5
$\begingroup$

As first shown by Gorini, Kossakowski, Sudarshan and Lindblad given some linear map $\mathcal L:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$, $e^{t\mathcal L}$ is a quantum channel for all $t\geq 0$ if and only if there exist $H$ Hermitian as well as $L_1,\ldots,L_m$, $m\in\mathbb N_0$ such that \begin{equation} \mathcal{L}(\rho) = -i[H,\rho]+\sum_{k=1}^m\Big( L_k \rho L_k^{\dagger} - \frac{1}{2} L_k^{\dagger}L_k \rho - \frac{1}{2}\rho L_k^{\dagger}L_k \Big)\tag1 \end{equation} for all $\rho$; see also this, this, or this post.

To pass to a more restricted version of this problem imagine we only knew that $e^{t\mathcal L}$ is a channel for certain, but not all times $t$. Note that this does not necessarily mean that $e^{t\mathcal L}$ fails to be a channel for some of the other times, just that we do not have any information about these times. In the limiting case where we're guaranteed to have CPTP for only one time $t_0$ (i.e. all we know is that $e^{t_0\mathcal L}$ is CPTP for some $t_0>0$) this does not automatically imply that $\mathcal L$ is of form (1). The reasoning is simple: take any bijective channel $\Phi$ (so one can define $\mathcal L:=\log\Phi$ using some branch of the complex logarithm) such that $\Phi$ has an odd number of negative eigenvalue—e.g., the qubit Holevo-Werner channel $\Phi:=\frac{{\rm tr}(\cdot){\bf1}+(\cdot)^T}3$ which has three positive and one negative eigenvalues. Then $e^{t\mathcal L}$ is a channel if and only if $t\in\mathbb N_0$, and for all other times the resulting map is not even Hermitian preserving: for all $t\in\mathbb R\setminus\mathbb N_0$ there is an odd number of complex eigenvalues which is at odds with the fact that complex eigenvalues of Hermitian-preserving maps always come in complex conjugate pairs. This begs the following question:

Given some $\mathcal L:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ linear and some subset $\mathcal T:=\{t_j\}_{j\in J}\subset(0,\infty)$ such that $e^{t_j\mathcal L}$ is completely positive and trace preserving under what assumptions on $\mathcal T$ can we conclude that $\mathcal L$ can be expressed as in Eq. (1)?

First observe the following sufficient condition: If $0\in\overline{\mathcal T}$, i.e. there exist times $t_j\in\mathcal T$ such that $t_j\to 0$ as $j\to\infty$, then $\mathcal L=\lim_{j\to\infty}\frac{e^{t_j\mathcal L}-{\rm id}}{t_j}$ is a limit of Lindblad generators${}^1$ $\frac1{t_j}(e^{t_j\mathcal L}-{\rm id})$, hence a Lindblad generator itself (as the latter, like every Lie wedge, form a closed set).

As seen previously it is not enough to know the channel property for just one time $t$. Another example—which will further motivate our question—is based on the example of a non-Hermitian matrix $A$ such that $e^{iA}$ is still unitary; there the corresponding generator $\mathcal L:=-i[A,\cdot]$ is not of Lindblad form (because not Hermitian-preserving) but $e^{\mathcal L}=e^{-iA}(\cdot)e^{iA}$ is a unitary channel. This example can be modified to the case where $\mathcal T$ is a finite, rationally dependent set. Interestingly, however, this example breaks down once we know that there exist $t_1,t_2\in\mathcal T$ rationally independent such that both $e^{iAt_1},e^{iAt_2}$ are unitary. In other words for the case of closed systems it is necessary and sufficient to have access to two (rationally independent) times.

Thus, in general, it is necessary that $\mathcal T$ contains two rationally independent elements if we want to conclude Lindblad form from the CPTP property; this is also in line with the initially discussed Holevo-Werner example. However, because the above example is tailored quite heavily to unitary matrices it is not clear to me whether this criterion is sufficient (beyond the special case of unitary channels); for example, already the implication that because $e^{iA}$ is unitary it is diagonalizable, $A$ must be diagonalizable does not have any counterpart for Lindbladians, cf. also the section "Quantum channels and dynamics" in the list of counterexamples in quantum information. In other words the following seems to be the simplest non-trivial—as well as a concise—version of my question:

Can it happen that, say, $e^{\mathcal L},e^{\sqrt2\mathcal L}$ are channels but the linear map $\mathcal L$ is still not of Lindblad form (1)?


1: Given any channel $\Phi$ it is known that $\Phi-{\rm id}$ (and any non-negative multiple thereof) is a Lindblad generator. This is also known as "Markovian approximation" (arXiv).

$\endgroup$
1
  • 1
    $\begingroup$ Nice question. As a quick shot, I would try the "pancake channel" as a potential counterexample, i.e. channel which contracts the Bloch sphere much in one direction, and less in the other two. This one is not created by a Liouvillian (in fact, it is unphysical if it does not contract enough), yet I would suspect that starting from a certain amount of contraction it remains physical. I'll try to check in the next days. $\endgroup$ Commented May 29 at 12:18

1 Answer 1

2
$\begingroup$

Consider the "pancake channel" $$ \mathcal P:\rho \mapsto \mathrm{tr}\rho\,I + \sum_{\alpha=x,y,z} \lambda_\alpha\mathrm{tr}(\rho\sigma_\alpha)\sigma_\alpha\ , $$ where we choose $\lambda_{x}=\lambda_{y} = e^{-at}$ and $\lambda_z = e^{-bt}$, and where $a<b$. (It maps the Bloch sphere to a pancake, as it flattens is more in the $z$ direction.)

This channel is diagonal when written in the Pauli basis (including identity), with representation $$ T=\begin{pmatrix} 1 \\ & e^{-at} \\ && e^{-at}\\ &&& e^{-bt} \end{pmatrix}\ , \tag{1} $$ and is thus generated by the linear map $$ \mathcal L = \begin{pmatrix} 0 \\ & -a \\ && -a\\ &&& -b \end{pmatrix}\ . $$

Channels of the form (1) are completely positive if and only if $\lambda_1^\downarrow + \lambda_2^\downarrow \le 1+ \lambda_3^\downarrow$ for descendingly ordered eigenvalues of the lower $3\times 3$ block, see e.g. arXiv:math-ph/0611057, Eq. (33) in the arXiv version (alternatively, this directly follows by inspection of the Choi state).

Specifically, this means that the map $\mathcal P$ is completely positive if and only if $$ 2 e^{-at} \le 1 + e^{-bt}\ .\tag{2} $$ If you now choose $2a<b$, e.g., $a=1$, and $b=4$, it is immediate to check using (2) that that

  1. For small enough $t$, is $\mathcal P$ is not completely positive -- that is, it cannot be generated by a Liouvillian, and
  2. For large enough $t$, $\mathcal P$ is completely positive (in particular, for $a=1$, $b=4$, it is completely positive for all $t\ge1$).
$\endgroup$
3
  • $\begingroup$ If fact, there are pancake channels which cannot even be subdivided into other channels, let alone generated continuously (see arxiv.org/abs/1708.00880, Example 14). Indeed, flat pancakes must be rather small to be physical. $\endgroup$ Commented May 30 at 20:32
  • 1
    $\begingroup$ A simple, yet elegant counterexample! What I also find interesting about this example is that for small times—while the dynamics fail to be CP because $\mathcal L$ is not conditionally completely positive—the dynamics are at least positive trace-preserving. Thus even the stronger information that $e^{t\mathcal L}$ is PTP for small enough $t$ and CPTP for all $t$ large enough this is still not enough to conclude Lindblad form for $\mathcal L$. $\endgroup$ Commented May 31 at 7:48
  • 2
    $\begingroup$ @Frederik Indeed. I still find it puzzling that a map which contracts the Bloch sphere only (or primarily) along one direction is not CP, while one which does so in two directions is. (Of course, one can just check it ... but it still puzzles me.) $\endgroup$ Commented May 31 at 13:00

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.