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Clean and dirty ancilla identities extend to references and mixed states

Statement

Let $U$ and $V$ satisfy either the clean or dirty identity in Clean and dirty ancillas, and let $R$ be any finite ordered reference register. Put $P_{0} = R_{0_{A}, 0_{A}}$ , where the rank-one operator notation used here is defined by $R_{u, v} (z) = ⟨ z, v ⟩_{r o o t} u$ . In the clean case, for every matrix $X$ on $R \otimes D$ ,

(I_{R} \otimes U) (X \otimes P_{0}) (I_{R} \otimes U^{†}) = ((I_{R} \otimes V) X (I_{R} \otimes V^{†})) \otimes P_{0} .

In the dirty case, for every matrix $Y$ on $R \otimes D \otimes A$ ,

(I_{R} \otimes U) Y (I_{R} \otimes U^{†}) = (I_{R} \otimes V \otimes I_{A}) Y (I_{R} \otimes V^{†} \otimes I_{A}) .

Thus the stated identities apply in particular to density matrices, meaning positive semidefinite matrices of trace one.

Facts & Assumptions

Given: Finite-dimensional registers $R, D, A$ , ordered so the operation acts on $D \otimes A$ and as the identity on $R$ . Prove the stated extension in the clean and dirty cases.

[F1]

Clean and dirty ancillas: In the clean case, $U (∣ ψ ⟩_{D} \otimes ∣ 0_{A} ⟩) = V ∣ ψ ⟩_{D} \otimes ∣ 0_{A} ⟩$ for every $∣ ψ ⟩_{D}$ . In the dirty case, $U = V \otimes I_{A}$ .

[F2]

Tensor products of finitely many linear maps: For a finite ordered family of linear maps, the tensor map is the unique linear map whose action on each elementary tensor is the tensor of the factor-map actions.

[F3]

Ordered tensor registers, their inner product, and wire permutations: The tensor inner product on elementary tensors is the product of the factor inner products, with the displayed register order retained.

[F4]

The product basis is orthonormal and tensor dimensions multiply: For finite ordered registers with specified orthonormal factor bases, the lexicographically ordered elementary tensors form an orthonormal basis and therefore span the product register.

Proof

1.1givenF1F2F4algebra

In the clean case, for every $w \in H_{R} \otimes H_{D}$ , write $w$ as a finite sum of elementary tensors. The clean identity in [F1] and the tensor-map rule [F2] give $(I_{R} \otimes U) (w \otimes ∣ 0_{A} ⟩) = ((I_{R} \otimes V) w) \otimes ∣ 0_{A} ⟩$ .

2.1step 1.1F2F3F4algebra

For any linear map $A$ with adjoint and vectors $u, v$ , conjugate symmetry and the defining adjoint relation give $⟨ A^{†} q, v ⟩_{r o o t} = \overline{⟨ v, A^{†} q ⟩_{r o o t}} = \overline{⟨ A v, q ⟩_{r o o t}} = ⟨ q, A v ⟩_{r o o t}$ for every $q$ . Hence $A R_{u, v} A^{†} = R_{A u, A v}$ . By [F3], $R_{w, z} \otimes P_{0} = R_{w \otimes ∣ 0_{A} ⟩, z \otimes ∣ 0_{A} ⟩}$ . Apply the rank-one identity with $A = I_{R} \otimes U$ , $u = e_{i} \otimes ∣ 0_{A} ⟩$ , and $v = e_{j} \otimes ∣ 0_{A} ⟩$ , where $e_{i}, e_{j}$ range over the product orthonormal basis of $R \otimes D$ . Step 1.1 gives the clean conjugation formula for each $R_{e_{i}, e_{j}} \otimes P_{0}$ . Every matrix $X$ on $R \otimes D$ equals $\sum_{i, j} ⟨ X e_{j}, e_{i} ⟩_{r o o t} R_{e_{i}, e_{j}}$ , since the right side agrees with $X$ on each basis vector; hence these matrix units span and linearity proves the clean identity for every $X$ .

3.1givenF1F2F3F4algebra∎

In the dirty case, [F1] and [F2] give $I_{R} \otimes U = I_{R} \otimes V \otimes I_{A}$ on every elementary tensor and hence on the whole product space. The product inner product [F3] makes its adjoint $I_{R} \otimes V^{†} \otimes I_{A}$ : this follows first on elementary tensors from the defining adjoint relation for $V$ , then on all vectors because they span by [F4]. Substitution yields the dirty identity for every matrix $Y$ , including every density matrix; for a product $Y = Z_{R D} \otimes S_{A}$ , the output is $((I_{R} \otimes V) Z_{R D} (I_{R} \otimes V^{†})) \otimes S_{A}$ , so the dirty factor remains unchanged and uncorrelated in this product case.

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Sources

John Preskill, Quantum Computation, Chapter 5: Classical and Quantum Circuits (updated July 2015) (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)