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Example conventions, energy units and constructed/formal data — Examples

1 · Prerequisites

2 · Summary

The companion page gives two worked checks against the A-page suppliers. The finite two-level example chooses times where the exact unitary matrix element square and the first-order squared coefficient separate visibly; it also compares the exact survival value with a positive-rate exponential. This makes the finite-time truncation warning numerical without claiming that a continuum unstable particle has periodic survival.

The oscillator example evaluates the one-mode time-ordered correlator at a negative time and compares it with the explicitly defined kernel $G_{R}$ . At the chosen time the correlator is nonzero and $G_{R}$ vanishes; this is a comparison of two functions in the stated model and makes no claim about response to an external source. Both examples are exact calculations in the stated mathematical models. No experimental datum is added, and no probabilistic interpretation is assigned to either matrix-element square.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Worked example B01 — exact finite dynamics and its first coefficient

Example

Use the two-level matrix model of Finite Dyson remainder, coefficient unitarity, and regulator counterexamples with $H = g_{E} σ_{x}$ , where $g_{E} > 0$ is in joules. At $t_{1} = \frac{π ℏ}{2 g_{E}} > 0,$ the exact off-diagonal matrix-element square is $1$ , while its first-order squared coefficient is $π^{2} / 4 > 1$ . At $t_{2} = π ℏ / g_{E}$ , the exact survival matrix-element square is $1$ , whereas $e^{- Γ t_{2}} < 1$ for every positive rate $Γ$ in $s^{- 1}$ . These are finite-matrix identities. This mathematical example assigns no probabilistic interpretation to the amplitudes or their matrix-element squares.

Facts & Assumptions

Given: ${A C}_{ω}$ , the two-level Hamiltonian $H = g_{E} σ_{x}$ with $g_{E} > 0$ in joules, the positive $ℏ$ convention, and the times $t_{1}, t_{2}$ above.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is assumed by the finite-dynamics supplier.

[F1]

QFT example units, metric, gauge, and pairing conventions: $ℏ$ is in joule-seconds and matrix-element modulus squares are unchanged between the first-linear Hilbert pairing and coordinate convention.

[F2]

Finite Dyson remainder, coefficient unitarity, and regulator counterexamples: $e^{- i t H / ℏ} = \cos (g_{E} t / ℏ) I - i \sin (g_{E} t / ℏ) σ_{x}$ ; the exact transition square is $\sin^{2} (g_{E} t / ℏ)$ , the first-order squared coefficient is $(g_{E} t / ℏ)^{2}$ , and the survival square is not $e^{- Γ t}$ for $Γ > 0$ .

Verification

1.1givenA1F1F2algebra

At $t_{1}$ , the dimensionless argument is $g_{E} t_{1} / ℏ = π / 2$ . Applying [F2], the exact off-diagonal entry is $- i$ and its modulus square is $1$ . The first-order term is $- i π / 2$ , whose modulus square is $π^{2} / 4 > 1$ .

2.1givenF1F2step 1.1algebra

At $t_{2}$ , the argument is $π$ , so the diagonal entry of the exact evolution is $- 1$ and its modulus square is $1$ . Since $t_{2} > 0$ and $Γ > 0$ in $s^{- 1}$ , $0 < e^{- Γ t_{2}} < 1$ . Hence these exact finite-model survival values cannot equal one positive-rate exponential for all times. This is a matrix calculation, not a continuum decay conclusion.

3.1givenF2step 1.1step 2.1∎

At $t_{1}$ , the first-order coefficient square is $π^{2} / 4 > 1$ , while the corresponding exact matrix-element square is $1$ . They are distinct mathematical quantities in this finite-matrix example; no probabilistic interpretation is made.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Worked example B03 — nonretarded one-mode correlator

Example

For the one-mode oscillator in One-mode oscillator, Feynman boundary prescription, and retarded contrast, choose $t_{*} = - π / (2 ω)$ , where $ω > 0$ is in $s^{- 1}$ . Then the normalized Feynman correlator is $- i / (2 ω)$ , while the explicitly defined kernel $G_{R}$ is zero. This compares the two functions in the stated one-mode model and makes no claim about a response to an external source.

Facts & Assumptions

Given: ${A C}_{ω}$ , $ω > 0$ in $s^{- 1}$ , $t_{*} = - π / (2 ω) < 0$ , and the fixed-mode definitions of the Feynman correlator and retarded kernel.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is assumed by the oscillator supplier.

[F1]

QFT example units, metric, gauge, and pairing conventions: $ω t$ is dimensionless, and these formulas carry no spatial-locality claim.

[F2]

One-mode oscillator, Feynman boundary prescription, and retarded contrast: $C_{F} (t) = e^{- i ω ∣ t ∣} / (2 ω)$ and $G_{R} (t) = 1_{[0, \infty)} (t) \sin (ω t) / ω$ .

Verification

1.1givenA1F1F2algebra

Since $ω ∣ t_{*} ∣ = π / 2$ , the oscillator formula gives $C_{F} (t_{*}) = e^{- i π / 2} / (2 ω) = - i / (2 ω)$ . The units are seconds on both sides.

2.1givenF2step 1.1algebra

Since $t_{*} < 0$ , the support indicator in $G_{R}$ is zero, so $G_{R} (t_{*}) = 0$ . Thus the correlator is nonzero at negative time while the retarded response vanishes there.

3.1givenF1F2step 2.1∎

This distinguishes the two kernels in this single oscillator mode only; it proves no spatial signal property or interacting-field statement.

Example conventions, energy units and constructed/formal data — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Worked example B01 — exact finite dynamics and its first coefficient

Example

Facts & Assumptions

Verification

Worked example B03 — nonretarded one-mode correlator

Example

Facts & Assumptions

Verification

Sources