7.11 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 mixtures142 (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 decomposition143 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 sense144) 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}). 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_{ij} B_i\rho B^\dagger_j \left(\sum _k c_{ki}c^\star_{kj}\right) \\=& \sum_{ij} 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.


  1. The number of vectors contributing to each mixture (and hence the number of corresponding Kraus operators) may be different, but we can always simply extend the smaller set to the required size by adding zero operators.↩︎

  2. We talk about spectral decomposition in more detail in Section 11.10.1.↩︎

  3. Recall that the Hilbert–Schmidt product (A|B) of two operators A and B is defined by (A|B)=\frac{1}{2}\operatorname{tr}A^\dagger B.↩︎