$\newcommand{\Y}{\mathcal{Y}}\newcommand{\X}{\mathcal{X}}\newcommand{\rmL}{\mathrm{L}}$As explained for example in Watrous' book (chapter 2, p. 79), given an arbitrary linear map $\Phi\in\rmL(\rmL( \X),\rmL(\Y))$, for every linear operator $X\in\rmL(\X)$ we can write the Kraus representation of $\Phi(X)$ as $$\Phi(X)=\sum_a A_a X B_a^\dagger,\tag1$$ where $A_a,B_a\in\rmL(\X,\Y)$.

As far as I understand it, the essential step in showing this is given in Corollary 2.21 of the above book, when we write the Choi representation as $J(\Phi)=\sum_a u_a v_a^\dagger$.

I've been trying to understand how the Kraus representation can be obtained more directly in the natural representation of $\Phi$, which essentially means to think of $\Phi$ as a matrix acting on the vectorized versions of the operators, via the equivalence $\rmL(\X)\sim\X\otimes \X$. In this representation, we can write $\Phi(X)_{ij}=\Phi_{ijk\ell}X_{k\ell}$, where $\Phi_{ijk\ell}$ are the components of the natural representation.

One can apply SVD to the four-index object $\Phi_{ijk\ell}$ separating $i,k$ from $j,\ell$, thus obtaining a decomposition of the form $$\Phi_{ijk\ell}=\sum_\alpha s_\alpha A^\alpha_{ik}\bar B^{\alpha}_{j\ell}.\tag2$$ This looks very close to the typical Kraus representation for a general map as given in (1). Being $s_\alpha\ge0$, one can also write $s_\alpha=\sqrt{s_\alpha}\sqrt{s_\alpha}$ and redefine the operators $A, B$ to contain $\sqrt{s_\alpha}$, thus getting an even more similar form.

What I'm wondering is: is this way of decomposing the channel via SVD equivalent to the Kraus representation? If so, we would also know additional properties for the $A_{ik}^\alpha, B_{j\ell}^\alpha$ operators, such as their orthogonality: $\sum_{ik}A^\alpha_{ik} \bar A^\beta_{ik}=\delta_{\alpha\beta}$ etc. In this sense, it seems to me that absorbing the singular values $s_\alpha$ into the operators, like is done in (1), is actually hiding information, because then we lose the orthogonality between the $A_a, B_a$. Does one approach have any advantage over the other (assuming they are both equally correct)?


1 Answer 1


As matrices, the natural representation and Choi representation of a map $\Phi$ have exactly the same entries, but arranged into matrices in different ways. One way to express this is like this: $$ \langle k,\ell | K(\Phi) | i,j \rangle = \langle k | \Phi(|i\rangle\langle j|) | \ell \rangle = \langle k,i | J(\Phi) | \ell,j \rangle, $$ where $K(\Phi)$ denotes the natural representation, $J(\Phi)$ denotes the Choi representation, $i$ and $j$ index into the input system, and $k$, $\ell$ index into the output system. It appears to me that applying the singular value decomposition to $K(\Phi)$ in the particular way the question describes is equivalent to applying the singular value decomposition to $J(\Phi)$ for this reason. Of course, we do not need to use the singular value decomposition to write $J(\Phi) = \sum_a u_a v_a^{\dagger}$, but it is one way to obtain this form (and as a bonus to get $\{u_a\}$ and $\{v_a\}$ to be orthogonal sets, in case that is useful).

I do not understand the statement in the question about information being hidden if the singular values are absorbed into the Kraus operators. The sets $\{A_a\}$ and $\{B_a\}$ are both orthogonal sets, and this does not change when their elements are multiplied by scalars.

In my view, the main advantage in starting from the Choi representation is that it works better for completely positive maps. In this case, $J(\Phi)$ is positive semidefinite, so we can use the spectral decomposition rather than the singular value decomposition to obtain $A_a = B_a$ for all $a$. Indeed, equation (1) in the question would be considered by many to be a generalized type of Kraus representation, whereas a proper Kraus representation would require $A_a = B_a$, and therefore only exists when $\Phi$ is completely positive.

In general, the Choi representation $J(\Phi)$ is nice because its properties as an operator translate to properties of $\Phi$ (particularly $J(\Phi)\geq 0$ if and only if $\Phi$ is completely positive). In some sense the natural representation $K(\Phi)$ ignores the operator structure of the inputs and outputs of $\Phi$, which makes it a less useful representation.


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