1
$\begingroup$

Does the unitary freedom in the choice of Kraus operators for a given quantum channel just come from the unitary freedom in choice of purification of a quantum state?

Here's what I'm thinking. If I have two representations of the same quantum channel $$\Lambda(\rho) = \sum_k A_k\rho A_k^\dagger = \sum_k B_k \rho B_k^\dagger = \Lambda'(\rho),$$ then we also have $$(\Lambda \otimes id) \Omega = (\Lambda' \otimes id)\Omega,$$ where $\Omega = \sum_i \frac{1}{d}\vert{ii}\rangle\langle ii \vert.$ Each of these results in different representations of the same density matrix. So, they can be purified and there is unitary freedom in the choice of purification. Tracing out the auxiliary state seems to explain why there is unitary freedom in the choice of Kraus operators.

My question is twofold: is this correct? Is there a better (or another) way to see why we have unitary freedom in Kraus operators?

$\endgroup$

2 Answers 2

2
$\begingroup$

Short answer: yes, tracing out the auxiliary system gives the unitary freedom in the choice of the Kraus operators.

Longer answer: Given an initial system state $\rho_{sys}$ and bath/auxillary state $|B_0\rangle\langle B_0|$, the initial total density matrix $\rho_{total} = \rho_{sys} \otimes |B_0\rangle\langle B_0|$ evolves according to some propagator $U_{total}$ that defines the total system evolution as $U_{tot}\rho_{tot}U_{tot}^\dagger$.

If we now only care about the state of the system we care about, we can trace out the bath WRT some basis for it $\{|j\rangle\}$ as $\Lambda(\rho_{sys}) = \sum_j \langle j| U_{tot}\rho_{tot}U_{tot}^\dagger |j\rangle = \sum_j \langle j| U_{tot}(\rho_{sys} \otimes |B_0\rangle\langle B_0|)U_{tot}^\dagger |j\rangle = \sum_j K_j \rho_{sys} K_j^\dagger$ when we define $K_j$ as $\langle j| U_{tot}|B_0\rangle$.

This means we can get a valid set of Kraus operators $\{K_j\}$ that define our define our channel for any choice of bath basis $\{|j\rangle\}$, which gives the unitary freedom since we can unitarily map from one basis to another.

$\endgroup$
1
$\begingroup$
  1. The unitary freedom in the choice of Kraus operators for a given channel $\Phi$ can be traced back directly to the freedom in the choice of decomposition of the associated Choi $J(\Phi)$. More precisely, given $\Phi:\operatorname{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal Y)$ and its Choi $$J(\Phi)\equiv(\Phi\otimes\operatorname{Id})(|\Omega\rangle\!\langle\Omega|)\in\mathrm{Lin}(\mathcal Y\otimes \mathcal X),$$ which is positive semidefinite iff $\Phi$ is a channel, any decomposition of $J(\Phi)$ in terms of unit-rank semidefinite operators: $$J(\Phi) = \sum_a v_a v_a^\dagger, \qquad v_a\in\mathcal Y\otimes\mathcal X,\tag1$$ corresponds to the Kraus decomposition $$\Phi(X) = \sum_a A_a X A_a^\dagger,\qquad A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y),$$ where $A_a$ and $v_a$ are directly related by $(A_a)_{ij} \equiv (v_a)_{ij}$ (that is, $v_a$ is the vectorisation of $A_a$). Vice versa, any Kraus decomposition with operators $A_a$ provides a decomposition for the Choi in terms of vectors $v_a$ related to $A_a$ as above.

  2. The reason this observation helps to answer the question, is that we know how to fully characterise the possible ways in which a positive semidefinite operator can be decomposed as in (1). More specifically, if we assume $v_a$ to be orthogonal vectors (i.e. we are doing an eigendecomposition of $J(\Phi)$), then this freedom is tightly related to the degeneracy of the eigenvalues of $J(\Phi)$. If $J(\Phi)$ is nondegenerate, then there is a unique such decomposition. For any degenerate eigenspace, there is unitary freedom in the choice of associated eigenvectors.

  3. It is also possible to consider decompositions in (1) in terms of vectors that are not orthogonal. Also in this case, we know what kinds of decompositions are possible, because then the question becomes equivalent to asking the possible pure state decompositions of a given mixed state, and we now a state $\rho$ can be written as $$\rho=\sum_k p_k |\varphi_k\rangle\!\langle\varphi_k|\tag2$$ for some choice of (not necessarily orthogonal) states $|\varphi_k\rangle$ iff $\boldsymbol p\preceq\boldsymbol{\lambda}(\rho)$ (that is, iff the probability vector $\boldsymbol p$ is majorised by the vector of eigenvalues of $\rho$).

  4. Finally, regarding the connection with the freedom in choosing purifications: yes, I think you can kinda put it that way as well, because the freedom in choosing pure state decompositions of a density matrix is tightly related to the freedom in choosing its purifications: given any decomposition as in (2), there is a set of corresponding purifications $$|\psi_\rho\rangle = \sum_k \sqrt{p_k} (|\varphi_k\rangle\otimes|u_k\rangle),$$ for an arbitrary set of orthonormal auxiliary states $|u_k\rangle$. Clearly, however, this additional freedom in choosing the auxiliary states in the purification of the states does not affect in any way the associated pure state decomposition of the state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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