I know that the Choi operator is a useful tool to construct the Kraus representation of a given map, and that the vectorization map plays an important role in such construction.

How exactly does the vectorization map work in this context, and how does it relate the Choi and Kraus representations of a given map?


1 Answer 1


One way to understand the relationship between the Choi representation of a channel and its possible Kraus representations is to use the vectorization map.

Suppose that we have two finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{Y}$, and that we have fixed a standard basis $\{|1\rangle,\ldots,|n\rangle\}$ of $\mathcal{X}$ and a standard basis $\{|1\rangle,\ldots,|m\rangle\}$ of $\mathcal{Y}$. The vectorization map is the linear map of the form $$ \text{vec}:\text{L}(\mathcal{X},\mathcal{Y})\rightarrow\mathcal{Y}\otimes\mathcal{X}, $$ where $\text{L}(\mathcal{X},\mathcal{Y})$ is the space of all linear maps from $\mathcal{X}$ to $\mathcal{Y}$, that is defined by the following action on standard basis elements: $$ \text{vec}\bigl( | j\rangle \langle k |\bigr) = |j\rangle |k\rangle. $$ For other elements of $\text{L}(\mathcal{X},\mathcal{Y})$ the action of the mapping is determined by linearity.

If we view elements of $\text{L}(\mathcal{X},\mathcal{Y})$ as $m\times n$ matrices that act by left-multiplication on column vectors of dimension $n$, the action of the vectorization mapping is to take the rows of the matrix, turn them into column vectors, and then stack them on top of one another like this (in the case $n=m=2$): $$ \text{vec}\begin{pmatrix} \alpha & \beta\\ \gamma & \delta \end{pmatrix} = \begin{pmatrix} \alpha\\ \beta\\ \gamma\\ \delta \end{pmatrix}. $$

Now the relationship between all of the possible Kraus representations and the Choi representation of a channel $\Phi$ having the form $\Phi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Y})$ can be easily stated: for any choice of a positive integer $N$ and operators $A_1,\ldots,A_N\in\text{L}(\mathcal{X},\mathcal{Y})$, it is the case that $$ \Phi(X) = \sum_{k=1}^N A_k X A_k^{\dagger} $$ is a Kraus representation of $\Phi$ if and only if the Choi representation of $\Phi$ satisfies the equality $$ J(\Phi) = \sum_{k=1}^N \text{vec}(A_k) \text{vec}(A_k)^{\dagger}. $$

This relationship allows you to convert between the two relationships pretty easily. To get the Choi representation from a Kraus representation, it is just a matter of computing. To go from a Choi representation to a Kraus representation, you can use the spectral theorem to write $$ J(\Phi) = \sum_{k=1}^r v_k v_k^{\dagger} $$ for vectors $v_1,\ldots,v_r\in\mathcal{Y}\otimes\mathcal{X}$, where $r=\text{rank}(J(\Phi))$, then take your Kraus operators to be the operators $A_1,\ldots,A_r\in\text{L}(\mathcal{X},\mathcal{Y})$ that satisfy $\text{vec}(A_k) = v_k$ for each $k\in\{1,\ldots,r\}$. This is why $\text{rank}(J(\Phi))$ is always the minimum possible number of Kraus operators you need to specify a given channel.

One warning about this relationship: I am assuming that the Choi representation of a channel $\Phi$ of the form described above is defined as $$ J(\Phi) = \sum_{1\leq j,k \leq n} \Phi\bigl(|j\rangle \langle k|\bigr) \otimes |j\rangle \langle k|. $$ Some people choose to define the Choi representation with the tensor factors in the reverse order, like this: $$ J'(\Phi) = \sum_{1\leq j,k \leq n} |j\rangle \langle k| \otimes \Phi\bigl(|j\rangle \langle k|\bigr). $$ If you do that, you can still get a similar relationship, but you will have to define your vectorization mapping in a different way, as $\text{vec}'(|j\rangle\langle k|) = |k\rangle|j\rangle$. (You will find the vectorization map defined in these two different ways by different authors, so you just need to be aware of which definition is being used whenever you see this mapping.)

Concerning references, you can find this relationship in Choi's 1975 paper:

Man-Duen Choi. Completely positive linear maps on complex matrices. Linear Algebra and Its Applications 10: 285-290, 1975.

Choi's proof of this relationship is terse, but certainly comprehensible. You can also find more detail in Section 4.4 of Mark Wilde's book:

Mark Wilde. Quantum Information Theory, second edition. Cambridge University Press, 2017.

Assuming you'll forgive me for a self-citation, I also cover this in Section 2.2.2 of my book, using a similar notation to the description above:

John Watrous. The Theory of Quantum Information. Cambridge University Press, 2018. (Available at https://cs.uwaterloo.ca/~watrous/TQI/ .)

  • 3
    $\begingroup$ wish I could give more than one upvote.. :) $\endgroup$ Commented Jul 14, 2020 at 3:09
  • $\begingroup$ I'm a little bit confused, shouldn't vec stand for stacks of columns of matrix, i.e. $$\mathrm{vec}\left( \begin{matrix} \alpha& \beta\\ \gamma& \delta\\ \end{matrix} \right) =\left( \begin{array}{c} \alpha\\ \gamma\\ \beta\\ \delta\\ \end{array} \right) ?$$ $\endgroup$
    – narip
    Commented Sep 15, 2022 at 0:48
  • $\begingroup$ Since for example:$$\left( \begin{matrix} a& b\\ c& d\\ \end{matrix} \right) |0\rangle \langle 1|\left( \begin{matrix} \bar{a}& \bar{c}\\ \bar{b}& \bar{d}\\ \end{matrix} \right) \otimes |0\rangle \langle 1|=\left( \begin{array}{c} a\\ c\\ 0\\ 0\\ \end{array} \right) \left( \begin{matrix} \begin{matrix} \begin{matrix} 0& 0\\ \end{matrix}& \bar{b}\\ \end{matrix}& \bar{d}\\ \end{matrix} \right) $$ $\endgroup$
    – narip
    Commented Sep 15, 2022 at 0:49
  • 1
    $\begingroup$ Do you agree that $\vert 1\rangle\vert 0\rangle = \begin{pmatrix}0\\0\\1\\0\end{pmatrix}$? $\endgroup$ Commented Sep 15, 2022 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.