[Timescale bound] Let $u : = ψ (t_{0})$ with $∥ u ∥ = 1$ and $f^{'} (t_{0}) \neq 0$ . By hypotheses 1-3, $u \in D (K) \cap D (A)$ and $K u \in D (A)$ , so step 2.1 gives $∣ f^{'} (t_{0}) ∣ = (2 / ℏ) ∣ Im ⟨ A u, K u ⟩ ∣$ . The weak bound in [F3] gives $Δ K (t_{0}) Δ A (t_{0}) \geq ∣ Im ⟨ A u, K u ⟩ ∣ = (ℏ / 2) ∣ f^{'} (t_{0}) ∣$ . Dividing by $∣ f^{'} (t_{0}) ∣ > 0$ proves (c), without assuming $A u \in D (K)$ .

4.1givenstep 3.2algebra∎

[Scope and vanishing derivative] If $f^{'} (t_{0}) = 0$ , the quotient $τ_{A} (t_{0}) = Δ A (t_{0}) / ∣ f^{'} (t_{0}) ∣$ is undefined, so no timescale is supplied; and nothing in the argument produces a time operator, a measurement error or an energy-time observable, since all quantities were formed from the specified curve and the two operators. This is (d).

Graph-domain Ehrenfest identities and the conditional timescale bound

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources