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Linear Combinations of Unitaries and Taylor Simulation — Examples

1 · Prerequisites

2 · Summary

These examples verify cancelling identity padding, show why a preserved unknown state cannot program its own exact reflection, and calculate a two-term LCU block and its input-dependent postselection weight. The counterexample concerns only an exact program-state interface and leaves explicitly declared reflection access unaffected. This page is a logical leaf.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Cancelling identity padding fixes the LCU scale

Example

For the truncated Taylor LCU in Finite Taylor LCU polynomial, suppose its listed coefficient norm is $s_{K} = \sum_{k = 0}^{K} \frac{(Λ ∣ h ∣)^{k}}{k!} \leq 2.$ Set $q = (2 - s_{K}) / 2$ and append two distinct identity entries with coefficients $+ q$ and $- q$ . The represented polynomial is unchanged while its listed coefficient norm becomes exactly $2$ .

For example, take the one-term Hamiltonian $H = 1 \cdot Z$ , $h = \log 2$ , and $K = 0$ . Then $M_{0} = I$ , $s_{0} = 1$ , and the padded list $(1 I, \frac{1}{2} I, - \frac{1}{2} I)$ still sums to $I$ and has absolute coefficient sum $2$ .

Facts & Assumptions

Given: A truncated Taylor LCU whose listed coefficient norm satisfies $s_{K} \leq 2$ .

[F1]

Finite Taylor LCU polynomial: the $k = 0$ term is the empty tuple with coefficient $1$ and unitary $I$ , and the listed coefficient norm is $s_{K} (h) = \sum_{k = 0}^{K} (Λ ∣ h ∣)^{k} / k!$ .

[F2]

The Pauli matrices: in the ordered qubit basis, $Z = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ .

[F3]

The natural logarithm as the inverse of the exponential function: for $x > 0$ , $\log x$ is the real number whose exponential is $x$ .

Verification

Given: As in Facts & Assumptions.

1.1givenF1algebra

The hypothesis $s_{K} \leq 2$ gives $q = (2 - s_{K}) / 2 \geq 0$ .

2.1givenF1step 1.1algebra

The two appended terms add $q I - q I = 0$ to the matrix and add $∣ q ∣ + ∣ - q ∣ = 2 q = 2 - s_{K}$ to the coefficient norm. The new norm is therefore $s_{K} + (2 - s_{K}) = 2$ , including $q = 0$ .

3.1givenF1F2F3step 2.1algebra∎

In the stated example, [F2] gives $Z^{†} = Z$ and $Z^{†} Z = I$ , so $H = 1 \cdot Z$ is a valid one-term Hamiltonian with $Λ = 1$ ; [F3] makes $h = \log 2$ a real parameter. Since $K = 0$ , [F1] gives $M_{0} = I$ and $s_{0} = 1$ . Hence $q = 1 / 2$ ; the three coefficients sum to $1$ and their absolute values sum to $1 + 1 / 2 + 1 / 2 = 2$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A preserved unknown state cannot program its own reflection

Statement

Let $P$ and $T$ be qubit registers. There is no fixed unitary $U$ on $P \otimes T$ that, for every unit program vector $∣ ψ ⟩ \in P$ and target vector $∣ ϕ ⟩ \in T$ , implements $U (∣ ψ ⟩ \otimes ∣ ϕ ⟩) = ∣ ψ ⟩ \otimes (I - 2 ∣ ψ ⟩ ⟨ ψ ∣) ∣ ϕ ⟩,$ while preserving the program register. This only rules out that exact program-state interface; it does not rule out a declared circuit for a known reflection or a separate oracle interface.

Facts & Assumptions

Given: The two-dimensional register with ordered orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ and the reflection convention $S_{ψ} = I - 2 ∣ ψ ⟩ ⟨ ψ ∣$ .

[F1]

Finite registers and their ordered computational bases and The computational basis is orthonormal: $∣ 0 ⟩, ∣ 1 ⟩$ are an orthonormal basis of the qubit register.

[F2]

Dirac Notation Conversion: $∣ ψ ⟩ ⟨ ψ ∣$ sends $z$ to $⟨ z, ψ ⟩_{r o o t} ψ$ ; Rank-One Reflection Is Ray-Invariant and Its Signs Differ Under Control proves that for unit $ψ$ it is an orthogonal rank-one projector.

[F3]

Orthogonal-Projector Reflection and Access Contract and Projector Reflections Are Hermitian Unitary Involutions: $S_{ψ} = I - 2 ∣ ψ ⟩ ⟨ ψ ∣$ is unitary.

[F4]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces and Finite complex linear isometries preserve inner products: every unitary preserves inner products.

[F5]

Ordered tensor registers, their inner product, and wire permutations: the product inner product satisfies $⟨ x \otimes y, x^{'} \otimes y^{'} ⟩ = ⟨ x, x^{'} ⟩ ⟨ y, y^{'} ⟩$ .

[F6]

Oblivious ancilla reflection: the reflection and its controlled form are declared access operations; a state or matrix description alone supplies neither call.

Proof

Given: As in Facts & Assumptions.

1.1givenF1F5F6assume-contraconstructalgebra

Assume for contradiction that such a unitary $U$ exists as the one fixed implementation replacing the explicitly declared reflection access of [F6]. Compare the unit inputs $∣ 0 ⟩_{P} \otimes ∣ 0 ⟩_{T}$ and $∣ + ⟩_{P} \otimes ∣ 0 ⟩_{T}$ , where $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ . Their inner product is $1 / \sqrt{2}$ by [F1] and [F5].

2.1givenF2F3step 1.1algebra

The required output for the first input is $- ∣ 0 ⟩_{P} \otimes ∣ 0 ⟩_{T}$ , since $S_{0} ∣ 0 ⟩ = - ∣ 0 ⟩$ . For the second, $2 ∣ + ⟩ ⟨ + ∣ 0 ⟩ = ∣ 0 ⟩ + ∣ 1 ⟩$ , so $S_{+} ∣ 0 ⟩ = - ∣ 1 ⟩$ and the output is $- ∣ + ⟩_{P} \otimes ∣ 1 ⟩_{T}$ . These computations use [F2]–[F3].

3.1F1F4F5step 1.1step 2.1discharge-contradiction∎

The two required outputs have inner product zero because $⟨ 0 ∣ 1 ⟩ = 0$ on $T$ , by [F1] and [F5]. But [F4] says $U$ preserves the input inner product $1 / \sqrt{2}$ . This contradiction proves that no such fixed unitary exists.

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Two-term LCU postselection is input dependent

Example

On a qubit let $M = I + 2 Z$ , with LCU terms $(1, I)$ and $(2, Z)$ , so $α = 3$ . Use the complete PREPARE unitary $P = (\begin{array}{cc} 1 / \sqrt{3} & - \sqrt{2 / 3} \\ \sqrt{2 / 3} & 1 / \sqrt{3} \end{array}),$ whose first column prepares $3^{- 1 / 2} ∣ 0 ⟩ + \sqrt{2 / 3} ∣ 1 ⟩$ , and explicitly supply inverse access $P^{†}$ . Let SELECT apply $I$ and $Z$ on the corresponding indices. The projected block is $M / 3 = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 / 3 \end{array}) .$ For $∣ ψ ⟩ = a ∣ 0 ⟩ + b ∣ 1 ⟩$ with $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , the clean-ancilla branch vector is $a ∣ 0 ⟩ - (b / 3) ∣ 1 ⟩$ and its weight is $p_{0} = ∣ a ∣^{2} + ∣ b ∣^{2} / 9.$ The branch is always nonzero, with $1 / 9 \leq p_{0} \leq 1$ ; its conditional state is the branch vector divided by $\sqrt{p_{0}}$ . The least weight occurs at $a = 0$ , and the greatest at $b = 0$ .

Facts & Assumptions

Given: A qubit system, the LCU list $(1, I), (2, Z)$ , the complete PREPARE unitary and its explicitly supplied inverse access, SELECT, and a unit input $a ∣ 0 ⟩ + b ∣ 1 ⟩$ .

[F1]

The Pauli matrices: $Z ∣ 0 ⟩ = ∣ 0 ⟩$ and $Z ∣ 1 ⟩ = - ∣ 1 ⟩$ .

[F2]

Finite registers and their ordered computational bases and The computational basis is orthonormal: the qubit basis is orthonormal.

[F3]

LCU projected-block and postselection correctness: a supplied LCU block has clean branch vector $M ∣ ψ ⟩ / α$ and weight $∥ M ∣ ψ ⟩ ∥^{2} / α^{2}$ .

[F4]

Clean-ancilla zero-branch postselection and Clean-ancilla postselection has normalized branch weights: a positive branch weight gives the normalized conditional system vector.

Verification

Given: As in Facts & Assumptions.

1.1givenF1F2F3constructalgebra

The absolute coefficient sum is $α = ∣ 1 ∣ + ∣ 2 ∣ = 3$ . The specified real rotation $P$ is unitary because its columns have norm one and inner product zero by [F2]; its first column is the required prepared index state, and the inverse access is explicitly supplied. SELECT is unitary because it is block diagonal with blocks $I$ and $Z$ , and direct multiplication of the matrices in [F1] gives $I^{†} I = Z^{†} Z = I$ . Thus the access satisfies [F3].

2.1givenF1F2F3step 1.1algebra

By [F1], $M = I + 2 Z$ is diagonal with entries $3, - 1$ , so [F3] gives the projected block $M / 3 = diag (1, - 1 / 3)$ and branch vector $a ∣ 0 ⟩ - (b / 3) ∣ 1 ⟩$ . Orthonormality in [F2] makes its squared norm $p_{0} = ∣ a ∣^{2} + ∣ b ∣^{2} / 9$ .

3.1givenF2F4step 2.1algebra∎

Since $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , $p_{0} = 1 - (8 / 9) ∣ b ∣^{2}$ . The bounds $0 \leq ∣ b ∣^{2} \leq 1$ give $1 / 9 \leq p_{0} \leq 1$ ; equality occurs at $a = 0$ and $b = 0$ , respectively. Thus the branch is nonzero for every unit input and [F4] gives the stated conditional vector.

Linear Combinations of Unitaries and Taylor Simulation — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Cancelling identity padding fixes the LCU scale

Example

Facts & Assumptions

Verification

A preserved unknown state cannot program its own reflection

Statement

Facts & Assumptions

Proof

Two-term LCU postselection is input dependent

Example

Facts & Assumptions

Verification

Sources