Can the Kraus decomposition always be chosen to be a statistical mixture of unitary evolutions?


If $\mathcal{E}$ is a CPTP map between hermitian operators on two Hilbert spaces, then we can find a set of operators $\{K_j\}_j$ such that

$$\mathcal{E}(\rho)=\sum_j K_j\rho K_j^\dagger $$

in the same spirit as any density matrix $\rho$ can be decomposed in a pure states ensemble

$$\rho=\sum_k p_k |\psi_k\rangle\langle\psi_k| $$

with $\sum_k p_k=1$ and be intepreted as a classical statistical mixture of pure states, can I always find a Kraus decomposition such that $K_j= \sqrt{p_j} U_j$ with $U_j U_j^\dagger=\mathbb{1}$ and $\sum_j p_j=1$ and interpret it as a classical statistical mixture of unitary evolutions?


Posted 2019-08-06T16:15:21.107

Reputation: 845



You cannot always find such a Kraus decomposition. Notice that any CPTP map $\mathcal E$ which does have a decomposition as a probabilistic mixture unitaries is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state: $$ \mathcal E(\tfrac{1}{d} \mathbf 1) = \tfrac{1}{d} \mathbf 1 . $$ This is true because $U \cdot \mathbf 1 \cdot U^\dagger = \mathbf 1$ for each unitary $U$, and taking a mixture over several unitaries $U$ does not change the result for the operator $\mathbf 1$. But it is easy to find maps that don't have this property — so for such maps, there is no interpretation as a probabilistic mixture of unitary processes. For instance, the map $\mathcal R$ that resets a qubit to $\lvert 0 \rangle$ does not preserve the maximally mixed state on a qubit: $$ \mathcal R(\rho) \,=\, \mathrm{tr}(\rho) \cdot \lvert 0 \rangle\!\langle 0 \rvert . $$ Of course, this map can be expressed using Kraus operators: for instance, you could take $K_0 = \lvert 0 \rangle\!\langle 0 \rvert$ and $K_1 = \lvert 0 \rangle\!\langle 1 \rvert$, which would suffice for $\mathcal R(\rho) = K_0^{\phantom\dagger} \rho K_0^\dagger + K_1^{\phantom\dagger} \rho K_1^\dagger$.

Edited to add: Following up on Norbert Schuh's comment below, not all unital channels are mixtures of unitaries either. An example for qutrits can be found in John Watrous' Theory of Quantum Information (which is freely available online here), in Example 4.3: $$ \begin{gathered} \Phi(\rho) = A_1^{\phantom\dagger} \rho A_1^\dagger + A_2^{\phantom\dagger} \rho A_2^\dagger + A_3^{\phantom\dagger} \rho A_3^\dagger, \\[1ex] A_1 = \tfrac{1}{\sqrt 2}\mbox{$\scriptstyle\begin{bmatrix} 0 \!&\! 0 \!&\! 0 \\ 0 \!&\! 0 \!&\! 1 \\ 0 \!&\! -1 \!&\! 0 \end{bmatrix}$}\,, \quad A_2 = \tfrac{1}{\sqrt 2}\mbox{$\scriptstyle\begin{bmatrix} 0 \!&\! 0 \!&\! 1 \\ 0 \!&\! 0 \!&\! 0 \\ -1 \!&\! 0 \!&\! 0 \end{bmatrix}$}\,, \quad A_3 = \tfrac{1}{\sqrt 2}\mbox{$\scriptstyle\begin{bmatrix} 0 \!&\! 1 \!&\! 0 \\ -1 \!&\! 0 \!&\! 0 \\ 0 \!&\! 0 \!&\! 0 \end{bmatrix}$}\,. \end{gathered} $$ As Watrous observes, it is possible to show that this is an extreme point among the channels on a qutrit (i.e., not a convex combination of any other channels) by showing that the operators $A_j^\dagger A_k^{\phantom\dagger}$ are a linearly independent set of operators; then as $\Phi$ is not itself unitary, it isn't a mixture of unitaries either.

Watrous also notes the following characterisation of channels which are mixtures of unitaries:

Theorem 4.8 (paraphrased) — Let $\Phi$ be a channel on a Hilbert space $\mathcal X$, characterised by a Stinespring dilation $$\Phi(\rho) = \mathrm{Tr}_{\mathcal Z}(A \rho A^\dagger)$$ for some unitary embedding $A: \mathcal X \to \mathcal X \otimes \mathcal Z$. Then $\Phi$ is a mixture of unitaries if and only if there exists a POVM $\{E_k\}_{k \in K}$ on $\mathcal Z$ with outcomes labeled by elements of some set $K$, and some channels $\Psi_k$ on $\mathcal X$ for $k \in K$, such that $$ \rho = \sum_{k \in K} \Psi_k\bigl(\tilde \rho_k\bigr), $$ where $\tilde \rho_k = \mathrm{Tr}_{\mathcal Z}\bigl(\bigl[\mathbf 1 \!\otimes\! E_k\bigr]\bigl[A \rho A^\dagger\bigr]\bigr)$.

That is to say: if (and only if) $\Phi$ can be characterised by an interaction of an input system $\mathcal X$ with an environment $\mathcal Z$, in such a way that there is a way in principle that you could measure the environment and use the outcome to control a 'correction' operation to recover the original input, then $\Phi$ is a mixture of unitaries.

Niel de Beaudrap

Posted 2019-08-06T16:15:21.107

Reputation: 9 858

It might be worth noting that even for unital channels, this is not always possible. – Norbert Schuch – 2019-08-07T00:42:28.920

@NorbertSchuch Does it says anything about the channel whether this can be found or not? (A part from the obvious "it is not a statistical mixture of unitary evolutions"). It seemed like a very natural idea and it strikes me as if a (unital) quantum channel cannot be represented in such a way, it should imply something is different about that channel – user2723984 – 2019-08-07T09:37:33.397

2@user2723984 This is (was?) known as the "Quantum Birkhoff conjecture". You might want to check the corresponding papers. – Norbert Schuch – 2019-08-07T09:48:13.197