Choi's theorem on completely positive maps

'''Choi's theorem.''' Let \Phi : \mathbb{C}^{n\times n} \to \mathbb{C}^{m\times m} be a linear map. The following are equivalent: :(i) is -positive (i.e. \left (\operatorname{id}_n\otimes\Phi \right )(A)\in\mathbb{C}^{n\times n}\otimes\mathbb{C}^{m\times m} is positive whenever A\in\mathbb{C}^{n\times n}\otimes\mathbb{C}^{n\times n} is positive). :(ii) The matrix with operator entries ::C_\Phi= \left (\operatorname{id}_n\otimes\Phi \right ) \left (\sum_{ij}E_{ij}\otimes E_{ij} \right ) = \sum_{ij}E_{ij}\otimes\Phi(E_{ij}) \in \mathbb{C} ^{nm \times nm} :is positive semi-definite (PSD), where E_{ij} \in \mathbb{C}^{n\times n} is the matrix with 1 in the -th entry and 0s elsewhere. (The matrix CΦ is sometimes called the Choi matrix of .) :(iii) is completely positive. == Proof ==

(i) implies (ii) We observe that if :E=\sum_{ij}E_{ij}\otimes E_{ij}, then E=E* and E2=nE, so E=n−1EE* which is positive. Therefore CΦ =(In ⊗ Φ)(E) is positive by the n-positivity of Φ. (iii) implies (i) This holds trivially. (ii) implies (iii) This mainly involves chasing the different ways of looking at Cnm×nm: : \mathbb{C}^{nm\times nm} \cong\mathbb{C}^{nm}\otimes(\mathbb{C}^{nm})^* \cong\mathbb{C}^n\otimes\mathbb{C}^m\otimes(\mathbb{C}^n\otimes\mathbb{C}^m)^* \cong\mathbb{C}^n\otimes(\mathbb{C}^n)^*\otimes\mathbb{C}^m\otimes(\mathbb{C}^m)^* \cong\mathbb{C}^{n\times n}\otimes\mathbb{C}^{m\times m}. Let the eigenvector decomposition of CΦ be :C_\Phi = \sum _{i = 1} ^{nm} \lambda_i v_i v_i ^*, where the vectors v_i lie in Cnm . By assumption, each eigenvalue \lambda_i is non-negative so we can absorb the eigenvalues in the eigenvectors and redefine v_i so that : \; C_\Phi = \sum _{i = 1} ^{nm} v_i v_i ^* . The vector space Cnm can be viewed as the direct sum \textstyle \oplus_{i=1}^n \mathbb{C}^m compatibly with the above identification \textstyle\mathbb{C}^{nm}\cong\mathbb{C}^n\otimes\mathbb{C}^m and the standard basis of Cn. If Pk ∈ Cm × nm is projection onto the k-th copy of Cm, then Pk* ∈ Cnm×m is the inclusion of Cm as the k-th summand of the direct sum and : \; \Phi (E_{kl}) = P_k \cdot C_\Phi \cdot P_l^* = \sum _{i = 1} ^{nm} P_k v_i ( P_l v_i )^*. Now if the operators Vi ∈ Cm×n are defined on the k-th standard basis vector ek of Cn by :\; V_i e_k = P_k v_i, then :\Phi (E_{kl}) = \sum _{i = 1} ^{nm} P_k v_i ( P_l v_i )^* = \sum _{i = 1} ^{nm} V_i e_k e_l ^* V_i ^* = \sum _{i = 1} ^{nm} V_i E_{kl} V_i ^*. Extending by linearity gives us :\Phi(A) = \sum_{i=1}^{nm} V_i A V_i^* for any A ∈ Cn×n. Any map of this form is manifestly completely positive: the map A \to V_i A V_i^* is completely positive, and the sum (across i) of completely positive operators is again completely positive. Thus \Phi is completely positive, the desired result. The above is essentially Choi's original proof. Alternative proofs have also been known. == Consequences ==

Kraus operators In the context of quantum information theory, the operators {Vi} are called the Kraus operators (after Karl Kraus) of Φ. Notice, given a completely positive Φ, its Kraus operators need not be unique. For example, any "square root" factorization of the Choi matrix gives a set of Kraus operators. Let :B^* = [b_1, \ldots, b_{nm}], where bi*'s are the row vectors of B, then :C_\Phi = \sum _{i = 1} ^{nm} b_i b_i ^*. The corresponding Kraus operators can be obtained by exactly the same argument from the proof. When the Kraus operators are obtained from the eigenvector decomposition of the Choi matrix, because the eigenvectors form an orthogonal set, the corresponding Kraus operators are also orthogonal in the Hilbert–Schmidt inner product. This is not true in general for Kraus operators obtained from square root factorizations. (Positive semidefinite matrices do not generally have a unique square-root factorizations.) If two sets of Kraus operators {Ai}1nm and {Bi}1nm represent the same completely positive map Φ, then there exists a unitary operator matrix :\{U_{ij}\}_{ij} \in \mathbb{C}^{nm^2 \times nm^2} \quad \text{such that} \quad A_i = \sum _{j = 1} U_{ij} B_j. This can be viewed as a special case of the result relating two minimal Stinespring representations. Alternatively, there is an isometry scalar matrix {uij}ij ∈ Cnm × nm such that :A_i = \sum _{j = 1} u_{ij} B_j. This follows from the fact that for two square matrices M and N, M M* = N N* if and only if M = N U for some unitary U. Completely copositive maps It follows immediately from Choi's theorem that Φ is completely copositive if and only if it is of the form :\Phi(A) = \sum _i V_i A^T V_i ^* . Hermitian-preserving maps Choi's technique can be used to obtain a similar result for a more general class of maps. Φ is said to be Hermitian-preserving if A is Hermitian implies Φ(A) is also Hermitian. One can show Φ is Hermitian-preserving if and only if it is of the form :\Phi (A) = \sum_{i=1} ^{nm} \lambda_i V_i A V_i ^* where λi are real numbers, the eigenvalues of CΦ, and each Vi corresponds to an eigenvector of CΦ. Unlike the completely positive case, CΦ may fail to be positive. Since Hermitian matrices do not admit factorizations of the form B*B in general, the Kraus representation is no longer possible for a given Φ. == See also ==