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Golden-rule coefficient approximate identity

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Let $ℏ > 0$ .

(a) The golden-rule kernel. For $t > 0$ define $K_{t} (x) : = \frac{1}{ℏ^{2} t} ∣ \int_{0}^{t} e^{i x s / ℏ} d s ∣^{2}, x \in R .$ Then $K_{t}$ is continuous, nonnegative and even, and for $x \neq 0$ $K_{t} (x) = \frac{4 \sin^{2} (x t / (2 ℏ))}{x^{2} t}, K_{t} (0) = \frac{t}{ℏ^{2}};$ moreover for every $δ > 0$ and every $x$ with $∣ x ∣ \geq δ$ , $K_{t} (x) \leq \frac{4}{x^{2} t} \leq \frac{4}{δ^{2} t}, and \int_{R} K_{t} (x) d x = \frac{2 π}{ℏ} .$

(b) Approximate identity. Let $J \in L^{1} (R; C)$ (including real-valued functions) (Integrable real and complex functions, and their integrals, The space $L^{p} (μ)$ as the quotient by null functions) and let $Δ \in R$ . Choose a measurable representative $J : R \to C$ . Continuity at $Δ$ means that for every $ϵ > 0$ there is $δ > 0$ such that $∣ J (y) - J (Δ) ∣ < ϵ$ whenever $∣ y - Δ ∣ < δ$ ; for real-valued $J$ this is Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, and for complex-valued $J$ it is equivalent to continuity of its real and imaginary parts. If this condition holds, then $\lim_{t \to \infty} \int_{R} K_{t} (x) J (Δ + x) d x = \frac{2 π}{ℏ} J (Δ)$ (For a complex-valued function $Q : (0, \infty) \to C$ , $Q (t) \to L$ means that for every $ϵ > 0$ there is $T > 0$ with $∣ Q (t) - L ∣ < ϵ$ for all $t > T$ ; equivalently its real and imaginary parts have the real limits in Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point.)

(c) First-order emission coefficient. Adopt the objects and hypotheses of Cutoff dipole relative bound: $α > 0$ , the transverse space $h = ran M_{Π}$ of Transverse continuum free Fock Hamiltonian with $ε (k) = α ∣ k ∣$ , the cutoff $Λ > 0$ , the vectors $g_{1}, \dots, g_{J} \in h$ ( $J \geq 1$ ) with $supp g_{j} \subseteq {k : 0 < ε (k) \leq Λ}, \frac{g_{j}}{\sqrt{ε}} \in h,$ the Hermitian $d \times d$ matrices $H_{a}, D_{1}, \dots, D_{J}$ and the coupling $η_{0} \in R$ . Let $v_{a}, v_{b}$ be orthonormal eigenvectors of $H_{a}$ with real eigenvalues $E_{a}, E_{b}$ satisfying $H_{a} v_{a} = E_{a} v_{a}$ , $H_{a} v_{b} = E_{b} v_{b}$ , and let $Δ : = E_{a} - E_{b} > 0$ . Put $c_{j} : = ⟨ D_{j} v_{a}, v_{b} ⟩ (C^{d} inner product), M : = \sum_{j = 1}^{J} c_{j} g_{j} \in h,$ and define the finite Borel measure $ν_{b} (A) : = \int_{R^{3}} 1_{A} (ε (k)) ∥ M (k) ∥^{2} d^{3} k, A \subseteq R Borel,$ and for $t \geq 0$ the amplitude $A_{b} (k, t) : = - \frac{i η_{0}}{ℏ} M (k) \int_{0}^{t} e^{i (ε (k) - Δ) s / ℏ} d s .$ Then:

(i) $ν_{b}$ is a finite Borel measure on $R$ with $ν_{b} (R ∖ (0, Λ]) = 0$ , and $\int_{R} f d ν_{b} = \int_{R^{3}} f (ε (k)) ∥ M (k) ∥^{2} d^{3} k$ for every Borel $f \geq 0$ ;

(ii) $A_{b} (\cdot, t) \in h$ for every $t \geq 0$ , and for $t > 0$ , $∥ A_{b} (\cdot, t) ∥_{h}^{2} = η_{0}^{2} t \int_{R} K_{t} (E - Δ) d ν_{b} (E) .$ At $t = 0$ the amplitude is zero, so its squared norm is zero; no $K_{0}$ is used.

(iii) if $ν_{b} = J_{b} λ$ for some $J_{b} \in L^{1} (λ)$ with $J_{b} = 0$ $λ$ -a.e. on $R ∖ (0, Λ]$ and the representative $J_{b}$ continuous at $Δ$ , then for $0 < Δ < Λ$ $\lim_{t \to \infty} \frac{1}{t} ∥ A_{b} (\cdot, t) ∥_{h}^{2} = \frac{2 π η_{0}^{2}}{ℏ} J_{b} (Δ);$ if instead $Δ > Λ$ , then the same limit exists and equals $0$ . The boundary case $Δ = Λ$ is not asserted.

Motivation and scope. In the cutoff dipole model of Cutoff dipole relative bound the vector $A_{b} (\cdot, t)$ is the explicit single-vertex amplitude with the atom passing from $v_{a}$ to $v_{b}$ and one photon emitted; $∥ A_{b} (\cdot, t) ∥^{2}$ is the corresponding first-order photon coefficient, and (iii) is the golden-rule limit of that coefficient divided by $t$ . This item computes only the limit of the explicit coefficient, under the stated hypotheses on the spectral density $J_{b}$ . It asserts no approximation of the exact model evolution, no all-time or weak-coupling limit, no Markov property and no exact exponential decay, and it reports no measurement or empirical rate. The parameter $ℏ$ , the coupling $η_{0}$ and the matrices $D_{j}$ enter as mathematical normalisation data of the model; their physical identification is a separate assumption.

Facts & Assumptions

Given: the coupling data $(ℏ, α, Λ, η_{0}, g_{j}, H_{a}, D_{j}, v_{a}, v_{b})$ of the cutoff model, the kernel $K_{t}$ , a function $J \in L^{1} (λ)$ continuous at $Δ$ , and the pushforward measure $ν_{b}$ with its possible density $J_{b}$ .

[F1]

For real $θ$ , $∣ e^{i θ} - 1 ∣^{2} = 4 \sin^{2} (θ / 2)$ , $∣ e^{i θ} ∣ = 1$ , and $∣ \sin ∣ \leq 1$ ; the derivative of $\sin$ at $0$ gives $\lim_{a \to 0} \sin (a) / a = 1$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The derivatives of sine and cosine are cosine and minus sine).

[F2]

Fourier/Plancherel package: for $g \in L^{1} (R)$ the integral transform $\hat{g} (ξ) = \int g (u) e^{- 2 π i ξ u} d u$ is defined, on $L^{1} \cap L^{2}$ it agrees with the $L^{2}$ transform, and $F_{2} : L^{2} \to L^{2}$ is a unitary isometry preserving the norm (Fourier transform on complex L1 classes, Agreement of the integral and L2 transforms, Plancherel theorem).

[F3]

Every interval $[0, T]$ , $T \geq 0$ , has Lebesgue measure $T$ , so $∥ 1_{[0, T]} ∥_{2}^{2} = T$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F4]

Change of variables: if $U, V \subseteq R$ are open, $T : U \to V$ is a $C^{1}$ diffeomorphism and $f \in L^{1} (λ)$ , then $\int_{V} f d λ = \int_{U} (f \circ T) ∣ T^{'} ∣ d λ$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions); this applies in particular to the translations $x \mapsto Δ + x$ and the dilations $x \mapsto - 2 π x$ of $R$ (both with $∣ T^{'} ∣$ constant). The primitive $- 1 / x$ and monotone convergence on bounded intervals give $\int_{∣ x ∣ \geq δ} x^{- 2} d x = 2 / δ$ for $δ > 0$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , Monotone convergence for the integral).

[F5]

Measure and integral calculus: monotone convergence; additivity and linearity of the Lebesgue integral ( $L^{1}$ linearity, nonnegative additivity); for a nonnegative measurable $f$ the set function $A \mapsto \int_{A} f d μ$ is a measure; a measure with density $h$ is given by $(h μ) (A) = \int_{A} h d μ$ (Monotone convergence for the integral, Additivity of the nonnegative Lebesgue integral, The Lebesgue integral is linear on $L^{1} (μ)$ , The indefinite integral of a nonnegative measurable function is a measure, The measure with density $f$ relative to $μ$ ).

[F6]

Model objects: $h = {f \in L^{2} (R^{3}; C^{3}) : k \cdot f (k) = 0 a.e.}$ is a closed linear subspace, $ε (k) = α ∣ k ∣$ is continuous, $g_{j} \in h$ with $supp g_{j} \subseteq {0 < ε \leq Λ}$ and $g_{j} / \sqrt{ε} \in h$ ; $H_{a}, D_{j}$ are Hermitian with $v_{a}, v_{b}$ orthonormal eigenvectors (Transverse continuum free Fock Hamiltonian, Cutoff dipole relative bound).

[F7]

Cauchy-Schwarz in the finite-dimensional Hilbert space $C^{J}$ gives $∣ \sum_{j} c_{j} g_{j} (k) ∣^{2} \leq (\sum_{j} ∣ c_{j} ∣^{2}) (\sum_{j} ∥ g_{j} (k) ∥^{2})$ for a.e. $k$ , hence $∥ M ∥_{L^{2}}^{2} \leq (\sum_{j} ∣ c_{j} ∣^{2}) (\sum_{j} ∥ g_{j} ∥^{2}) < \infty$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, The space $L^{p} (μ)$ as the quotient by null functions).

[F8]

Use the modulus definitions of continuity and limits stated in (b); they agree with Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point and Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point for real functions. For complex functions the inequalities $∣ Re z ∣, ∣ Im z ∣ \leq ∣ z ∣ \leq ∣ Re z ∣ + ∣ Im z ∣$ prove equivalence with the two real-component conditions.

Proof

1.1F1algebra

[Explicit kernel and pointwise bounds] For $x \neq 0$ , integrating the exponential gives $\int_{0}^{t} e^{i x s / ℏ} d s = ℏ (e^{i x t / ℏ} - 1) / (i x)$ , so by [F1] $K_{t} (x) = ℏ^{2} ∣ e^{i x t / ℏ} - 1 ∣^{2} / (x^{2} ℏ^{2} t) = 4 \sin^{2} (x t / (2 ℏ)) / (x^{2} t)$ ; at $x = 0$ the defining integral equals $t$ and $K_{t} (0) = t / ℏ^{2}$ . With $a = x t / (2 ℏ)$ the identity $4 \sin^{2} (x t / (2 ℏ)) / (x^{2} t) = (t / ℏ^{2}) (\sin a / a)^{2}$ and [F1] give continuity at $0$ and on $R$ ; $K_{t} \geq 0$ and $K_{t}$ is even by [F1]. For $∣ x ∣ \geq δ > 0$ the same formula and $∣ \sin ∣ \leq 1$ give $K_{t} (x) \leq 4 / (x^{2} t) \leq 4 / (δ^{2} t)$ .

1.2F1F2F3F4algebra

[Kernel mass] Put $T : = t / ℏ$ and $g : = 1_{[0, T]} \in L^{1} \cap L^{2}$ , $∥ g ∥_{2}^{2} = T$ by [F3]. By [F2], $\hat{g} (ξ) = \int_{0}^{T} e^{- 2 π i ξ u} d u$ satisfies $\int_{R} ∣ \hat{g} ∣^{2} = ∥ g ∥_{2}^{2} = T$ . For every real $x$ , $\int_{0}^{T} e^{i x u} d u = \hat{g} (- x / (2 π))$ ; applying [F4] to the dilation $ξ \mapsto - 2 π ξ$ of the $L^{1}$ function $ξ \mapsto ∣ \hat{g} (ξ) ∣^{2}$ gives $\int_{R} ∣ \hat{g} (- x / (2 π)) ∣^{2} d x = 2 π \int_{R} ∣ \hat{g} (ξ) ∣^{2} d ξ = 2 π T$ . Substituting $s = ℏ u$ in the elementary integral gives $\int_{0}^{t} e^{i x s / ℏ} d s = ℏ \int_{0}^{T} e^{i x u} d u$ , hence $\int_{R} K_{t} d x = (1 / (ℏ^{2} t)) \cdot ℏ^{2} \cdot 2 π T = 2 π / ℏ$ .

1.3F5F6F7algebra

[The pushforward measure] With $c_{j} : = ⟨ D_{j} v_{a}, v_{b} ⟩$ and $M = \sum_{j} c_{j} g_{j}$ , [F7] gives $∥ M ∥_{L^{2}}^{2} < \infty$ and $M \in h$ by [F6], and $supp M \subseteq {0 < ε \leq Λ}$ . Hence for every Borel $A \subseteq R$ the function $k \mapsto 1_{A} (ε (k)) ∥ M (k) ∥^{2}$ is nonnegative and measurable, so $A \mapsto ν_{b} (A) : = \int_{R^{3}} 1_{A} (ε (k)) ∥ M (k) ∥^{2} d^{3} k$ is a finite measure by [F5], with $ν_{b} (R) = ∥ M ∥_{L^{2}}^{2} < \infty$ and $ν_{b} (R ∖ (0, Λ]) = 0$ . The identity $\int f d ν_{b} = \int f (ε (k)) ∥ M (k) ∥^{2} d^{3} k$ holds for indicators by definition, for nonnegative simple $f$ by additivity, and for all Borel $f \geq 0$ by monotone convergence [F5]. This proves (i).

2.1F4F5F8step 1.1step 1.2algebra

[Approximate identity] Let $J \in L^{1} (λ; C)$ have the representative continuous at $Δ$ specified in (b). For fixed $t > 0$ , $K_{t} \leq t / ℏ^{2}$ by the defining time integral, so $K_{t} (x) J (Δ + x)$ is integrable by translation invariance, and the subtraction below is licensed by $L^{1}$ linearity [F5]. Fix $δ > 0$ . By step 1.2, $\int K_{t} (x) J (Δ + x) d x - (2 π / ℏ) J (Δ) = \int K_{t} (x) (J (Δ + x) - J (Δ)) d x$ , and we split the right side at ${∣ x ∣ \geq δ}$ . On $∣ x ∣ < δ$ the bound $K_{t} \geq 0$ and step 1.2 give at most $(2 π / ℏ) \sup_{∣ x ∣ < δ} ∣ J (Δ + x) - J (Δ) ∣$ ; on $∣ x ∣ \geq δ$ , separate the two terms to obtain at most $(4 / (δ^{2} t)) \int_{∣ x ∣ \geq δ} ∣ J (Δ + x) ∣ d x + ∣ J (Δ) ∣ \int_{∣ x ∣ \geq δ} 4 / (x^{2} t) d x \leq 4 ∥ J ∥_{1} / (δ^{2} t) + 8 ∣ J (Δ) ∣ / (δ t)$ , since $\int_{∣ x ∣ \geq δ} x^{- 2} d x = 2 / δ$ , where $\int_{R} ∣ J (Δ + x) ∣ d x = ∥ J ∥_{1}$ by [F4] applied to the translation $x \mapsto Δ + x$ . Given $ε > 0$ , continuity of $J$ at $Δ$ [F8] yields $δ > 0$ with $(2 π / ℏ) \sup_{∣ x ∣ < δ} ∣ J (Δ + x) - J (Δ) ∣ \leq ε / 3$ , and then $t$ large enough that $4 ∥ J ∥_{1} / (δ^{2} t) + 8 ∣ J (Δ) ∣ / (δ t) < ε / 3$ ; hence the modulus of the displayed difference is $< ε$ for all large $t$ , which is the asserted limit $\frac{2 π}{ℏ} J (Δ)$ [F8].

2.2step 1.1step 1.3F6algebra

[Coefficient computation] For $t = 0$ the defining time integral is zero and $A_{b} (\cdot, 0) = 0$ . For $t > 0$ let $φ_{t} (E) : = \int_{0}^{t} e^{i (E - Δ) s / ℏ} d s$ , so that $A_{b} (k, t) = - (i η_{0} / ℏ) M (k) φ_{t} (ε (k))$ and $∣ φ_{t} ∣ \leq t$ . The function $k \mapsto - (i η_{0} / ℏ) φ_{t} (ε (k))$ is bounded and measurable, so $A_{b} (\cdot, t) \in L^{2} (R^{3}; C^{3})$ with $∣ A_{b} (k, t) ∣ \leq (∣ η_{0} ∣ t / ℏ) ∥ M (k) ∥$ , and pointwise scalar multiplication preserves the transversality condition $k \cdot f (k) = 0$ a.e.; hence $A_{b} (\cdot, t) \in h$ by [F6]. Then $∥ A_{b} (\cdot, t) ∥_{h}^{2} = (η_{0}^{2} / ℏ^{2}) \int_{R^{3}} ∥ M (k) ∥^{2} ∣ φ_{t} (ε (k)) ∣^{2} d^{3} k = (η_{0}^{2} / ℏ^{2}) \int_{R} ∣ φ_{t} (E) ∣^{2} d ν_{b} (E)$ by step 1.3 with $f = ∣ φ_{t} ∣^{2}$ , and $∣ φ_{t} (E) ∣^{2} = ∣ \int_{0}^{t} e^{i (E - Δ) s / ℏ} d s ∣^{2} = ℏ^{2} t K_{t} (E - Δ)$ for every $E$ by the definition of $K_{t}$ in (a) and step 1.1; this proves (ii).

3.1F4F5step 1.1step 2.1step 1.3step 2.2algebra

[Density and the limit] Assume $ν_{b} = J_{b} λ$ with $J_{b} \in L^{1} (λ)$ , $J_{b} = 0$ $λ$ -a.e. on $R ∖ (0, Λ]$ , and $J_{b}$ continuous at $Δ$ . As in step 1.3, the identity $\int f d ν_{b} = \int f J_{b} d λ$ holds for every Borel $f \geq 0$ (indicators are the hypothesis, simple functions additivity, general $f$ monotone convergence, [F5]). Hence by step 2.2, for every $t > 0$ , $\frac{1}{t} ∥ A_{b} (\cdot, t) ∥_{h}^{2} = η_{0}^{2} \int K_{t} (E - Δ) d ν_{b} (E) = η_{0}^{2} \int K_{t} (E - Δ) J_{b} (E) d E = η_{0}^{2} \int K_{t} (x) J_{b} (Δ + x) d x$ , the last equality by the change of variables $E = Δ + x$ [F4], the integrand $K_{t} (\cdot - Δ) J_{b}$ being bounded times $L^{1}$ . If $0 < Δ < Λ$ , then step 2.1 gives $\lim_{t} \frac{1}{t} ∥ A_{b} (\cdot, t) ∥^{2} = η_{0}^{2} \frac{2 π}{ℏ} J_{b} (Δ) = \frac{2 π η_{0}^{2}}{ℏ} J_{b} (Δ)$ . If instead $Δ > Λ$ , then $J_{b} (Δ + x) \neq 0$ for $λ$ -a.e. $x$ only when $Δ + x \leq Λ$ , i.e. $x \leq Λ - Δ < 0$ ; on the support up to a null set one has $∣ x ∣ \geq Δ - Λ = : δ > 0$ , so step 1.1 and $\int ∣ J_{b} (Δ + x) ∣ d x = ∥ J_{b} ∥_{1}$ [F4] give $∣ \int K_{t} (x) J_{b} (Δ + x) d x ∣ \leq (4 / (δ^{2} t)) ∥ J_{b} ∥_{1} \to 0$ , and the limit is $0$ . This proves (iii); the case $Δ = Λ$ is excluded, consistently with the statement.

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

[Conclusion and scope] Steps 1.1 and 1.2 prove (a), step 2.1 proves (b), and steps 1.3, 2.2 and 3.1 prove (i)-(iii). The scope paragraph of the statement records that only the explicit single-vertex coefficient is treated and that no all-time, Markov, exponential-decay or empirical claim is made. This completes the proof.

Golden-rule coefficient approximate identity

Statement

Facts & Assumptions

Proof

Depends on

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