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Golden-rule kernel and its quantified approximate-identity limit

Statement

Assume the Countable Choice axiom ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $ℏ > 0$ in joule-seconds, $T > 0$ in seconds, an energy $E_{0} \in R$ in joules, a dimensionless $λ \in R$ , and a bounded nonnegative Borel function $w : R \to [0, \infty)$ continuous at $E_{0}$ . Here $w$ is a specified transition-density weight in joules; no state density or matrix element is inferred from the name.

For angular frequency $ω \in R$ in $s^{- 1}$ , define $K_{T} (ω) = \frac{1}{T} {∣ \int_{0}^{T} e^{i ω t} d t ∣}^{2} = {\begin{cases} \frac{4 \sin^{2} (ω T / 2)}{T ω^{2}}, & ω \neq 0, \\ T, & ω = 0. \end{cases}$ Then $K_{T} \geq 0$ , $\int_{R} K_{T} (ω) d ω = 2 π$ , and for every $r > 0$ , $\int_{∣ ω ∣ > r} \frac{K_{T} (ω)}{2 π} d ω \leq \frac{4}{π T r} .$ For bounded Borel $f : R \to C$ continuous at $0$ , define $ω_{f} (r) = \sup_{∣ ω ∣ \leq r} ∣ f (ω) - f (0) ∣ .$ Then $∣ \int_{R} \frac{K_{T} (ω)}{2 π} f (ω) d ω - f (0) ∣ \leq ω_{f} (r) + \frac{8 ∥ f ∥_{\infty}}{π T r},$ so the left-hand integral tends to $f (0)$ as $T \to \infty$ .

Define the finite-time second-order coefficient quantity $P_{2} (T) = \frac{λ^{2}}{ℏ^{2}} \int_{R} w (E) {∣ \int_{0}^{T} e^{i (E - E_{0}) t / ℏ} d t ∣}^{2} d E .$ It is finite and $\lim_{T \to \infty} \frac{P_{2} (T)}{T} = \frac{2 π λ^{2}}{ℏ} w (E_{0}) .$ This is a theorem about the stated integral coefficient, not an exact long-time transition probability, an exponential decay law, or a product of distributions such as $δ^{2}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , $T > 0$ , $ℏ > 0$ , $λ \in R$ , $E_{0} \in R$ , and the bounded nonnegative weight $w$ continuous at $E_{0}$ .

[F1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ), Fourier transform on complex L1 classes, Plancherel theorem, Agreement of the integral and L2 transforms: ${A C}_{ω}$ is the stated countable-choice principle; the one-dimensional Fourier transform is $\hat{g} (ν) = \int g (t) e^{- 2 π i t ν} d t$ ; under that axiom Plancherel preserves the $L^{2}$ norm, and for $g \in L^{1} \cap L^{2}$ the integral transform represents the $L^{2}$ transform almost everywhere.

[F2]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : $∣ e^{i x} ∣ = 1$ and $e^{i x} - 1 = 2 i e^{i x / 2} \sin (x / 2)$ for real $x$ .

[F3]

Borel change of variables from the compact-support formula and Radon uniqueness: under ${A C}_{ω}$ , a $C^{1}$ diffeomorphism between open subsets of $R^{m}$ changes nonnegative Borel Lebesgue integrals by its absolute Jacobian determinant.

[F4]

QFT example units, metric, gauge, and pairing conventions: $E, ℏ$ use SI joules and joule-seconds, angular frequency is in $s^{- 1}$ , and $λ$ is dimensionless.

Proof

1.1givenF2algebra

For $ω \neq 0$ , integration of the exponential gives $\int_{0}^{T} e^{i ω t} d t = (e^{i ω T} - 1) / (i ω)$ . Fact [F2] gives its squared modulus $4 \sin^{2} (ω T / 2) / ω^{2}$ . At $ω = 0$ the integral is $T$ , so the continuation is $K_{T} (0) = T$ . For any integrable complex $q$ , put $I = \int q$ . If $I = 0$ , $∣ I ∣ \leq \int ∣ q ∣$ is immediate; otherwise $α = \overline{I} / ∣ I ∣$ gives $∣ I ∣ = \int Re (α q) \leq \int ∣ q ∣$ . Applied to $q (t) = e^{i ω t}$ on $[0, T]$ , with $∣ q ∣ = 1$ by [F2], this gives $K_{T} \leq T$ . Also $K_{T} \leq 4 / (T ω^{2})$ for $ω \neq 0$ . These bounds show $K_{T}$ is integrable and establish the tail estimate by integrating $ω^{- 2}$ on both rays.

1.2givenF1F3

Let $g = 1_{[0, T]}$ . It belongs to $L^{1} (R) \cap L^{2} (R)$ and $∥ g ∥_{2}^{2} = T$ . For $ν = - ω / (2 π)$ , its integral transform is $\hat{g} (ν) = \int_{0}^{T} e^{i ω t} d t$ . By [F1], this integral represents the $L^{2}$ transform almost everywhere and Plancherel gives $\int_{R} ∣ \hat{g} (ν) ∣^{2} d ν = T$ . The linear substitution $ω = - 2 π ν$ [F3] gives $\int_{R} K_{T} (ω) d ω = \frac{2 π}{T} \int_{R} ∣ \hat{g} (ν) ∣^{2} d ν = 2 π .$ The affine substitution is the one-dimensional case of [F3].

2.1givenstep 1.1step 1.2algebra

Since $\int K_{T} / (2 π) = 1$ , subtracting $f (0)$ writes the error as $\int (K_{T} / (2 π)) (f - f (0))$ . On $∣ ω ∣ \leq r$ the absolute difference is at most $ω_{f} (r)$ and the kernel mass there is at most one. On $∣ ω ∣ > r$ it is at most $2 ∥ f ∥_{\infty}$ , and the mass is at most $4 / (π T r)$ by step 1.1. Adding these bounds proves the estimate.

3.1givenstep 2.1

Given $ε > 0$ , continuity at zero gives $r > 0$ with $ω_{f} (r) < ε / 2$ . For this fixed $r$ , choose $T$ large enough that $8 ∥ f ∥_{\infty} / (π T r) < ε / 2$ . Step 2.1 then gives error below $ε$ . Thus the approximate-identity limit is $f (0)$ .

4.1givenF3F4step 1.1step 3.1algebra∎

The bounds in step 1.1 make $K_{T} (ω) w (E_{0} + ℏ ω)$ integrable because $w$ is bounded. In the definition of $P_{2}$ , substitute $E = E_{0} + ℏ ω$ ; this is an affine $C^{1}$ diffeomorphism with Jacobian $ℏ > 0$ , so [F3] applies to the nonnegative integrand. Since ${∣ \int_{0}^{T} e^{i ω t} d t ∣}^{2} = T K_{T} (ω),$ we obtain $\frac{P_{2} (T)}{T} = \frac{λ^{2}}{ℏ} \int_{R} K_{T} (ω) w (E_{0} + ℏ ω) d ω = \frac{2 π λ^{2}}{ℏ} \int_{R} \frac{K_{T} (ω)}{2 π} f (ω) d ω,$ where $f (ω) = w (E_{0} + ℏ ω)$ is bounded and continuous at zero. Step 3.1 gives the claimed limit. The units are $s^{- 1}$ because $w (E_{0}) / ℏ$ has units $J / (J s)$ . No exact-dynamics or probability claim follows from this coefficient calculation.

Golden-rule kernel and its quantified approximate-identity limit

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources