# Which differential equations correspond to the Lindblad equation?

Victor Kasatkin<EMAIL_ADDRESS>Center for Quantum Information Science &
Technology, University of Southern California, Los Angeles, CA 90089, USA
Larry Gu<EMAIL_ADDRESS>Department of Mathematics, University of Southern
California, Los Angeles, CA 90089, USA  Daniel A. Lidar<EMAIL_ADDRESS>Departments of Electrical Engineering, Chemistry, Physics and Astronomy, and
Center for Quantum Information Science & Technology, University of Southern
California, Los Angeles, CA 90089, USA

(September 3, 2024)

###### Abstract

The Lindblad master equation can always be transformed into a first-order
linear ordinary differential equation (1ODE) for the coherence vector. We pose
the inverse problem: given a finite-dimensional, non-homogeneous 1ODE, does a
corresponding Lindblad equation exist? If so, what are the corresponding
Hamiltonian and Lindblad operators? We provide a general solution to this
problem, including a complete positivity test in terms of the parameters of
the 1ODE. We also derive a host of properties relating the two representations
(master equation and 1ODE) of independent interest.

##  I Introduction

The Gorini-Kossakowski-Lindblad-Sudarshan (GKLS) master equation [1, 2] is
widely used in modeling the evolution of open quantum systems subject to
Markovian dynamics [3, 4, 5, 6, 7]. It can be written in the following form:

$\displaystyle\dot{\rho}$ $\displaystyle=\mathcal{L}\rho\
,\quad\mathcal{L}=\mathcal{L}_{H}+\mathcal{L}_{a}$ (1a)
$\displaystyle\mathcal{L}_{H}[\cdot]$ $\displaystyle=-i[H,\cdot]$ (1b)
$\displaystyle\mathcal{L}_{a}[\cdot]$
$\displaystyle=\sum_{i,j=1}^{d^{2}-1}a_{ij}\big{(}F_{i}\cdot
F_{j}-\frac{1}{2}\\{F_{j}F_{i},\cdot\\}\big{)}\ ,$ (1c)

commonly known simply as the Lindblad equation.111Sadly, Prof. Lindblad passed
away while this work was in preparation. We dedicate this work to his memory.
Here, the dot denotes a time-derivative, $\rho$ is the state (density matrix)
of the open system whose Hilbert space is $d$-dimensional, $\mathcal{L}$ is
the Lindbladian, $H=H^{\dagger}$ is the system Hamiltonian, $\\{F_{i}\\}$ is a
Hermitian operator basis, and the constants $a_{ij}\in\mathbb{C}$ are the
elements of a positive semidefinite matrix of rates $a$ (we provide complete
definitions below). The positivity property of $a$ is crucial since it is
necessary and sufficient for the superoperator $\mathcal{L}$ to generate a
completely positive (CP) map. This is known as the GKLS theorem.222This is
actually a pair of theorems that were discovered independently and nearly
simultaneously for the finite-dimensional case by Gorini, Kossakowski, and
Sudarshan [1], and the infinite-dimensional case by Lindblad [2]; for a
detailed account of this history see [8].

When $a\not\geq 0$, Eq. 1 still describes a time-independent Markovian quantum
master equation but no longer generates a CP map. This situation arises, e.g.,
when the initial system-bath state is sufficiently correlated [9, 10, 11, 12,
13, 14].

The Lindblad equation is a first-order linear differential equation, and it
can easily be transformed to make this explicit. By expanding the density
matrix $\rho$ in the operator basis $\\{F_{i}\\}$ one straightforwardly
derives the equivalent nonhomogeneous form

$\dot{\vec{v}}=(Q+R)\vec{v}+\vec{c}\ ,$ (2)

where the “coherence vector” $\vec{v}$ collects the real-valued expansion
coefficients of $\rho$, $Q=-Q^{T}$ is an antisymmetric matrix determined by
$H$, while $R$ and $\vec{c}$ are determined by $a$ [3].

The problem we address in this work is the inverse question:

_When does a general, nonhomogeneous linear first-order differential equation_

$\dot{\vec{v}}=G\vec{v}+\vec{c}$ (3)

_describe a Markovian quantum master equation, and in particular a Lindblad
equation?_

An elementary necessary condition is that $G$ must have eigenvalues whose real
parts are non-positive (otherwise $\lim_{t\to\infty}\|\vec{v}(t)\|$ is
unbounded). Here we go well beyond this observation and give a complete
solution to the problem of inverting both $H$ and $a$ from $G$ and $\vec{c}$.
We also reformulate the condition for complete positivity in terms of $G$ and
$\vec{c}$.

Our results apply in the finite-dimensional setting, and we leave the inverse
problem for the infinite-dimensional or unbounded settings open.

We start by stating our main result – the solution of the inverse problem
along with the complete positivity condition – in Section II. We devote the
rest of this work to providing a general background, a derivation and proof of
the solution, and examples. In more detail, the background is given in Section
III, where we introduce the representation of general superoperators over a
“nice” operator basis such as the Gell-Mann matrices, use it to express the
Markovian quantum master equation over real vector spaces, and derive a
variety of general properties of $Q$, $R$, and $\vec{c}$. In Section IV, we
derive the solution of the inverse problem and provide its proof. In Section
V, we illustrate a few aspects of the general theory in terms of examples.
Given the question we pose in the title of this work, it is natural to ask
about the probability that a randomly selected pair $(G,\vec{c})$ will give
rise to a valid Lindblad equation. We provide a partial answer in Section VI,
where we point out that Lindbladians are extremely rare in the space of random
matrices. In Section VII we discuss the relationship between our work and
previous results. We conclude in Section VIII and provide additional
supporting material in the appendices, including the analytical solution of
Eq. 3 in Appendix A.

##  II Main result

Consider a finite-dimensional Hilbert space $\mathcal{H}$, with
$d=\dim(\mathcal{H})<\infty$, and the Banach space (a complete, normed vector
space) $\mathcal{B}(\mathcal{H})$ of operators acting on $\mathcal{H}$
equipped with the Hilbert-Schmidt inner product $\langle
A,B\rangle\equiv\operatorname{Tr}(A^{\dagger}B)$. Throughout this work, we use
the following definitions.

######  Definition 1.

A “nice operator basis” for $\mathcal{H}$ is a set
$\\{F_{j}\\}_{j=0}^{J}\in\mathcal{B}(\mathcal{H})$, $J\equiv d^{2}-1$, such
that $F_{0}=\frac{1}{\sqrt{d}}I$ ($I$ denotes the identity operator in
$\mathcal{B}(\mathcal{H})$), $F_{j}=F_{j}^{\dagger}$ and
$\operatorname{Tr}F_{j}=0$ for $1\leq j\leq J$, and [15]:

$\langle F_{j},F_{k}\rangle=\operatorname{Tr}(F_{j}F_{k})=\delta_{jk}\ ,\quad
0\leq j\leq J\ .$ (4)

######  Definition 2.

Any equation of the form Eq. 1, where $H=H^{\dagger}$ is a Hermitian operator
in $B(\mathcal{H})$ and $a=a^{\dagger}$ is a Hermitian $J\times J$ matrix, is
a “Markovian quantum master equation”. In this case, the superoperator
$\mathcal{L}$ in Eq. 1 is called a “Liouvillian”.

When, in addition, $a\geq 0$, Eq. 1 is a “Lindblad master equation”, “Lindblad
equation”, or “completely positive Markovian quantum master equation”. In this
case, the superoperator $\mathcal{L}$ is called a “Lindbladian” or “completely
positive Liouvillian”.

The superoperator $\mathcal{L}_{a}$ is called the _dissipator_.

Whenever we refer to Eq. 1 or the superoperator $\mathcal{L}$ without
explicitly specifying complete positivity, we assume the former case, i.e., we
only assume the conditions $H=H^{\dagger}$, $a=a^{\dagger}$.

Our main result is the following:

######  Theorem 1.

For any $J\times J$ matrix $G$ and any vector $\vec{c}$ of length $J$ with
real coefficients there is a pair $(H,a)$ describing a Markovian quantum
master equation Eq. 1 which transforms to Eq. 3 with these $G$ and $\vec{c}$.
If, in addition to the above, we require $H$ to be traceless, then such a pair
$(H,a)$ is unique and can be computed as follows:

$\displaystyle H$
$\displaystyle=\frac{1}{2id}\sum_{m,n=1}^{J}G_{nm}\left[F_{m},F_{n}\right],$
(5a) $\displaystyle a_{mn}$
$\displaystyle=\sum_{i=1}^{J}\operatorname{Tr}\bigg{[}\bigg{(}\sum_{j=1}^{J}G_{ij}F_{j}+c_{i}I\bigg{)}F_{m}F_{i}F_{n}\bigg{]}.$
(5b)

Moreover, the matrix $a=\\{a_{mn}\\}$ is positive semidefinite [i.e., Eq. 1 is
a Lindblad master equation] if and only if (iff) for all traceless
$B\in\mathcal{B}(\mathcal{H})$:

$\sum_{i=1}^{J}\operatorname{Tr}\bigg{[}\bigg{(}\sum_{j=1}^{J}G_{ij}F_{j}+c_{i}I\bigg{)}B^{\dagger}F_{i}B\bigg{]}\geq
0\ .$ (6)

Condition (6) is equivalent to the positive semidefiniteness of $a$, which is
simpler to check. Thus, in practice, it is preferable to first compute $a$
using Eq. 5b and check the sign of its smallest eigenvalue, rather than to
work directly with Eq. 6.

Note that adding a constant, i.e., a term of the form $cI$ with
$c\in\mathbb{R}$, to $H$ does not change the Liouvillian $\mathcal{L}$.
Therefore, if the requirement that $H$ is traceless is not imposed, $H$ can
only be recovered from $G$ up to an additive constant.

##  III Background

Let $M(d,F)$ denote the vector space of $d\times d$ matrices with coefficients
in $F$, where $F\in\\{\mathbb{R},\mathbb{C}\\}$. For our purposes it suffices
to identify $\mathcal{B}(\mathcal{H})$ with $M(d,\mathbb{C})$. Quantum states
are represented by density operators $\rho\in\mathcal{B}_{+}(\mathcal{H})$
(positive trace-class operators acting on $\mathcal{H}$) with unit trace:
$\operatorname{Tr}\rho=1$. Elements of
$\mathcal{B}[\mathcal{B}(\mathcal{H})]$, i.e., linear transformations
$\mathcal{E}:\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H})$, are called
superoperators, or maps.

Complete positivity of a superoperator $\mathcal{E}$ is equivalent to the
statement that $\mathcal{E}$ has a Kraus representation [16]: $\forall
X\in\mathcal{B}(\mathcal{H})$,

$\mathcal{E}(X)=\sum_{i}K_{i}XK_{i}^{\dagger}\ ,$ (7)

where the $\\{K_{i}\\}$ are called Kraus operators. When they satisfy
$\sum_{i}K_{i}^{\dagger}K_{i}=I$, the map $\mathcal{E}$ is trace preserving.

A quantum dynamical semigroup is defined as a one-parameter family of strongly
continuous CP maps, satisfying $\Lambda_{0}=I$ (the identity operator), and
the Markov property $\Lambda_{s}\Lambda_{t}=\Lambda_{s+t}$ for all $s,t\geq
0$.333Strong continuity means that $\forall X\in\mathcal{B}(\mathcal{H})$,
$\|\Lambda_{t}X-X\|\to 0$ as $t\searrow 0$, i.e., the map $\Lambda_{t}$ is
continuous in the strong operator topology.

We can more formally state the GKLS theorem(s) [1, 2] as follows: a
superoperator
$\mathcal{L}:\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H})$ is the
generator of a quantum dynamical semigroup iff it is of the form given in Eq.
1, with $a\geq 0$. I.e., when $\mathcal{L}$ is in the form of Eq. 1, the
solution of $\dot{\rho}=\mathcal{L}\rho$ is $\rho(t)=\Lambda_{t}\rho(0)$,
where $\Lambda_{t}=e^{\mathcal{L}t}$ is an element of a quantum dynamical
semigroup.

###  III.1 General superoperators over a nice operator basis

Given a nice operator basis, let us “coordinatize” the operator
$X\in\mathcal{B}(\mathcal{H})$ in the nice operator basis as

$X=\sum_{j=0}^{J}\boldsymbol{X}_{j}F_{j},$ (8)

where we use the notation $\boldsymbol{X}=\\{\boldsymbol{X}_{j}\\}$ for the
vector of coordinates of the operator $X$ (we interchangeably use the
$\vec{X}$ and $\boldsymbol{X}$ notations to denote a vector). I.e., for any
$X\in\mathcal{B}(\mathcal{H})$:

$\boldsymbol{X}_{i}=\langle F_{i},X\rangle=\operatorname{Tr}(F_{i}X)\ .$ (9)

In particular, applying Eqs. 8 and 9 to $|{k}\rangle\\!\langle l|$ we obtain

$|{k}\rangle\\!\langle l|=\sum_{i=0}^{J}(F_{i})_{lk}F_{i},$ (10)

or

$\sum_{i=0}^{J}F_{i\,lk}F_{i\,k^{\prime}l^{\prime}}=\delta_{ll^{\prime}}\delta_{kk^{\prime}}.$
(11)

######  Proposition 1.

The operator $X\in\mathcal{B}(\mathcal{H})$ is Hermitian iff
$\boldsymbol{X}\in\mathbb{R}^{d^{2}}$ in a nice operator basis.

###### Proof.

If $X$ is Hermitian, then $\boldsymbol{X}_{i}^{*}=\langle
F_{i},X\rangle^{*}=\langle
X,F_{i}\rangle=\operatorname{Tr}(XF_{i})=\operatorname{Tr}(F_{i}X)=\boldsymbol{X}_{i}$.
Vice-versa, if all the coefficients $\boldsymbol{X}_{i}$ are real, then the
operator $X=\sum_{j=0}^{J}\boldsymbol{X}_{j}F_{j}$ is Hermitian because it is
a sum of Hermitian operators $F_{j}$ with real coefficients. ∎

More generally, for Hermitian matrices, the inner product becomes $\langle
A,B\rangle=\operatorname{Tr}(AB)$. Such matrices have $d^{2}=J+1$ real
parameters [$d$ real diagonal elements plus $(d^{2}-d)/2$ independent complex
off-diagonal elements], so they can be represented in terms of vectors in
$\mathbb{R}^{d^{2}}$. Moreover, if $A$ and $B$ are both Hermitian operators
then:

$\langle
A,B\rangle=\sum_{{j,k=0}}^{J}\boldsymbol{A}_{j}\boldsymbol{B}_{k}\operatorname{Tr}(F_{j}F_{k})=\sum_{{j,k=0}}^{J}\boldsymbol{A}_{j}\boldsymbol{B}_{k}\delta_{jk}=\boldsymbol{A}^{t}\boldsymbol{B}\
,$ (12)

so that in these coordinates, the inner product $\langle A,B\rangle$
corresponds to the standard inner product in $\mathbb{R}^{d^{2}}$.

The matrix elements of a superoperator
$\mathcal{E}:\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H})$ are given by

$\boldsymbol{\mathcal{E}}_{ij}=\langle
F_{i},\mathcal{E}(F_{j})\rangle=\operatorname{Tr}[F_{i}\mathcal{E}(F_{j})]\ .$
(13)

The action of a general (not necessarily CP) superoperator
$\mathcal{E}\in\mathcal{B}[\mathcal{B}(\mathcal{H})]$ can always be
represented for any $A\in\mathcal{B}(\mathcal{H})$ as

$\mathcal{E}(A)=\sum_{{i,j=0}}^{J}c_{ij}F_{i}AF_{j},$ (14)

where $c_{ij}\in\mathbb{C}$ (see Appendix D). Thus the matrix $c=\\{c_{ij}\\}$
specifies $\mathcal{E}$ in the given basis $\\{F_{i}\\}$. CP maps are a
special case of Eq. (14), where $c$ is positive semidefinite.

For superoperators we denote the Hilbert-Schmidt adjoint of $\mathcal{E}$ by
$\mathcal{E}^{\dagger}$, which is defined via

$\langle\mathcal{E}^{\dagger}(A),B\rangle\equiv\langle
A,\mathcal{E}(B)\rangle\quad\forall A,B\in\mathcal{B}(\mathcal{H}).$ (15)

######  Definition 3.

$\mathcal{E}$ is _Hermiticity-preserving_ iff:

$\mathcal{E}^{\dagger}\left(A^{\dagger}\right)=\mathcal{E}(A)\ \quad\forall
A\in\mathcal{B}(\mathcal{H}).$ (16)

$\mathcal{E}$ is _Hermitian_ iff

$\mathcal{E}^{\dagger}\left(A\right)=\mathcal{E}(A)\ \quad\forall
A\in\mathcal{B}(\mathcal{H}).$ (17)

Let us find explicit formulas for $[\mathcal{E}(A^{\dagger})]^{\dagger}$ and
$\mathcal{E}^{\dagger}(A)$ in terms of the general representation of a
superoperator given by Eq. 14. First, from Eq. (14) we have

$[\mathcal{E}(A^{\dagger})]^{\dagger}=\left(\sum_{ij}c_{ij}F_{i}A^{\dagger}F_{j}\right)^{\dagger}=\sum_{ij}c^{*}_{ij}F_{j}AF_{i}.$
(18)

Second, substituting Eq. 14 into the complex conjugate of Eq. 15 yields:

$\displaystyle\left<B,\mathcal{E}^{\dagger}(A)\right>=\left<\mathcal{E}(B),A\right>=\operatorname{Tr}\left(\sum_{ij}(c_{ij}F_{i}BF_{j})^{\dagger}A\right)$
(19a)
$\displaystyle=\operatorname{Tr}\left(\sum_{ij}c_{ij}^{*}F_{j}B^{\dagger}F_{i}A\right)=\operatorname{Tr}\left(B^{\dagger}\sum_{ij}c_{ij}^{*}F_{i}AF_{j}\right)$
(19b) $\displaystyle=\langle B,\sum_{ij}c_{ij}^{*}F_{i}AF_{j}\rangle.$ (19c)

Since this equality holds for every $B$, we have:

$\mathcal{E}^{\dagger}(A)=\sum_{ij}c_{ij}^{*}F_{i}AF_{j}.$ (20)

It will turn out to be useful to have another representation of
$\mathcal{E}^{\dagger}(A)$. Consider two sets of arbitrary operators
$\\{L_{p}\\}$ and $\\{M_{q}\\}$, which we can expand in the nice operator
basis as $L_{p}=\sum_{i=0}^{J}l_{ip}F_{i}$ and
$M_{q}=\sum_{i=0}^{J}m_{iq}F_{i}$. In addition, let $b$ be some arbitrary
matrix in $M(d^{2},\mathbb{C})$ and let $c=lbm^{\dagger}$. Then, using Eq. 14:

$\displaystyle\mathcal{E}(A)$
$\displaystyle=\sum_{i,j=0}^{J}[\sum_{p,q}l_{ip}b_{pq}m^{*}_{jq}]F_{i}AF_{j}$
(21a)
$\displaystyle=\sum_{p,q}b_{pq}\left(\sum_{i=0}^{J}l_{ip}F_{i}\right)A\left(\sum_{j=0}^{J}m_{jq}^{*}F_{j}\right)$
(21b) $\displaystyle=\sum_{p,q}b_{pq}L_{p}AM_{q}^{\dagger}\ ,$ (21c)

and, using Eq. 20:

$\displaystyle\mathcal{E}^{\dagger}(A)$
$\displaystyle=\sum_{i,j=0}^{J}[\sum_{p,q}l_{ip}b_{pq}m^{*}_{jq}]^{*}F_{i}AF_{j}$
(22a)
$\displaystyle=\sum_{p,q}b_{pq}^{*}\left(\sum_{i=0}^{J}l_{ip}^{*}F_{i}\right)A\left(\sum_{j=0}^{J}m_{jq}F_{j}\right)$
(22b) $\displaystyle=\sum_{p,q}b_{pq}^{*}L_{p}^{\dagger}AM_{q}\ .$ (22c)

######  Proposition 2.

$\mathcal{E}$ is simultaneously Hermiticity-preserving and Hermitian iff $c$
in Eq. 14 is real-symmetric, i.e., $c_{ij}=c_{ji}=c^{*}_{ij}$ $\forall i,j$.

###### Proof.

This follows immediately by equating the right-hand sides of Eqs. 18 and 20,
both of which are then equal to $\mathcal{E}(A)$ by 3. ∎

######  Proposition 3.

$\mathcal{E}$ is Hermiticity-preserving iff $\boldsymbol{\mathcal{E}}\in
M(d^{2},\mathbb{R})$.

###### Proof.

Using Eq. (13), we obtain:

$\boldsymbol{\mathcal{E}}_{ij}^{*}=\operatorname{Tr}\left(\left[\mathcal{E}\left(F_{j}\right)\right]^{\dagger}F_{i}^{\dagger}\right)=\operatorname{Tr}\left(\mathcal{E}\left(F_{j}\right)F_{i}\right)=\boldsymbol{\mathcal{E}}_{ij}\
.$ (23)

In the other direction, assume $\boldsymbol{\mathcal{E}}\in
M(d^{2},\mathbb{R})$. $\mathcal{E}(A)$ is represented in a nice operator basis
as

$\displaystyle\mathcal{E}(A)$
$\displaystyle=\sum_{{i,j=0}}^{J}\boldsymbol{\mathcal{E}}_{ij}\boldsymbol{A}_{j}F_{i}\
,$ (24)

so that

$\displaystyle[\mathcal{E}(A)]^{\dagger}$
$\displaystyle=\left(\,\,\sum_{{i,j=0}}^{J}\boldsymbol{\mathcal{E}}_{ij}\boldsymbol{A}_{j}F_{i}\right)^{\dagger}=\sum_{{i,j=0}}^{J}\boldsymbol{\mathcal{E}}_{ij}^{*}\boldsymbol{A}_{j}^{*}F_{i}^{\dagger}$
(25a)
$\displaystyle=\sum_{{i,j=0}}^{J}\boldsymbol{\mathcal{E}}_{ij}\boldsymbol{A}_{j}^{*}F_{i}=\mathcal{E}(A^{\dagger})\
.$ (25b)

∎

Thus, after coordinatization, a Hermiticity-preserving superoperator
$\mathcal{E}$ can be seen as a real-valued $d^{2}\times d^{2}$-dimensional
matrix $\boldsymbol{\mathcal{E}}$.

###  III.2 General properties of a Liouvillian

Since $a$ in Eq. 1 is Hermitian, it can be written as
$a=u^{{\dagger}}\tilde{\gamma}u$ where $u$ is unitary and $\tilde{\gamma}$ is
diagonal with the eigenvalues $\\{\gamma_{\alpha}\\}_{\alpha=1}^{J}$ of $a$ on
its diagonal. The $\gamma_{\alpha}$’s are always real when Eq. 1 is a
Markovian quantum master equation. When $a\geq 0$, i.e., when Eq. 1 is a
Lindblad equation, they are non-negative. Defining

$L_{\alpha}=\sum_{j=1}^{J}u^{*}_{\alpha j}F_{j}\ ,$ (26)

we have $F_{i}=\sum_{\alpha=1}^{J}u_{\alpha i}L_{\alpha}$. Then:

$\displaystyle\sum_{{i,j=1}}^{J}a_{ij}F_{i}\rho F_{j}^{\dagger}$
$\displaystyle=\sum_{{\alpha,\beta=1}}^{J}L_{\alpha}\rho
L_{\beta}^{\dagger}\sum_{{i,j=1}}^{J}u_{\alpha i}a_{ij}u^{\dagger}_{\beta l}$
$\displaystyle=\sum_{{\alpha=1}}^{J}{\gamma}_{\alpha}L_{\alpha}\rho
L_{\alpha}^{\dagger}\ ,$ (27a)
$\displaystyle\sum_{{i,j=1}}^{J}a_{ij}F_{j}^{\dagger}F_{i}$
$\displaystyle=\sum_{{\alpha,\beta=1}}^{J}L_{\beta}^{\dagger}L_{\alpha}\sum_{{i,j=1}}^{J}u_{\alpha
i}a_{ij}u^{\dagger}_{\beta l}$
$\displaystyle=\sum_{{\alpha=1}}^{J}{\gamma}_{\alpha}L_{\alpha}^{\dagger}L_{\alpha}\
.$ (27b)

Therefore the Markovian quantum master equation [Eq. 1] becomes:

$\displaystyle\dot{\rho}={\cal L}\rho$
$\displaystyle=\mathcal{L}_{H}\rho+\mathcal{L}_{a}\rho$ (28a)
$\displaystyle\mathcal{L}_{H}[\cdot]$ $\displaystyle=-i[H,\cdot]\ ,$ (28b)
$\displaystyle\mathcal{L}_{a}[\cdot]$
$\displaystyle=\sum_{\alpha=1}^{J}\gamma_{\alpha}\left(L_{\alpha}\cdot
L^{{\dagger}}_{\alpha}-\frac{1}{2}\\{L^{{\dagger}}_{\alpha}L_{\alpha},\cdot\\}\right)\
.$ (28c)

The $L_{\alpha}\in\mathcal{B}(\mathcal{H})$ are called _Lindblad operators_.

We remark that the representation of the dissipator $\mathcal{L}_{a}$ in the
form (28c) need not be unique: for example, if $a=I$ (the identity matrix), it
can be written as $u^{\dagger}Iu$ for any unitary $u$, but [as can be seen
from Eq. 26] different choices of $u$ lead to different Lindblad operators.
More generally, $\tilde{\gamma}$ is unique only up to permutations, and any
permutation redefines the Lindblad operators by modifying the diagonalizing
unitary matrix $u$.

######  Proposition 4.

The Liouvillian is Hermiticity-preserving.

###### Proof.

To show this we can directly compare $[\mathcal{L}(X)]^{\dagger}$ with
$\mathcal{L}(X^{\dagger})$ using Eqs. 28b and 28c. For the Hamiltonian part,
we have the following:

$\displaystyle[\mathcal{L}_{H}(X)]^{{\dagger}}$ $\displaystyle=i(HX-
XH)^{{\dagger}}=i(X^{{\dagger}}H^{{\dagger}}-H^{{\dagger}}X^{{\dagger}})$
(29a)
$\displaystyle=-i(HX^{{\dagger}}-X^{{\dagger}}H)=\mathcal{L}_{H}(X^{{\dagger}})\
,$ (29b)

while for the dissipative part,

$\displaystyle[\mathcal{L}_{a}(X)]^{{\dagger}}$
$\displaystyle=\sum_{\alpha=1}^{J}\gamma_{\alpha}\left((L_{\alpha}XL_{\alpha}^{{\dagger}})^{{\dagger}}-\frac{1}{2}\\{L_{\alpha}^{{\dagger}}L_{\alpha},X\\}^{\dagger}\right)$
(30a)
$\displaystyle=\sum_{\alpha=1}^{J}\gamma_{\alpha}\left((L_{\alpha}X^{{\dagger}}L_{\alpha}^{{\dagger}})-\frac{1}{2}\\{L_{\alpha}^{{\dagger}}L_{\alpha},X^{{\dagger}}\\}\right)$
(30b) $\displaystyle=\mathcal{L}_{a}(X^{{\dagger}})\,.$ (30c)

Thus,

$\mathcal{L}\left(X^{\dagger}\right)=[\mathcal{L}\left(X\right)]^{\dagger}\ .$
(31)

∎

######  Corollary 1.

The Markovian quantum master equation $\dot{\rho}=\mathcal{L}\rho$ in a nice
operator basis is a real-valued linear ordinary differential equation (ODE)
for the vector $\boldsymbol{\rho}$ whose coordinates are
$\boldsymbol{\rho}_{j}=\operatorname{Tr}(\rho F_{j})$:

$\dot{\boldsymbol{\rho}}=\boldsymbol{\mathcal{L}}\boldsymbol{\rho}\
,\quad\boldsymbol{\rho}\in\mathbb{R}^{d^{2}}\
,\quad\boldsymbol{\mathcal{L}}\in M(d^{2},\mathbb{R})\ .$ (32)

###### Proof.

This follows directly from Propositions 1, 3 and 4 with $X=\rho$ and
$\mathcal{E}=\mathcal{L}$. ∎

######  Proposition 5.

The Liouvillian maps any operator $X$ to a traceless operator. That is, for
any operator $X$

$\operatorname{Tr}[\mathcal{L}(X)]=0.$ (33)

###### Proof.

From Eq. 1 we have

$\displaystyle\operatorname{Tr}[\mathcal{L}_{H}(X)]$
$\displaystyle=-i\operatorname{Tr}([H,X]){}$
$\displaystyle=-i[\operatorname{Tr}(HX)-\operatorname{Tr}(XH)]=0,$ (34)

and, using $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$:

$\operatorname{Tr}(\mathcal{L}_{a}(X))=\sum_{i,j=1}^{J}a_{ij}\operatorname{Tr}\left(F_{i}XF_{j}-\frac{1}{2}\left\\{F_{j}F_{i},X\right\\}\right)\\\
=\sum_{i,j=1}^{J}a_{ij}\left(F_{j}F_{i}X-\frac{1}{2}F_{j}F_{i}X-\frac{1}{2}F_{j}F_{i}X\right)=0.$
(35)

∎

######  Proposition 6.

Suppose the dissipator $\mathcal{L}_{a}$ is of the form (1c) with a Hermitian
matrix $a$. Then the following conditions are equivalent.

  1. 1.

$\mathcal{L}_{a}$ is Hermitian.

  2. 2.

$\mathcal{L}_{a}$ can be written in the form (28c) s.t.
$\forall\alpha\;L_{\alpha}=L_{\alpha}^{\dagger}$.

  3. 3.

$a$ is symmetric, i.e., $a=a^{T}$.

  4. 4.

All the matrix elements $a_{ij}$ are real.

###### Proof.

$2\implies 1$. If all the Lindblad operators $L_{\alpha}$ are Hermitian then
$\mathcal{L}_{a}^{\dagger}=\mathcal{L}_{a}$. Indeed, from Eqs. 28c and 22c,
for any operator $A$ we have

$\displaystyle\mathcal{L}_{a}^{\dagger}(A)$
$\displaystyle=\sum_{\alpha=1}^{J}\gamma_{\alpha}\left(L_{\alpha}^{{\dagger}}XL_{\alpha}-\frac{1}{2}\\{L^{{\dagger}}_{\alpha}L_{\alpha},X\\}\right)=\mathcal{L}_{a}(A).$
(36)

$3\Leftrightarrow 4$. Conditions 3 and 4 mean that for each $i,j$ we have
$a_{ji}=a_{ij}$ and $a_{ji}=a_{ji}^{*}$ respectively. Since $a=a^{\dagger}$,
the right-hand sides of these two equations are equal for all $i,j$.

$3,4\implies 2$. We know that $a$ is real-symmetric; hence it can be
diagonalized by an orthogonal matrix. Therefore, $u$ in Eq. 26 can be chosen
to have real matrix elements. For that choice, we have
$\forall\alpha\;L_{\alpha}=L_{\alpha}^{\dagger}$.

$1\implies 4$. According to Eq. 22c,
$\mathcal{L}_{a}^{\dagger}=\mathcal{L}_{a^{*}}$. From
$\mathcal{L}_{a}=\mathcal{L}_{a}^{\dagger}$ we have

$\mathcal{L}_{a}=(\mathcal{L}_{a}+\mathcal{L}_{a^{*}})/2=\mathcal{L}_{\tilde{a}},$
(37)

where $\tilde{a}=(a+a^{*})/2$ — a matrix with real matrix elements. Later in 2
we will show that $a$ is uniquely determined by $\mathcal{L}_{a}$, hence
$a=\tilde{a}$ and $a=a^{*}$.

$2\implies 3$. If $L_{\alpha}=L_{\alpha}^{\dagger}$ then it can be expanded in
the given basis $\\{F_{j}\\}$ with real coefficients:
$L_{\alpha}=\sum_{j=1}^{J}w_{\alpha j}F_{j}$, with $w_{\alpha
j}\in\mathbb{R}$. Substituting this into Eq. 28c we obtain Eq. 1c with
$\tilde{a}=w^{T}\tilde{\gamma}w$ in place of $a$ and
$\tilde{\gamma}=\text{diag}(\gamma_{1},\dots,\gamma_{J})$. Thus
$\tilde{a}=\tilde{a}^{T}$. Later in 2 (see also Propositions 18 and 2) we will
show that $a$ is uniquely determined by $\mathcal{L}_{a}$; hence,
$a=\tilde{a}$ and $a=a^{T}$. ∎

By definition, the generator $\mathcal{L}_{a}$ is unital if
$\mathcal{L}_{a}(I)=0$ [17] [a CPTP map is unital if it maps $I$ to itself:
$\mathcal{E}(I)=I$]. Another general property of the Liouvillian is the
following:

######  Proposition 7.

The Liouvillian is unital if any of the conditions in Proposition 6 are
satisfied.

###### Proof.

It suffices to assume that $a$ is symmetric; then using Eq. 1c:

$\displaystyle\mathcal{L}_{a}(I)$
$\displaystyle=\sum_{i,j=1}^{J}a_{ij}(F_{i}F_{j}-F_{j}F_{i})$ (38a)
$\displaystyle=\sum_{i,j=1}^{J}a_{ij}F_{i}F_{j}-\sum_{i,j=1}^{J}a_{ji}F_{j}F_{i}=0\
.$ (38b)

∎

Note that the converse is false: unitality does not imply that $a$ must be
symmetric (or any of the other conditions in Proposition 6). As a
counterexample, consider any nice operator basis containing two or more
commuting operators [e.g., the last two matrices in Eq. 170 below]. Then, if
$a$ contains a non-zero matrix element only between these two operators [e.g.,
$a_{78}=1$, all the rest zero, for the nice operator basis in Eq. 170], Eq.
38a gives $\mathcal{L}_{a}(I)=0$, but $a$ is not symmetric.

###  III.3 Markovian quantum master equation for the coherence vector: from
$H$ and $a$ to $G=Q+R$ and $\vec{c}$

By 1, we can expand the density matrix in the nice operator basis as:

$\rho=\vec{F}^{\prime}\cdot\boldsymbol{\rho}=\frac{1}{d}I+\vec{F}\cdot\vec{v}\
,$ (39)

where $\vec{F}^{\prime}=(F_{0},F_{1},\dots,F_{J})$,
$\boldsymbol{\rho}=(1/\sqrt{d},v_{1},\dots,v_{J})^{T}$,
$\vec{F}=(F_{1},\dots,F_{J})$ collects the traceless basis-operators into a
vector, and the corresponding coordinate vector
$\vec{v}=(v_{1},\dots,v_{J})^{T}\in\mathbb{R}^{J}$ is called the _coherence
vector_. We have:

$v_{j}=\operatorname{Tr}(\rho F_{j})=\boldsymbol{\rho}_{j}\ .$ (40)

We note that the conditions guaranteeing that a given vector represents a
valid (i.e., non-negative) density matrix are nontrivial. A simple necessary
condition follows from the condition that the purity
$\operatorname{Tr}(\rho^{2})\leq 1$:

$\displaystyle\operatorname{Tr}(\rho^{2})$
$\displaystyle=\operatorname{Tr}\left[\left(\frac{1}{d}I+\vec{F}\cdot\vec{v}\right)^{2}\right]$
(41a)
$\displaystyle=\frac{1}{d}+\sum_{i,j=1}^{J}\operatorname{Tr}(F_{i}F_{j})v_{i}v_{j}=\frac{1}{d}+\left\|\vec{v}\right\|^{2}\
,$ (41b)

i.e.,

$\|\vec{v}\|\leq\left(1-\frac{1}{d}\right)^{1/2}\ ,$ (42)

which is saturated for pure states. Consequently, as mentioned in Section I,
$G$’s eigenvalues are constrained to have non-positive real parts. This is a
necessary condition any candidate Eq. 3 must satisfy to represent a Lindblad
equation. Additional inequalities have been derived in Ref. [18].

######  Proposition 8.

The coherence vector satisfies Eq. 3. Moreover, the decomposition of
$\mathcal{L}$ as $\mathcal{L}=\mathcal{L}_{H}+\mathcal{L}_{a}$ with
$\mathcal{L}_{H}$ and $\mathcal{L}_{a}$ given by Eqs. (1b) and (1c) induces
the decomposition of $G$ in Eq. 3 into $G=Q+R$, where
$\mathcal{L}_{H}[\rho]\leadsto Q\vec{v}$, $\mathcal{L}_{a}[\rho]\leadsto
R\vec{v}+\vec{c}$, $\vec{c}\in\mathbb{R}^{J}$, $R,Q\in M(J,\mathbb{R})$, and
$Q=-Q^{T}$ (antisymmetric).

###### Proof.

After separating out $v_{0}=\boldsymbol{\rho}_{0}=1/\sqrt{d}$, Eq. (32)
becomes

$\dot{v}_{i}=\sum_{j=1}^{J}\boldsymbol{\mathcal{L}}_{ij}v_{j}+\frac{1}{\sqrt{d}}\boldsymbol{\mathcal{L}}_{i0}\
,\quad 1\leq i\leq J\ .$ (43)

This already establishes that $\vec{v}$ satisfies Eq. 2, with

$\displaystyle Q_{ij}+R_{ij}$
$\displaystyle\equiv\boldsymbol{\mathcal{L}}_{ij}=\operatorname{Tr}[F_{i}\mathcal{L}(F_{j})]\equiv
G_{ij}$ (44a) $\displaystyle c_{j}$
$\displaystyle\equiv\frac{1}{\sqrt{d}}\boldsymbol{\mathcal{L}}_{j0}=\frac{1}{\sqrt{d}}(Q_{j0}+R_{j0})\
,$ (44b)

for $1\leq i,j\leq J$.

To prove the remaining claims, recall that
$\boldsymbol{\mathcal{L}}_{ij}=\boldsymbol{\mathcal{L}}_{ij}^{*}$ by 1.
Therefore, by linearity, for $1\leq i,j\leq J$:

$\displaystyle Q_{ij}$
$\displaystyle\equiv\operatorname{Tr}[F_{i}\mathcal{L}_{H}(F_{j})]=Q_{ij}^{*}$
(45a) $\displaystyle R_{ij}$
$\displaystyle\equiv\operatorname{Tr}[F_{i}\mathcal{L}_{a}(F_{j})]=R_{ij}^{*}\
,$ (45b)

i.e., $R,Q\in M(J,\mathbb{R})$. In addition,

$\displaystyle Q_{ji}$
$\displaystyle=\operatorname{Tr}[F_{j}\mathcal{L}_{H}(F_{i})]=-i\operatorname{Tr}\left(F_{j}[H,F_{i}]\right)$
(46a)
$\displaystyle=i\operatorname{Tr}\left(F_{i}[H,F_{j}]\right)=-\operatorname{Tr}[F_{i}\mathcal{L}_{H}(F_{j})]=-Q_{ij}\
,$ (46b)

i.e., $Q$ is antisymmetric. Moreover, by expanding $H$ in the nice operator
basis as

$H=\sum_{m=0}^{J}h_{m}F_{m}\ ,$ (47)

we can write $Q$’s matrix elements as:

$\displaystyle Q_{ij}$
$\displaystyle=-i\operatorname{Tr}\left(F_{i}[H,F_{j}]\right)$ (48a)
$\displaystyle=-i\sum_{m=1}^{J}h_{m}\operatorname{Tr}\left(F_{i}[F_{m},F_{j}]\right)\
.$ (48b)

As for $\vec{c}$, since $F_{0}=I/\sqrt{d}$, we have

$Q_{j0}=\frac{-i}{\sqrt{d}}\operatorname{Tr}\left(F_{j}[H,I]\right)=0,$ (49)

so that for $1\leq j\leq J$, Eq. 44b is replaced by:

$\displaystyle
c_{j}=\frac{1}{d}\operatorname{Tr}[F_{j}\mathcal{L}_{a}(I)]=\frac{1}{\sqrt{d}}R_{j0}=c_{j}^{*}\
,$ (50)

i.e., $\vec{c}\in\mathbb{R}^{J}$. ∎

Let $\mathrm{spec}(A)$ denote the spectrum of the operator $A$, i.e., the set
of its eigenvalues.

######  Proposition 9.

$\mathrm{spec}(\mathcal{L})=\\{0\\}\cup\mathrm{spec}(G).$ (51)

###### Proof.

Note that, as follows from proposition 5,

$\boldsymbol{\mathcal{L}}_{0j}=0\quad\text{i.e.},\quad\boldsymbol{\mathcal{L}}_{0}=\vec{0}\
.$ (52)

Combining this with Eq. 45, we see that in the basis $\\{F_{j}\\}_{j=0}^{J}$:

$\boldsymbol{\mathcal{L}}=\left(\begin{array}[]{cc}0&0\cdots 0\\\
\sqrt{d}\vec{c}&G\end{array}\right)\,.$ (53)

Computing the spectrum of the superoperator $\mathcal{L}$ is equivalent to
finding the eigenvalues of $\boldsymbol{\mathcal{L}}$, which are the solutions
of the characteristic equation

$\det\left(\lambda
I_{J+1}-\boldsymbol{\mathcal{L}}\right)=\lambda\det\left(\lambda
I_{J}-G\right)=0,$ (54)

where $I_{n}$ denotes the $n\times n$ identity matrix. These solutions are
$\lambda=0$ and $\mathrm{spec}(G)$. ∎

###  III.4 When does $\vec{c}$ vanish?

######  Proposition 10.

$\vec{c}=\vec{0}$ iff $\mathcal{L}_{a}$ is unital.

###### Proof.

The proof is immediate from Eq. 53, which shows that $\vec{c}$ is the vector
representing the matrix $\mathcal{L}_{a}(I)/d$. Thus $\vec{c}=\vec{0}$ iff
$\mathcal{L}_{a}(I)=0$.

A more explicit proof is the following calculation. First, if
$\mathcal{L}_{a}$ is unital then, using Eqs. (45b) and (50), we have:

$c_{k}=\frac{1}{d}\operatorname{Tr}[F_{k}\mathcal{L}_{a}(I)]=0\quad(\mathcal{L}_{a}\text{
unital})\ .$ (55)

On the other hand, if $\vec{c}=0$,

$\displaystyle\operatorname{Tr}[F_{k}\mathcal{L}_{a}(I)]=dc_{k}=0,$ (56)

hence $\mathcal{L}_{a}(I)=0$. ∎

Note that using Eq. 1c we can write down the general formula for $\vec{c}$’s
elements in a given nice operator basis:

$c_{k}=\frac{1}{d}\sum_{i,j=1}^{J}a_{ij}\operatorname{Tr}(\big{[}F_{i},F_{j}\big{]}F_{k})\,.$
(57)

###  III.5 When is $R$ symmetric?

Here by $R$ we mean the $J\times J$ matrix with elements
$\\{R_{kl}\\}_{k,l=1}^{J}$. Using Eq. 45b:

$\displaystyle R_{kl}$
$\displaystyle=\operatorname{Tr}\big{[}F_{k}\mathcal{L}_{a}(F_{l})\big{]}$
(58a)
$\displaystyle=\sum_{{i,j=1}}^{J}a_{ij}\operatorname{Tr}\big{[}F_{k}\big{(}F_{i}F_{l}F_{j}-\frac{1}{2}\left\\{F_{j}F_{i},F_{l}\right\\}\big{)}\big{]}\
.$ (58b)

Note that this is the general formula for $R$’s elements in a given nice
operator basis. $R$ is not symmetric in general. However, we have two special
cases presented in the following Propositions.

######  Proposition 11.

$R$ is symmetric in the single qubit ($d=2$) case.

###### Proof.

By direct calculation, it follows that if we choose the traceless elements of
the nice operator basis as the normalized Pauli matrices
$\\{\sigma^{x},\sigma^{y},\sigma^{z}\\}/\sqrt{2}$, then $R_{kl}=R_{lk}$ for
arbitrary $a$.

An alternative way to see this is from 2 below: the dimension of the space of
possible antisymmetric components of $R$ is $J(J-3)/2$, which is $0$ iff
$J\in\\{0,3\\}$, i.e., when $d\in\\{1,2\\}$. ∎

######  Proposition 12.

The following statements are equivalent:

  1. 1.

$R$ is symmetric and $\vec{c}=\vec{0}$;

  2. 2.

Any of the conditions in Proposition 6 are satisfied.

###### Proof.

If $\mathcal{L}_{a}^{{\dagger}}=\mathcal{L}_{a}$ (one of the four equivalent
conditions in Proposition 6) then, using Eq. 45b:

$\displaystyle R_{lk}$ $\displaystyle=\langle
F_{l},\mathcal{L}_{a}(F_{k})\rangle=\langle\mathcal{L}_{a}(F_{l}),F_{k}\rangle$
(59a) $\displaystyle=\langle
F_{k},\mathcal{L}_{a}(F_{l})\rangle^{*}=R_{kl}^{*}=R_{kl}\ ,$ (59b)

where we also used that the matrix elements of $R$ are real (Proposition 8).
This proves that 2 $\implies$ 1\. Similarly,

$\displaystyle c_{k}$
$\displaystyle=\frac{1}{d}\left<F_{k},\mathcal{L}_{a}(I)\right>=\frac{1}{d}\left<\mathcal{L}_{a}(F_{k}),I\right>$
(60a)
$\displaystyle=\frac{1}{d}\operatorname{Tr}[\mathcal{L}_{a}(F_{k})]^{*}=0,$
(60b)

where in the last equality we used Proposition 5.

The proof that 1 $\implies$ 2 is almost identical:

$\displaystyle\langle
F_{l},\mathcal{L}_{a}(F_{k})\rangle=R_{lk}=R_{kl}=R_{kl}^{*}$
$\displaystyle=\langle F_{k},\mathcal{L}_{a}(F_{l})\rangle^{*}$ (61a)
$\displaystyle=\langle\mathcal{L}_{a}(F_{l}),F_{k}\rangle\ ,$ (61b)

and

$\left<F_{k},\mathcal{L}_{a}(F_{0})\right>=\sqrt{d}c_{k}=0=\left<\mathcal{L}_{a}(F_{k}),F_{0}\right>,$
(62)

which means $\mathcal{L}_{a}^{{\dagger}}=\mathcal{L}_{a}$. ∎

In particular, $a=a^{T}$ implies that $R$ is symmetric. Note that, consistent
with the comment following Proposition 7, the converse is false, i.e., $R$
being symmetric does not imply that $a$ is real-symmetric: any $a$ yielding a
symmetric $R$ and non-zero $\vec{c}$ is a counterexample. As we describe in 2,
from any such pair $(R,c)$ one could recover $a$ via Eq. 5b, which would be a
counterexample. For a specific counterexample, see Section V.1.2.

According to 2 below, the dimension of the space of possible antisymmetric
components of $R$ is $J(J-3)/2$, which is non-zero if and only if $d\geq 3$.
Thus, for $d\geq 3$ the matrix $R$ is not always symmetric. An explicit
example of this for $d=3$ is the following. Consider a qutrit subject to
amplitude damping (spontaneous emission) involving just two of the three
levels:

$\displaystyle a=\left(\begin{array}[]{ccc}1&-i&\\\ i&1&\\\ &&0_{6\times
6}\end{array}\right)\ ,$ (66)

which is a non-negative matrix with eigenvalues $0$ ($7$-fold degenerate) and
$2$. Note that in all the examples given in this work, we use a specific
choice of a nice operator basis $\\{F_{i}\\}$: the generalized Gell-Mann
matrices normalized to satisfy the normalization condition Eq. (4); see
Section III.7 and Ref. [19]. By Eq. 58 we have

$\displaystyle
R=\left(\begin{array}[]{cccccccc}-\frac{1}{4}&0&0&0&-\frac{1}{4}&0&0&0\\\
0&-\frac{1}{4}&0&\frac{1}{4}&0&0&0&0\\\ 0&0&-\frac{1}{2}&0&0&0&0&0\\\
0&-\frac{3}{4}&0&-\frac{5}{4}&0&0&0&0\\\
\frac{3}{4}&0&0&0&-\frac{5}{4}&0&0&0\\\
0&0&0&0&0&-\frac{1}{2}&\frac{1}{4}&\frac{1}{4\sqrt{3}}\\\
0&0&0&0&0&-\frac{3}{4}&-\frac{5}{4}&-\frac{\sqrt{3}}{4}\\\
0&0&0&0&0&-\frac{\sqrt{3}}{4}&-\frac{\sqrt{3}}{4}&-\frac{3}{4}\\\
\end{array}\right)$ (75)

which is not symmetric. We revisit this example in Section V.2.1.

###  III.6 Properties of the linear map $a\mapsto(R,\vec{c})$

Consider a map $\mathcal{F}:a\mapsto(R,\vec{c})$ defined by Eqs. 58 and 57.
This is a linear map of real vector spaces
$\mathcal{F}:M_{\textrm{sa}}(J,\mathbb{C})\to
M(J,\mathbb{R})\oplus\mathbb{R}^{J}$, where $M_{\textrm{sa}}(J,\mathbb{C})$ is
the $\mathbb{R}$-vector space of Hermitian (or self-adjoint; hence the “sa”
subscript) $J\times J$ matrices over $\mathbb{C}$.

Then, as will follow later from 2 and 14, this map is injective, i.e., $a\neq
a^{\prime}\implies\mathcal{F}(a)\neq\mathcal{F}(a^{\prime})$. In other words,
if the pairs $(R,\vec{c})$ and $(R^{\prime},\vec{c}^{\prime})$ are equal, then
also the corresponding rate matrices $a$ and $a^{\prime}$ are equal. A direct
way to prove the injectivity of $\mathcal{F}$, which does not rely on 2 (but
in part uses similar ideas), is presented in Appendix B.

The following Lemma, together with 2 below, describes the image of
$\mathcal{F}$.

######  Lemma 1.

For all symmetric $R$, $R_{\text{sym}}$:
$(R_{\text{sym}},\vec{0})\in\text{Im}(\mathcal{F})$. In particular,
$(R_{\text{sym}},\vec{0})$ is the image of some real-valued symmetric $a$.

###### Proof.

The subspace of real-valued symmetric $a$ in $M(J,\mathbb{R})$ is of dimension
$J(J+1)/2$. We know by Proposition 12 that $\mathcal{F}$ maps real-symmetric
$a$ to $\vec{c}=\vec{0}$ and $R$ real-symmetric. The subspace
$V=\\{(R_{\text{sym}},\vec{0})\\}$ has dimension $J(J+1)/2$ in the codomain of
$\mathcal{F}$. Since, as explained above, $\mathcal{F}$ is injective, the
image of all real-symmetric $a$ is the subspace $V$. ∎

###  III.7 Properties of a nice operator basis

In this subsection, we define structure constants corresponding to a nice
operator basis and provide alternative forms of Eqs. 57 and 48b using these
structure constants. We note that the structure constants are often explicitly
defined for the case of (normalized and generalized) Gell-Mann matrices [20].
As mentioned above, this is our choice in all the examples in this work.
However, the theory presented here applies to any choice of a nice operator
basis $\\{F_{i}\\}_{i=0}^{J}$ (see 1).

For any such choice, the elements $\\{F_{i}\\}_{i=1}^{J}$ form a generator set
of the Lie algebra $\mathfrak{su}(d)$.444Not every generator set of a Lie
algebra is a nice operator basis. For example, the set
$\\{\sigma^{x},\sigma^{y},\sigma^{z}\\}$ is a generator set of
$\mathfrak{su}(2)$ but it violates the normalization conditions Eq. 4; indeed,
with this choice we have $\left<F_{i},F_{j}\right>=2\delta_{ij}$. We can
define structure constants $f_{ijk}$ via

$[F_{i},F_{j}]=i\sum_{k=1}^{J}f_{ijk}F_{k}=if_{ijk}F_{k}\ ,$ (76)

where in the second equality we used the Einstein summation convention of
summing over repeated indices, which we use henceforth when convenient.

The structure constants are totally antisymmetric, i.e.,
$f_{jkl}=-f_{kjl}=f_{klj}$ and themselves satisfy a type of orthogonality
relation [21, 22, 23]:

$f_{jkl}f_{jkm}=2d\delta_{lm}\ .$ (77)

Using Eqs. 4 and 76 we can then further simplify Eq. 57:

$\displaystyle c_{k}=\frac{1}{d}ia_{ij}f_{ijk}\ .$ (78)

We may also derive an explicit expression relating the coordinates $h_{m}$ of
the Hamiltonian in the expansion (47) to the matrix elements of $Q$. Namely,
inserting Eq. 76 into Eq. 48, we have:

$\displaystyle Q_{jk}$
$\displaystyle=-ih_{l}\operatorname{Tr}(F_{j}[F_{l},F_{k}])=-ih_{l}if_{lkj}=-f_{jkl}h_{l}\
.$ (79)

On the other hand, using

$f_{jkm}Q_{jk}=-f_{jkl}f_{jkm}h_{l}=-2d\delta_{lm}h_{l}=-2dh_{m}\ ,$ (80)

we have

$h_{m}=-\frac{1}{2d}f_{jkm}Q_{jk}\ .$ (81)

This implies that $H$ can be computed given $Q$ as

$H=-\frac{1}{2d}f_{jkm}Q_{jk}F_{m}\ .$ (82)

We discuss the consistency between this expression and Eq. 5a in Proposition
15.

##  IV Solution of the inverse problem

We now set up the necessary mathematical framework to solve the inverse
problem in full generality. Since some of the results after Proposition 5 used
2 and its corollaries, in this section we only use the results up to (and
including) Proposition 5, to avoid any circular references.

###  IV.1 Forward and inverse transformations

For a field $F\in\\{\mathbb{R},\mathbb{C}\\}$ let $M_{0}(d,F)$ denote the
subspace of traceless matrices. Let $M_{\textrm{sa}}(d,F)$ denote the
$\mathbb{R}$-subspace of Hermitian matrices. E.g.,
$M_{\textrm{sa}}(d,\mathbb{R})$ is a vector space of real-valued $d\times d$
symmetric matrices. Finally, let $M_{0,\textrm{sa}}(d,F)$ be the
$\mathbb{R}$-subspace of matrices which are both traceless and Hermitian.

######  Theorem 2.

For any nice operator basis $\\{F_{n}\\}_{n=1}^{J}$ there is a linear
bijective correspondence between the following objects.

  1. 1.

Pairs $(H,a)$ where $H\in M_{0,\textrm{sa}}(d,\mathbb{C})$, $a\in
M_{\textrm{sa}}(J,\mathbb{C})$.

  2. 2.

$d\times d\times d\times d$ tensors $x=\\{x_{ijkl}\\}_{i,j,k,l=1..d}$ over
$\mathbb{C}$ satisfying

$x_{ijkl}=x_{lkji}^{*},\quad\sum_{k=1}^{d}(x_{ijkk}+x_{kkij})=0\ .$ (83)

  3. 3.

$d\times d\times d\times d$ tensors
$\tilde{x}=\\{\tilde{x}_{ijkl}\\}_{i,j,k,l=1..d}$ over $\mathbb{C}$ satisfying

$\tilde{x}_{ijkl}=\tilde{x}_{lkji}^{*}\
,\quad\sum_{i=1}^{d}\tilde{x}_{ijki}=0\ .$ (84)

  4. 4.

Linear Hermiticity-preserving superoperators $\mathcal{L}$ acting on
$M(d,\mathbb{C})$ satisfying

$\mathcal{L}\left(X^{\dagger}\right)=[\mathcal{L}(X)]^{\dagger}\
,\quad\operatorname{Tr}[\mathcal{L}(X)]=0$ (85)

for each $X\in M(d,\mathbb{C})$.

  5. 5.

Linear superoperators

$\mathcal{L}:M_{sa}(d,\mathbb{C})\to M_{0,sa}(d,\mathbb{C})\ .$ (86)

  6. 6.

Pairs $(G,\vec{c})$ where $G\in M(J,\mathbb{R})$, $c\in\mathbb{R}^{J}$.

The corresponding transformations are given as follows:

  * •

$1\to 2$:

$x_{ijkl}=-iH_{ij}\delta_{kl}+i\delta_{ij}H_{kl}+\sum_{m,n=1}^{J}a_{mn}(F_{m})_{ij}(F_{n})_{kl}\
.$ (87)

  * •

$2\to 1$:

$\displaystyle H_{ij}$
$\displaystyle=\frac{1}{2id}\sum_{k=1}^{d}(x_{kkij}-x_{ijkk})\ ,$ (88a)
$\displaystyle a_{mn}$
$\displaystyle=\sum_{i,j,k,l=1}^{d}(F_{m})_{ji}x_{ijkl}(F_{n})_{lk}\ .$ (88b)

  * •

$2\to 4,5$:

$[\mathcal{L}(X)]_{il}=\sum_{j,k=1}^{d}(x_{ijkl}X_{jk}-\frac{1}{2}x_{jkij}X_{kl}-\frac{1}{2}X_{ij}x_{kljk})\
.$ (89)

  * •

$4\to 5$ is simply the restriction to Hermitian operators $X$.

  * •

$5\to 4$:

$\mathcal{L}(X)=\mathcal{L}[\operatorname{Re}(X)]+i\mathcal{L}[\operatorname{Im}(X)]\
,$ (90)

where

$\operatorname{Re}(X)=\frac{X+X^{\dagger}}{2}\
,\qquad\operatorname{Im}(X)=\operatorname{Re}(X/i)=\frac{X-X^{\dagger}}{2i}\
.$ (91)

  * •

$1\to 4$ is given by Eqs. (1a)-(1c).

  * •

$5\to 6$:

$\displaystyle G_{nm}$
$\displaystyle=\operatorname{Tr}[F_{n}\mathcal{L}(F_{m})]\ ,$ (92a)
$\displaystyle c_{n}$
$\displaystyle=\frac{1}{\sqrt{d}}\operatorname{Tr}[F_{n}\mathcal{L}(F_{0})]\
.$ (92b)

  * •

$6\to 4,5$:

$\mathcal{L}(X)=\sum_{n=1}^{J}\left(\sum_{m=1}^{J}G_{nm}\operatorname{Tr}(F_{m}X)+c_{n}\operatorname{Tr}(X)\right)F_{n}\
.$ (93)

  * •

$4\to 3$:

$\tilde{x}_{ijkl}=\left[\mathcal{L}(\ket{j}\\!\\!\bra{k})\right]_{il}\ .$ (94)

  * •

$3\to 1$ is given by Eq. (88a) and Eq. (88b) with $\tilde{x}$ in place of $x$.

  * •

$3\to 4$ is given by Eq. (89) with $\tilde{x}$ in place of $x$.

  * •

$3\to 2$ is given by

$x_{ijkl}=\tilde{x}_{ijkl}-b_{ij}\delta_{kl}-\delta_{ij}b_{kl}\ ,$ (95)

where

$b_{ij}=\frac{1}{2d}\sum_{k=1}^{d}\left(\tilde{x}_{ijkk}+\tilde{x}_{kkij}-\frac{1}{d}\sum_{l=1}^{d}\tilde{x}_{llkk}\delta_{ij}\right)\
.$ (96)

We denote the spaces in 2 as $\mathcal{V}_{1},\dots,\mathcal{V}_{6}$ and the
corresponding maps as $\varphi_{ij}:\mathcal{V}_{j}\to\mathcal{V}_{i}$. 2 and
Lemma 2 below imply that the diagram in Fig. 1 is commutative and all its
arrows are $\mathbb{R}$-linear bijections (given by the formulas in 2 and
Lemma 2). Once the theorem is proven, we will use the maps $\varphi_{ij}$ for
any indices $i,j=1,\dots,6$. Such maps are defined as the compositions of the
maps in Fig. 1; any such composition will provide the same result due to the
commutativity of Fig. 1.

$\mathcal{V}_{1}$$(H,a)$$\mathcal{V}_{2}$$x$$\mathcal{V}_{3}$$\tilde{x}$$\mathcal{V}_{4}$$\mathcal{L}$$\mathcal{V}_{5}$$\mathcal{L}$$\mathcal{V}_{6}$$(G,\vec{c})$
Figure 1: A commutative diagram representing the transformations between the
various spaces described in 2 (solid lines) and Lemma 2 (dotted lines). These
transformations capture the equivalence between the algebraic objects
appearing in Markovian quantum master equations and first-order differential
equations. The maps to $\mathcal{V}_{5}$ from $\mathcal{V}_{2}$ and
$\mathcal{V}_{6}$ are represented by dashed lines as they are represented by
the same formulas (89) and (93) as the corresponding maps to
$\mathcal{V}_{4}$: indeed, these maps to $\mathcal{V}_{5}$ trivially coincide
with the composition of the corresponding maps to $\mathcal{V}_{4}$ with the
restriction $\mathcal{V}_{4}\to\mathcal{V}_{5}$.

###### Proof.

First, note that the dimension of each of the $\mathbb{R}$-vector spaces
$\mathcal{V}_{i}$ is $(d^{2}-1)d^{2}=J(J+1)$.

We would like to prove that all maps are well defined and that the diagram is
commutative, i.e., if we start at any object in any node and apply the maps in
the diagram until we end up in the same node, we will have obtained the same
object as the one we started with.

For $\varphi_{21}\colon\mathcal{V}_{1}\to\mathcal{V}_{2}$ we need to show that
the right-hand side of Eq. (87) satisfies Eq. (83). We check this by direct
computation:

$\displaystyle x_{lkji}^{*}$
$\displaystyle=iH_{lk}^{*}\delta_{ji}-i\delta_{lk}H_{ji}^{*}+\sum_{m,n=1}^{J}a_{mn}^{*}(F_{m})_{lk}^{*}(F_{n})_{ji}^{*}$
$\displaystyle=-iH_{ij}\delta_{kl}+i\delta_{ij}H_{kl}+\sum_{m,n=1}^{J}a_{mn}(F_{m})_{ij}(F_{n})_{kl}$
$\displaystyle=x_{ijkl}\ .$ (97)

Here we used that $H,a,\\{F_{n}\\}$ are Hermitian, swapped the order of two
terms with $H$, and renamed the indices $n\leftrightarrow m$ in the summation.
To check the second property of $x$ we note that the $\\{F_{n}\\}$ are
traceless. Hence, the second term in Eq. (87) does not contribute and

$\displaystyle\sum_{k=1}^{d}(x_{ijkk}+x_{kkij})=$ (98)
$\displaystyle\quad\sum_{k=1}^{d}\Bigl{(}-iH_{ij}\delta_{kk}+i\delta_{ij}H_{kk}-iH_{kk}\delta_{ij}+i\delta_{kk}H_{ij}\Bigr{)}=0\
.$

For $\varphi_{12}\colon\mathcal{V}_{2}\to\mathcal{V}_{1}$ the fact that $H$
and $a$ are Hermitian follows from the first property of $x$ in Eq. 83 and
Eqs. 88a and 88b for $H$ and $a$. The trace of the right-hand side of Eq.
(88a) evaluates to 0.

If we start with $(H,a)$ and apply the maps
$\mathcal{V}_{1}\to\mathcal{V}_{2}\to\mathcal{V}_{1}$ given by Eqs. 87, 88a
and 88b we obtain some
$(H^{\prime},a^{\prime})=\varphi_{12}(\varphi_{21}((H,a)))$. Let us prove that
these match the original $(H,a)$, i.e., that
$\varphi_{12}\circ\varphi_{21}=\operatorname{id}_{\mathcal{V}_{1}}$ Since
$F_{m}$ and $F_{n}$ are traceless, the term corresponding to $a$ in the
expression for $H^{\prime}$ is zero. Using the fact that
$\operatorname{Tr}(H)=0$ and $\operatorname{Tr}(I)=d$ we obtain

$\displaystyle H^{\prime}_{ij}$
$\displaystyle=\frac{1}{2id}\sum_{k=1}^{d}(x_{kkij}-x_{ijkk})$ (99a)
$\displaystyle=\frac{1}{2id}\Bigl{(}-i\operatorname{Tr}(H)\delta_{ij}+i\operatorname{Tr}(I)H_{ij}+iH_{ij}\operatorname{Tr}(I)$
$\displaystyle\qquad-i\delta_{ij}\operatorname{Tr}(H)\Bigr{)}$ (99b)
$\displaystyle=\frac{1}{2id}\left(idH_{ij}+iH_{ij}d\right)=H_{ij}\ .$ (99c)

Similarly, we can observe that the term with $H$ from Eq. (87) does not
contribute to $a^{\prime}$ since the $\\{F_{n}\\}$ are traceless. Using the
fact that the $\\{F_{m}\\}$ are Hermitian and form an orthonormal basis we
obtain

$a^{\prime}_{mn}=\sum_{i,j=1}^{J}a_{ij}\operatorname{Tr}(F_{m}F_{i})\operatorname{Tr}(F_{n}F_{j})=a_{mn}\
.$ (100)

It follows that
$\varphi_{21}\circ\varphi_{12}=\operatorname{id}_{\mathcal{V}_{2}}$. Indeed,
it is a general fact that a left inverse of a linear map between 2 vector
spaces of the same dimension is necessarily its right inverse as well [24,
Proposition 1.20]. We also use this argument for other loops, proving the
commutativity of each loop in the diagram for only a single starting point.

By direct comparison of Eqs. (1a)-(1c) with Eq. (87) and Eq. (89) one can
conclude that they yield the same operator $\mathcal{L}$. To check that
$\varphi_{41}$ is well defined, one would need to check that the operator
$\mathcal{L}$ defined by Eqs. (1a)-(1c) satisfies Eq. (85). This was already
done in Proposition 4 and Proposition 5. Therefore,
$\varphi_{42}\circ\varphi_{21}=\varphi_{41}$. Note, that $\varphi_{42}$ is
well defined because for any $x\in\mathcal{V}_{2}$ we have

$\varphi_{42}(x)=(\varphi_{42}\circ\varphi_{21}\circ\varphi_{12})(x)=\varphi_{41}(\varphi_{12}(x)).$
(101)

The maps
$\varphi_{45},\,\varphi_{54}:\mathcal{V}_{5}\leftrightarrow\mathcal{V}_{4}$
are clearly well-defined. Also
$\varphi_{54}\circ\varphi_{45}=\operatorname{id}_{\mathcal{V}_{5}}$ and,
hence, $\varphi_{45}\circ\varphi_{54}=\operatorname{id}_{\mathcal{V}_{4}}$.

The fact that the elements of $G$ and $\vec{c}$ are real follows from Eq.
(92), the fact that the $\\{F_{n}\\}$ form an orthonormal basis, and the fact
that $\mathcal{L}$ maps Hermitian operators to Hermitian operators. Let us
check that the composition
$\varphi_{46}\circ\varphi_{65}\circ\varphi_{54}=\operatorname{id}_{\mathcal{V}_{4}}$.
Denote the resulting operator given by Eq. (93) as
$\mathcal{L}^{\prime}=(\varphi_{46}\circ\varphi_{65}\circ\varphi_{54})(\mathcal{L})$.
Since $\\{F_{n}\\}_{n=0,\dots,J}$ form a basis in $M(d,\mathbb{C})$ and
$\mathcal{L}$ and both $\mathcal{L}^{\prime}$ are $\mathbb{C}$-linear, it is
sufficient to check that $\mathcal{L}(F_{l})=\mathcal{L}^{\prime}(F_{l})$ for
$l=0,\dots,J$. For $l=0$ we have, using Eq. (93):

$\displaystyle\mathcal{L}^{\prime}(F_{0})$
$\displaystyle=\sum_{n=1}^{J}c_{n}\operatorname{Tr}(I/\sqrt{d})F_{n}=\sum_{n=1}^{J}c_{n}\sqrt{d}F_{n}$
(102a)
$\displaystyle=\sum_{n=1}^{J}\operatorname{Tr}[F_{n}\mathcal{L}(F_{0})]F_{n}=\mathcal{L}(F_{0}).$
(102b)

And for $n>0$ we have:

$\displaystyle\mathcal{L}^{\prime}(F_{l})$
$\displaystyle=\sum_{n,m=1}^{J}G_{nm}\operatorname{Tr}(F_{m}F_{l})F_{n}=$
(103a)
$\displaystyle=\sum_{n=1}^{J}G_{nl}F_{n}=\sum_{n=1}^{J}\operatorname{Tr}[F_{n}\mathcal{L}(F_{l})]F_{n}$
(103b) $\displaystyle=\mathcal{L}(F_{l})\ .$ (103c)

Here we again used that $\\{F_{n}\\}_{n=0,\dots,J}$ are Hermitian and form an
orthonormal basis, and that $\operatorname{Tr}[F_{0}\mathcal{L}(X)]=0$ for any
$X$ [Eq. 85].

Now consider the map $\varphi_{34}:\mathcal{L}\mapsto\tilde{x}$. Eq. 84
follows from Eq. 85 applied to $X=\ket{j}\\!\\!\bra{k}$. Let us prove that
$\varphi_{23}\circ\varphi_{34}\circ\varphi_{42}=\operatorname{id}_{\mathcal{V}_{2}}$.
Start with $x\in\mathcal{V}_{2}$ and apply the above maps to obtain
$\mathcal{L}=\varphi_{42}(x)\in\mathcal{V}_{4}$,
$\tilde{x}=\varphi_{34}(\mathcal{L})\in\mathcal{V}_{3}$,
$x^{\prime}=\varphi_{23}(\tilde{x})\in\mathcal{V}_{2}$. From Eqs. 89 and 94 we
have:

$\displaystyle\tilde{x}_{ijkl}$
$\displaystyle=x_{ijkl}-\frac{1}{2}\sum_{n=1}^{d}(x_{njin}\delta_{lk}+x_{nlkn}\delta_{ij})$
(104a)
$\displaystyle=x_{ijkl}+\tilde{b}_{ij}\delta_{kl}+\delta_{ij}\tilde{b}_{kl}\
,$ (104b)

where

$\tilde{b}_{ij}=-\frac{1}{2}\sum_{n=1}^{d}x_{njin}.$ (105)

In order to compute $x^{\prime}=\varphi_{23}(\tilde{x})$ we compute $b$ given
by Eq. 96 by substituting Eq. 104b into Eq. (96) and using Eq. (83).

$\displaystyle b_{ij}$
$\displaystyle=\frac{1}{2d}\sum_{k=1}^{d}\biggl{(}x_{ijkk}+\tilde{b}_{ij}\delta_{kk}+\delta_{ij}\tilde{b}_{kk}$
(106a)
$\displaystyle\quad+x_{kkij}+\tilde{b}_{kk}\delta_{ij}+\delta_{kk}\tilde{b}_{ij}-\frac{1}{d}\sum_{l=1}^{d}(x_{kkll}+2\tilde{b}_{kk}\delta_{ll})\delta_{ij}\biggr{)}{}$
$\displaystyle=\tilde{b}_{ij}+(1-1)\frac{1}{d}\delta_{ij}\sum_{k=1}^{d}\tilde{b}_{kk}=\tilde{b}_{ij}\
.$ (106b)

Therefore, Eq. (95) subtracts the same term as the one added in Eq. (104).
Hence, $x=x^{\prime}$. Note that the computation we performed also implies
that $\varphi_{23}$ is well-defined by a similar argument as used for
$\varphi_{42}$ above.

Finally, concerning $\varphi_{13}$ and $\varphi_{43}$, one can check that the
term of the form

$\tilde{b}_{ij}\delta_{kl}+\delta_{ij}\tilde{b}_{kl}={b}_{ij}\delta_{kl}+\delta_{ij}{b}_{kl}\
,$ (107)

which appears in Eq. (95), evaluates to $0$ when substituted into Eqs. (88a),
(88b), and (89). Specifically,

$\displaystyle H_{ij}$
$\displaystyle=\frac{1}{2id}\sum_{k=1}^{d}[(\tilde{x}_{kkij}-b_{kk}\delta_{ij}-b_{ij}\delta_{kk})$
$\displaystyle\qquad-(\tilde{x}_{ijkk}-b_{ij}\delta_{kk}-\delta_{ij}b_{kk})]$
(108a)
$\displaystyle=\frac{1}{2id}\sum_{k=1}^{d}(\tilde{x}_{kkij}-\tilde{x}_{ijkk})\
.$ (108b)

Substituting just the term ${b}_{ij}\delta_{kl}+\delta_{ij}{b}_{kl}$ into Eq.
(88b), we have:

$\displaystyle\sum_{i,j,k,l=1}^{d}(F_{m})_{ji}(b_{ij}\delta_{kl}+\delta_{ij}b_{kl})(F_{n})_{lk}$
(109a)
$\displaystyle\quad=\sum_{i,j=1}^{d}b_{ij}[(F_{m})_{ji}\operatorname{Tr}(F_{n})+(F_{n})_{ji}\operatorname{Tr}(F_{m})]$
(109b) $\displaystyle\quad=0\ ,$ (109c)

which means that

$a_{mn}=\sum_{i,j,k,l=1}^{d}(F_{m})_{ji}\tilde{x}_{ijkl}(F_{n})_{lk}\ .$ (110)

Likewise, substituting the same Eq. (107) term into Eq. (89), we have:

$\displaystyle\sum_{j,k=1}^{d}[b_{ij}\delta_{kl}X_{jk}+\delta_{ij}b_{kl}X_{jk}$
$\displaystyle\qquad-\frac{1}{2}(b_{jk}\delta_{ij}X_{kl}+\delta_{jk}b_{ij}X_{kl})$
$\displaystyle\qquad-\frac{1}{2}(X_{ij}b_{kl}\delta_{jk}+X_{ij}\delta_{kl}b_{jk})]$
(111a)
$\displaystyle\quad=\frac{1}{2}\sum_{j=1}^{d}(2b_{ij}X_{jl}+2b_{jl}X_{ij}-b_{ij}X_{jl}-b_{ij}X_{jl}$
(111b) $\displaystyle\qquad-X_{ij}b_{jl}-X_{ij}b_{jl})$ $\displaystyle\quad=0\
,$ (111c)

which means that

$[\mathcal{L}(X)]_{il}=\sum_{j,k=1}^{d}(\tilde{x}_{ijkl}X_{jk}-\frac{1}{2}\tilde{x}_{jkij}X_{kl}-\frac{1}{2}X_{ij}\tilde{x}_{kljk})\
.$ (112)

This proves the remaining statements
$\varphi_{43}\circ\varphi_{34}=\operatorname{id}_{\mathcal{V}_{4}}$ and
$\varphi_{13}\circ\varphi_{34}\circ\varphi_{41}=\operatorname{id}_{\mathcal{V}_{1}}$.
∎

###  IV.2 Explicit formulas for $H$ and $a$ from $G$ and $\vec{c}$

We are now finally prepared to complete the solution of the inverse problem
and provide explicit formulas for $H$ and $a$ given $G$ and $\vec{c}$. We
first solve the problem more generally: assume we have
$(G,\vec{c})\in\mathcal{V}_{6}$ and would like to know the corresponding
object from one of $\mathcal{V}_{1},\dots,\mathcal{V}_{5}$. If we are
interested in $\mathcal{L}\in\mathcal{V}_{4}$ or
$\mathcal{L}\in\mathcal{V}_{5}$ we could directly apply Eq. (93) from 2. For
$\mathcal{V}_{1}$, $\mathcal{V}_{2}$, or $\mathcal{V}_{3}$ we would, however,
have to compose multiple maps from that theorem.

Here we provide explicit formulas for these compositions.

######  Lemma 2.

Suppose we have $(G,\vec{c})\in\mathcal{V}_{6}$. Then the objects from
$\mathcal{V}_{1},\mathcal{V}_{2},\mathcal{V}_{3}$ corresponding to
$(G,\vec{c})$ in the sense of Theorem 2 are given by the following formulas:

  * •

$(H,a)\in\mathcal{V}_{1}$ are given by Eqs. 5a and 5b.

  * •

$\tilde{x}\in\mathcal{V}_{3}$ is given by

$\tilde{x}_{ijkl}=\sum_{n=1}^{J}\left(\sum_{m=1}^{J}G_{nm}(F_{m})_{kj}+c_{n}\delta_{kj}\right)(F_{n})_{il}.$
(113)

  * •

$x\in\mathcal{V}_{2}$ is given by Eq. (95) with $b$ given by

$b=\frac{1}{2d}\sum_{n=1}^{J}\left(\sum_{m=1}^{J}G_{nm}(\\{F_{m},F_{n}\\}-\delta_{mn}I/d)+2c_{n}F_{n}\right).$
(114)

###### Proof.

To simplify the computation, we introduce

$\tilde{G}_{n}=\sum_{m=1}^{J}G_{nm}F_{m}+c_{n}I\ .$ (115)

Eq. (113) is obtained by substituting Eq. (93) into Eq. (94). Eq. (114) is
obtained by substituting Eq. (113) into Eq. (96):

$\displaystyle b_{ij}$
$\displaystyle=\frac{1}{2d}\sum_{k=1}^{d}(\tilde{x}_{ijkk}+\tilde{x}_{kkij})$
(116a)
$\displaystyle=\frac{1}{2d}\sum_{k=1}^{d}\bigg{[}\sum_{n=1}^{J}(\tilde{G}_{n})_{kj}(F_{n})_{ik}+(\tilde{G}_{n})_{ik}(F_{n})_{kj}$
$\displaystyle\qquad-\frac{1}{d}\delta_{ij}(\tilde{G}_{n})_{lk}(F_{n})_{kl}\bigg{]}$
(116b)
$\displaystyle=\frac{1}{2d}\bigg{[}\sum_{m,n=1}^{J}G_{nm}\Bigl{[}(F_{n}F_{m})_{ij}+(F_{m}F_{n})_{ij}$
$\displaystyle\qquad-\operatorname{Tr}(F_{m}F_{n})\delta_{ij}/d\Bigr{]}+\sum_{n=1}^{J}2c_{n}(F_{n})_{ij}\bigg{]}\
.$ (116c)

Eqs. (5a) and (5b) are obtained by substituting $\tilde{x}$ from Eq. (113)
into Eq. (108b) and Eq. (110) respectively:

$\displaystyle H_{ij}$
$\displaystyle=\frac{1}{2id}\sum_{k=1}^{d}\sum_{n=1}^{J}\bigg{[}(\tilde{G}_{n})_{ik}(F_{n})_{kj}-(\tilde{G}_{n})_{kj}(F_{n})_{ik}\bigg{]}$
(117a)
$\displaystyle=\frac{1}{2id}\sum_{n=1}^{J}([\tilde{G}_{n},F_{n}])_{ij}=\frac{1}{2id}\sum_{m,n=1}^{J}G_{nm}(\left[F_{m},F_{n}\right])_{ij},$
(117b)

and

$\displaystyle a_{mn}$
$\displaystyle=\;\sum_{{i,j,k,l=1}}^{d}\;(F_{m})_{ji}\left(\sum_{n^{\prime}=1}^{J}(\tilde{G}_{n^{\prime}})_{kj}(F_{n^{\prime}})_{il}\right)(F_{n})_{lk}$
(118a)
$\displaystyle=\sum_{n^{\prime}=1}^{J}\operatorname{Tr}(\tilde{G}_{n^{\prime}}F_{m}F_{n^{\prime}}F_{n})$
(118b)
$\displaystyle=\sum_{i=1}^{J}\operatorname{Tr}\left[\left(\sum_{j=1}^{J}G_{ij}F_{j}+c_{i}I\right)F_{m}F_{i}F_{n}\right]\
.$ (118c)

∎

Eqs. 5a and 5b of 1 are now a special case (the first item) of Lemma 2.

Once $a$ is computed from Eq. (5b), one can check if the given ODE generates a
Lindblad equation (and not just a Markovian quantum master equation) by
testing whether $a$ is positive semi-definite. What remains is to formulate a
complete-positivity condition directly in terms of $G$ and $\vec{c}$; we do
this next.

###  IV.3 Complete positivity

We now complete the proof of 1.

######  Lemma 3 (Positive-semidefiniteness condition of 1).

A pair $(G,\vec{c})$ generates a (completely positive) Lindblad master
equation iff Eq. 6 holds for all traceless $B\in\mathcal{B}(\mathcal{H})$.

###### Proof.

The condition $a\geq 0$ is equivalent to the condition that for any vector
$b=\\{b_{m}\\}_{m=1}^{J}$ one has

$\sum_{m,n=1}^{J}b_{m}^{*}a_{mn}b_{n}\geq 0\ .$ (119)

Such vectors are in one-to-one correspondence with $B\in M_{0}(d,\mathbb{C})$
[i.e., with traceless $B\in B(\mathcal{H})$] via expansion in the basis
$\\{F_{m}\\}$:

$B=\sum_{m=1}^{J}b_{m}F_{m},\qquad b_{m}=\operatorname{Tr}(F_{m}B).$ (120)

Substituting Eq. 5b into the right hand side of Eq. 119 and simplifying it
using Eq. 120 we obtain the right-hand side of Eq. 6:

$\displaystyle\sum_{m,n=1}^{J}b_{m}^{*}a_{mn}b_{n}={}$
$\displaystyle\quad\sum_{m,n=1}^{J}b_{m}^{*}\sum_{i=1}^{J}\operatorname{Tr}\bigg{[}\bigg{(}\sum_{j=1}^{J}G_{ij}F_{j}+c_{i}I\bigg{)}F_{m}F_{i}F_{n}\bigg{]}b_{n}={}$
$\displaystyle\quad\sum_{i=1}^{J}\operatorname{Tr}\bigg{[}\bigg{(}\sum_{j=1}^{J}G_{ij}F_{j}+c_{i}I\bigg{)}B^{\dagger}F_{i}B\bigg{]}\
.$ (121)

∎

###  IV.4 Complete positivity and convex geometry

There is a fruitful complementary description of our complete positivity
results using convex geometry, which we explain in this subsection. For the
necessary background in convex geometry, see Appendix C.

The last part of 1, i.e., the just-proven Lemma 3, describes the set of pairs
$(G,\vec{c})$ corresponding to positive semidefinite $a$ (or, equivalently, to
Lindblad master equations). We denote this set by $\mathcal{V}_{6{+}}$ (a
subset of $\mathcal{V}_{6}$). Moreover, we introduce $\mathcal{V}_{i{+}}$
($i=1,\dots,6$) to be images of $\mathcal{V}_{6{+}}$ under the maps
$\varphi_{i6}$ described by the commutative diagram Fig. 1.
$\mathcal{V}_{1{+}}$ consists of pairs $(H,a)$ where $H$ is an arbitrary
Hermitian matrix and $a$ is positive semidefinite.

The set $M_{+}(d,\mathbb{C})$ of positive semidefinite matrices is a convex
cone with compact support (see Lemma 6 in Appendix C). The complete positivity
part of 1 uses the description of the cone $M_{+}(d,\mathbb{C})$ along with a
set of linear inequalities Eq. 119, i.e., the set of supporting hyperplanes.

An alternative is to note that $M_{+}(d,\mathbb{C})$ is a convex hull of its
extreme rays (Lemma 8). The extreme rays of $M_{{+}}(J,\mathbb{C})$ are the
rays generated by rank 1 matrices (Lemma 7), i.e., matrices $a$ of the form
$bb^{\dagger}$, where $b\in\mathbb{C}^{J}$.

Using the map $\varphi_{61}$ this provides a description for
$\mathcal{V}_{6{+}}$ given by the following:

######  Proposition 13.

The cone $\mathcal{V}_{6+}$ of pairs $(G,\vec{c})$ generating a (completely
positive) Lindblad master equation coincides with the convex hull of the
following elements:

  * •

Elements of the form $(Q,\vec{0})$, where $Q$ is given by Eq. 48 for $H\in
M_{0,\textrm{sa}}(d,\mathbb{C})$;

  * •

Elements of the form $(R,\vec{c})$, where

$\displaystyle R_{kl}$
$\displaystyle=\operatorname{Tr}\big{[}F_{k}\big{(}BF_{l}B^{\dagger}-\frac{1}{2}\left\\{B^{\dagger}B,F_{l}\right\\}\big{)}\big{]}$
(122a) $\displaystyle c_{k}$
$\displaystyle=\frac{1}{d}\operatorname{Tr}(\big{[}B,B^{\dagger}\big{]}F_{k})$
(122b) $\displaystyle B$ $\displaystyle\in M_{0}(d,\mathbb{C})\ .$ (122c)

###### Proof.

According to the discussion above, $\mathcal{V}_{6+}$ is generated by elements
of the form $\varphi_{61}((H,a=0))$ and $\varphi_{61}((H=0,a=bb^{\dagger}))$.
The former are given by Eq. 48. The image of the latter can be computed using
Eqs. 58b and 57, after identifying $b\in\mathbb{C}^{J}$ with elements $B\in
M_{0}(d,\mathbb{C})$ via Eq. 120:

$\displaystyle R_{kl}$
$\displaystyle=\sum_{i,j=1}^{J}b_{i}b_{j}^{*}\operatorname{Tr}\big{[}F_{k}\big{(}F_{i}F_{l}F_{j}-\frac{1}{2}\left\\{F_{j}F_{i},F_{l}\right\\}\big{)}\big{]}$
$\displaystyle=\operatorname{Tr}\big{[}F_{k}\big{(}BF_{l}B^{\dagger}-\frac{1}{2}\left\\{B^{\dagger}B,F_{l}\right\\}\big{)}\big{]}$
(123a) $\displaystyle c_{k}$
$\displaystyle=\frac{1}{d}\sum_{i,j=1}^{J}b_{i}b_{j}^{*}\operatorname{Tr}(\big{[}F_{i},F_{j}\big{]}F_{k})=\frac{1}{d}\operatorname{Tr}(\big{[}B,B^{\dagger}\big{]}F_{k})\
.$ (123b)

∎

###  IV.5 Consistency

By construction, the forward maps defined in 2 are consistent with our
previously defined maps. More explicitly, this is stated in the following
proposition.

######  Proposition 14.

Let $H^{\prime}$ be a Hermitian operator in $\mathcal{B}(\mathcal{H})$ and let
$a$ be a Hermitian $J\times J$ matrix. Let $Q,R,\vec{c}$ be obtained from
$(H^{\prime},a)$ using Eqs. 45a, 45b and 44b, and $G=Q+R$. Let
$H=H^{\prime}-\frac{1}{d}\operatorname{Tr}(H^{\prime})I$ be the traceless
component of $H^{\prime}$, and let $\varphi_{ij}$ be the maps from 2. Then

$\displaystyle(Q,0)$ $\displaystyle=\varphi_{61}((H,0)),$ (124a)
$\displaystyle(R,c)$ $\displaystyle=\varphi_{61}((0,a)),$ (124b)
$\displaystyle(G,c)$ $\displaystyle=\varphi_{61}((H,a)).$ (124c)

###### Proof.

First, note that Eq. 1b yields the same operator $\mathcal{L}$ if we change
$H$ by a constant. Hence, we could apply Eq. 45a to $H$ instead of
$H^{\prime}$ and get the same $Q$. One of the equivalent ways to describe
$\varphi_{16}$ is
$\varphi_{61}=\varphi_{65}\circ\varphi_{54}\circ\varphi_{41}$. Note that

$(\varphi_{54}\circ\varphi_{41})((H,0))=\mathcal{L}_{H}+\left.\mathcal{L}_{a}\right|_{a=0}=\mathcal{L}_{H}.$
(125)

Thus, Eq. 124a follows from comparison of Eq. 45a with Eq. 92. Similarly,

$(\varphi_{54}\circ\varphi_{41})((0,a))=\mathcal{L}_{a},$ (126)

and Eq. 124b follows from comparison of Eqs. 45b and 44b with Eq. 92. Finally,
Eq. 124c is the sum of Eqs. 124a and 124b. ∎

In the following Proposition, we show that Eqs. 5a and 82 are consistent,
i.e., yield the same result when used to recover $H$.

######  Proposition 15.

For any $J\times J$ matrix $G$, the r.h.s. of Eq. 5a and Eq. 82 (when $Q$ is
replaced with $G$) are equal to each other.

###### Proof.

Comparing Eq. 5a to Eq. 82 one can see that for a given $G$ the claim is
equivalent to

$\frac{1}{2id}G_{nm}[F_{m},F_{n}]=-\frac{1}{2d}f_{nml}G_{nm}F_{l}$ (127)

where we use the Einstein summation notation. Eq. 127 follows directly from
the definition of the structure constants Eq. 76. ∎

Next, we show that the inverse maps [not only Eqs. 5a and 82] also yield the
same result when used to recover $H$, and in fact one may interchange $G$ and
$Q$ when using these maps, or subtract any matrix from $G$ which can be
obtained using the formula for $R$ (even from a different $\mathcal{L}_{a}$;
e.g., any real symmetric matrix).

######  Proposition 16.

Let $H,a,Q,R,G,\vec{c}$ be the same as in Proposition 14. Then all of the
following methods recover the same $H$:

  1. 1.

Applying Eq. 5a to $G$ (as is);

  2. 2.

Applying Eq. 5a to $Q$ in place of $G$;

  3. 3.

Computing $Q^{\prime}$ to be the antisymmetric part of $G$ and applying Eq. 5a
to $Q^{\prime}$ in place of $G$;

  4. 4.

Computing $Q^{\prime\prime}=G-R^{\prime}$ where
$(R^{\prime},\vec{c}^{\prime})=\varphi_{61}((0,a^{\prime}))$ for any Hermitian
$J\times J$ matrix $a^{\prime}$ (possibly different from $a$) and applying Eq.
5a to $Q^{\prime\prime}$ in place of $G$.

###### Proof.

According to 2, the diagram Fig. 1 is commutative. Since Eq. 5 is is the
formula for $\varphi_{16}$ and $\varphi_{61}((H,a))=(G,c)$, applying Eq. 5a to
$G$ recovers the original $H$ (statement #1).

Since $\varphi_{61}((H,0))=(Q,\vec{0})$, applying Eq. 5a to $Q$ also recovers
the original $H$ (statement #2).

Since Eq. 5a contracts $G_{nm}$ with a commutator, which is antisymmetric with
respect to the order of indices, the result of that application is independent
of adding any symmetric matrix to $G$ (statement #3).

The final statement follows from
$\varphi_{61}((H,a-a^{\prime}))=(Q^{\prime\prime},c-c^{\prime})$. ∎

###  IV.6 Decomposing $G$ into $G=Q+R$, and the space of possible matrices $R$

######  Proposition 17.

The decomposition of $G$ into $Q$ and $R$ can be obtained by the following
formulas:

$\displaystyle Q_{ij}$
$\displaystyle=-\frac{1}{2d}\sum_{m,n=1}^{J}G_{nm}\operatorname{Tr}\left([F_{i},F_{j}],[F_{n},F_{m}]\right)$
(128a) $\displaystyle=\frac{1}{2d}\sum_{m,n,k=1}^{J}G_{nm}f_{knm}f_{kij}.$
(128b) $\displaystyle R_{ij}$ $\displaystyle=G_{ij}-Q_{ij}.$ (128c)

###### Proof.

Substituting Eq. 5a into Eq. 48a, we obtain:

$Q_{ij}=\frac{1}{2d}\sum_{m,n=1}^{J}G_{nm}\operatorname{Tr}\left(F_{i}\left[F_{j},\left[F_{m},F_{n}\right]\right]\right),$
(129)

which is equivalent to Eq. 128a. Eq. 128b is then obtained using the
definition of the structure constants in Eq. 76. Finally, Eq. 128c follows
from $G=R+Q$.

Note that an alternative way to obtain the formula for $R$ is to substitute
Eq. 5b into Eq. 58b (which results in a significantly more complex computation
giving the same final result). ∎

Naively, one might be tempted to set $Q$ to be the antisymmetric part of $G$,
or $R$ to be the symmetric part of $G$, instead of using the expressions given
in Proposition 17. This, however, only works for $d\leq 2$. This follows from
the dimension counting done in 2 below: the space of antisymmetric matrices
$R$ has dimension $J(J-3)/2$, which is non-vanishing unless $d\in\\{1,2\\}$
(recall that $J=d^{2}-1$). We give an example illustrating this for $d=3$ in
Section V.2.2.

However, as stated in Propositions 15 and 16 above, taking the antisymmetric
part of $G$ would still result in the correct $H$ being recovered.

######  Proposition 18.

For any $R\in M(J,\mathbb{R})$, $\vec{c}\in\mathbb{R}^{J}$ a pair
$(R,\vec{c})$ can be obtained from some Hermitian matrix $a$ if and only if

$\sum_{m,n=1}^{J}R_{mn}[F_{n},F_{m}]=0.$ (130)

In particular, this condition is independent of $\vec{c}$, i.e., for any
$\vec{c}^{\prime}\in\mathbb{R}^{J}$ the pair $(R,\vec{c})$ can be obtained
from some Hermitian matrix $a$ if and only if $(R,\vec{c}^{\prime})$ can (from
a possibly different Hermitian matrix $a^{\prime}$).

###### Proof.

According to Proposition 14 a pair $(R,\vec{c})$ can be obtained from some
Hermitian matrix $a$ if and only if $(R,\vec{c})\in\varphi_{61}(\\{(0,a):a\in
M_{\textrm{sa}}(J,\mathbb{C})\\})$. According to 2 this is equivalent to

$H=0\quad\textrm{ where }\quad(H,a)=\varphi_{16}((R,\vec{c})).$ (131)

From Lemma 2, this is equivalent to Eq. 5a evaluating to 0 when $R$ is used
instead of $G$, i.e. when Eq. 130 holds. ∎

######  Corollary 2.

The space of matrices $R$ which can be obtained from some Hermitian matrix $a$
is an $\mathbb{R}$-linear subspace of $M(J,\mathbb{R})$ of dimension $J^{2}-J$
and includes the space of symmetric matrices.

The space of antisymmetric matrices $R$, which can be obtained from some
Hermitian matrix $a$ has dimension $J(J-3)/2$.

###### Proof.

As explained in the proof of Proposition 18, the space of possible $(R,c)$ is
a bijective image of a $J^{2}$ dimensional space. According to Proposition 18
it contains all possible pairs of the form $(0,\vec{c})$ —- a $J$-dimensional
subspace, hence the space of possible values for $R$ has dimension $J^{2}-J$.

Furthermore, any symmetric matrix satisfies Eq. 130 due to contraction with an
antisymmetric commutator there.

The final statement follows from the direct calculation:

$J^{2}-J-J(J+1)/2=J(J-3)/2\ ,$ (132)

where $J(J+1)/2$ is the dimension of the space of symmetric matrices. ∎

##  V Examples

This section provides several examples to illustrate the general theory
developed above. Various supporting calculations can be found in Ref. [19].

###  V.1 A qubit

Consider a single qubit, i.e., $\mathcal{H}=\mathbb{C}^{2}$. As the nice
operator basis, we choose the normalized Pauli matrices
$\frac{1}{\sqrt{2}}\\{I,\sigma^{x},\sigma^{y},\sigma^{z}\\}$.

####  V.1.1 A qubit with pure dephasing

Consider the Lindblad equation for pure dephasing:
$\dot{\rho}=\mathcal{L}_{a}(\rho)=\gamma(\sigma^{z}\rho\sigma^{z}-\rho)$, for
which

$\displaystyle a=\begin{bmatrix}0&0&0\\\ 0&0&0\\\ 0&0&2\gamma\\\
\end{bmatrix}\ .$ (133)

Here $Q=0$ (since $H=0$) and, by Propositions 6, 7 and 10, $\vec{c}=0$, since
$a$ is real-symmetric. Using Eq. 58 we find

$\displaystyle G=R=\begin{bmatrix}-2\gamma&0&0\\\ 0&-2\gamma&0\\\ 0&0&0\\\
\end{bmatrix}\ .$ (134)

The inverse problem retrieves $H$ from $G$ and $a$ from $(G,\vec{c})$. Using
Eqs. 5a and 5b, we indeed find $H=0$ and $a$ as given in Eq. 133.

####  V.1.2 A qubit with amplitude damping

Now consider the Lindblad equation for spontaneous emission:

$\displaystyle\dot{\rho}$
$\displaystyle=\mathcal{L}_{H}(\rho)+\mathcal{L}_{a}(\rho)$ (135a)
$\displaystyle\mathcal{L}_{H}(\rho)$ $\displaystyle=-i\omega[\sigma^{z},\rho]$
(135b) $\displaystyle\mathcal{L}_{a}(\rho)$
$\displaystyle=\gamma(\sigma^{-}\rho\sigma^{+}-\frac{1}{2}\\{\sigma^{+}\sigma^{-},\rho\\})\
,$ (135c)

where $\sigma^{\pm}=\frac{1}{2}(\sigma^{x}\mp i\sigma^{y})$. This yields

$\displaystyle a=\left[\begin{array}[]{ccc}\gamma&-i\gamma&0\\\
i\gamma&\gamma&0\\\ 0&0&0\\\ \end{array}\right]\ ,$ (139)

i.e., $a$ is not real-symmetric, and the Lindbladian is non-unital (in
particular, $\vec{c}$ is non-zero).

Then, using Eqs. 48 and 58 we find

$\displaystyle Q$ $\displaystyle=\left[\begin{array}[]{ccc}0&-2\omega&0\\\
2\omega&0&0\\\ 0&0&0\\\ \end{array}\right]$ (140d) $\displaystyle R$
$\displaystyle=\begin{bmatrix}-\gamma&0&0\\\ 0&-\gamma&0\\\
0&0&-2\gamma\end{bmatrix}\ ,$ (140e)

and using Eq. 57

$\displaystyle\vec{c}=(0,0,\sqrt{2}\gamma)^{T}\ .$ (141)

Using Eqs. 5a and 5b with $G=Q+R$ we then indeed find $H=\omega\sigma^{z}$ and
and $a$ as given in Eq. 139.

####  V.1.3 Example of $G$ giving rise to a non-CP map

Let $\vec{c}=\vec{0}$ and

$\displaystyle G=\begin{bmatrix}0&1&0\\\ 0&0&0\\\ 0&0&0\\\ \end{bmatrix}\ ,$
(142)

Using Eq. 5b this yields:

$a=\left[\begin{array}[]{ccc}0&\frac{1}{2}&0\\\ \frac{1}{2}&0&0\\\ 0&0&0\\\
\end{array}\right]\ ,$ (143)

whose eigenvalues are $\\{-1/2,1/2,0\\}$. I.e., $a$ is not positive
semidefinite, which means that the corresponding Markovian quantum master
equation does not generate a CP map. Hence, the pair $(G,\vec{c})$ with matrix
$G$ given in Eq. 142 and $\vec{c}=0$ does not correspond to a Lindblad
equation, only to a Markovian quantum master equation.

###  V.2 A qutrit

As the nice operator basis for this $d=3$-dimensional case, we choose the $8$
normalized Gell-Mann matrices, including the normalized identity matrix:

$\displaystyle F_{0}=\frac{1}{\sqrt{3}}I,\
\\{F_{m}\\}_{m=1}^{d}=\left\\{\left(\begin{array}[]{ccc}0&\frac{1}{\sqrt{2}}&0\\\
\frac{1}{\sqrt{2}}&0&0\\\ 0&0&0\\\
\end{array}\right),\left(\begin{array}[]{ccc}0&0&\frac{1}{\sqrt{2}}\\\
0&0&0\\\ \frac{1}{\sqrt{2}}&0&0\\\ \end{array}\right),\right.$ (150)
$\displaystyle\left.\left(\begin{array}[]{ccc}0&0&0\\\
0&0&\frac{1}{\sqrt{2}}\\\ 0&\frac{1}{\sqrt{2}}&0\\\
\end{array}\right),\left(\begin{array}[]{ccc}0&-\frac{i}{\sqrt{2}}&0\\\
\frac{i}{\sqrt{2}}&0&0\\\ 0&0&0\\\
\end{array}\right),\left(\begin{array}[]{ccc}0&0&-\frac{i}{\sqrt{2}}\\\
0&0&0\\\ \frac{i}{\sqrt{2}}&0&0\\\ \end{array}\right),\right.$ (160)
$\displaystyle\left.\left(\begin{array}[]{ccc}0&0&0\\\
0&0&-\frac{i}{\sqrt{2}}\\\ 0&\frac{i}{\sqrt{2}}&0\\\
\end{array}\right),\left(\begin{array}[]{ccc}\frac{1}{\sqrt{2}}&0&0\\\
0&-\frac{1}{\sqrt{2}}&0\\\ 0&0&0\\\
\end{array}\right),\left(\begin{array}[]{ccc}\frac{1}{\sqrt{6}}&0&0\\\
0&\frac{1}{\sqrt{6}}&0\\\ 0&0&-\sqrt{\frac{2}{3}}\\\
\end{array}\right)\right\\}$ (170)

####  V.2.1 A qutrit with amplitude damping

Consider the example of $a$ given in Eq. 66, i.e., a qutrit undergoing
amplitude damping between its two lowest levels (similar to the qubit example
given in Section V.1.2). The corresponding $R$ matrix was given in Eq. 75;
assuming $H=0$ we have $G=R$. In addition, we find

$\vec{c}=(0,0,0,0,0,\frac{\sqrt{2}}{3},0,0)\ .$ (171)

Computing $a$ using Eq. 5b, we find that it is indeed identical to Eq. 66.

####  V.2.2 Illustration of Propositions 17 and 16

Note that for symmetric $G$, the decomposition is trivial: $Q=0$, $R=G$.
Therefore, to illustrate Proposition 17, let us choose an arbitrary
antisymmetric $G$; for example

$\displaystyle G=\left(\begin{array}[]{cccccccc}0&0&0&0&-\frac{1}{4}&0&0&0\\\
0&0&0&\frac{1}{4}&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&-\frac{1}{4}&0&0&0&0&0&0\\\
\frac{1}{4}&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&\frac{1}{4}&-\frac{\sqrt{3}}{4}\\\
0&0&0&0&0&-\frac{1}{4}&0&\frac{\sqrt{3}}{4}\\\
0&0&0&0&0&\frac{\sqrt{3}}{4}&-\frac{\sqrt{3}}{4}&0\\\ \end{array}\right)\ .$
(180)

Computing the Hamiltonian $H$ that arises from this $G$ using Eq. 82 [or,
equivalently, Eq. 5a] yields:

$H=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&\frac{1}{6}\\\ 0&\frac{1}{6}&0\\\
\end{array}\right)\ .$ (181)

When computing $Q$ from this Hamiltonian using Eq. 48, we observe that $Q$
does not equal the antisymmetric part of $G$, as anticipated by 2 and the
comment after Proposition 17. Instead, we find:

$\displaystyle Q=\left(\begin{array}[]{cccccccc}0&0&0&0&\frac{1}{6}&0&0&0\\\
0&0&0&\frac{1}{6}&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&-\frac{1}{6}&0&0&0&0&0&0\\\
-\frac{1}{6}&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&\frac{1}{6}&-\frac{1}{2\sqrt{3}}\\\
0&0&0&0&0&-\frac{1}{6}&0&0\\\ 0&0&0&0&0&\frac{1}{2\sqrt{3}}&0&0\\\
\end{array}\right)\ .$ (190)

As expected, if we then use Eq. 128a to decompose $G$ into $Q$ and $R$, we get
the same $Q$ as the one in Eq. 190. Also, as guaranteed by Proposition 16, if
we use that $Q$ to compute $H$ again, we will get the same $H$ as in Eq. 181.

##  VI The rarity of Lindbladians

As posed in the title, the question that motivated this work can be
interpreted as the probability that a given pair $(G,\vec{c})$ will give rise
to a Lindbladian (given a natural choice of a distribution over the pairs
$(G,\vec{c})$). Alternatively, one can use a natural choice of a distribution
for $(H,a)$ in order to induce the distribution on the pairs $(G,\vec{c})$ and
compute a probability that a given pair $(G,\vec{c})$ with that distribution
gives rise to a Lindbladian. In the remainder of this section, we provide
partial answers to these questions and describe the difference between these
two distributions.

###  VI.1 Natural distribution on the pairs of $(G,\vec{c})$

Suppose all elements of $G$ and $\sqrt{d}\vec{c}$ are picked independently
from a standard normal distribution. Such a distribution is known as the
Ginibre Orthogonal Ensemble (GinOE) [25]. One might be interested in the
following question: “How likely is it that these $G$ and $\vec{c}$ generate a
Lindbladian, i.e. $a\geq 0$?”. Let us denote this probability as
$\tilde{P}^{\text{GinOE}}_{J}$. While we do not know the asymptotics of
$\tilde{P}^{\text{GinOE}}_{J}$, here we provide a non-rigorous attempt at
deriving the asymptotics of an upper bound. Recall that a necessary condition
for $a\geq 0$ is that all eigenvalues of $G$ have a non-positive real part.
Denoting the probability of the latter by $P^{\text{GinOE}}_{J}$, this implies

$\tilde{P}^{\text{GinOE}}_{J}\leq P^{\text{GinOE}}_{J}.$ (191)

The distribution of eigenvalues in the GinOE is known: see, e.g. [25, Eq.
(1.7)]. One can use either the methods of [26] or some rigorous alternative to
those methods to estimate the asymptotics of $P^{\text{GinOE}}_{J}$, which
would be an upper bound for $\tilde{P}^{\text{GinOE}}_{J}$ due to Eq. 191.

More specifically, assuming the same (non-rigorous) argument as in Ref. [26]
applies to the GinOE, we can state that

$P^{\text{GinOE}}_{J}=\exp(-\theta^{\text{GinOE}}J^{2}+O(J)).$ (192)

We discuss the estimation of the positive constant $\theta^{\text{GinOE}}$ in
Appendix E. The important point about Eq. 192 is that the probability decays
rapidly in the Hilbert space dimension $d$ (recall that $J=d^{2}-1$).

###  VI.2 Natural distribution on $a$

In this subsection, we examine the prevalence of Lindbladians within the set
of all Liouvillians.

Specifically, we attempt to answer the following question: “Given a random $H$
and $a$, what is the probability that $\mathcal{L}_{H,a}$ is a Lindbladian?”
Since the condition $a\geq 0$ that guarantees a Liouvillian is a Lindbladian
depends only on $a$ and not on $H$, defining the distribution in the space of
Hermitian matrices $a$ is sufficient to answer this question. Additionally,
the scale of $a$ is irrelevant; changing $a$ to $\alpha a$ (where $\alpha>0$
can be randomly selected from a distribution which may depend on $a$) does not
alter the answer to $a\geq 0$.

One natural choice for the distribution is the Gaussian Unitary Ensemble
(GUE), also known as the $\beta=2$ Gaussian Ensemble. In this ensemble, the
real and imaginary parts of each component of a $J\times J$ matrix $A$ are
independently selected from a normal distribution with a mean of $0$ and555We
choose the normalization to be consistent with Ref. [27]. An alternative
normalization used in other works is variance $=1$, i.e., the standard normal
distribution. a variance of $1/\beta=1/2$. The matrix $a$ is then computed as
$a=(A+A^{\dagger})/2$. The condition $a\geq 0$ is equivalent to
$\lambda_{j}\geq 0$ for $j=1,\dots,J$, where
$\lambda=\\{\lambda_{j}\\}_{j=1}^{J}$ is the vector of eigenvalues of $a$. Let
us denote the probability of this event as $P^{\text{GUE}}_{J}$.

The joint probability density function (PDF) of eigenvalues of a matrix from
the GUE is well known to be:

$\displaystyle p(\lambda)$
$\displaystyle=\frac{1}{G_{\beta,J}}e^{-\frac{\beta}{2}\sum_{j=1}^{J}\lambda_{j}^{2}}\prod_{1\leq
j<k\leq J}\\!\\!\\!\\!\left|\lambda_{k}-\lambda_{j}\right|^{\beta}$ (193a)
$\displaystyle=\frac{1}{G_{\beta,J}}e^{-\beta E(\lambda)}\ ,$ (193b)

where $E(\lambda)=\frac{1}{2}\sum_{j=1}^{J}\lambda_{j}^{2}-\sum_{1\leq j<k\leq
J}\ln\left|\lambda_{j}-\lambda_{k}\right|$ and $G_{\beta,J}$ is the
normalization factor (partition function); see [27, Eq. (1.4)] for further
details. Thus, the distribution of $\lambda$ can be interpreted as the
canonical ensemble of a two-dimensional gas (in $\mathbb{C}$) with $J$ charged
particles (each of unit charge with the same sign), constrained to the real
line in a harmonic potential at temperature $1/\beta=1/2$, also known as a
log-gas. The probability $P^{\text{GUE}}_{J}$ in this interpretation is the
probability that, by random chance, all particles happen to be to the right of
the origin. $P^{\text{GUE}}_{J}$ is known as the probability of atypically
large fluctuations of extreme value statistics of the GUE, and was estimated
in Ref. [26] as:

$P^{\text{GUE}}_{J}=3^{-J^{2}/2+O(J)}\ .$ (194)

The estimate uses non-rigorous techniques from statistical mechanics,
including the replacement of particle density (sum of delta functions) with a
smooth function and the use of functional integrals and functional Fourier
transforms. While rigorously proving this estimate is left to future research,
Eq. 194 shows that the probability of a randomly picked Liouvillian being a
Lindbladian decays rapidly with $d$, just like Eq. 192 for the GinOE. For
example, already for $d=4$, the estimate gives $P^{\text{GUE}}_{J}\sim
10^{-54}$, demonstrating that it is extremely unlikely to find a Lindbladian
for $d=4$ by picking a random $a$ from the GUE.

###  VI.3 Comparison between these distributions

Given the distribution on the pairs $(G,\vec{c})$ described in Section VI.1,
one can derive a distribution for $a$ or a matrix related to the tensor $x$
from 2. $\tilde{P}^{\text{GinOE}}$ is then equal to the probability that
$a\geq 0$ (or, equivalently, $x\geq 0$) given that distribution.

Due to the linearity of $\varphi_{16}$, the components of $a$ also follow a
multivariate Gaussian distribution with mean $0$. Using Eq. 5b one can compute
the covariances:

$\mathbb{E}(a_{mn}a_{m^{\prime}n^{\prime}})=\delta_{mn^{\prime}}\delta_{nm^{\prime}}-\frac{1}{d}\operatorname{Tr}(F_{m^{\prime}}F_{n^{\prime}}F_{m}F_{n})\
.$ (195)

This distribution does not match the GUE investigated in Section VI.2: for
comparison covariances in the GUE are given by

$\mathbb{E}(a_{mn}a_{m^{\prime}n^{\prime}})=\frac{1}{2}\delta_{mn^{\prime}}\delta_{nm^{\prime}}.$
(196)

The normalization difference ($1$ vs $1/2$) is due to a difference in
conventions and is irrelevant to the question of whether $a\geq 0$. The
presence of the second term in Eq. 195, however, is significant and, for
$d>2$, introduces a dependence of the probability distribution given by Eq.
195 on the choice of the nice operator basis.

Nevertheless, the distribution of $x$ computed from $(H=0,a)$ is independent
of both the nice operator basis and the choice of the orthonormal basis in the
Hilbert space $\mathcal{H}$:

$\displaystyle\mathbb{E}(x_{ijkl}x_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}})=\delta_{jk^{\prime}}\delta_{j^{\prime}k}\delta_{li^{\prime}}\delta_{l^{\prime}i}$
$\displaystyle+d^{-1}(-\delta_{jk^{\prime}}\delta_{j^{\prime}i^{\prime}}\delta_{lk}\delta_{l^{\prime}i}-\delta_{jk^{\prime}}\delta_{j^{\prime}k}\delta_{li}\delta_{l^{\prime}i^{\prime}}-\delta_{ji}\delta_{j^{\prime}k}\delta_{li^{\prime}}\delta_{l^{\prime}k^{\prime}})$
$\displaystyle+d^{-2}(\delta_{jk^{\prime}}\delta_{j^{\prime}i}\delta_{lk}\delta_{l^{\prime}i^{\prime}}+\delta_{ji}\delta_{j^{\prime}k}\delta_{lk^{\prime}}\delta_{l^{\prime}i^{\prime}}$
$\displaystyle\quad+\delta_{jk^{\prime}}\delta_{j^{\prime}i^{\prime}}\delta_{li}\delta_{l^{\prime}k}+\delta_{ji^{\prime}}\delta_{j^{\prime}k}\delta_{li}\delta_{l^{\prime}k^{\prime}})$
$\displaystyle+d^{-3}(-\delta_{jk^{\prime}}\delta_{j^{\prime}i^{\prime}}\delta_{lk}\delta_{l^{\prime}i}-\delta_{ji}\delta_{j^{\prime}k^{\prime}}\delta_{lk}\delta_{l^{\prime}i^{\prime}}-\delta_{ji}\delta_{j^{\prime}i^{\prime}}\delta_{lk^{\prime}}\delta_{l^{\prime}k}$
$\displaystyle\quad-\delta_{jk}\delta_{j^{\prime}i^{\prime}}\delta_{li}\delta_{l^{\prime}k^{\prime}}-\delta_{ji}\delta_{j^{\prime}k}\delta_{li^{\prime}}\delta_{l^{\prime}k^{\prime}}-\delta_{ji^{\prime}}\delta_{j^{\prime}i}\delta_{lk}\delta_{l^{\prime}k^{\prime}})$
$\displaystyle+4d^{-4}\delta_{ji}\delta_{j^{\prime}i^{\prime}}\delta_{lk}\delta_{l^{\prime}k^{\prime}}+Jd^{-4}\delta_{ji}\delta_{j^{\prime}i^{\prime}}\delta_{lk}\delta_{l^{\prime}k^{\prime}}.$
(197)

This $x$ can be interpreted as a matrix with indices $(ij)$ and $(lk)$:
$x_{(ij)(lk)}=x_{ijkl}$. With this interpretation, such $x$ is positive
semidefinite if and only if $a$ is.

While beyond the scope of this work, one can use Eq. 195 or Section VI.3 to
attempt to estimate $\tilde{P}^{\text{GinOE}}_{J}$. One approach might be to
first compute the joint distribution of eigenvalues of $a$ or of $x$ similarly
to how they are computed for the GUE (see, e.g., Ref. [27]), i.e. by
integrating out unitary symmetry. One may attempt to integrate out the
unitaries in $U(\mathcal{H})$ first, and then integrate over the quotient
space $U(B_{0}(\mathcal{H}))/U(\mathcal{H})$, which consists of the unitaries
in $B_{0}(\mathcal{H})$ modulo those in $U(\mathcal{H})$, where
$U(\mathcal{H})$ denotes the manifold of unitaries acting on a finite-
dimensional Hilbert space $\mathcal{H}$. That second integration appears to be
non-trivial because while Section VI.3 is symmetric with respect to
$U(\mathcal{H})$, it is no longer symmetric with respect to
$U(B_{0}(\mathcal{H}))/U(\mathcal{H})$. Once the joint distribution of
eigenvalues is computed (or estimated), one can attempt to use it to estimate
the probability that all of them are non-negative.

##  VII Overlap with prior work

After this work was completed, it came to our attention that our main results
could be reconstructed by combining several earlier results, in particular
using Refs. [5, Section 3.2.2], [28, Section 7.1.2], and [29]. An aspect that
is common to these three works and distinguishes them from ours is that they
do not represent $\mathcal{L}$ by its action on the basis elements $F_{i}$,
i.e., they do not have $(G,\vec{c})$. In this sense, their motivation is
different from ours, given that our starting point is the title question
“which (nonhomogeneous linear first-order) differential equations correspond
to the Lindblad equation?”, formulated in terms of $(G,\vec{c})$ [Eq. 3].
Nevertheless, it is possible to use these earlier results to prove 1: one can
show that the map $\varphi_{64}$ given by Eq. 92 is a bijective linear map
with inverse given by Eq. 93. With this established, one can substitute Eq. 93
instead of $\mathcal{L}$ to apply the above results to obtain 1.

The overlap is described in more detail below. We first summarize it in the
following list, mentioning the parts of the 1 and the overlaps we have found.

  1. 1.

Bijectivity of the map $\varphi_{61}$.

     * •

Surjectivity follows from [5, Section 3.2.2].

     * •

Injectivity follows from [28, Section 7.1.2].

  2. 2.

Explicit formulas [Eq. 5] for the inverse ($\varphi_{16}$).

     * •

Follows from [5, Section 3.2.2].

  3. 3.

Complete positivity condition Eq. 6 in terms of $G$ and $\vec{c}$.

     * •

Follows from [29].

###  VII.1 Overlap with [5, Section 3.2.2]

Item 2 and surjectivity in Item 1 follow from [5, Section 3.2.2]. To see this,
one needs to take $V(t)$ given by

$V(t)\rho=\rho+t\mathcal{L}(\rho)+O(t^{2})$ (198)

with $\mathcal{L}$ given by Eq. 93, and write it in the form [5, Eq. (3.50)]
(see, e.g., Appendix D on how to do that). Then one can follow [5, Eqs.
(3.54)–(3.63)] to recover $a$ and $H$ and write $\mathcal{L}$ in the form Eq.
1. After simplification, this would provide explicit formulas Eq. 5 for $a$
and $H$ representing that $\mathcal{L}$. Since the procedure works for any
$\mathcal{L}$ from $\mathcal{V}_{4}$ (i.e. any $(G,\vec{c})$), this shows the
surjectivity of $\varphi_{61}$ in Item 1.

###  VII.2 Overlap with [28, Section 7.1.2]

Reference [28, Theorem 7.1] asserts that a linear map is a Lindbladian if and
only if it can be represented in any of the four forms described in [28, Eqs.
(7.20)-(7.23)]. [28, Eq. (7.23)] matches Eq. 1, where [28] uses the matrix
$C=a^{T}/2$ referred to as the “Kossakowski matrix”.

Reference [28, Proposition 7.4] establishes the uniqueness of the
decomposition of the Lindbladian $\mathcal{L}$ into $\mathcal{L}_{H}$ and
$\mathcal{L}_{a}$, with $H$ being unique up to an additive constant. The
uniqueness of $a$ can be deduced from the second part of this proposition.

The complete positivity condition, Eq. 6, can be obtained from [28] in two
steps. First, by comparing [28, Eq. (7.20)] with [28, Eq. (7.14)], one may
note that that the condition we need is called “Conditional complete
positivity” in [28]. Second, one may transform [28, Proposition 7.2, Eq.
(7.15)] into Eq. 6.

It should be noted that the expression of the complete positivity condition in
Ref. [28] differs from the expression used in our paper, and thus, some work
would be required to derive one condition from the other.

###  VII.3 Overlap with [29]

The blog post [29] provides a complete positivity condition in a form more
similar to the one used in our work. [29] defines666“PT” stands for “partial
transpose”. $\mathcal{L}^{\textrm{PT}}$ as follows. First, introduce a matrix
notation for a superoperator $S$ acting on $M_{d}(\mathbb{C})$ where $S$ is
described by a matrix with elements $S_{(nn^{\prime})(mm^{\prime})}$ (matrix
rows and columns indexed by elements of $\\{1,\dots,d\\}^{2}$) such that:

$(S(A))_{nn^{\prime}}=\sum_{m,m^{\prime}=1}^{d}S_{(nn^{\prime})(mm^{\prime})}A_{mm^{\prime}}.$
(199)

Then, define $S^{\textrm{PT}}$ using

$S^{\textrm{PT}}_{(nm)(n^{\prime}m^{\prime})}=S_{(nn^{\prime})(mm^{\prime})}.$
(200)

An alternative way to describe $S^{\textrm{PT}}$ is

$\operatorname{Tr}(AS^{\textrm{PT}}(B))=\sum_{j,k=1}^{d}\left(AS(|{j}\rangle\\!\langle
k|)B\right)_{jk}.$ (201)

With this notation, the complete positivity condition becomes777In the blog
post, the condition used a “superprojector” removing the trace. The equivalent
alternative used here is to require that $B$ is traceless.

$\forall B\in
M_{0}(d,\mathbb{C})\quad\operatorname{Tr}(B^{\dagger}\mathcal{L}^{\textrm{PT}}(B))\geq
0\ .$ (202)

Using Eq. 201, this can be rewritten as

$\forall B\in
M_{0}(d,\mathbb{C})\quad\sum_{j,k=1}^{d}\left(B^{\dagger}\mathcal{L}(|{j}\rangle\\!\langle
k|)B\right)_{jk}\geq 0.$ (203)

This is the same as requiring Eq. 6 to hold for all traceless
$B\in\mathcal{B}(\mathcal{H})$, as in 1.

##  VIII Summary and Outlook

A standard approach to solving a (time-independent quantum) Markovian quantum
master equation is to vectorize it and solve the corresponding nonhomogeneous
first-order linear ODE for the coherence vector. Here we posed the inverse
problem: when does such a 1ODE of the form $\dot{\vec{v}}=G\vec{v}+\vec{c}$
correspond to a Markovian quantum master equation? When does it correspond to
a completely positive Markovian quantum master equation, i.e., a Lindblad
equation? What are the parameters, i.e., $a$ and $H$, of such master equations
in terms of the parameters $G$ and $\vec{c}$ of the 1ODE? Finally, how
ubiquitous are Lindbladians?

We have shown that the answer to the first question is “always”. We also
expressed the parameters of such Markovian quantum master equations using an
expansion in a nice operator basis, which yields explicit expressions for the
Hamiltonian and the matrix $a$ of coefficients of the dissipator in terms of
the parameters $(G,\vec{c})$ of the 1ODE; see 1. In essence, this means that
every 1ODE of the form $\dot{\vec{v}}=G\vec{v}+\vec{c}$ is directly
representable as a Markovian quantum master equation, Eq. 1. However, complete
positivity (i.e., whether the result is a Lindblad equation) is not guaranteed
and must be checked on a case-by-case basis. Toward this end, we have also
formulated the complete positivity condition directly in terms of
$(G,\vec{c})$; see Eq. 6. This condition is equivalent to the positive
semidefiniteness of $a$, which is simpler to check in practice after $a$ has
been computed from $G$ and $\vec{c}$ using Eq. 5b.

Our work assumed the setting of finite-dimensional Hilbert spaces. We left
open for future research the problem of connecting 1ODEs to quantum master
equations in the infinite-dimensional setting. We also left open the complete
answer to the question of the ubiquity of Lindbladians: namely, while we have
argued that the condition of positivity of $a$ makes Lindbladians extremely
rare when viewed from the perspective of random matrix theory, the answer to
what is the probability that $a\geq 0$ for all real $G$ (sampled from the
Ginibre Orthogonal Ensemble (GinOE) [25]) such that $G$ has non-positive
eigenvalues, is still open. Answering this question will quantify the
probability that a randomly selected 1ODE results in a Lindblad master
equation.

###### Acknowledgements.

We thank Dr. Evgeny Mozgunov for his valuable input and helpful discussions.
We also extend our appreciation to Dr. Frederik vom Ende for pointing out that
Eq. 14 was proven in [1] and for asking us to provide a simple proof (see
Proposition 19), and especially to Dr. Jess Riedel for pointing out the
connections to, and overlap with [28, 29], which led us to add Section VII to
the final version of this work. This research was sponsored by the Army
Research Office and was accomplished under Grant Number W911NF-20-1-0075. The
views and conclusions contained in this document are those of the authors and
should not be interpreted as representing the official policies, either
expressed or implied, of the Army Research Office or the U.S. Government. The
U.S. Government is authorized to reproduce and distribute reprints for
Government purposes, notwithstanding any copyright notation herein.

##  Appendix A Solution of Eq. 3 for the coherence vector

Eq. 3 is a linear, first-order, nonhomogeneous differential equation (1ODE).
We provide the solution in two parts, first for diagonalizable and invertible
$G$, then for general $G$.

###  A.1 Solution for diagonalizable and invertible $G$

Let us first assume that $G$ is diagonalizable over $\mathbb{R}^{J}$ and also
invertible. We look for a solution in the form

$\vec{v}(t)=\vec{v}^{(0)}(t)+\vec{v}^{(\infty)}\ ,$ (204)

where $\vec{v}^{(0)}(t)$ is the homogeneous part and $\vec{v}^{(\infty)}$ is
the nonhomogeneous (time-independent) part. Let $\vec{x}^{(k)}$ and
$\lambda_{k}$ represent the eigenvectors and (possibly degenerate and complex)
eigenvalues of $G$, i.e.,

$G\vec{x}^{(k)}=\lambda_{k}\vec{x}^{(k)}\ ,\qquad k=1,\dots,J\ .$ (205)

It is then straightforward to check by direct differentiation and substitution
that

$\displaystyle\vec{v}^{(0)}(t)$
$\displaystyle=\sum_{k=1}^{M}s_{k}e^{\lambda_{k}t}\vec{x}^{(k)}$ (206a)
$\displaystyle\vec{v}^{(\infty)}$ $\displaystyle=-G^{-1}\vec{c}$ (206b)

in the solution in Eq. 3. The coefficients $s_{k}$ are determined by the
initial condition $\vec{v}(0)$:

$\vec{v}^{(0)}(0)=\sum_{k=1}^{M}s_{k}\vec{x}^{(k)}=X\vec{s}\
,\qquad\text{col}_{k}(X)=\vec{x}^{(k)}\ ,$ (207)

i.e., $X$ is the matrix whose columns are the eigenvectors of $G$. Also,
$\vec{v}^{(0)}(0)=\vec{v}(0)-\vec{v}^{(\infty)}$. Thus

$\vec{s}=X^{-1}(\vec{v}(0)+G^{-1}\vec{c})\ .$ (208)

The eigenvalues can be decomposed as
$\lambda_{k}=\operatorname{Re}(\lambda_{k})+i\operatorname{Im}(\lambda_{k})$.
The imaginary part describes a rotation of $\vec{v}$. The real part is
constrained by complete positivity and trace preservation to be non-positive,
or else the norm of the coherence vector would not be bounded [recall Eq. 42].

###  A.2 Solution for general $G$

The general case is where $G$ is not diagonalizable over $\mathbb{R}^{J}$ and
may not be invertible. In this case, we can still use a similarity
transformation $S$ to transform $G$ into Jordan canonical form:

$G_{J}=SGS^{-1}=\left(\begin{array}[]{ccc}J_{1}&&\\\ &\ddots&\\\ &&J_{q}\\\
\end{array}\right)\ ,$ (209)

where the $q$ Jordan blocks have the form

$J_{j}=\left(\begin{array}[]{ccccc}\mu_{j}&1&&&\\\ &\mu_{j}&\ddots&&\\\
&&\ddots&\ddots&\\\ &&&\mu_{j}&1\\\ &&&&\mu_{j}\\\
\end{array}\right)=\mu_{j}I+K_{j}\ .$ (210)

The $\mu_{j}$’s are the (possibly degenerate, complex) eigenvalues, and
$K_{j}$ are nilpotent matrices: $K_{j}^{d_{j}}=0$, where $d_{j}$ is the
dimension of $J_{j}$. When all $d_{j}=1$, $G$ is diagonalizable, and $G_{J}$
reduces to the diagonalized form of $G$. When one or more of the $d_{j}>1$,
then $G$ is not diagonalizable, meaning that a similarity transformation does
not exist that transforms $G$ into a diagonal matrix. The eigenvalues are the
solutions of $\det{(G-\mu I)}=0$.

Applying $S$ from the left to Eq. 3 yields

$S\dot{\vec{v}}=SGS^{-1}S\vec{v}+S\vec{c}\quad\implies\quad\dot{\vec{w}}=G_{J}\vec{w}+\vec{c}^{\prime}\
,$ (211)

where $\vec{w}=S\vec{v}$ and we defined $\vec{c}^{\prime}=S\vec{c}$. We again
look for a solution in the form

$\vec{w}(t)=\vec{w}^{(0)}(t)+\vec{w}^{(\infty)}\ ,$ (212)

where now $\vec{w}^{(0)}(t)$ is the homogeneous part and $\vec{w}^{(\infty)}$
is the nonhomogeneous (time-independent) part.

First, let us solve for the homogeneous part. Since the Jordan blocks are
decoupled, the general solution of the homogenous part is

$\vec{w}^{(0)}(t)=\bigoplus_{j=1}^{q}\vec{w}_{j}^{(0)}(t)\ ,$ (213)

where we use the direct sum notation to denote that the summands must be
stacked into a single-column vector. Each of the $q$ summands satisfies a
differential equation of the form

$\dot{\vec{w}}_{j}^{(0)}=J_{j}{\vec{w}}_{j}^{(0)}\ .$ (214)

######  Claim 1.

The solution for a general $d_{j}$ dimensional Jordan block is a vector

$\vec{w}_{j}^{(0)}=({w}_{j,1}^{(0)},\dots,{w}_{j,k}^{(0)},\dots,{w}_{j,d_{j}}^{(0)})^{T}$
(215)

with components:

$\displaystyle{w}_{j,k}^{(0)}(t)$ $\displaystyle=e^{\mu_{j}t}p_{d_{j}-k}(t)$
(216a) $\displaystyle p_{d_{j}-k}(t)$
$\displaystyle=\sum_{n=k}^{d_{j}}{w}_{j,n}^{(0)}(0)\frac{t^{n-k}}{(n-k)!}\
,\quad k=1,\dots,d_{j}\ .$ (216b)

Here $p_{d_{j}-k}(t)$ denotes a polynomial of degree $d_{j}-k$. The highest
degree of such polynomials is $d_{j}-1$, of the last component
${w}_{j,d_{j}}^{(0)}$.

###### Proof.

From the structure of the $j$’th Jordan block $J_{j}$, it is clear that

$\displaystyle\dot{w}_{j,d_{j}}^{(0)}$
$\displaystyle=\mu_{j}{w}_{j,d_{j}}^{(0)}$ (217a)
$\displaystyle\dot{w}_{j,k}^{(0)}$
$\displaystyle=\mu_{j}{w}_{j,k}^{(0)}+{w}_{j,k+1}^{(0)}\ ,\qquad
k\in\\{1,\dots d_{j}-1\\}\ ,$ (217b)

and if we differentiate Eq. (216) we obtain, first for $k=d_{j}$:

$\dot{w}_{j,d_{j}}^{(0)}=\mu_{j}e^{\mu_{j}t}p_{0}=\mu_{j}{w}_{j,d_{j}}^{(0)}\
,$ (218)

and second, for $k<d_{j}$:

$\displaystyle\dot{w}_{j,k}^{(0)}$
$\displaystyle=\mu_{j}{w}_{j,k}^{(0)}+\sum_{n=k+1}^{d_{j}}{w}_{j,n}^{(0)}(0)\frac{t^{n-k-1}}{(n-k-1)!}$
(219a) $\displaystyle=\mu_{j}{w}_{j,k}^{(0)}+{w}_{j,k+1}^{(0)}\ ,\qquad
k\in\\{1,\dots d_{j}-1\\}\ .$ (219b)

∎

Note that we can be certain that for all $d_{j}>1$, the corresponding
$\operatorname{Re}(\mu_{j})<0$, since a positive or zero real part would
violate Eq. 42.

Next, let us solve for the nonhomogeneous part. As before, it is determined by
the solution form (212) subject to the boundary conditions

$\displaystyle\vec{w}(0)$ $\displaystyle=\vec{w}^{(0)}(0)+\vec{w}^{(\infty)}$
(220a) $\displaystyle\vec{w}(\infty)$ $\displaystyle=\vec{w}^{(\infty)}\,.$
(220b)

On the one hand, $\dot{\vec{w}}=\dot{\vec{w}}^{(0)}$, and on the other hand
$\dot{\vec{w}}=G_{J}(\vec{w}^{(0)}(t)+\vec{w}^{(\infty)})+\vec{c}^{\prime}=G_{J}\vec{w}^{(0)}(t)+G_{J}\vec{w}^{(\infty)}+\vec{c}^{\prime}$.
Since only the homogeneous part is time-dependent, we must have
$\dot{\vec{w}}^{(0)}=G_{J}\vec{w}^{(0)}(t)$, so that the particular solution
satisfies

$G_{J}\vec{w}^{(\infty)}=-\vec{c}^{\prime}\ .$ (221)

This linear system has a solution for $\vec{w}^{(\infty)}$ iff
$\text{rank}(G_{J})=\text{rank}(G_{J}|\vec{c}^{\prime})$ (the augmented
matrix).

The solution is unique only if, in addition, $G_{J}$ is full rank (equal to
$J$); otherwise, the solution for $\vec{w}^{(\infty)}$ is under-determined.
Since $G_{J}$ is in Jordan form, it is in full-rank row-echelon form unless
there are Jordan blocks with $\mu_{j}=0$. However, as a consequence of the
norm upper bound Eq. 42, any such Jordan blocks must have a dimension
$d_{j}=1$, since otherwise, the polynomials in Eq. 216 would grow unboundedly.
Therefore, the kernel dimension $\dim(G_{J})-\mathrm{rank}(G_{J})$ must be the
number of Jordan blocks with $\mu_{j}=0$ and $d_{j}=1$, and in these blocks,
Eq. 211 becomes the trivial scalar differential equation
$\dot{w}_{j}=c^{\prime}_{j}$, whose solution is
$w_{j}(t)=c^{\prime}_{j}t+w_{j}(0)$. But this means that $c^{\prime}_{j}=0$
since otherwise $w_{j}(t)$ is unbounded. Thus, in each block with $\mu_{j}=0$
we have $w_{j}(t)=w_{j}(0)$. Such Jordan blocks represent frozen degrees of
freedom that do not evolve in time.

##  Appendix B Injectivity of the map $a\mapsto(R,\vec{c})$

Here we provide an alternative proof of injectivity of the linear map
$\mathcal{F}\colon a\mapsto(R,\vec{c})$ introduced in Section III.6, which
repurposes the ideas of 2 to focus on injectivity alone. But first, we need a
Lemma separate from 2:

######  Lemma 4.

If $\mathcal{F}$ maps matrices $a_{1}$ and $a_{2}$ to the same $(R,\vec{c})$,
then $\mathcal{L}_{a_{1}}=\mathcal{L}_{a_{2}}$ where $\mathcal{L}_{a}$ is the
dissipative Liouvillian of Eq. 1c acting on arbitrary $d\times d$ operators
$X$, not just positive $\rho$ with $\operatorname{Tr}(\rho)=1$.

###### Proof.

Proposition 8 trivially holds for arbitrary $d\times d$ operators $X$ and
vectors $\vec{v}$ with complex-valued components computed using Eq. 40, with
$\rho$ replaced by $X$. It shows that after coordinatization the Markovian
quantum master equation is equivalent to a linear ordinary differential
equation for such vectors, so the solution to $\dot{X}=\mathcal{L}_{a}(X)$ is
unique up to initial condition for any $X\in\mathcal{B}(\mathcal{H})$. Suppose
$a_{1}$ and $a_{2}$ both map to the same $\vec{c}$ and $R$ by Eqs. 57 and 58.
Let $X_{1}(t)$ be the solution to
$\dot{X}_{1}(t)=\mathcal{L}_{a_{1}}[X_{1}(t)]$ and $X_{2}(t)$ be the solution
to $\dot{X}_{2}(t)=\mathcal{L}_{a_{2}}[X_{2}(t)]$. Furthermore, let them have
the same initial conditions: $X(0)=X_{1}(0)=X_{2}(0)$. Let $\vec{v}_{1}(t)$
and $\vec{v}_{2}(t)$ be the vectors associated with $X_{1}(t)$ and $X_{2}(t)$
via Eq. 40. Then we have

$\displaystyle\dot{\vec{v}}_{1}(t)$ $\displaystyle=R\vec{v}_{1}(t)+\vec{c}$
(222a) $\displaystyle\dot{\vec{v}}_{2}(t)$
$\displaystyle=R\vec{v}_{2}(t)+\vec{c}\ .$ (222b)

By uniqueness of the solution to $\dot{\vec{v}}=R\vec{v}+\vec{c}$ and that
$\vec{v}_{1}(0)=\vec{v}_{2}(0)$ since $X_{1}(0)=X_{2}(0)$, we have
$\vec{v}_{1}(t)=\vec{v}_{2}(t)$ for all $t$, which implies $X_{1}(t)=X_{2}(t)$
for all $t$. Thus $\dot{X}=\mathcal{L}_{a_{1}}(X)$ and
$\dot{X}=\mathcal{L}_{a_{2}}(X)$ have the same unique solution up to initial
conditions.

For any operator $X(0)$, let $X(t)$ be the unique solution to both
$\dot{X}=\mathcal{L}_{a_{1}}(X)$ and $\dot{X}=\mathcal{L}_{a_{2}}(X)$ with
initial condition $X(0)$. Then
$\dot{X}(t)|_{0}=\mathcal{L}_{a_{1}}[X(0)]=\mathcal{L}_{a_{2}}[X(0)]$ for
arbitrary $X(0)$. Thus $\mathcal{L}_{a_{1}}=\mathcal{L}_{a_{2}}$. ∎

######  Lemma 5.

$\mathcal{F}$ is injective (one-to-one).

###### Proof.

To prove injectivity we need to show that
$\mathcal{F}(a_{1})=\mathcal{F}(a_{2})\implies a_{1}=a_{2}$, but since
$\mathcal{F}$ is linear it suffices to show that
$\mathcal{F}(a)=(0,\vec{0})\implies a=0$.

Let us take a matrix $a$ such that $\mathcal{F}(a)=(0,\vec{0})$. From Lemma 4:

$\mathcal{L}_{a}=\mathcal{L}_{0}=0,$ (223)

where $\mathcal{L}_{a}$ is the dissipative Liouvillian in Eq. 1. Let $\ket{j}$
be the vector with the $j$-th component equal to $1$ and all other components
equal to $0$. Then, from Eq. 223, $\forall j,k=1,\dots,d$ we have
$\mathcal{L}_{a}(|{j}\rangle\\!\langle k|)=0$. In particular, using Eq. 1c:

$\displaystyle 0$
$\displaystyle=\sum_{k=1}^{d}\bra{i}\mathcal{L}_{a}(|{j}\rangle\\!\langle
k|)\ket{k}$ (224a)
$\displaystyle=\,\sum_{{m,n=1}}^{J}\,a_{mn}\big{[}\bra{i}F_{m}\ket{j}\sum_{k=1}^{d}\bra{k}F_{n}\ket{k}-\frac{1}{2}\bra{i}F_{n}F_{m}\ket{j}\sum_{k=1}^{d}\langle{k}|{k}\rangle$
$\displaystyle\qquad-\frac{1}{2}\langle{i}|{j}\rangle\sum_{k=1}^{d}\bra{k}F_{n}F_{m}\ket{k}\big{]}$
(224b)
$\displaystyle=\sum_{m,n=1}^{J}a_{mn}\big{[}\bra{i}F_{m}\ket{j}\operatorname{Tr}(F_{n})-\frac{d}{2}\bra{i}F_{n}F_{m}\ket{j}$
$\displaystyle\qquad-\frac{1}{2}\delta_{ij}\operatorname{Tr}(F_{n}F_{m})\big{]}$
(224c)
$\displaystyle=-\frac{1}{2}\delta_{ij}\operatorname{Tr}(a)-\frac{d}{2}\sum_{m,n=1}^{J}a_{mn}\bra{i}F_{n}F_{m}\ket{j}\
,$ (224d)

where we used $\operatorname{Tr}(F_{n})=0$ and
$\operatorname{Tr}(F_{n}F_{m})=\delta_{nm}$ (1). For the same reason,
$\sum_{i=1}^{d}\bra{i}F_{n}F_{m}\ket{i}=\operatorname{Tr}(F_{n}F_{m})=\delta_{nm}$,
so that after summing Eq. 224d over $i=j=1,\ldots,d$ we obtain

$0=-\frac{1}{2}\sum_{i=1}^{d}\operatorname{Tr}(a)-\frac{d}{2}\sum_{m,n=1}^{J}a_{mn}\delta_{nm}=-d\operatorname{Tr}(a)\
,$ (225)

i.e., $\operatorname{Tr}(a)=0$. From Eq. 224d it follows that
$\bra{i}A\ket{j}=0$ $\forall i,j$, where
$A\equiv\sum_{m,n=1}^{J}a_{mn}F_{n}F_{m}$, i.e., $A=0$. Hence, we have shown
that $\forall X$:

$\mathcal{L}_{a}(X)=\sum_{m,n=1}^{J}a_{mn}F_{n}XF_{n}\ .$ (226)

We can now express the matrix elements $a_{mn}$ in terms of $\mathcal{L}_{a}$,
as follows:

$\displaystyle a_{mn}$
$\displaystyle=\sum_{m^{\prime},n^{\prime}=1}^{J}a_{m^{\prime}n^{\prime}}\operatorname{Tr}(F_{n}F_{n^{\prime}})\operatorname{Tr}(F_{m^{\prime}}F_{m})$
(227a)
$\displaystyle=\sum_{j,k=1}^{d}\bra{j}F_{n}\big{[}\sum_{m^{\prime},n^{\prime}=1}^{J}a_{m^{\prime}n^{\prime}}F_{n^{\prime}}|{j}\rangle\\!\langle
k|F_{m^{\prime}}\big{]}F_{m}\ket{k}$ (227b)
$\displaystyle=\sum_{j,k=1}^{J}\bra{j}F_{n}\mathcal{L}_{a}(|{j}\rangle\\!\langle
k|)F_{m}\ket{k}=0\ ,$ (227c)

since $\mathcal{L}_{a}=0$. That is, $a=0$. ∎

##  Appendix C Facts from convex geometry

In this section, we summarize the facts from convex geometry needed for this
work.

Let $M_{{+}}(J,\mathbb{C})$ be the set of positive semidefinite $J\times J$
matrices with complex coefficients:

$M_{{+}}(J,\mathbb{C})=\left\\{a\in M_{\textrm{sa}}(J,\mathbb{C}):\forall
b\in\mathbb{C}^{J}\;b^{\dagger}ab\geq 0\right\\}.$ (228)

This equation describes $M_{{+}}(J,\mathbb{C})$ as the intersection of
infinitely many half-spaces in $M_{\textrm{sa}}(J,\mathbb{C})$, whose
boundaries all pass through the origin $a=0$. Therefore,
$M_{{+}}(J,\mathbb{C})$ is a closed convex cone in
$M_{\textrm{sa}}(J,\mathbb{C})$.

Let

$M_{{+},1}(J,\mathbb{C})=\\{a\in
M_{{+}}(J,\mathbb{C})\colon\;\operatorname{Tr}(a)=1\\},$ (229)

i.e., the set of unit-trace, positive-semidefinite matrices.

######  Lemma 6.

$M_{{+}}(J,\mathbb{C})$ is a closed convex cone with a compact base, which can
be chosen to be $M_{{+},1}(J,\mathbb{C})$.

###### Proof.

As noted above, $M_{{+}}(J,\mathbb{C})$ is a closed convex cone. By
definition, $M_{{+},1}(J,\mathbb{C})$ is an intersection of that cone and a
hyperplane not passing through the origin. Thus, it remains to prove that

  1. 1.

any of the rays in $M_{{+}}(J,\mathbb{C})$ passes through
$M_{{+},1}(J,\mathbb{C})$, i.e., every non-zero element of
$M_{{+}}(J,\mathbb{C})$ is proportional to an element of
$M_{{+},1}(J,\mathbb{C})$ with a positive coefficient;

  2. 2.

$M_{{+},1}(J,\mathbb{C})$ is compact.

To prove the first statement, let $a\neq 0,a\in M_{{+}}(J,\mathbb{C})$. Since
$a$ is Hermitian it can be diagonalized. Eq. 228 implies that all its
eigenvalues are non-negative, hence $\operatorname{Tr}(a)\geq 0$. In fact, at
least one of the eigenvalues has to be positive, otherwise $a=0$, hence
$\operatorname{Tr}(a)>0$. Therefore
$a=\operatorname{Tr}(a)(a/\operatorname{Tr}(a))$ where the coefficient
$\operatorname{Tr}(a)>0$ and the element $a/\operatorname{Tr}(a)\in
M_{{+},1}(J,\mathbb{C})$.

Since $M_{{+},1}(J,\mathbb{C})$ is an intersection of closed sets, it is
closed. Thus, it remains to prove it is bounded, e.g.:

$\forall a\in
M_{{+},1}(J,\mathbb{C})\quad\sum_{i,j=1}^{J}\left|a_{ij}\right|^{2}\leq 1\ .$
(230)

Let $\\{\lambda_{i}\\}_{i=1}^{J}$ be the eigenvalues of such an $a$. Then
$\sum_{i=1}^{J}\lambda_{i}=1$ and $0\leq\lambda_{i}\leq 1$. Therefore, indeed,

$\sum_{i,j=1}^{J}\left|a_{ij}\right|^{2}=\operatorname{Tr}(a^{\dagger}a)=\sum_{i=1}^{J}\lambda_{i}^{2}\leq\sum_{i=1}^{J}\lambda_{i}=1\
.$ (231)

∎

######  Lemma 7.

The extreme rays of $M_{{+}}(J,\mathbb{C})$ are the rays generated by rank 1
matrices.

Note: this is [30, Exercise 7.5.15]. Here we provide a proof for completeness.

###### Proof.

Let $\\{\alpha a:\alpha\in\mathbb{R}_{+}\\}$ be an extreme ray in
$M_{{+}}(J,\mathbb{C})$. Here $a\in M_{{+}}(J,\mathbb{C})\setminus\\{0\\}$.
Since $a$ is Hermitian, we can diagonalize it by a unitary, hence write it in
the form

$a=\sum_{i=1}^{k}b_{i}b_{i}^{\dagger}$ (232)

for some $b_{i}\in\mathbb{C}^{J}$ orthogonal to each other, where
$k=\operatorname{rank}(a)$. Since $a\neq 0$, $k\geq 1$. If $k\geq 2$ then Eq.
232 represents $a$ as a sum of non-proportional elements of
$M_{{+}}(J,\mathbb{C})$, contradicting the assumption that $a$ lies on an
extreme ray in $M_{{+}}(J,\mathbb{C})$. Thus, $k=1$.

Vice-versa, if $\operatorname{rank}(a)=1$, $a\in M_{{+}}(J,\mathbb{C})$, we
want to prove that $a$ lies on an extreme ray of $M_{{+}}(J,\mathbb{C})$.
Assume to the contrary that

$a=a_{1}+a_{2}$ (233)

where $a_{1},a_{2}\in M_{{+}}(J,\mathbb{C})$ and $a_{1}$ is not proportional
to $a$. We can write $a=\alpha bb^{\dagger}$ for some
$\alpha\in\mathbb{R}_{+}$, $b\in\mathbb{C}^{J}$ with $\left\|b\right\|=1$,
$\alpha>0$. Then for $i=1,2$, we can decompose

$a_{i}/\alpha=\beta_{i}bb^{\dagger}+\tilde{a}_{i}$ (234)

where $\beta_{i}=b^{\dagger}a_{i}b/\alpha$, $b^{\dagger}\tilde{a}_{i}b=0$. Eq.
233 implies that $\beta_{1}+\beta_{2}=1$, $\tilde{a}_{1}+\tilde{a}_{2}=0$. By
our assumption $\tilde{a}_{1}\neq 0$ (otherwise $a_{1}$ would be proportional
to $a$). Therefore, there is a vector $c$ s.t. $c^{\dagger}\tilde{a}_{1}c\neq
0$. Let $c=\gamma b+c_{\perp}$, where $b^{\dagger}c_{\perp}=0$. Note that
$c_{\perp}^{\dagger}\tilde{a}_{1}c_{\perp}=c_{\perp}^{\dagger}a_{1}c_{\perp}/\alpha$,
hence $c_{\perp}^{\dagger}\tilde{a}_{1}c_{\perp}\geq 0$. Similarly,
$c_{\perp}^{\dagger}\tilde{a}_{1}c_{\perp}=-c_{\perp}^{\dagger}a_{2}c_{\perp}/\alpha$,
hence $c_{\perp}^{\dagger}\tilde{a}_{1}c_{\perp}\leq 0$. Thus,
$c_{\perp}^{\dagger}\tilde{a}_{1}c_{\perp}=0$.

For any $\delta\in\mathbb{C}$ we have

$(b+\delta c_{\perp})^{\dagger}a_{1}(b+\delta c_{\perp})\geq 0\ ,$ (235)

and we have just shown that the quadratic term in this expression is zero.
Since the inequality Eq. 235 has to hold for all $\delta$, the linear term is
zero too, i.e., $b^{\dagger}a_{1}c_{\perp}=0$. Thus,
$c^{\dagger}\tilde{a}_{1}c=0$, contradicting the assumption. ∎

######  Lemma 8.

$M_{{+}}(J,\mathbb{C})$ is the convex hull of its extreme rays.

Here we present two alternative ways to see this.

###### Proof 1.

Any positive semidefinite matrix can be diagonalized. Such diagonalization
results [30, Theorem 7.5.2] in a decomposition of the matrix into rank-$1$
positive semidefinite matrices [as in Eq. 232], which – according to Lemma 7 –
lie on the extreme rays of $M_{{+}}(J,\mathbb{C})$. ∎

###### Proof 2.

In the context of finite dimensional vector spaces, Krein-Milman (also known
as Minkowski’s theorem, see, e.g., [31, Corollary 18.5.1]) states that any
compact convex set is the convex hull of its extreme points. As a corollary
any convex cone with a compact base is the convex hull of its extreme rays. As
follows from Lemma 6, $M_{{+}}(J,\mathbb{C})$ is such a cone. ∎

##  Appendix D Representation of a superoperator using a nice operator basis

######  Proposition 19.

Any superoperator $\mathcal{E}\in\mathcal{B}[\mathcal{B}(\mathcal{H})]$ can
always be represented as

$\mathcal{E}(A)=\sum_{{i,j=0}}^{J}c_{ij}F_{i}\bullet F_{j}\ .$ (236)

This is [1, Lemma 2.2].

It can also be seen directly by noting that any linear operator can be
represented by a matrix and, thus,
$\mathcal{E}(A)_{kn}=\sum_{l,m=1}^{d}\mathcal{E}_{klmn}A_{lm}$ for some tensor
$\mathcal{E}_{klmn}$. The tensor $\mathcal{E}_{klmn}$ can be seen as a matrix
in 2 ways: first, as a function of indices $k$ and $l$, and second, as a
function of indices $m$ and $n$. Applying “coordinatization” [Eqs. 8 and 9]
twice to $\mathcal{E}_{klmn}$ we get $c_{ij}$.

###### Proof.

Let $\mathcal{E}$ be any superoperator in
$\mathcal{B}[\mathcal{B}(\mathcal{H})]$. Denoting
$\mathcal{E}_{klmn}=\mathcal{E}(|{l}\rangle\\!\langle m|)_{kn}$ and applying
Eq. 11 to $\delta_{kk^{\prime}}\delta_{ll^{\prime}}$ and
$\delta_{mm^{\prime}}\delta_{nn^{\prime}}$ for any
$A\in\mathcal{B}(\mathcal{H})$ we get

$\displaystyle\mathcal{E}(A)_{kn}=\sum_{l,m=1}^{d}\mathcal{E}_{klmn}A_{lm}$
(237a)
$\displaystyle\quad=\sum_{k^{\prime},l^{\prime},m^{\prime},n^{\prime},l,m=1}^{d}\delta_{kk^{\prime}}\delta_{ll^{\prime}}\delta_{mm^{\prime}}\delta_{nn^{\prime}}\mathcal{E}_{k^{\prime}l^{\prime}m^{\prime}n^{\prime}}A_{lm}$
(237b)
$\displaystyle\quad=\sum_{k^{\prime},l^{\prime},m^{\prime},n^{\prime},l,m=1}^{d}\sum_{i,j=0}^{J}F_{i\,kl}F_{i\,l^{\prime}k^{\prime}}F_{j\,mn}F_{j\,n^{\prime}m^{\prime}}\mathcal{E}_{k^{\prime}l^{\prime}m^{\prime}n^{\prime}}A_{lm}$
(237c) $\displaystyle\quad=\sum_{i,j=0}^{J}c_{ij}(F_{i}AF_{j})_{kn},$ (237d)

where

$c_{ij}=\sum_{k^{\prime},l^{\prime},m^{\prime},n^{\prime}=1}^{d}F_{i\,l^{\prime}k^{\prime}}F_{j\,n^{\prime}m^{\prime}}\mathcal{E}_{k^{\prime}l^{\prime}m^{\prime}n^{\prime}}.$
(238)

∎

##  Appendix E Computation of the constant $\theta^{\text{GinOE}}$

The constant $\theta^{\text{GinOE}}$ in Eq. 192 can be computed as

$\theta^{\text{GinOE}}=\inf_{\rho\in\mathcal{S}_{{+}}}E(\rho)-\inf_{\rho\in\mathcal{S}}E(\rho),$
(239)

where

$\displaystyle E(\rho)$
$\displaystyle=\frac{1}{2}\int_{\mathbb{C}}\left(\left|z\right|^{2}-\int_{\mathbb{C}}\ln\left|z-w\right|\rho(w)d^{2}w\right)\rho(z)d^{2}z,$
(240a) $\displaystyle\mathcal{S}$ $\displaystyle=\Bigl{\\{}\rho\in
C_{0}(\mathbb{C}\to\mathbb{R}_{+}):\int_{\mathbb{C}}\rho(z)d^{2}z=1,$
$\displaystyle\qquad\forall z\;\rho(z)=\rho(z^{*})\Bigr{\\}},$ (240b)
$\displaystyle\mathcal{S}_{{+}}$
$\displaystyle=\left\\{\rho\in\mathcal{S}:\forall
z\;\operatorname{Re}(z)<0\Rightarrow\rho(z)=0\right\\},$ (240c) $\displaystyle
d^{2}z$
$\displaystyle=d\operatorname{Re}(z)d\operatorname{Im}(z)=\frac{i}{2}dzdz^{*},$
(240d)

where $C_{0}(X\to Y)$ is the set of continuous functions from $X$ to $Y$ with
compact support. One can then compute that
$\inf_{\rho\in\mathcal{S}}E(\rho)=5/8-\ln(2)/4$ and is achieved near
$\rho(z)=\frac{1}{2\pi}\mathbb{I}(\left|z\right|^{2}<2)$, where $\mathbb{I}$
is the indicator function. One can then try to argue that the
$\inf_{\rho\in\mathcal{S}_{{+}}}E(\rho)$ is achieved for $\rho$ of the
following form:

$\rho(x+iy)=g(y)\delta(x)+\frac{1}{2\pi}\mathbb{I}(0\leq x\leq f(y)).$ (241)

Note that while the function Eq. 241 is not continuous, it can be approximated
by a continuous function in $\mathcal{S}_{{+}}$ as soon as $f$ and $g$ are
non-negative continuous functions with compact support and

$\int_{\mathbb{R}}\left(g(y)+\frac{f(y)}{2\pi}\right)dy=1.$ (242)

For such $\rho$ we have

$E(\rho)=\int_{\mathbb{R}}dy_{1}\Biggl{(}\frac{f(y_{1})^{3}}{12\pi}+\frac{y_{1}^{2}f(y_{1})}{4\pi}+g(y_{1})\frac{y_{1}^{2}}{2}-\\\
\int_{\mathbb{R}}dy_{2}\biggl{(}g(y_{1})g(y_{2})\ln\left|y_{1}-y_{2}\right|/2+\\\
\frac{1}{2\pi}g(y_{2})I_{1}(f(y_{1}),\left|y_{2}-y_{1}\right|)+\\\
\frac{1}{8\pi^{2}}I_{2}(f(y_{1}),f(y_{2}),y_{2}-y_{1})\biggr{)}\Biggr{)},$
(243)

where

$\displaystyle I_{1}(x,y)$
$\displaystyle=\int_{0}^{x}dx_{1}\ln\left|x_{1}+iy\right|,$ (244a)
$\displaystyle I_{2}(x_{1},x_{2},y)$
$\displaystyle=\int_{0}^{x_{1}}dx_{3}\int_{0}^{x_{2}}dx_{4}\ln\left|x_{2}-x_{4}+iy\right|.$
(244b)

One can then solve this optimization problem numerically (or attempt to find
an analytical solution) to find $\theta^{\text{GinOE}}$.

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