I have a CPTP quantum channel $\mathcal{E}$ that I've characterized by an operator sum representation $\{E_i\}$ for $i=1, \dots, m$ which acts on an input state like $$ \mathcal{E}(\rho) = \sum_{i=1}^m E_i \rho E_i^\dagger $$ with $\sum_i E_i^\dagger E_i = I$. I would like to know if there is an alternative Kraus representation with operators $\{F_j\}$ for $j=1, \dots, n$ that satisfies similar properties, $\mathcal{E}(\rho) =\sum_{j=1}^n F_j \rho F_j^\dagger$ and $\sum_{j=1}^n F_j^\dagger F_j=I$ but also that each new operator is proportional to a unitary, $$ F^\dagger_j F_j = F_j F_j^\dagger = c_j I $$ for some real $c_j$. For example a choice of dephasing channel satisfies this property since it can be defined either by the operators $\{\frac{1}{\sqrt{2}}I, \frac{1}{\sqrt{2}}Z\}$ or $\{|0\rangle\langle 0 |, |1\rangle \langle 1|\}$.

I do understand that the two channels must be related by some unitary transformation $U$ with, $E_i = \sum_j U_{ij} F_j$ but this seems only to constrain the problem for specific choices of $\{E_i\}$.

  • 1
    $\begingroup$ related: quantumcomputing.stackexchange.com/a/6970/55 $\endgroup$
    – glS
    Jun 29, 2021 at 22:49
  • 2
    $\begingroup$ This is possible for some channels and not for others. Do you have a specific channel you are interested in, or are you asking more generally if this is always possible? $\endgroup$ Jun 30, 2021 at 0:00
  • $\begingroup$ @JohnWatrous I would like to know for any given channel whether this alternative representation will be possible $\endgroup$
    – forky40
    Jun 30, 2021 at 1:14
  • $\begingroup$ Did your formula missing $I$ in $\sum_{j=1}^n F_j^\dagger F_j(=I)$ ? $\endgroup$
    – narip
    Jun 30, 2021 at 6:14
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    $\begingroup$ (Lee and Watrous 2020) might be relevant here (and also possibly what @JohnWatrous was referring to?) $\endgroup$
    – glS
    Jul 1, 2021 at 9:13

2 Answers 2


As Adam Zalcman has stated in his answer, channels whose Kraus operators are proportional to unitary operators are called mixed-unitary channels (or, alternatively, random unitary channels).

Every mixed-unitary channel is unital (meaning that it maps the identity operator to itself), so if you want a channel that is not mixed unitary, just pick any non-unital channel. For example, the qubit channel $$ \Phi(X) = \operatorname{Tr}(X)\, |0\rangle \langle 0| $$ is a non-unital channel, so it is not mixed unitary.

Also as Adam has answered, qubit channels are mixed unitary if and only if they are unital. This is no longer true in dimension 3 and higher. For example, the $n$-dimensional anti-symmetric Werner-Holevo channel $$ \Phi(X) = \frac{\operatorname{Tr}(X)\mathbb{1}_n - X^{\scriptsize\mathsf{T}}}{n-1} $$ is unital but not mixed-unitary for every odd integer $n\geq 3$. In fact, for odd $n$ you cannot come up with any Kraus representation of this channel where even one of its Kraus operators is non-zero and proportional to a unitary operator.

Finally, as the reference gIS mentioned in a comment shows, the computational problem of deciding whether or not a given channel is mixed unitary is NP-hard, even under a pretty relaxed notion of approximation, so you should not expect a simple test to reveal whether or not such an expression is possible. This is why I asked if you had a particular channel in mind; for a specific channel, it may be possible to make the determination based on an analysis of that channel, but an easy-to-check general solution is unlikely.


I don't know the fully general answer, but have found a solution for channels acting on a single qubit.

Mixed-unitary channels

Quantum channels that admit a Kraus representation consisting solely of multiples of unitary operators are known as mixed-unitary channels, i.e. $\mathcal{E}$ is a mixed-unitary channel if there exists unitary operators $U_i$ and positive real numbers $c_i$ with $i=1,\dots,m$ such that

$$ \mathcal{E}(\rho) = \sum_{i=1}^mc_iU_i\rho U_i^\dagger $$

for all density matrices $\rho$. Note that $\sum_{i=1}^m c_i=1$ when $\mathcal{E}$ is trace-preserving.

Unital channels

It is easy to see that if $\mathcal{E}$ is a mixed-unitary channel then it is unital, i.e. it maps identity to identity $\mathcal{E}(I) = I$. It turns out that if $\mathcal{E}$ acts on density matrices of a single qubit, i.e. $\mathcal{E}:\mathcal{L}(\mathbb{C^2})\to \mathcal{L}(\mathbb{C^2})$ then this necessary condition is also sufficient.

In single-qubit case, $\mathcal{E}$ is mixed-unitary if and only if it is unital

Claim. Let $\mathcal{E}$ be a single-qubit quantum channel. If $\mathcal{E}$ is unital then it is mixed-unitary.

Proof. Recall that any single-qubit density matrix $\rho$ can be written as

$$ \rho = \frac{I + \vec{r}_\rho\cdot\vec{\sigma}}{2} $$

where $\vec{\sigma}=(\sigma_x, \sigma_y,\sigma_z)$ is the vector of Pauli matrices and $\vec{r}_\rho\in\mathbb{R}^3$ with $\|\vec{r}_\rho\| \le 1$ is the Bloch vector corresponding to $\rho$. Moreover, any single-qubit CPTP map $\mathcal{E}$ is equivalent to an affine map acting on Bloch vectors

$$ \vec{r}_\rho \xrightarrow{\mathcal{E}} \vec{r}_{\mathcal{E}(\rho)}=M\vec{r}_\rho + \vec{c}\tag1 $$

where $M$ is a $3\times 3$ real matrix and $\vec{c}$ a constant vector, c.f. equation $(8.89)$ on page 375 in Nielsen & Chuang. The maximally mixed state $I/2$ corresponds to the zero Bloch vector, so if $\mathcal{E}$ is unital then $\vec{c}=0$ and $(1)$ becomes

$$ \vec{r}_\rho \xrightarrow{\mathcal{E}} \vec{r}_{\mathcal{E}(\rho)}=M\vec{r}_\rho.\tag{1'} $$

Next, we show that any Bloch sphere transformation described by $(1')$ corresponds to a mixed-unitary quantum channel. To see this, write $M$ as

$$ M = OS\tag2 $$

for a real orthogonal matrix $O$ with unit determinant and a real symmetric matrix $S$, c.f. equation $(8.93)$ on page 375 in Nielsen & Chuang. Geometrically, $(2)$ describes a deformation of the Bloch sphere along the principal axes determined by the eigenvectors of $S$ followed by the rotation described by $O$.

Now, Bloch sphere rotations, such as $O$, correspond to unitary channels. Bloch sphere deformations along the $x$, $y$ and $z$ axes can be performed using bit-flip and phase-flip channels. Finally, Bloch sphere deformations along arbitrary axes can be implemented using deformations along the $x$, $y$ and $z$ axes and rotations.

In summary, any unital single-qubit quantum channel $\mathcal{E}$ corresponds to a Bloch sphere transformation which does not include translation, i.e. one given by $(1')$, and any such transformation can be effected by a composition of bit-flip, phase-flip and unitary channels. Any such composition is a mixed-unitary channel. $\square$


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