Kraus operators, revisited
One thing that is very important is that state-channel 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}.
From this viewpoint, 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 two mixtures \{p_k, |\psi_k\rangle\} and \{q_l, |\phi_l\rangle\} described by the same density operator
\widetilde{\mathcal{E}}
= \sum_k|\widetilde\psi_k\rangle\langle\widetilde\psi_k|
= \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) are related: 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.
How many Kraus operators do we really need?
State-channel duality tells us that the minimal number of Kraus operators needed to express \mathcal{E} in the operator-sum form is given by the rank of its Choi matrix \widetilde{\mathcal{E}}, and so we need no more than dd' such operators.
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_{ij}|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_{ij}|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_{ij} \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}).
In summary then, \langle\widetilde{v}_k|\widetilde{v}_l\rangle=0 implies that \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}.