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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.

Worked example B01 — exact finite dynamics and its first coefficient

Example

Facts & Assumptions

Verification

Depends on

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