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Finite Dyson remainder, coefficient unitarity, and regulator counterexamples

Statement

Assume the Countable Choice axiom ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $d \geq 1$ , $T > 0$ , $ℏ > 0$ , and $λ \in R$ . Let $H_{0} = H_{0}^{†}$ be a $d \times d$ Hermitian matrix with entries in joules, and let $V : [0, T] \to M_{d} (C)$ be norm-continuous, Hermitian-valued, and in joules. Define $V_{I} (t) = e^{i t H_{0} / ℏ} V (t) e^{- i t H_{0} / ℏ}, B = \sup_{0 \leq t \leq T} ∥ V_{I} (t) ∥, z = \frac{∣ λ ∣ B T}{ℏ} .$ The coupling $λ$ is dimensionless and $z$ is dimensionless. Put $S_{0} (t) = I$ and $S_{n + 1} (t) = - \frac{i}{ℏ} \int_{0}^{t} V_{I} (r) S_{n} (r) d r, S (t) = \sum_{n = 0}^{\infty} λ^{n} S_{n} (t) .$ Then the series converges in operator norm on $[0, T]$ , $S$ is the unique solution of $\dot{S} = - i λ V_{I} S / ℏ$ , $S (0) = I$ , and $S (t)$ is unitary. For every $N \geq 0$ , $∥ S (T) - \sum_{n = 0}^{N} λ^{n} S_{n} (T) ∥ \leq e^{z} \frac{z^{N + 1}}{(N + 1)!} .$

For a finite matrix $A$ , define its Hermitian imaginary part by ${Im}_{H} A : = (A - A^{†}) / (2 i)$ ; for a scalar entry $w \in C$ , $Im w$ has its ordinary scalar meaning.

Define $S (T; λ) = I + i T (λ)$ and $T (λ) = \sum_{n \geq 1} λ^{n} T_{n}$ . For every pair of basis indices $f, i$ , exact unitarity gives $2 Im T_{i i} (λ) = \sum_{f} ∣ T_{f i} (λ) ∣^{2},$ and coefficient comparison gives $T_{1} = T_{1}^{†}$ and $2 {Im}_{H} T_{2} = T_{1}^{†} T_{1}$ . These are identities for this finite matrix model.

For $H = g_{E} σ_{x}$ on $C^{2}$ , where $g_{E} > 0$ is in joules and $σ_{x} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , the evolution matrix is $e^{- i t H / ℏ} = \cos (g_{E} t / ℏ) I - i \sin (g_{E} t / ℏ) σ_{x} .$ Its exact off-diagonal matrix-element square is $\sin^{2} (g_{E} t / ℏ)$ ; its first-order squared coefficient is $(g_{E} t / ℏ)^{2}$ , which can exceed $1$ . For every positive rate $Γ$ in $s^{- 1}$ , the survival matrix-element square $\cos^{2} (g_{E} t / ℏ)$ is not $e^{- Γ t}$ .

For an integer $N \geq 0$ , on the finite basis $∣ 0 ⟩, \dots, ∣ N ⟩$ define $a_{N} ∣ n ⟩ = \sqrt{n} ∣ n - 1 ⟩$ (zero at $n = 0$ ) and $a_{N}^{†} ∣ n ⟩ = \sqrt{n + 1} ∣ n + 1 ⟩$ for $n < N$ , with $a_{N}^{†} ∣ N ⟩ = 0$ . Then $[a_{N}, a_{N}^{†}] = I - (N + 1) ∣ N ⟩ ⟨ N ∣ .$ No finite-dimensional matrices can satisfy $[A, B] = I$ , because the trace of a commutator is zero while the trace of $I$ is the dimension.

Facts & Assumptions

Given: $d \geq 1$ , $T > 0$ , $ℏ > 0$ , real dimensionless $λ$ , Hermitian $H_{0}$ , and a norm-continuous Hermitian matrix family $V (t)$ on $[0, T]$ , all with the units fixed in QFT example units, metric, gauge, and pairing conventions.

[F1]

QFT example units, metric, gauge, and pairing conventions: $ℏ$ has units $J s$ , energy matrices have units joules, $λ$ is dimensionless, and finite matrices use the operator norm.

[F2]

Norm Dyson series and uniqueness: for the stated norm-continuous bounded self-adjoint operator family, the Volterra equation has a unique continuous solution.

[F3]

Norm Dyson series and uniqueness: every propagator supplied by that theorem is unitary; its full statement also supplies the norm-Dyson convergence and factorial bound.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is the choice principle assumed by the exact controlled-dynamics supplier; no stronger choice principle is used.

Proof

1.1givenF1algebra

The matrix exponential is defined by its norm-convergent power series; termwise differentiation on compact time intervals gives $W^{'} (t) = i H_{0} W (t) / ℏ$ for $W (t) = e^{i t H_{0} / ℏ}$ . Hence $(W (t)^{†} W (t))^{'} = 0$ and $W (0)^{†} W (0) = I$ , so $W$ is unitary (in finite dimension a left inverse is an inverse). Therefore $V_{I} (t) = W (t) V (t) W (t)^{†}$ is Hermitian and norm-continuous. Its norm is bounded on compact $[0, T]$ , so $B < \infty$ .

1.2givenF1algebra

Recursion expresses $λ^{n} S_{n} (T)$ as an $n$ -fold ordered integral with integrand of norm at most $(∣ λ ∣ B / ℏ)^{n}$ . The volume of $0 < t_{n} < \dots < t_{1} < T$ is $T^{n} / n!$ : induction integrates the preceding volume $t^{n - 1} / (n - 1)!$ from $0$ to $T$ . Thus $∥ λ^{n} S_{n} (T) ∥ \leq z^{n} / n!$ and the same bound with $T$ replaced by any $t \in [0, T]$ holds uniformly. The scalar exponential series proves uniform absolute convergence.

1.3givenF1algebra

For $σ_{x}^{2} = I$ , separate the even and odd powers in the absolutely convergent exponential series: $e^{- i z σ_{x}} = \sum_{k \geq 0} \frac{(- 1)^{k} z^{2 k}}{(2 k)!} I - i \sum_{k \geq 0} \frac{(- 1)^{k} z^{2 k + 1}}{(2 k + 1)!} σ_{x} = \cos z I - i \sin z σ_{x},$ where $z = g_{E} t / ℏ$ . Its off-diagonal entry has modulus square $\sin^{2} z$ , and the first-order term has modulus square $z^{2}$ . Taking $∣ z ∣ > 1$ makes the latter exceed $1$ . Also $\cos^{2} z$ has derivative $0$ at $t = 0$ , whereas $e^{- Γ t}$ has derivative $- Γ < 0$ there; so they cannot agree for all $t$ when $Γ > 0$ .

2.1step 1.2algebra

For the tail, write $n = N + 1 + k$ . Then $\sum_{n = N + 1}^{\infty} \frac{z^{n}}{n!} = \frac{z^{N + 1}}{(N + 1)!} \sum_{k = 0}^{\infty} \frac{z^{k}}{(N + 2) \dots (N + 1 + k)} \leq \frac{z^{N + 1}}{(N + 1)!} \sum_{k = 0}^{\infty} \frac{z^{k}}{k!} = e^{z} \frac{z^{N + 1}}{(N + 1)!},$ where an empty product is $1$ and $(N + 2) \dots (N + 1 + k) \geq k!$ . The norm tail is bounded by this scalar tail.

2.2givenA1F2F3step 1.2

The family $H (t) = λ V_{I} (t)$ satisfies the exact hypotheses of [F2] and [F3] on the compact interval $[0, T]$ : it is bounded, norm-continuous and self-adjoint, and the required ${A C}_{ω}$ is [A1]. The recursion above is its Dyson recursion, so [F2] supplies the Volterra equation and uniqueness, [F3] supplies unitarity of $S (t)$ , and step 1.2 supplies the displayed sharper bound with $B$ .

3.1step 1.2step 2.2algebra

From $S = I + i T$ , $S^{†} S = I$ expands to $i (T - T^{†}) + T^{†} T = 0$ . Its $i i$ entry is $- 2 Im T_{i i} + \sum_{f} ∣ T_{f i} ∣^{2} = 0$ , proving the exact identity. The norm-convergent power series makes $T (λ)$ analytic near $0$ (indeed entire), so coefficient comparison is valid: the $λ$ term gives $T_{1} = T_{1}^{†}$ , and the $λ^{2}$ term gives $i (T_{2} - T_{2}^{†}) + T_{1}^{†} T_{1} = 0$ , hence $2 {Im}_{H} T_{2} = T_{1}^{†} T_{1}$ .

4.1step 3.1algebra

For the linear truncation $S_{t r} = I + i λ T_{1}$ with $T_{1} = T_{1}^{†}$ , multiplication gives $S_{t r}^{†} S_{t r} = I + λ^{2} T_{1}^{2}$ . If $λ T_{1} \neq 0$ , then $T_{1}^{2}$ is nonzero positive semidefinite because $⟨ T_{1}^{2} v, v ⟩ = ∥ T_{1} v ∥^{2}$ ; hence this truncation is not unitary. The exact coefficient relation in step 3.1 therefore requires the higher-order correction whenever its right side is nonzero.

5.1givenF1algebra∎

For $n < N$ , direct application gives $a_{N} a_{N}^{†} ∣ n ⟩ = (n + 1) ∣ n ⟩$ and $a_{N}^{†} a_{N} ∣ n ⟩ = n ∣ n ⟩$ , so the commutator equals $∣ n ⟩$ . At $n = N$ , the two terms are $0$ and $N ∣ N ⟩$ , so their difference is $- N ∣ N ⟩$ . These are exactly the values of $I - (N + 1) ∣ N ⟩ ⟨ N ∣$ on each basis vector. For arbitrary finite matrices $A, B$ , $tr (A B) = \sum_{i, j} A_{i j} B_{j i} = tr (B A)$ by swapping dummy indices, so $tr [A, B] = 0$ ; but $tr I = d > 0$ . Thus the unmodified canonical commutator cannot hold on a finite-dimensional space.

Finite Dyson remainder, coefficient unitarity, and regulator counterexamples

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources