## Kraus operators, revisited

Channel-state duality gives us more than just a one-to-one correspondence between states \widetilde{\mathcal{E}} and channels \mathcal{E} — it *also* gives a one-to-one correspondence between vectors in the statistical ensemble \widetilde{\mathcal{E}} and the Kraus operators in the decomposition of \mathcal{E}.

With the above in mind, we see that the freedom to choose the Kraus operators representing a channel in many different ways is really the same thing as the freedom to choose the ensemble of pure states representing a density operator in many different ways.

We already know that if two mixtures (p_k, |\psi_k\rangle) and (q_l, |\phi_l\rangle) are described by the same density operator
\sum_k|\widetilde\psi_k\rangle\langle\widetilde\psi_k|
= \widetilde{\mathcal{E}}
= \sum_l|\widetilde\phi_l\rangle\langle\widetilde\phi_l|
(where |\widetilde \psi_k\rangle=\sqrt{p_k}|\psi_k\rangle and |\widetilde \phi_l\rangle=\sqrt{q_l}|\phi_l\rangle) then they are related to one another: there exists some unitary R such that
|\widetilde \psi_k\rangle
= \sum_{l} R_{kl} |\widetilde \phi_l\rangle.
Using the aforementioned fact that any vector |\psi\rangle in \mathcal{H}\otimes\mathcal{H}' can be written as |\psi\rangle=\mathbf{1}\otimes V|\Omega\rangle, this implies the same unitary freedom in choosing the Kraus operators.

So how many Kraus operators do we really need?
Channel-state duality tells us that the *minimal* number of Kraus operators needed to express \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') in the operator-sum form is given by the rank of its Choi matrix \widetilde{\mathcal{E}}, i.e. we need *no more than* dd' such operators (where d=\dim\mathcal{H} and d'=\dim\mathcal{H}').
In fact, this minimal set of Kraus operators corresponds to the *spectral decomposition* of \widetilde{\mathcal{E}}.

Indeed, if \widetilde{\mathcal{E}}=\sum_k|\widetilde{v}_k\rangle\langle\widetilde{v}_k| and |\widetilde{v}_k\rangle=(\mathbf{1}\otimes E_k)|\Omega\rangle, then the orthogonality of |\widetilde{v}_k\rangle and |\widetilde{v}_l\rangle implies the orthogonality (in the Hilbert–Schmidt sense) of the corresponding Kraus operators E_k and E_l.
In order to see this, we write \langle\widetilde{v}_k|\widetilde{v}_l\rangle as
\begin{aligned}
\braket{\widetilde{v}_k |\widetilde{v}_l}
&= \langle\Omega|(\mathbf{1}\otimes E_k^\dagger)(\mathbf{1}\otimes E_l)|\Omega\rangle
\\&= \operatorname{tr}(\mathbf{1}\otimes E_k^\dagger E_l)|\Omega\rangle\langle\Omega|
\\&= \frac{1}{d}\operatorname{tr}\sum_{i,j}|i\rangle\langle j|\otimes E_k^\dagger E_l |i\rangle\langle j|
\end{aligned}
(using the fact that we can substitute \frac{1}{d}\sum_{i,j}|i\rangle\langle j|\otimes|i\rangle\langle j| for |\Omega\rangle\langle\Omega|).
Now, the trace of the tensor product of two matrices is the product of their traces, hence
\begin{aligned}
\langle\widetilde{v}_k|\widetilde{v}_l\rangle
&= \frac{1}{d}\sum_{i,j} \langle i|j\rangle\operatorname{tr}E_k^\dagger E_l |i\rangle\langle j|
\\&= \frac{1}{d} \operatorname{tr}E_k^\dagger E_l
\end{aligned}
(using the fact that \langle i|j\rangle=\delta_{ij} and \sum_i|i\rangle\langle i|=\mathbf{1}).
So we have shown that *if \langle\widetilde{v}_k|\widetilde{v}_l\rangle=0 then \operatorname{tr}E_k^\dagger E_l=0*.

A linear map \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') is completely positive if and only if it admits an operator-sum decomposition of the form
\mathcal{E}(\rho) = \sum_k E_k\rho E^\dagger_k.

If this is the case, then this decomposition has the following properties:

- \mathcal{E} is trace preserving if and only if \sum_k E^\dagger_kE_k=\mathbf{1}.
- Two sets of Kraus operators \{E_k\} and \{F_l\} represent the same map \mathcal{E} if and only if there exists a unitary R such that E_k =\sum_l R_{kl}F_l (where the smaller set of the Kraus operators is padded with zeros, if necessary).

Note that, for *any* \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}'), there always exists a representation with *at most* dd' mutually orthogonal Kraus operators: \operatorname{tr}E^\dagger_iE_j\propto\delta_{ij}.

For example, consider the simpler case where \dim\mathcal{H}=\dim\mathcal{H}'=d
Then the Kraus operators E_k are vectors in a d^2-dimensional Hilbert space, with the Hilbert–Schmidt inner product \operatorname{tr}E^\dagger_k E_l.
We can pick an orthonormal basis of operators \{B_i\} and express each Kraus vector in this basis as E_k=\sum c_{ki} B_i (where i=1,\ldots,d^2 and k=1,\ldots,n, with n possibly much larger than d^2).
This gives us
\begin{aligned}
\rho
\longmapsto
& \sum_{i,j} B_i\rho B^\dagger_j \left(\sum _k c_{ki}c^\star_{kj}\right)
\\=& \sum_{i,j} B_i\rho B^\dagger_j C_{ij}
\end{aligned}
The matrix C_{ij} is positive semidefinite, and hence unitarily diagonalisable: C_{ij}=\sum_k U_{ik} d_k U^\dagger_{kj} for some unitary U and some d_k\geqslant 0.
We can then unitarily “rotate” our operator basis and use the C_k=\sum_j U_{jk}B_j \sqrt{d_k} as our new Kraus operators.