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Finite-level truncation with bounded coupling error

Statement

Assume Countable Choice. Let $d \geq 1$ , $ℏ > 0$ , let $H = H^{*}$ on $C^{d}$ , and let $P = P^{*} = P^{2}$ be an orthogonal projection. Use the operator norm induced by the standard inner product. Put $P^{⊥} = I_{d} - P$ , $H_{D} = P H P + P^{⊥} H P^{⊥}$ , and $V = H - H_{D}$ . With finite matrix exponentials as defined on this page, for every $t \in R$ , $∥ e^{- i t H / ℏ} - e^{- i t H_{D} / ℏ} ∥ \leq \frac{∣ t ∣}{ℏ} ∥ V ∥, ∥ P^{⊥} e^{- i t H / ℏ} P ∥ \leq \frac{∣ t ∣}{ℏ} ∥ V ∥ .$ The second inequality bounds amplitude leakage from the retained subspace $ran P$ ; it asserts no convergence for an unbounded coupling.

Statement references

Finite-dimensional matrix exponential

Unitarity and differentiation of the finite Hermitian matrix exponential

Facts & Assumptions

Given: Countable Choice, $d \geq 1$ , $ℏ > 0$ , Hermitian $H$ , an orthogonal projection $P$ , and real $t$ .

[A1]

Countable Choice is the hypothesis of the Banach-valued fundamental theorem of calculus used to integrate the matrix derivative (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Fundamental theorem of calculus for Banach-valued continuous curves).

[F1]

Finite-dimensional matrix exponential defines the finite matrix exponential, and Unitarity and differentiation of the finite Hermitian matrix exponential gives its unitary property, group law, and operator-norm derivative for Hermitian matrices.

[F2]

The operator norm satisfies $∥ A B ∥ \leq ∥ A ∥ ∥ B ∥$ by its unit-ball definition (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces); a unitary matrix has norm one because it preserves vector norms and $d \geq 1$ . For a continuous matrix-valued curve, the integral of its derivative is the endpoint difference (Fundamental theorem of calculus for Banach-valued continuous curves), and its integral norm is bounded by the integral of the norm (Bochner integral norm inequality).

[F3]

For an endomorphism of a finite-dimensional complex inner product space, Hermitian means self-adjoint, $A = A^{*}$ (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

Proof

Given: [A1], [F1], [F2], and the matrices in the Statement.

1.1F1F3givenalgebra

By [F3], $H_{D}$ is Hermitian because $P = P^{*}$ , $P^{⊥} = (P^{⊥})^{*}$ , and $H = H^{*}$ . Also $P H_{D} P^{⊥} = P^{⊥} H_{D} P = 0$ , so $H_{D}$ preserves both $ran P$ and $ran P^{⊥}$ . Its power-series exponential therefore preserves those subspaces and commutes with $P$ , giving $P^{⊥} e^{- i t H_{D} / ℏ} P = 0$ . The vectors $P x$ and $P^{⊥} x$ are orthogonal because $P P^{⊥} = 0$ and $P = P^{*}$ ; hence $∥ x ∥^{2} = ∥ P x ∥^{2} + ∥ P^{⊥} x ∥^{2}$ , so $∥ P ∥, ∥ P^{⊥} ∥ \leq 1$ .

1.2A1F1F2algebra

For $s$ between $0$ and $t$ , set $F (s) = e^{- i (t - s) H / ℏ} e^{- i s H_{D} / ℏ}$ . By [F1] and the product rule, $F^{'} (s) = \frac{i}{ℏ} e^{- i (t - s) H / ℏ} (H - H_{D}) e^{- i s H_{D} / ℏ} = \frac{i}{ℏ} e^{- i (t - s) H / ℏ} V e^{- i s H_{D} / ℏ} .$ Integrating this continuous matrix curve from $0$ to $t$ by [F2] gives $e^{- i t H_{D} / ℏ} - e^{- i t H / ℏ} = \frac{i}{ℏ} \int_{0}^{t} e^{- i (t - s) H / ℏ} V e^{- i s H_{D} / ℏ} d s .$ This oriented integral identity holds also for $t < 0$ .

2.1F1F2step 1.2

The two exponentials in the integrand of step 1.2 are unitary by [F1], so its integrand has norm at most $∥ V ∥$ by [F2]. Taking norms in that identity and using the interval length $∣ t ∣$ proves the first bound.

3.1step 1.1step 2.1algebra∎

Step 1.1 gives $P^{⊥} e^{- i t H_{D} / ℏ} P = 0$ . Hence $P^{⊥} e^{- i t H / ℏ} P = P^{⊥} (e^{- i t H / ℏ} - e^{- i t H_{D} / ℏ}) P$ . The contraction bounds for $P$ and $P^{⊥}$ from step 1.1 and the first estimate in step 2.1 yield the claimed leakage bound.

Finite-level truncation with bounded coupling error

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources