(i) implies (ii)

We observe that if

${\displaystyle E=\sum _{ij}E_{ij}\otimes E_{ij},}$

then E=E^* and E²=nE, so E=n⁻¹EE^* which is positive. Therefore C_Φ =(I_n ⊗ 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 C^nm×nm:

${\displaystyle \mathbb {C} ^{nm\times nm}\cong \mathbb {C} ^{nm}\otimes (\mathbb {C} ^{nm})^{*}\cong \mathbb {C} ^{n}\otimes \mathbb {C} ^{m}\otimes (\mathbb {C} ^{n}\otimes \mathbb {C} ^{m})^{*}\cong \mathbb {C} ^{n}\otimes (\mathbb {C} ^{n})^{*}\otimes \mathbb {C} ^{m}\otimes (\mathbb {C} ^{m})^{*}\cong \mathbb {C} ^{n\times n}\otimes \mathbb {C} ^{m\times m}.}$

Let the eigenvector decomposition of C_Φ be

${\displaystyle C_{\Phi }=\sum _{i=1}^{nm}\lambda _{i}v_{i}v_{i}^{*},}$

where the vectors ${\displaystyle v_{i}}$ lie in C^nm . By assumption, each eigenvalue ${\displaystyle \lambda _{i}}$ is non-negative so we can absorb the eigenvalues in the eigenvectors and redefine ${\displaystyle v_{i}}$ so that

${\displaystyle \;C_{\Phi }=\sum _{i=1}^{nm}v_{i}v_{i}^{*}.}$

The vector space C^nm can be viewed as the direct sum ${\displaystyle \textstyle \oplus _{i=1}^{n}\mathbb {C} ^{m}}$ compatibly with the above identification ${\displaystyle \textstyle \mathbb {C} ^{nm}\cong \mathbb {C} ^{n}\otimes \mathbb {C} ^{m}}$ and the standard basis of Cⁿ.

If P_k ∈ C^{m × nm} is projection onto the k-th copy of C^m, then P_k^* ∈ C^nm×m is the inclusion of C^m as the k-th summand of the direct sum and

${\displaystyle \;\Phi (E_{kl})=P_{k}\cdot C_{\Phi }\cdot P_{l}^{*}=\sum _{i=1}^{nm}P_{k}v_{i}(P_{l}v_{i})^{*}.}$

Now if the operators V_i ∈ C^m×n are defined on the k-th standard basis vector e_k of Cⁿ by

${\displaystyle \;V_{i}e_{k}=P_{k}v_{i},}$

then

${\displaystyle \Phi (E_{kl})=\sum _{i=1}^{nm}P_{k}v_{i}(P_{l}v_{i})^{*}=\sum _{i=1}^{nm}V_{i}e_{k}e_{l}^{*}V_{i}^{*}=\sum _{i=1}^{nm}V_{i}E_{kl}V_{i}^{*}.}$

Extending by linearity gives us

${\displaystyle \Phi (A)=\sum _{i=1}^{nm}V_{i}AV_{i}^{*}}$

for any A ∈ C^n×n. Any map of this form is manifestly completely positive: the map ${\displaystyle A\to V_{i}AV_{i}^{*}}$ is completely positive, and the sum (across ${\displaystyle i}$) of completely positive operators is again completely positive. Thus ${\displaystyle \Phi }$ is completely positive, the desired result.

The above is essentially Choi's original proof. Alternative proofs have also been known.

Kraus operators

In the context of quantum information theory, the operators {V_i} 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 C_Φ = B^∗B gives a set of Kraus operators.

Let

${\displaystyle B^{*}=[b_{1},\ldots ,b_{nm}],}$

where b_i*'s are the row vectors of B, then

${\displaystyle C_{\Phi }=\sum _{i=1}^{nm}b_{i}b_{i}^{*}.}$

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 {A_i}₁^nm and {B_i}₁^nm represent the same completely positive map Φ, then there exists a unitary operator matrix

${\displaystyle \{U_{ij}\}_{ij}\in \mathbb {C} ^{nm^{2}\times nm^{2}}\quad {\text{such that}}\quad A_{i}=\sum _{j=1}U_{ij}B_{j}.}$

This can be viewed as a special case of the result relating two minimal Stinespring representations.

Alternatively, there is an isometry scalar matrix {u_ij}_ij ∈ C^{nm × nm} such that

${\displaystyle A_{i}=\sum _{j=1}u_{ij}B_{j}.}$

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

${\displaystyle \Phi (A)=\sum _{i}V_{i}A^{T}V_{i}^{*}.}$

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

${\displaystyle \Phi (A)=\sum _{i=1}^{nm}\lambda _{i}V_{i}AV_{i}^{*}}$

where λ_i are real numbers, the eigenvalues of C_Φ, and each V_i corresponds to an eigenvector of C_Φ. Unlike the completely positive case, C_Φ 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.