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Canonical Choi eigenvectors, Kraus rank and branch weights

Statement

Let $H$ and $K$ be finite-dimensional complex inner-product spaces with $d = \dim H > 0$ , and fix an ordered orthonormal basis $(e_{i})_{i = 1}^{d}$ of $H$ . Let $N : L (H) \to L (K)$ be a quantum channel: a completely positive, trace-preserving linear map. Its input-first, unnormalized Choi matrix is $J = \sum_{i, j} ∣ e_{i} ⟩ ⟨ e_{j} ∣ \otimes N (∣ e_{i} ⟩ ⟨ e_{j} ∣)$ . Write its positive-eigenvalue decomposition as $J = \sum_{k = 1}^{r} λ_{k} ∣ v_{k} ⟩ ⟨ v_{k} ∣$ , where $λ_{k} > 0$ and the $v_{k}$ are orthonormal. Define $K_{k} : H \to K$ by $\sum_{i} e_{i} \otimes K_{k} e_{i} = \sqrt{λ_{k}} v_{k}$ .

Then $N (X) = \sum_{k = 1}^{r} K_{k} X K_{k}^{†}$ , and $Tr (K_{k}^{†} K_{l}) = λ_{k} δ_{k l}$ . In particular, the weights $w_{k} = Tr (K_{k}^{†} K_{k})$ equal $λ_{k}$ , and the minimum number of operators in a Kraus representation of $N$ is $r = rank J$ . The maps $X \mapsto K_{k} X K_{k}^{†}$ form a finite instrument, whose branch probability on the maximally mixed input $I_{H} / d$ is $w_{k} / d$ . Within a degenerate eigenspace the orthonormal eigenbasis may be rotated, so the individual operators are not unique; these mathematical branches do not specify a unique measured trajectory.

Statement references

Finite support square roots and generalized inverses

Choi matrix with input factor first

Finite channel, Kraus, Choi, and Stinespring equivalence

Finite instruments and conditional states

The product basis is orthonormal and tensor dimensions multiply

Facts & Assumptions

Given: The spaces, basis, channel, Choi matrix and spectral vectors in the Statement.

Goal: Establish the Kraus representation, orthogonality, minimal number, and instrument weights stated above.

[F1]

A finite-dimensional positive semidefinite operator has an orthonormal eigenbasis with nonnegative eigenvalues; its support is the span of the positive-eigenvalue vectors. Finite support square roots and generalized inverses

[F2]

The Choi matrix uses the fixed input basis and input-first tensor order displayed in the Statement, with the inner product linear in its first argument. Choi matrix with input factor first

[F3]

For a finite-dimensional channel, the Choi matrix is positive semidefinite and a finite Kraus representation exists with $\sum_{a} A_{a}^{†} A_{a} = I_{H}$ . Finite channel, Kraus, Choi, and Stinespring equivalence

[F4]

For a nonzero finite-dimensional input space, a finite instrument is a family of completely positive maps indexed by a finite nonempty outcome set whose sum preserves trace; its outcome weight at a density operator $ρ$ is the trace of the corresponding output. Finite instruments and conditional states

[F5]

For finite-dimensional complex inner-product spaces with specified orthonormal bases, the elementary tensors of the basis vectors form an orthonormal basis of their tensor product, including when one factor is zero-dimensional. The product basis is orthonormal and tensor dimensions multiply

Proof

1.1givenF1F2F3F5algebra

By [F3], $J$ is positive semidefinite, and [F1] gives the stated decomposition with $r = rank J$ . Choose an orthonormal basis $(f_{j})$ of $K$ , available by applying [F1] to $I_{K}$ . By [F5], each tensor vector has a unique expansion $\sum_{i, j} c_{i j} e_{i} \otimes f_{j}$ , hence a unique expression $\sum_{i} e_{i} \otimes y_{i}$ with $y_{i} = \sum_{j} c_{i j} f_{j}$ . Assigning $K_{k} e_{i} = y_{i}$ for the vector $\sqrt{λ_{k}} v_{k}$ then defines a unique linear map $K_{k} : H \to K$ . For any map $A : H \to K$ , write $∣ A ⟩ ⁣ ⟩ = \sum_{i} e_{i} \otimes A e_{i}$ . Expanding [F2] for $X \mapsto A X A^{†}$ gives its Choi matrix as $∣ A ⟩ ⁣ ⟩ ⟨ ⁣ ⟨ A ∣$ . Thus the Choi matrix of $Φ (X) = \sum_{k} K_{k} X K_{k}^{†}$ is $J$ . Equality of Choi matrices implies equality of maps: the $(i, j)$ input block is the image of $∣ e_{i} ⟩ ⟨ e_{j} ∣$ , and these matrix units form a basis of $L (H)$ . Hence $Φ = N$ .

2.1givenF2step 1.1algebra

With the first-argument-linear convention in [F2], the tensor inner product gives $Tr (K_{k}^{†} K_{l}) = \sum_{i} ⟨ K_{l} e_{i}, K_{k} e_{i} ⟩ = ⟨ \sqrt{λ_{l}} v_{l}, \sqrt{λ_{k}} v_{k} ⟩ = λ_{k} δ_{k l}$ , since the $v_{k}$ are orthonormal. This proves the asserted weights. For any Kraus representation with $m$ operators $A_{a}$ , the same expansion yields $J = \sum_{a = 1}^{m} ∣ A_{a} ⟩ ⁣ ⟩ ⟨ ⁣ ⟨ A_{a} ∣$ ; its range lies in the span of these $m$ vectors, so $r \leq m$ . Step 1.1 supplies a representation with $r$ operators, proving minimality.

3.1givenF1F4step 1.1step 2.1algebra∎

If $r = 0$ , step 1.1 would give $N (∣ e_{1} ⟩ ⟨ e_{1} ∣) = 0$ , contradicting trace preservation because $Tr (∣ e_{1} ⟩ ⟨ e_{1} ∣) = 1$ . Thus $r > 0$ , so ${1, \dots, r}$ is a finite nonempty outcome set. Each branch $X \mapsto K_{k} X K_{k}^{†}$ is completely positive: after adjoining any finite-dimensional identity map it sends a positive operator $Y$ to $(I \otimes K_{k}) Y (I \otimes K_{k})^{†}$ , which is positive. Their sum is the trace-preserving channel $N$ , so [F4] makes them an instrument. Its branch weight at $I_{H} / d$ is $Tr (K_{k} K_{k}^{†}) / d = Tr (K_{k}^{†} K_{k}) / d = w_{k} / d$ . An orthonormal change of basis within a repeated-eigenvalue eigenspace leaves its spectral projection unchanged but changes its individual vectors and reshaped operators. The instrument constructed from a chosen basis is therefore one possible decomposition, with no uniqueness claim about a measured trajectory.

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