Do the Kraus operators of a CPTP channel need to be orthogonal?

Question

Let $\Phi\in\mathrm T(\mathcal X,\mathcal Y)$ be a CPTP map. Any such channel admits a Kraus decomposition of the form $$\Phi(X)=\sum_a A_a X A_a^\dagger,$$ for a set of operators $A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y)$ satisfying $\sum_a A_a^\dagger A_a=I_{\mathcal X}$.

The standard way to prove this passes through the Choi representation $J(\Phi)$ of the channel, showing that CP is equivalent to $J(\Phi)$ being a positive operator, and therefore $J(\Phi)$ admits a spectral decomposition with positive eigenvalues, and finally realise that the eigenvectors of $J(\Phi)$ are essentially equivalent to the Kraus operators $A_a$ (upon some reinterpretation of the indices). This is shown for example at pag. 83 (theorem 2.22) of Watrous' TQI book, and in some form also in this other answer here, as well as in a slightly different formalism in this other answer of mine.

What puzzles me about this is the following. The components in the spectral decomposition of the Choi operator $J(\Phi)$ will also have to satisfy an additional property, one that I haven't seen discussed in this context: the orthogonality of the eigenvectors.

If $J(\Phi)=\sum_a v_a v_a^\dagger$, then we also know that the vectors $v_a$ are orthogonal. More specifically, we can always write $J(\Phi)=\sum_a p_a v_a v_a^\dagger$ for some $p_a\ge0$ and $\langle v_a,v_b\rangle=\delta_{ab}$. Remembering that here $v_a\in\mathcal Y\otimes\mathcal X$, these vectors are essentially the Kraus operators of the channel in the sense that $(v_a)_{\alpha i}=(A_a)_{\alpha i}$ (using greek and latin letters to denote indices in $\mathcal Y$ and $\mathcal X$, respectively).

The orthogonality of the $v_a$ is thus equivalent to the fact that Kraus operators must satisfy $$\operatorname{Tr}(A_a^\dagger A_b)\equiv \sum_{i\alpha}(A_a^*)_{\alpha i} (A_b)_{\alpha i}=p_a\delta_{ab}.\tag A$$

However, this property doesn't seem to be usually remarked. Moreover, people often refer to Kraus operators that do not satisfy this orthogonality condition. An example is the Kraus operators used for the dephasing channel in this answer.

The question is therefore as follows: should the property (A) be considered as a necessary condition for a set $\{A_a\}_a$ to be called a set of Kraus operators of a channel? Moreover, regardless of the terminology that one chooses to use, is there any advantage in choosing a "Kraus decomposition" for the channel that is made out of orthogonal operators, rather than non-orthogonal ones?

Answer 1

There is an ambiguity in the choice of Kraus operators: If $\{E_a\}$ is a set of Kraus operators for a channel $\mathcal E$, so is $\{F_b\}$ with $F_b=\sum_a v_{ab} E_a$, with $(v_{ab})$ an isometry.

In particular, you can choose a $(v)$ which diagonalizes the matrix $X_{ac}=\mathrm{tr}[E_a^\dagger E_b]$, in which case $\{F_b\}$ satisfies your orthogonality property.

However, this is an ambiguity in the Kraus representation, and all such representations are called Kraus representations. That's the way it is, without a condition (A). If you think condition (A) is good to have, then you should give it a different name.

It is likely that this representation has certain advantages, just as an eigenvalue decomposition. Off the top of my head, it should e.g. mean that the different errors introduced by the different Kraus operators are orthogonal, so it should be a convenient representation in terms of error correction. Also, it is one representation with the minimal number of Kraus operators. I'm sure there are others.

Answer 2

To answer your two questions separately:

1) The property (A) is not a necessary condition for a set to be a Kraus decomposition. Your derivation is right - Kraus operators obtained from the eigenvectors of the Choi matrix (or the process matrix, for that matter), are almost always orthogonal to each other. Almost, because if the Choi matrix is degenerate, you can find a non-orthogonal eigenbasis $\{c_{\lambda_{i}}\}$ for an eigenspace $\lambda_{i}$ with $a.m. > 1$. This results in non-orthogonal Kraus operators as well.

2) There is yet another option: Kraus operators may be linear combinations of each other (e.g. the operators in the other answer you linked). That also means that they never can all be orthogonal to each other. It also introduces some form of ambiguity, because there are then multiple ways to express the action of the channel.

For instance, the canonical way of expressing the dephasing channel is with Kraus operators $A_{1} = \sqrt{1-p}I$ and $A_{2} = \sqrt{p}Z$ (which are, in fact, orthogonal). By limiting your answer to orthogonal Kraus operators, you omit this ambiguity.