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Where To Apply
Wisconsin residents must apply in the county where at least one applicant has resided for the last 30 days.
If neither applicant is a resident of Wisconsin, the couple must apply in the Wisconsin county where the marriage will be performed.
Both applicants must apply in person in the County Clerk's office. We suggest that applicants arrive no later than 3:15 p.m. on any given day.
Fee, Waiting Period, & License Expiration
The cost is $60 payable by check, cash or by credit card. We accept Mastercard, Discover, VISA and American Express.
The marriage license is issued after a five-day waiting period which starts counting on the day after the license was applied for. The license is valid for 30 days.
Under special circumstances, the waiting period may be waived on approval of the County Clerk. The waiver fee is $10.
Required Documents Needed
In Wisconsin, applicants who were previously married will have to show how the last marriage ended. If the applicants were divorced, in Wisconsin, they MAY NOT marry for six (6) months from the date of the final divorce. A document of how the marriage ended must be provided. This document will be either a Judgment of Divorce or an Annulment, it must be signed and stamped by the Judge of that Circuit Court.
If the spouse of an applicant is deceased, the applicant will have to provide a certified copy of the death certificate which then shows how that marriage ended.
Other documents that will also be needed when you are applying for a marriage license are:
Valid driver's license or government issued ID card showing applicant's residence. If the residence address is incorrect, also bring one of the following showing the new residence:
Utility bill (not less than 90 days' old)
Credit card statement
University, college or technical college ID card
Property tax statement.
Certified copy of Birth Certificate - this information is very important. This document is issued by the state where the applicant was born. It must be a certified that contains a raised seal on it. Many confuse this with the original birth record from the hospital where the applicant was born. If it says anything at all describing a hospital, it is the wrong document - check and double check.
A photocopy will not be accepted.
A certification of birth registration or a certificate issued by a hospital will not be accepted.
If you were previously married please bring proof of how the last marriage ended, such as the following:
Certified copy of a death certificate
Court stamped and signed copy of the divorce decree (judgment of divorce) or annulment.
Wedding Information Needed
Date of Wedding
Location (county, city, village or township)
Officiant name, address and telephone number
The age requirement is 18 years or order. Applicants 16 or 17 years old will need a signed consent form is required. The consent form must be signed by the parent(s), legal guardian, custodian (appointed by a court), or person having actual care, custody and control of the applicant.
The consent form is an official oath; it states the person signing swears he/she does in fact have the legal authority to give consent and that the consent is given. It must be signed in the presence of the County Clerk or in the presence of a Notary Public and it must be presented to the County Clerk at the time of license application.
If applicant is 16 or 17 years old the age requirement terms will apply.
If applicant is 18 years or older, the terms of each specific guardianship will apply.
returning a marriage license
The marriage license must be filed with the Register of Deeds Office within three (3) days following the wedding ceremony. A marriage license may be mailed or delivered in person.
Wisconsin residents that obtain a marriage license in the county they reside in should follow one of these two examples:
If applicant lives in Bayfield County and gets married in a different Wisconsin County, they will get their license at the Bayfield County Clerk's Office and will return the license to the Register of Deeds Office in the County where the marriage was held.
If applicant lives in Bayfield County and gets married in Bayfield County, they will get their license at the Bayfield County Clerk's Office and will return the license to the Bayfield County Register of Deeds Office.
Out of state applicants will apply and receive their marriage license in the county they are marrying in and must return it to the Register of Deeds Office in the same county.
For example: Any out of state applicants that are getting married in Bayfield County will apply for their marriage license at the Bayfield County Clerk's Office and return the license to the Bayfield County Register of Deeds Office.
Documents Not In English
If any of the above needed documents are not in English, documents need to be translated into English, prior to the filing for the marriage license. A Certificate of Translation must then accompany the original documents when applying for the marriage license.
Lynn Divine
Officiants for Civil Marriage


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text:
\section{Introduction}

We study spectral decomposition of a general quantum Lindblad equation \cite{lindblad,gorini,alicki} with quadratic generators \cite{fannes1,fannes2} for an open system of $n$ fermions. The physical picture and the motivation for the study of such a non-equilibrium quantum many-body problem have been outlined in Ref. \cite{njp} (see also \cite{prl}), on which we shall rely extensively.  However, in this note we show that an important assumption of diagonalizability made in Ref. \cite{njp} can be removed and that one can prove the spectral theorem for a general quadratic many-body Liouvillean and construct its canonical Jordan form.

The main results presented bellow, are: (i) computationally efficient\footnote{Computational problem of diagonalizing $4n\times 4n$ complex matrix of Ref.\cite{njp} has been reduced to a problem of diagonalizing a $2n\times 2n$ real matrix (or possibly finding its Jordan canonical form if it cannot be diagonalized) and/or
solving a $2n\times 2n$ linear matrix equation of Sylvester type \cite{bartels}.} canonical representation of the structure matrix of the Liouvillean quadratic form as summarized in lemmas \ref{lemma:zerodriving}, \ref{lemma:similar}, (ii) writing the general (possibly non-diagonalizable) {\em normal form} of the quadratic Lioivilleans in terms of normal master mode maps [formula (\ref{eq:dLL})],  (iii) constructing the full many-body Liouvillean spectrum, the invariant subspaces, and the largest Jordan block in each of them (theorem \ref{spctheorem}), and (iv) algebraic construction of the convex subspaces of possibly degenerate (multiple) non-equlibrium steady states (theorem \ref{uniqueness}).

A general master equation governing time evolution of the density matrix $\rho(t)$ of an open quantum system, preserving the trace and positivity of $\rho$, can be written in the Lindblad form
 \cite{lindblad} as
\begin{equation}
\frac{ {\rm d} \rho }{ {\rm d}  t} = {\hat {\cal L}}\rho :=
- {\rm i}  [H,\rho] + \sum_{\mu} \left(2 L_\mu \rho L_\mu^\dagger - \{L_\mu^\dagger L_\mu,\rho\} \right)
\label{eq:lind}
\end{equation}
where $H$ is a Hermitian operator (Hamiltonian), $[x,y]:=xy-yx$, $\{x,y\}:=xy+yx$, and $L_\mu$ are arbitrary operators, physically representing couplings to various baths.

We shall treat open quantum systems of a finite number, say $n$ fermions, with a {\em quadratic} generator ${\hat {\cal L}}$. Mathematically speaking, we shall restrict our discussion to the case where a set of $2n$ anti-commuting, Hermitian, bounded operators exist, say $w_j$, $j=1,\ldots,2n$,
\begin{equation}
\{w_j,w_k\} = 2\delta_{j,k}, \qquad j,k =1,\ldots, 2n,
\label{eq:CARw}
\end{equation}
so that the Hamiltonian, and the Lindblad operators, can be simultaneously expressed in terms of a {\em quadratic form}, and {\em linear forms}, respectively
\begin{eqnarray}
H &=& \sum_{j,k=1}^{2n} w_j H_{jk} w_k = \un{w} \cdot \mm{H}\, \un{w}, \label{eq:hamil}  \\
L_\mu &=& \sum_{j=1}^{2n} l_{\mu,j} w_j = \un{l}_\mu \cdot \un{w}.\  \label{eq:lindb}
\end{eqnarray}
Throughout this paper $\un{x}=(x_1,x_2,\ldots)^T$ will designate a vector (column)
of appropriate scalar valued or operator valued symbols $x_k$ and upright bold letter (e.g. $\mm{H}$) will designate
a complex or real matrix.

The main conceptual tool of our analysis is a Fock space ${\cal K}$ of operators (which may be referred to also as {\em Liouville-Fock} space) which the density operator $\rho(t)$ (\ref{eq:lind}) is a member of.
${\cal K}$ is in fact a $2^{2n} = 4^n$ dimensional linear vector space spanned by a canonical basis
 $\xket{P_{\un{\alpha}}}$ with
\begin{equation}
P_{\alpha_1,\alpha_2,\ldots,\alpha_{2n}} := 2^{-n/2} w_1^{\alpha_1}w_2^{\alpha_2}\cdots w_{2n}^{\alpha_{2n}}, \qquad
\alpha_j\in\{ 0,1\},
\end{equation}
which is  {\em orthonormal} with respect to Hilbert-Schmidt inner product
\begin{equation}
\xbraket{x}{y} = {\,{\rm tr}\,} x^\dagger y.
\label{eq:inner}
\end{equation}
Introducing  {\em creation and annihilation linear
maps} $\hat{c}_j$ over ${\cal K}$
\begin{equation}
 {\hat c} ^\dagger_j \xket{P_{\un{\alpha}}} = \delta_{\alpha_j,0} \xket{w_j P_{\un{\alpha}}} \quad
 {\hat c} _j \xket{P_{\un{\alpha}}} = \delta_{\alpha_j,1} \xket{w_j P_{\un{\alpha}}}
\label{eq:defanih}
\end{equation}
the space ${\cal K}$ becomes a fermionic Fock space. {\em Canonical anticommutation relations} (CAR) among operator maps
$\{ {\hat c} _j, {\hat c} _k\} = 0,\; \{ {\hat c} _j, {\hat c} ^\dagger_k\}=\delta_{j,k}$ and the fact that $ {\hat c} ^\dagger_j$ is indeed a Hermitian adjoint of $ {\hat c} _j$ w.r.t. (\ref{eq:inner}) can be checked straightforwardly.

As shown in Ref.\cite{njp} the Lindblad equation (\ref{eq:lind}) in the Liouville-Fock picture
\begin{equation}
\frac{ {\rm d} }{ {\rm d}  t}\ket{\rho} = {\hat {\cal L}} \ket{\rho}
\label{eq:liouville}
\end{equation}
takes manifestly quadratic form, where the Liouvillean map is expressed as \footnote{Note also a technical issue of a {\em parity symmetry} of
the Liouvillean and the operator space w.r.t. even and odd number of fermionic operator-excitations \cite{njp} which we may ignore here without affecting any of the
results.}
\begin{equation}
{\hat {\cal L}} = \un{ {\hat a} }\cdot\mm{A}\un{ {\hat a} } - A_0 \hat{\mathbbm{1}}
\label{eq:Amatrix}
\end{equation}
in terms of $4n$ Hermitian Majorana fermionic maps over the operator space ${\cal K}$,
\begin{equation}
 {\hat a} _{1,j} := ( {\hat c} _j +  {\hat c} _j^\dagger)/\sqrt{2},\quad
 {\hat a} _{2,j} :=  {\rm i}  ( {\hat c} _j -  {\hat c} _j^\dagger)/\sqrt{2}
\end{equation}
satisfying CAR
\begin{equation}
\{ {\hat a} _{\nu,j}, {\hat a} _{\mu,k}\} = \delta_{\nu,\mu}\delta_{j,k},\quad \nu,\mu=1,2,\;\; j,k=1,\ldots,2n.
\end{equation}
We note that physically interesting {\em non-equilibrium steady state} (NESS), $\xket{{\rm NESS}} \in {\cal K}$, can be defined as the (possibly degenerate, and then non-unique)
right zero-eigenvalue eigenvector of the Liouvillean ${\hat {\cal L}} \xket{{\rm NESS}} = 0$.
The antisymmetric $4n \times 4n$ {\em structure matrix} $\mm{A}$ (\ref{eq:Amatrix}) has the general form \cite{njp}
\begin{eqnarray}
\mm{A} &=&
\pmatrix{
-2 {\rm i} \mm{H} + 2 {\rm i}  \mm{M}_{\rm i} & 2 {\rm i}  \mm{M} \cr
-2 {\rm i} \mm{M}^T & -2 {\rm i} \mm{H} - 2 {\rm i}  \mm{M}_{\rm i}}  \nonumber \\
&=& -2 {\rm i}  \mathbbm{1}_2 \otimes \mm{H} - 2 \sigma^2 \otimes \mm{M}_{\rm r} - 2(\sigma^1 -  {\rm i}  \sigma^3)\otimes \mm{M}_{\rm i}
\label{eq:AA}
\end{eqnarray}
where $\sigma^\alpha,\alpha=1,2,3$ are the Pauli matrices, $\mm{H}=-\mm{H}^T$ is an antisymmetric (due to (\ref{eq:CARw})) $2n\times 2n$ Hermitean (purely imaginary) matrix, whereas
\begin{equation}
\mm{M}_{\rm r} := \frac{1}{2}(\mm{M}+ \bar{\mm{M}}) = \mm{M}_{\rm r}^T,\qquad
\mm{M}_{\rm i} := \frac{1}{2 {\rm i} }(\mm{M} - \bar{\mm{M}}) = -\mm{M}_{\rm i}^T
\end{equation}
are the real and imaginary parts of the positive semidefinite Hermitian $2n\times 2n$ Lindblad {\em bath matrix} $\mm{M} := \sum_\mu \un{l}_\mu \otimes \bar{\un{l}}_\mu$, $\mm{M} \ge 0$.
Note that, physically speaking,
$\mm{H}$ encodes the Hamiltonian, $\mm{M}_{\rm r}$ encodes the {\em dissipation} and $\mm{M}_{\rm i}$ encodes the {\em driving}.
The scalar $A_0$ (\ref{eq:Amatrix}) has a value
\begin{equation}
A_0 = 2{\,{\rm tr}\,} \mm{M} = 2{\,{\rm tr}\,} \mm{M}_{\rm r}.
\label{eq:A0}
\end{equation}
Writing the skew unit
\begin{equation}
\mm{J} := \sigma^1 \otimes \mathbbm{1}_{2n}
\end{equation}
we observe that the structure matrix satisfies a particular self-conjugation property
\begin{equation}
\bar{\mm{A}} = \mm{J}  \mm{A} \mm{J}, \quad {\rm or\; equivalently} \quad \mm{A} \mm{J} = \mm{J} \bar{\mm{A}}.
\end{equation}

\section{Preliminaries}

A series of useful lemmas on the spectral decomposition of the structure matrix $\mm{A}$ shall now be straightforwardly proven, which explore
the structure of the problem and go beyond the general spectral theorems for the anti-symmetric complex matrices \cite{matrixtheory}.

\begin{lemma} Structure matrix (\ref{eq:AA}) is unitarily equivalent to a block-triangular matrix
\begin{equation}
\mm{\tilde A} = -2 {\rm i} \mathbbm{1}_2\otimes \mm{H} - 2\sigma^3 \otimes \mm{M}_{\rm r} +
4 {\rm i}  \sigma^+ \otimes \mm{M}_{\rm i} = \pmatrix{
-\mm{X}^T  & 4 {\rm i}  \mm{M}_{\rm i}\cr
\mm{0} & \mm{X}} \label{eq:tA}
\end{equation}
where
\begin{equation}
\mm{X} := -2 {\rm i}  \mm{H} + 2\mm{M}_{\rm r}
\label{eq:X}
\end{equation}
is a {\rm real} $2n \times 2n$ matrix.
\label{lemmatilde}
\end{lemma}
\begin{proof}
The similarity transformation $\mm{A}\to \tilde{\mm{A}}$
\begin{equation}
\tilde{\mm{A}} = \mm{U}\mm{A}\mm{U}^\dagger
\label{eq:Asim}
\end{equation}
is trivially provided by the cyclic permutation $\sigma^1 \to \sigma^2, \sigma^2 \to \sigma^3, \sigma^3 \to \sigma^1$,
\begin{equation}
\mm{U} = \frac{1}{\sqrt{2}}\pmatrix{1 & - {\rm i}  \cr 1 &  {\rm i} } \otimes \mathbbm{1}_{2n}.
\end{equation}
\qed
\end{proof}

We shall refer to the representation in which the structure matrix has a form (\ref{eq:tA}) as the tilde-representation.
As an immediate consequence of lemma \ref{lemmatilde} we have:

\begin{corollary}
The characteristic polynomial of the structure matrix $\mm{A}$
factorizes in terms of the characteristic polynomial of $\mm{X}$ (\ref{eq:X})
\begin{equation}
p_\mm{A}(\beta) = p_\mm{X}(\beta) p_\mm{X}(-\beta),
\end{equation}
where $p_\mm{A}(\beta) := {\,{\rm det}\,}(\mm{A}-\beta \mathbbm{1}_{4n}), p_\mm{X}(\beta) := {\,{\rm det}\,}(\mm{X}-\beta \mathbbm{1}_{2n})$
and hence, if $\sigma(\mm{A})$ denotes the spectrum of a matrix,
\begin{equation}
\sigma(\mm{A}) = \sigma(\mm{X}) \cup (-\sigma(\mm{X})),
\end{equation}
where the union $\cup$ should keep track of (algebraic) multiplicities.
\end{corollary}
Curiously enough, this implies that the spectrum of the structure matrix is insensitive to the driving matrix $\mm{M}_{\rm i}$.
Note also that the real matrix $\mm{X}$ (\ref{eq:X}) has no general structure apart from the fact that $\mm{X} + \mm{X}^T = 4\mm{M}_r \ge 0$,
since $\mm{M}_r \ge 0$ is implied by $\mm{M} \ge 0$. Sometimes the eigenvalues $\{ \beta_j \}$ of $\mm{X}$ shall be referred to as {\em rapidities}.

\begin{lemma}
\label{lemma:stability}
Let $\mm{X}$ be a  {\em real} square matrix, such that $\mm{X} + \mm{X}^T \ge 0$.
Then:
\begin{enumerate}
\item Any eigenvalue $\beta$ of $\mm{X}$ satisfies ${\,{\rm Re}\,} \beta \ge 0$.
\item For any eigenvalue $\beta$ of $\mm{X}$ on the imaginary line, ${\,{\rm Re}\,} \beta= 0$, its algebraic and geometric multiplicities coincide.
\end{enumerate}
\end{lemma}
\begin{proof}
(i) Take an eigenvalue $\beta$ and an eigenvector $\un{u}$, write $\mm{X}\un{u} = \beta \un{u}$, and the complex conjugate of this equation
$\mm{X}\bar{\un{u}} = \bar{\beta} \bar{\un{u}}$. Then take a dot product of the first equation with $\bar{\un{u}}$ and the dot product of the second
equation with $\un{u}$ and sum up:
\begin{equation}
\bar{\un{u}}\cdot (\mm{X}+\mm{X}^T)\un{u} = (2 {\,{\rm Re}\,} \beta) \bar{\un{u}}\cdot \un{u}.
\end{equation}
Strict positivity of the eigenvector norm, $\bar{\un{u}} \cdot \un{u} > 0$, and non-negativity, $\bar{\un{u}}\cdot(\mm{X} + \mm{X}^T)\un{u} \ge 0$
(by definition of positive semidefiniteness of $\mm{X}+\mm{X}^T$), imply ${\,{\rm Re}\,} \beta \ge 0$.

(ii) Consider a linear system of differential equations,
\begin{equation}
( {\rm d} / {\rm d}  t) \un{u}(t) = -\mm{X} \un{u}(t).
\label{eq:ode}
\end{equation}
Positive semidefiniteness of $\mm{X} + \mm{X}^T$ is then equivalent to Lyapunov stability in control theory, namely $( {\rm d} / {\rm d}  t)||\un{u}||^2_2 = -\bar{\un{u}}\cdot (\mm{X} + \mm{X}^T)\un{u} \le 0$ iff $\mm{X} + \mm{X}^T \ge 0$. Then, the converse (equivalent) version of the statement (ii) of the above lemma says that:
If there exists an imaginary (or vanishing) eigenvalue $\beta$, $\beta =  {\rm i}  b$, and a corresponding Jordan block of dimension $k > 1$ in the Jordan canonical form of $\mm{X}$, then $\mm{X}+\mm{X}^T$ is not positive semidefinite.
This is indeed obvious, since if we take the initial vector $\un{u}(0)$ for (\ref{eq:ode}) from
${\rm ker\,}(\mm{X} - \beta \mathbbm{1})^k \ominus {\rm ker\,}(\mm{X} - \beta \mathbbm{1})^{k-1}$ then
$\un{u}(t) \propto t^{k-1} e^{- {\rm i}  b t}$, and so $\mm{X} + \mm{X}^T \not\ge 0$.
 \qed
\end{proof}

\begin{lemma}
\label{lemma:zerodriving}
Let us write the Jordan canonical form of the matrix $\mm{X}$ (\ref{eq:X}) as
\begin{equation}
\mm{X} = \mm{P} \mm{\Delta} \mm{P}^{-1}
\label{eq:XJ}
\end{equation}
where $\mm{P}$ is a non-singular $2n\times 2n$ matrix (columns of which consist of generalized eigenvectors of $\mm{X}$) and
$\mm{\Delta} = \bigoplus_{j,k} \mm{\Delta}_{\ell_{j,k}}(\beta_j)$ is a direct sum of Jordan blocks
\begin{equation}
\Delta_\ell(\beta) :=
\pmatrix{
\beta & 1 & & \cr
  & \beta & \ddots & \cr
 & & \ddots & 1 \cr
 & & & \beta},
 \label{eq:canJB}
 \end{equation}
 for the distinct eigenvalues - rapidities $\beta_j$, and the block sizes $\ell_{j,k}$ satisfying the completeness sum
 \begin{equation}
 \sum_{j,k} \ell_{j,k} = 2n.
 \label{eq:sumrule}
 \end{equation}
Then, the manifestly antisymmetric Jordan canonical form of the structure matrix (\ref{eq:AA}),
specialized to the case of zero driving $\mm{A}_0 := \mm{A}\vert_{\mm{M}_{\rm i} = 0}$,
\begin{equation}
\mm{A}_0 =
\pmatrix{
-2 {\rm i} \mm{H}  & 2 {\rm i}  \mm{M}_{\rm r} \cr
-2 {\rm i} \mm{M}_{\rm r} & -2 {\rm i} \mm{H}},
\label{eq:AA0}
\end{equation}
reads
\begin{equation}
\mm{A}_0 =
\mm{V}_0^T \pmatrix{ \mm{0} & \mm{\Delta} \cr -\mm{\Delta}^T & \mm{0}} \mm{V}_0
\label{eq:A0J}
\end{equation}
where
\begin{equation}
\mm{V}_0= (\mm{P}^T\oplus\mm{P}^{-1})\mm{U}
\label{eq:defV}
\end{equation}
is a
$4n\times 4n$ matrix (rows of which store the generalized eigenvectors of $\mm{A}_0$) which satisfies normalization
identity
\begin{equation}
\mm{V}_0\mm{V}_0^T = \mm{J}.
\label{eq:V0n}
\end{equation}
\end{lemma}
\begin{proof}
Specializing lemma \ref{lemmatilde} to the case of $\mm{M}_{\rm i}=0$, we have
$\tilde{\mm{A}}_0 = (-\mm{X}^T)\oplus \mm{X}$. Using the Jordan decomposition  (\ref{eq:XJ}) and the similarity transformation (\ref{eq:Asim}) we find
\begin{equation}
\mm{A}_0 = \mm{U}^\dagger (\mm{P}^{-T}\oplus \mm{P})[(-\mm{\Delta})^T\oplus \mm{\Delta}] (\mm{P}^T \oplus \mm{P}^{-1})\mm{U}.
\end{equation}
Introducing $\mm{V}_0:= (\mm{P}^T \oplus \mm{P}^{-1})\mm{U}$, i.e. equation (\ref{eq:defV}), and noting $\mm{U}^\dagger = \mm{U}^T \mm{J}$, we write
\begin{equation}
\mm{V}_0^{-1} = \mm{U}^\dagger (\mm{P}^{-T}\oplus \mm{P}) =
\mm{U}^T (\mm{P}\oplus\mm{P}^{-T}) \mm{J} = \mm{V}^T_0 \mm{J}
\end{equation}
which immediately implies (\ref{eq:A0J}) and (\ref{eq:V0n}).
\qed
\end{proof}

\begin{lemma}
\label{lemma:similar}
The general structure matrix $\mm{A}$ (\ref{eq:AA}) is always similar to its zero-driving counterpart $\mm{A}_0$ (\ref{eq:AA0})
\begin{equation}
\mm{A} = \mm{W}^{-1} \mm{A}_0 \mm{W}.
\label{eq:AWsim}
\end{equation}
The similarity transformation is provided by a complex orthogonal matrix
\begin{equation}
\mm{W} = \mathbbm{1}_{4n} +2 (\sigma^1 -  {\rm i}  \sigma^3)\otimes\mm{Z},\qquad \mm{W} \mm{W}^T = \mathbbm{1}_{4n},
\end{equation}
where $\mm{Z}$ is an antisymmetric $2n \times 2n$ real matrix, $\mm{Z}^T=-\mm{Z}$, obtained as a solution
of the continuous Lyapunov equation \cite{lyapunov}
\begin{equation}
\mm{X}^T \mm{Z} + \mm{Z} \mm{X} = \mm{M}_i
\label{eq:Lyap}
\end{equation}
which always exists (for the straightforward method of solution, see the proof).
\end{lemma}
\begin{proof}
Working in the tilde-representation (\ref{eq:Asim}), $\tilde{\mm{W}} := \mm{U} \mm{W} \mm{U}^\dagger = \mathbbm{1}_{4n} -4 {\rm i}  \sigma^+ \otimes \mm{Z}$, we have
$\tilde{\mm{W}}^{-1} =  \mathbbm{1}_{4n} + 4 {\rm i}  \sigma^+ \otimes \mm{Z}$ and
\begin{equation}
\tilde{\mm{W}}^{-1} \tilde{\mm{A}}_0 \tilde{\mm{W}} = \pmatrix{ -\mm{X}^T & 4 {\rm i}  (\mm{X}^T\mm{Z}+ \mm{Z}\mm{X}) \cr \mm{0} & \mm{X}},
\end{equation}
so (\ref{eq:AWsim}) is indeed consistent with (\ref{eq:AA}) iff $\mm{Z}$ is antisymmetric and solves the continous Lyapunov equation (\ref{eq:Lyap}).
\footnote{All admisible structure matrices (\ref{eq:AA}) form a non-semisimple $n (6n-1)$ dimensional
{\em Lie algebra} spanned -- in a tilde-representation --
by $4n\times 4n$ matrices of the form
\begin{equation}
\pmatrix{-\mm{X}^T &  {\rm i}  \mm{Y} \cr \mm{0} & \mm{X}}
\end{equation}
where $\mm{X}$ is an arbitrary $2n \times 2n$ real matrix and $\mm{Y}$ is an antisymmetric real matrix $\mm{Y}^T=-\mm{Y}$.
The similarity transformation (\ref{eq:AWsim}) is generated by a nilpotent subalgebra with $\mm{X}=0$.}

We shall now show by construction that under condition of positive semidefiniteness of $\mm{M} = \mm{M}_{\rm r} +  {\rm i}  \mm{M}_{\rm i}$, the solution of (\ref{eq:Lyap}) always exists, but may not be unique if some of the eigenvalues $\beta_j$ of $\mm{X}$ (\ref{eq:X}) lie on the imaginary line.

Writing a trivial mapping ${\rm vec} : \mathbb{R}^{2n \times 2n} \to \mathbb{R}^{4n^2}$,  which vectorizes a matrix $({\rm vec\,} X)_{2n (i-1)+j} := X_{i,j}$,
and obeys ${\rm vec\,} (\mm{X} \mm{Z} \mm{Y}) = (\mm{X}\otimes \mm{Y}^T) {\rm vec\,}\mm{Z}$,
the continuous Lyapunov equation (\ref{eq:Lyap}) reads
\begin{equation}
(\mm{X}^T \otimes \mathbbm{1}_{2n} + \mathbbm{1}_{2n} \otimes \mm{X}^T)\,{\rm vec\,} \mm{Z} = {\rm vec\,} \mm{M}_{\rm i}.
\end{equation}
This equation is solved efficiently (in ${\cal O}(n^3)$ steps) using the Jordan canonical form of $\mm{X}$ (\ref{eq:XJ})
\begin{equation}
(\mm{\Delta}^T \otimes \mathbbm{1}_{2n} + \mathbbm{1}_{2n} \otimes \mm{\Delta}^T)\,{\rm vec\,}(\mm{P}^T \mm{Z} \mm{P}) = {\rm vec\,}(\mm{P}^T \mm{M}_{\rm i} \mm{P})
\label{eq:omegaeq}
\end{equation}
Note that the $(2n)^2 \times (2n)^2$ matrix
$\mm{\Omega}:=\mm{\Delta}^T \otimes \mathbbm{1}_{2n} + \mathbbm{1}_{2n} \otimes \mm{\Delta}^T$ is {\em lower triangular} with diagonal entries
$\Omega_{r,r}$ of the form $\beta_j + \beta_{j'}$, so ${\rm vec\,}(\mm{P}^T \mm{Z} \mm{P})$ can be solved for uniquely if no such pairs $\beta_j$, $\beta_{j'}$ exist that $\beta_j+\beta_{j'}=0$ \cite{bartels,sylvester}.
Since ${\,{\rm Re}\,} \beta_j \ge 0$ the problem ($\Omega_{r,r} = 0$) may arise in either of the two cases:
 (i) $\beta_j=0,\beta_{j'}=0$, or (ii) $\beta_j= {\rm i}  b,\beta_{j'}=- {\rm i}  b$, $b\in\mathbb{R}\setminus\{0\}$.
It follows from lemma \ref{lemma:stability} that in either of these cases (i,ii) the corresponding Jordan blocks are trivial (of dimension 1), so there are no non-vanishing elements in the entire $r$-th row or column, $\Omega_{r,s}=\Omega_{s,r} = 0$.
Thus there will be a solution of (\ref{eq:omegaeq}) only if the corresponding coefficient on the RHS  vanishes
\begin{equation}
\omega_r:=\left\{{\rm vec\,}(\mm{P}^T\mm{M}_{\rm i}\mm{P})\right\}_r=0,
\label{eq:omr}
\end{equation}
and then, the solution may be {\em non-unique} as the coefficient $\{ {\rm vec\,}(\mm{P}^T\mm{Z}\mm{P})\}_r$ is {\em arbitrary}
(unless it is fixed to zero by the antisymmetry of $\mm{Z}$ like in the case $j=j', \beta_j=0$).
Let us now study these two singular cases:
\begin{enumerate}
\item $\beta_j=0$ and $\beta_{j'}=0$. Suppose eigenvalue $0$ has multiplicity $m={\rm dim\, ker\,}\mm{X}$, and denote the corresponding basis vectors
$\mm{\Delta} \un{e}_l = 0$, as $\un{e}_l$, $l=1,\ldots,m$.
The RHS coefficient $\omega_r$ (\ref{eq:omr}) can be any of the matrix elements of an {\em antisymmetric} $m \times m$ matrix
$K_{l,l'} := (\un{e}_l\otimes \un{e}_{l'})\cdot {\rm vec\,}(\mm{P}^T\mm{M}_{\rm i}\mm{P}) = (\mm{P}\un{e}_{l}) \cdot \mm{M}_{\rm i}(\mm{P}\un{e}_{l'})=-K_{l',l}.$
Note that the corresponding eigenvectors of $\mm{X}$, $\mm{P}\un{e}_l$, are {\em real} (since $\mm{X}$ is a real matrix), and hence the matrix
$\mm{K}$ is {\rm real} as well.
Positive semidefiniteness of $\mm{M}$ implies positive semidefiniteness of its counterpart reduced on the null space
$Q_{l,l'} := (\mm{P}\un{e}_l) \cdot \mm{M} (\mm{P}\un{e}_{l'}) =  (\mm{P}\un{e}_l) \cdot \{\frac{1}{4}(\mm{X} + \mm{X}^T) + {\rm i}  \mm{M}_{\rm i}\} (\mm{P}\un{e}_{l'}) =  {\rm i}  K_{l,l'}$, $\mm{Q} \ge 0$.
Since $\mm{K}^T=-\mm{K}$, this is only possible if $K_{l,l'} \equiv 0$.
 \item $\beta_j= {\rm i}  b,\beta_{j'}=- {\rm i}  b$, $b\in\mathbb{R}\setminus\{0\}$. Suppose eigenvalue $ {\rm i}  b$ has multiplicity $m={\rm dim\, ker\,}(\mm{X} \pm  {\rm i}  b \mathbbm{1}_{2n})$,
 and denote the corresponding basis vectors
$\mm{\Delta} \un{f}^{\pm}_l = \pm {\rm i}  b \un{f}^{\pm}_l$, as $\un{f}^{\pm}_l$, $l=1,\ldots,m$.
Since $\mm{X}$ is real, the corresponding eigenvectors $\un{u}^{\pm}_l:=\mm{P} \un{f}^{\pm}_l$,
$\mm{X} \un{u}^{\pm}_l = \pm {\rm i}  b \un{u}^{\pm}_l$, can be chosen to be conjugate paired
$\un{u}^{-}_l = \overline{\un{u}^{+}_l}$ (i.e. the corresponding columns of $\mm{P}$ can be chosen to be conjugate paired).
The RHS coefficient $\omega_r$ (\ref{eq:omr}) can be any of the matrix elements of an {\em antihermitian} $m \times m$ matrix
$K_{l,l'} := (\un{f}^+_l\otimes \un{f}^-_{l'})\cdot {\rm vec\,}(\mm{P}^T\mm{M}_{\rm i}\mm{P}) = \un{u}^+_l \cdot \mm{M}_{\rm i}\un{u}^-_{l'} = -\bar{K}_{l',l}$.
Positive semidefiniteness of $\mm{M}^+:=\mm{M}$ and $\mm{M}^-:=\bar{\mm{M}}$
implies positive semidefiniteness of two Hermitian $m \times m$ matrices
$Q^{\pm}_{l,l'} := \un{u}^+_l \cdot \mm{M}^\pm \un{u}^-_{l'} =  \un{u}^+_l \cdot \{\frac{1}{4}(\mm{X} + \mm{X}^T) \pm  {\rm i}  \mm{M}_{\rm i}\} \un{u}^-_{l'} =
\pm  {\rm i}  K_{l,l'}$, $\mm{Q}^\pm \ge 0$. $ {\rm i} \mm{K}$ and $- {\rm i} \mm{K}$ can be simultaneously positive semideifinite only if $K_{l,l'} \equiv 0$.
\qed
\end{enumerate}
\end{proof}

\section{Jordan canonical form on the tensor product spaces}

To best of the author's knowledge the theory of Jordan canonical form on tensor product spaces has not been very much developed. However, being able to manipulate Jordan decompositions on tensor product spaces seems crucial for discussing the general non-diagonalizable situation of quantum Liouvillean dynamics.
In this section we are demonstrating two simple results, namely one on Jordan decomposition of a sum of Jordan blocks on tensor product spaces, and the other one on Jordan decomposition of a many-body nilpotent map on the fermionic Fock space whose structure matrix is a simple Jordan block. Both results will be
needed for the proof of the spectral theorem later in section \ref{theorems}.

\subsection{Jordan canonical form of the sum of Jordan blocks on a tensor product space}

Let us consider a tensor product space $\mathbb{C}^k \otimes \mathbb{C}^l$ and ask for the Jordan canonical form of the following $kl \times kl$ matrix
\begin{equation}
\mm{\Delta}_{k,l} = \mm{\Delta}_{k}(\alpha)\otimes \mathbbm{1}_l + \mathbbm{1}_k \otimes\mm{\Delta}_{l}(\beta)
\label{eq:Deltakl}
\end{equation}
where the Jordan block matrix is defined in (\ref{eq:canJB}).
\begin{lemma}
\label{2nilp}
The matrix (\ref{eq:Deltakl}) has the following Jordan canonical form
\begin{equation}
\mm{\Delta}_{k,l} \simeq \bigoplus_{r=1}^{\min\{k,l\}} \mm{\Delta}_{k+l-2r+1}(\alpha+\beta).
\label{eq:tensorsum}
\end{equation}
\end{lemma}
The proof will be constructive, so we shall also show how to compute the corresponding generalized eigenvectors.
\begin{proof}
Without loss of generality we can assume $k \ge l$.
Let $\ket{i,j},i=1,\ldots,k,j=1,\ldots,l$ denote the basis of $\mathbb{C}^k \otimes \mathbb{C}^l$. Clearly, since $\mm{\Delta}_r(\gamma) = \mathbbm{1}_r + \gamma\mm{\Delta}_r $
(we write $\mm{\Delta}_r := \mm{\Delta}_r(0)$), it is enough to show (\ref{eq:tensorsum}) for the nilpotent case $\alpha=\beta=0$, and then at the end
add $(\alpha+\beta)\mathbbm{1}_{kl}$.
The crucial tool will be the Newton's binomial formula on the tensor product sum
\begin{equation}
\mm{\Delta}_{k,l}^p = \sum_{r=0}^p \left({p\atop r}\right) \mm{\Delta}_k^r \otimes \mm{\Delta}_l^{p-r}.
\end{equation}
From the above formula, and nilpotency $\mm{\Delta}^r_r = \mm{0}$, $\mm{\Delta}^{r-1}_r \neq \mm{0}$,
it follows that $\mm{\Delta}_{k,l}^{k+l-1} = \mm{0}$ but $\mm{\Delta}_{k,l}^{k+l-2}= \left({k+l-2\atop l-1}\right) \mm{\Delta}^{k-1}_k\otimes\mm{\Delta}^{l-1}_l \neq \mm{0}$,
so the {\em largest} Jordan block in the decomposition of $\mm{\Delta}_{k,l}$ has size $k+l-1$, with the first nilpotent series of
generalized eigenvectors
\begin{equation}
\ket{p;1} := \mm{\Delta}^p_{k,l} \ket{0;1}, \qquad p = 0,\ldots, k+l-2,
\label{eq:maximal}
\end{equation}
and a `seed' vector (null vector of $\mm{\Delta}_{k,l}^{k+l-2}$), $\ket{0;1}:=\ket{k,l}$.

Let us introduce the diagonal number matrix $\mm{N} := \sum_{i=1}^k\sum_{j=1}^l (i+j)\ket{i,j}\bra{i,j}$ which slices the vector space to a direct sum of eigenspaces of the following dimensions
\begin{equation}
\dim\ker (\mm{N}-(q+1)\mathbbm{1}_{kl}) = \dim\ker(\mm{N}-(k+l+1-q)\mathbbm{1}_{kl}) = q,
\end{equation}
which change exactly by 1 when changing the eigenvalue by 1.
Further, note that $ \mm{N}\mm{\Delta}_{k,l}= \mm{\Delta}_{k,l} (\mm{N}-\mathbbm{1}_{kl})$, or equivalently $\mm{\Delta}_{k,l}$ always decreases the eigenvalue of $\mm{N}$ exactly by $1$
\begin{equation}
\mm{\Delta}_{k,l} \ker (\mm{N}- (q+1)\mathbbm{1}_{kl}) \subseteq  \ker (\mm{N}- q\mathbbm{1}_{kl}).
\end{equation}

Then we make the following inductive argument. Suppose that for all $q < r$, the series of generalized eigenvectors
\begin{equation}
\ket{p;q}:=\mm{\Delta}^p_{k,l}\ket{0;q},\qquad p = 0,\ldots, k+l-2q,
\end{equation}
explore all the eigenspaces of $\mm{N}$ with eigenvalues $\le r$ and $\ge k+l+2-r$. This is trivially true for $r=2$, or the maximal series with $q=1$ which we have
constructed above (\ref{eq:maximal}).

We need to show now that the $r$-th series explores the other two extremal eigenvalues, $r+1$ and $k+l+1-r$ and is of length $k+l-2r+1$.
Clearly, the seed vector for the $r-$th series $\ket{0;r}$ should have the number eigenvalue $k+l+1-r$, i.e. it  should be of the form
\begin{equation}
\ket{0;r} = \sum_{q=1}^r c^{(r)}_q \ket{k-q+1,l-r+q},
\end{equation}
and the size of the corresponding Jordan block can not be longer than $k+l-2r+1$
\begin{equation}
\mm{\Delta}_{k,l}^{k+l-2r+1}\ket{0;r} =
\sum_{j=1}^{r-1} \ket{r-j,j} \sum_{q=1}^r  \left({ k+l-2r+1\atop l-r-j+q}\right) c^{(r)}_q = 0.
\end{equation}
This results, for the $r$th Jordan block, in a homogeneous system of $r-1$ equations for $r$ unknown coefficients $c^{(r)}_j$,
\begin{equation}
\sum_{q=1}^r  \left({ k'+l'+1\atop l'-j+q}\right) c^{(r)}_q = 0,\quad j \in \{1,\ldots, r-1\},
\label{eq:iden}
\end{equation}
where on RHS and below we write $k':=k-r, l':=l-r$.
The above system has (up to a prefactor) a unique solution
\begin{equation}
c^{(r)}_q = (-1)^{r-q}  \left({ k'+r-q\atop r-q}\right)  \left({l'+q-1\atop q-1}\right).
\label{eq:crq}
\end{equation}
What is left to show is that a nilpotent series starting from a seed $\ket{0;r}$ is indeed not shorter than $k+l-2r+1$, i.e.
$\mm{\Delta}_{k,l}^{k+l-2r} \ket{0;r} \neq 0$, or equivalently
\begin{equation}
\sum_{q=1}^r  \left({ k'+l'\atop l'-j+q}\right) c^{(r)}_q \ne 0,
\label{eq:noniden}
\end{equation}
for at least some $j \in\{1,\ldots,r\}$.
Verification of (\ref{eq:iden}) and (\ref{eq:noniden}) with ansatz (\ref{eq:crq}) is an easy case for a computer-assisted proving system based on Zeilberger algorithm \cite{paule}. \qed
\end{proof}

\subsection{Jordan canonical form of the many-body nilpotent map}

\label{mbnilp}

Here we study the following abstract  many body problem. Let $ {\hat b} _{k}, {\hat b} '_{k}, k=1,\ldots,\ell $ be fermionic maps satisfying CAR,
$\{ {\hat b} _j, {\hat b} _k\}=\{ {\hat b} '_j, {\hat b} '_k\}=0,\{ {\hat b} _j, {\hat b} '_k\}=\delta_{j,k}$, and
defining $\ell $ fermionic modes, i.e. spanning a $2^\ell$ dimensional vector space ${\cal V}$ with the Fock basis and its dual defined, respectively, as
\begin{eqnarray}
\ket{\nu_1,\ldots,\nu_{\ell}} :=  {( {\hat b} '_1)}^{\nu_1} \cdots {( {\hat b} '_{\ell})}^{\nu_{\ell}}\ket{{\rm right}},\\
\bra{\nu_1,\ldots,\nu_{\ell}} := \bra{{\rm left}}( {\hat b} _{\ell})^{\nu_{\ell}} \cdots ( {\hat b} _1)^{\nu_1},\qquad \nu_k\in\{0,1\}.
\end{eqnarray}
The vector $\ket{\rm right}$ and its dual $\bra{\rm left}$, $\braket{\rm left}{\rm right}=1$, are defined via $\bra{\rm left} {\hat b} '_k \equiv 0,  {\hat b} _k \ket{\rm right}\equiv 0$, hence the biorthogonality $\braket{\un{\nu}}{\un{\nu}'} = \delta_{\un{\nu},\un{\nu}'}$ simply follows from the CAR.

We shall study the following {\em nilpotent} linear operator over ${\cal V}$
\begin{equation}
{\hat {\cal M}} = \sum_{k=1}^{\ell-1}  {\hat b} '_{k+1}  {\hat b} _k.
\label{eq:nilpotent}
\end{equation}
We observe immediately that ${\hat {\cal M}}$ commutes with a {\em number map} ${\hat {\cal N}} := \sum_{k=1}^\ell  {\hat b} '_k  {\hat b} _k$, $[{\hat {\cal M}},{\hat {\cal N}}]=0$, which foliates the linear space to a direct sum ${\cal V} = \bigoplus_{m=0}^\ell {\cal V}_m$, of ${\rm dim\, }{\cal V}_m = \left({\ell\atop m}\right) = \frac{\ell!}{m!(\ell-m)!}$ dimensional spaces ${\cal V}_n$, which are  spanned by Fock vectors $\ket{\un{\nu}}$ with exactly $|\un{\nu}|:=\sum_{k=1}^\ell \nu_k = m$ `particle' excitations ($m$ being the eigenvalue of ${\hat {\cal N}}$).

We shall now {\em fix} the number of particles $m$ and restrict the nilpotent operator ${\hat {\cal M}}$ (\ref{eq:nilpotent}) to the space ${\cal V}_m$, ${\hat {\cal M}}_m := {\hat {\cal M}}|_{{\cal V}_m}$. For $m=0$ (and $m=\ell$) ${\hat {\cal M}}$ acts trivially, ${\hat {\cal M}}_0 = {\hat {\cal M}}_m = 0$, and for $m=1$ (and $m=\ell-1$),
${\hat {\cal M}}_1 \simeq {\hat {\cal M}}_{\ell-1} \simeq \mm{\Delta}_\ell$. However, the interesting question now is to find a Jordan canonical form of ${\cal M}_m$,
for arbitrary $m$.

Let us define a {\em weight} of a basis Fock vector $\ket{\un{\nu}}$ as
\begin{equation}
\omega_{\un{\nu}} = \sum_{k=1}^\ell k \nu_k - \frac{1}{2}m(m+1).
\end{equation}
Having $\ell,m$ fixed, there are two extremal states, with minimum and maximum weight, namely
\begin{eqnarray}
\ket{\rm min} &=& \ket{1,\ldots,1,0,\ldots,0},\quad \omega_{\rm min} = 0, \nonumber \\
\ket{\rm max} &=& \ket{0,\ldots,0,1,\ldots,1},\quad \omega_{\rm max} = (\ell-m)m.
\end{eqnarray}
Denoting the linear spans ${\cal V}_m^r = L\{ \ket{\un{\nu}}; |\un{\nu}|=m, \omega_{\un{\nu}} = r\}$, we have a direct sum
\begin{equation}
{\cal V}_m = \bigoplus_{r=0}^{(\ell-m)m} {\cal V}^r_m.
\end{equation}
Dimensionality ${\rm dim}{\cal V}^r_m =: \left({\ell\atop m}\right)_{r}$ has a clear combinatorial-physical meaning, namely $\left({\ell\atop m}\right)_r$ is the number of distributions (combinations) of $m$ particles on $\ell$ exclusive equidistant vertical levels, such that the total potential energy is exactly $r$
steps larger than the minimal. Such a {\em restricted binomial symbol} can be calculated from a straightforward recursion relation
\begin{equation}
\left({\ell\atop m}\right)_r =  \left({\ell-1\atop m}\right)_r + \left({\ell-1\atop m-1}\right)_{r-\ell+m}, \qquad
\left({0\atop m}\right)_r = \delta_{m,0} \delta_{r,0}
\end{equation}
and clearly satisfies the sum rule
\begin{equation}
\sum_{r=0}^{(\ell-m)m}  \left({\ell\atop m}\right)_r =  \left({\ell\atop m}\right)
\label{eq:sumrule1}
\end{equation}
and the symmetry
\begin{equation}
\left({\ell\atop m}\right)_r = \left({\ell\atop m}\right)_{(\ell-m)m-r}.
\end{equation}
Note that $ \left({\ell\atop m}\right)_r = 0$ if $r < 0$ or $r > m(\ell-m)$.
Crucial observation for the rest of our discussion is the fact that ${\hat {\cal M}}_m$ maps ${\cal V}_m^r$ to ${\cal V}_m^{r+1}$,
i.e. that the map ${\hat {\cal M}}$ (\ref{eq:nilpotent}) increases the weight of the state exactly by $1$,
\begin{equation}
{\hat {\cal M}} {\cal V}_m^r \subseteq {\cal V}_m^{r+1}.
\label{eq:incl}
\end{equation}

Clearly, one can now state the following:
\begin{lemma}
\label{nilplemma}
The largest Jordan block in the canonical form of ${\hat {\cal M}}_m$ has size $(\ell-m)m + 1$.
\end{lemma}
\begin{proof}
Taking $\ket{0;1}:=\ket{\rm min}$ as a `seed' for the first nilpotent series, we see that, due to (\ref{eq:incl}),
\begin{equation}
\ket{p;1} := {\hat {\cal M}}^p \ket{\rm min} \in {\cal V}_{m}^{p}.
\end{equation}
All these vectors for $p \le (m-\ell)m$ should be non-vanishing since only positive integer coefficients can appear in expressing the series
${\hat {\cal M}}^p \ket{\rm min}$ in the canonical basis $\ket{\un{\nu}}$, and the only canonical basis vector $\ket{\nu}$ of ${\cal V}_m$ which is annihilated by
${\hat {\cal M}}$ is $\ket{\rm max}$. So we have $\ket{(\ell - m)m;1} = q\ket{\rm max}$, where $q$ is a positive integer since $\dim {\cal V}_m^{(\ell - m)m} = 1$, and
${\hat {\cal M}}\ket{(\ell-m)m;1} = 0$. \qed
\end{proof}

The rest of Jordan decomposition of ${\hat {\cal M}}_n$ we are only able to state if the following conjecture is true:
\begin{conjecture}
\label{conjecture}
The map ${\hat {\cal M}} : {\cal V}^r_m \to {\cal V}^{r+1}_m$
is {\em injective} for $r \le \lfloor ((\ell-m)m-1)/2\rfloor $ and {\em surjective} for
$r \ge \lfloor (\ell-m)m/2\rfloor.$
\end{conjecture}
Conjecture has been verified explicitly using computer algebra code for all $\ell \le 12$, and for any $\ell$ and $m \le 3$ \cite{semrl}, however its general proof remains open due to
a rather involved combinatorics.
The conjecture of course also implies the ordering on the restricted binomial sequence (dimensions of ${\cal V}^r_m$), namely
\begin{equation}
\left({\ell\atop m}\right)_0 \le \left({\ell\atop m}\right)_1 \le \ldots \le \left({\ell\atop m}\right)_{\lfloor (\ell-m)m/2 \rfloor}
\end{equation}
which can be verified directly.

Omitting a straightforward proof \cite{semrl} we state that the conjecture implies the following simple general result
\begin{equation}
{\hat {\cal M}}_n \simeq \bigoplus_{r=0}^{\lfloor (\ell-m)m/2\rfloor} \left[ \left({\ell\atop m}\right)_r - \left({\ell\atop m}\right)_{r-1}\right] \mm{\Delta}_{(\ell - m)m+1-2r}
\end{equation}
where the notation $[k]\mm{\Delta}$ means $\mm{\Delta}\oplus\mm{\Delta}\ldots\oplus\mm{\Delta}(k {\rm\; times})$, and $[0]\mm{\Delta} := 0$.
In other words, the number of Jordan blocks of size $(\ell-m)m+1-2r$ (it can only decrease from the maximal in steps of $2$) is
$\left({\ell\atop m}\right)_r - \left({\ell\atop m}\right)_{r-1}$ (which can also be zero for some instances $r$).
The total number of Jordan blocks and thus the total number of proper eigenvectors of ${\hat {\cal M}}_m$ is then simply equal to
$\left({\ell\atop m}\right)_{\lfloor (\ell-m)m/2\rfloor}$.

\section{The spectral theorem}

\label{theorems}

With the lemmas \ref{lemma:zerodriving},\ref{lemma:similar} we have shown that any Liouvillean structure matrix of the form (\ref{eq:AA}) admits a canonical
representation in terms of the Jordan canonical form (\ref{eq:XJ},\ref{eq:canJB}) of the real matrix $\mm{X}$ (\ref{eq:X}), and the anti-symmetric real matrix $\mm{Z}=-\mm{Z}^T$ which solves the Lyapunov equation (\ref{eq:Lyap}), namely
\begin{equation}
\mm{A} =
\mm{V}^T \pmatrix{ \mm{0} & \mm{\Delta} \cr -\mm{\Delta}^T & \mm{0}} \mm{V}.
\label{eq:AJ}
\end{equation}
The eigenvector matrix (rows of which store the generalized eigenvectors of $\mm{A}$) reads
\begin{eqnarray}
\mm{V} &=& \mm{V}_0 \mm{W} = (\mm{P}^T \oplus \mm{P}^{-1} )\mm{U}
(\mathbbm{1}_{4n} +2 (\sigma^1 -  {\rm i}  \sigma^3)\otimes\mm{Z}) \nonumber \\
&=& \frac{1}{\sqrt{2}}\pmatrix{\mm{P}^T(\mathbbm{1}_{2n}-4 {\rm i} \mm{Z})  &  - {\rm i}  \mm{P}^T (\mathbbm{1}_{2n}-4 {\rm i} \mm{Z}) \cr \mm{P}^{-1} &  {\rm i}  \mm{P}^{-1}}
\label{eq:V}
\end{eqnarray}
and satisfies the canonical normalization condition
\begin{equation}
\mm{V}\mm{V}^T = \mm{J}.
\label{eq:VJ}
\end{equation}

Let us name the first $2n$ rows of $\mm{V}$ as $\un{v}_{j,k,l}$, and the last $2n$ rows as $\un{v}'_{j,k,l}$, which are exactly the generalized eigenvectors pertaining to $k$-th Jordan block of the eigenvalue (rapidity)
$\beta_j$, and $-\beta_j$ respectively, and $l=1,\ldots,\ell_{j,k}$ ($l=1$ designates the {\em proper} eigenvector) where $\ell_{j,k}$ is the size of the Jordan block $(j,k)$.
Then we introduce the {\em normal master mode} (NMM) maps as
\begin{equation}
 {\hat b} _{j,k,l} := \un{v}_{j,k,l} \cdot \un{ {\hat a} },\qquad  {\hat b} '_{j,k,l} := \un{v}'_{j,k,l}\cdot \un{ {\hat a} },
\label{eq:bfroma}
\end{equation}
satisfying the almost-CAR (simply following from (\ref{eq:VJ}))
\begin{equation}
\{  {\hat b} _{j,k,l},  {\hat b} _{j',k',l'} \} = 0,\;\; \{  {\hat b} _{j,k,l},  {\hat b} '_{j',k',l'} \} = \delta_{j,j'} \delta_{k,k'} \delta_{l,l'},\;\; \{  {\hat b} '_{j,k,l},  {\hat b} '_{j',k',l'} \} = 0.
\label{eq:almostCAR}
\end{equation}
In terms of decomposition (\ref{eq:AJ}) and NMM maps (\ref{eq:bfroma}) the Liouvillean (\ref{eq:Amatrix}) gets almost-diagonal form
\begin{eqnarray}
{\hat {\cal L}} &=& (\mm{V}\un{ {\hat a} }) \cdot \left[\mm{\Delta} \oplus (-\mm{\Delta}^{T})\right]\mm{J} (\mm{V}\un{ {\hat a} }) - A_0 \hat{\mathbbm{1}} \\
&=& \sum_{j,k} \Bigl\{ \beta_j \sum_{l=1}^{\ell_{j,k}} ( {\hat b} _{j,k,l}  {\hat b} '_{j,k,l} -  {\hat b} '_{j,k,l}  {\hat b} _{j,k,l}) - \ell_{j,k} \beta_j \hat{\mathbbm{1}} \nonumber \\
&& \quad\;\;\; + \sum_{l=1}^{\ell_{j,k}-1} ( {\hat b} _{j,k,l}  {\hat b} '_{j,k,l+1} -  {\hat b} '_{j,k,l+1}  {\hat b} _{j,k,l}) \Bigr\}
\end{eqnarray}
where the scalar (\ref{eq:A0}) has been expressed as $A_0 = {\,{\rm tr}\,} \mm{X} = \sum_{j,k} \ell_{j,k} \beta_j$.
After exercising (\ref{eq:almostCAR}) the Liouvillean finally simplifies to a {\em normal form}
\begin{equation}
{\hat {\cal L}} = -2\sum_{j,k} \left\{ \beta_j \sum_{l=1}^{\ell_{j,k}}   {\hat b} '_{j,k,l}  {\hat b} _{j,k,l} + \sum_{l=1}^{\ell_{j,k}-1}  {\hat b} '_{j,k,l+1}  {\hat b} _{j,k,l} \right\}.
\label{eq:dLL}
\end{equation}
Note that the second (non-diagonal) sum vanishes for the trivial Jordan blocks with $\ell_{j,k}=1$.

Following Ref. \cite{njp} a {\em unique} pair\footnote{Unique with respect to a chosen,
possibly non-unique (see lemma \ref{lemma:similar}) decomposition of the structure matrix (\ref{eq:AJ}).} of operator Fock states exist, satisfying
\begin{equation}
 {\hat b} _{j,k,l} \xket{{\rm NESS}} = 0,\qquad \bra{1}  {\hat b} '_{j,k,l} = 0.
\label{eq:vacua}
\end{equation}
The bra (dual) vector $\bra{1}$ in fact represents a trivial operator $1$, dual to $\xket{{\rm NESS}}$, $\braket{1}{\rm NESS}={\rm tr}\rho_{\rm NESS} = 1$. Indeed $\xket{{\rm NESS}}$, which is an empty Fock state w.r.t. creation NMM maps $ {\hat b} '_{j,k,l}$, is a {\em non-equilibrium steady state} (NESS) since straightforward inspection following from almost-diagonal form of ${\hat {\cal L}}$ (\ref{eq:dLL}) shows that ${\hat {\cal L}} \xket{{\rm NESS}} = 0$, as well as the dual relation $\bra{1}{\hat {\cal L}} = 0$.

The complete spectrum of eigenvalues and eigenvectors of $4^n \times 4^n$ matrix of ${\hat {\cal L}}$ over ${\cal K}$ is trivial to construct in case that all Jordan blocks of $\mm{X}$ are trivial ($\ell_{j,k}=1$), i.e.
if $\mm{A}$ is diagonalizable, the case which has been assumed in Ref.\cite{njp}. However, here we treat the general case.

\subsection{Spectral decomposition and Jordan canonical form of the Liouvillean}

\begin{theorem}
\label{spctheorem}
(i) The complete spectrum of Liouvillean ${\hat {\cal L}}$ is given by the following integer linear combinations
\begin{equation}
\lambda_{\un{m}} = -2\sum_{j,k} m_{j,k}\beta_j\,,\quad m_{j,k} \in \{0,1,\ldots,\ell_{j,k}\}.
\end{equation}
(ii) The $4^n$ dimensional operator space, and its dual (the bra-space), admit the following decomposition
\begin{equation}
{\cal K} = \bigoplus_{\un{m}} {\cal K}_{\un{m}}\, , \qquad {\cal K}' = \bigoplus_{\un{m}} {\cal K}'_{\un{m}}
\label{eq:direct}
\end{equation}
in terms of $\dim {\cal K}_\un{m} = \prod_{j,k}\left({\ell_{j,k}\atop m_{j,k}}\right)$ dimensional
invariant subspaces
\begin{equation}
{\hat {\cal L}}{\cal K}_{\un{m}} \subseteq {\cal K}_{\un{m}}\, ,\qquad{\cal K}'_{\un{m}}{\hat {\cal L}} \subseteq {\cal K}'_{\un{m}}
\label{eq:inv}
\end{equation}
spanned by
\begin{eqnarray}
{\cal K}_{\un{m}} = L\left\{ \prod_{j,k}\prod_{\eta=1}^{m_{j,k}}  {\hat b} '_{j,k,l_\eta}  \xket{{\rm NESS}};\; 1\le l_1 < \ldots < l_{m_{j,k}} \le \ell_{j,k} \right\}, \nonumber\\
{\cal K}'_{\un{m}}= L\left\{ \bra{1}\prod_{j,k}\prod_{\eta=1}^{m_{j,k}}  {\hat b} _{j,k,l_\eta};\; 1\le l_1 < \ldots < l_{m_{j,k}} \le \ell_{j,k}\right\}. \label{eq:spaces}
\end{eqnarray}
(iii) However, the dimension of the eigenspace (the number of proper eigenvectors corresponding to $\lambda_\un{m}$) is smaller than $\dim {\cal K}_\un{m}$ in the nontrivial case
when at least one $\ell_{j,k} > 1$. The size of the largest Jordan block corresponding to $\lambda_\un{m}$ is
\begin{equation}
1+\sum_{j,k} (\ell_{j,k}-m_{j,k})m_{j,k} .
\label{eq:maxblock}
\end{equation}
\end{theorem}
\begin{proof}
We start by showing (ii). The fact that direct sums (\ref{eq:direct}) span the entire space follows from dimension counting $\sum_{\un{m}}\dim{\cal K}_{\un{m}}=\sum_{\un{m}}  \prod_{j,k}\left({\ell_{j,k}\atop m_{j,k}}\right)=\prod_{j,k}2^{\ell_{j,k}} = 2^{2n}$ and non-singularity of the complex orthogonal transformation to NMM maps (\ref{eq:bfroma}).
The invariance conditions (\ref{eq:inv}) are equivalent to
\begin{equation}
\bra{y}{\hat {\cal L}}\ket{x} = 0, \quad \forall  \ket{x} \in {\cal K}_{\un{m}}, \bra{y} \in {\cal K}'_{\un{m}'},\; {\rm such\; that}\; \un{m} \ne \un{m}'\,,
\label{eq:orth}
\end{equation}
which in turn follows from the normal form (\ref{eq:dLL}). More explicitly, one may write (\ref{eq:dLL}) as ${\hat {\cal L}} = \sum_{j,k} {\hat {\cal L}}_{j,k}$ where
\begin{eqnarray}
{\hat {\cal L}}_{j,k} &:=& -2(\beta_j\, {\hat {\cal N}}_{j,k} + {\hat {\cal M}}_{j,k}),\\
{\hat {\cal N}}_{j,k} &=& \sum_{l=1}^{\ell_{j,k}}   {\hat b} '_{j,k,l}  {\hat b} _{j,k,l},\quad
{\hat {\cal M}}_{j,k} = \sum_{l=1}^{\ell_{j,k}-1}  {\hat b} '_{j,k,l+1}   {\hat b} _{j,k,l} \nonumber
\end{eqnarray}
and show (\ref{eq:orth}) separately for each term
$\bra{y}{\hat {\cal L}}_{j,k}\ket{x}=0$, and considering only $\ket{x} = \prod_{\eta=1}^{m_{j,k}}  {\hat b} '_{j,k,l'_\eta}  \xket{{\rm NESS}},
\bra{y} = \bra{1}\prod_{\eta=1}^{m_{j,k}}  {\hat b} _{j,k,l_\eta}$ for some $\{l_\eta\},\{l'_\eta\}$, since the other NMM maps $\hat{b}_{j',k',l'}$ for $(j',k')\neq (j,k)$ entering $\bra{x}$ and $\ket{y}$ (\ref{eq:spaces})
all annihilate as they commute with ${\hat {\cal L}}_{j,k}$. The rest is just trivial CAR algebra.

To show (i) we write the Liouvillean as ${\hat {\cal L}} = {\hat {\cal L}}_0 + {\hat {\cal M}}$ where ${\hat {\cal L}}_0 := -2\sum_{j,k} \beta_j\,{\hat {\cal N}}_{j,k}$ and ${\hat {\cal M}} = \sum_{j,k} {\hat {\cal M}}_{j,k}$.
Since ${\hat {\cal M}}_{j,k}$ are nilpotent (lemma \ref{nilplemma}), and $[{\hat {\cal M}}_{j,k},{\hat {\cal M}}_{j',k'}]=0$, ${\hat {\cal M}}$ is nilpotent as well.
 Clearly $[{\hat {\cal L}}_0,{\hat {\cal M}}]=0$ as all the terms commute $[{\hat {\cal N}}_{j,k},{\hat {\cal M}}_{j',k'}]=0$, so ${\hat {\cal L}}$ and ${\hat {\cal L}}_0$ should have identical spectra.
 But the eigenspaces of ${\hat {\cal L}}_0$ are ${\cal K}_{\un{m}}$, ${\hat {\cal L}}_0 {\cal K}_{\un{m}} = \lambda_{\un{m}} {\cal K}_{\un{m}}$, so $\{ \lambda_\un{m}\}$ is also the spectrum of ${\hat {\cal L}}$.

 As for (iii) we observe that the invariant space ${\cal K}_\un{m}$ admits a tensor product decomposition
 \begin{equation}
 {\cal K}_{\un m} \sim \bigotimes_{j,k} {\cal K}_{j,k,m_{j,k}}
 \end{equation}
where ${\cal K}_{j,k,m}$ is a $\left( \ell_{j,k}\atop m\right)$ dimensional space spanned by all $\prod_{\eta=1}^m  {\hat b} '_{j,k,l_\eta}\xket{{\rm NESS}}$, with  $1\le l_1 <\ldots< l_m \le \ell_{j,k}$.
Let us consider the restriction of the map on the invariant space
${\hat {\cal L}}|_{{\cal K}_{\un{m}}} = \lambda_{\un{m}} \mathbbm{1} + {\hat {\cal M}}_{\un{m}}$ where the nilpotent ${\hat {\cal M}}_{\un{m}}$ acts independently on tensor factors
\begin{equation}
{\hat {\cal M}}_{\un{m}} = \sum_{j,k} \left( \bigotimes_{j',k'}^{(j',k') \neq (j,k)} \mathbbm{1}_{{\cal K}_{j',k',m_{j',k'}}} \right) \otimes {\hat {\cal M}}^{(\ell_{j,k})}_{m_{j,k}}.
\end{equation}
${\hat {\cal M}}^{(\ell_{j,k})}_{m_{j,k}}$ is exactly the abstract nilpotent many-body map studied in subsection \ref{mbnilp} where $\ell = \ell_{j,k}$.
Noting the lemma \ref{nilplemma} each factor map has a nontrivial Jordan canonical form with a maximal block of size $(\ell_{j,k}-m_{j,k})m_{j,k}+1$.
Then we use lemma \ref{2nilp} recursively, noting at each step always only the largest Jordan block, to arrive at (\ref{eq:maxblock}).
\qed
\end{proof}
Writing down the complete Jordan canonical form of Liouvillean ${\hat {\cal L}}$ seems at this point a very difficult combinatorial problem even assuming the conjecture \ref{conjecture} were true.
We stress however an interesting non-trivial aspect of the result that we have just proven. Namely, in spite of the fact that the dimension of the invariant subspace pertaining to
some possibly highly degenerate rapidity $\beta_j$ is {\em exponentially large} in the (algebraic) multiplicity of the rapidity, the largest Jordan block is {\em only polynomial} (at most quadratic) in the multiplicity. This may have interesting physical consequences in studying Liouvillean time evolution (relaxation) where all terms of the form $t^{p-1} \exp( \lambda_{\un{m}}t)$ can appear
with integer power $p$ not larger than (\ref{eq:maxblock}).

\subsection{Uniqueness of NESS}

\begin{theorem}
\label{uniqueness}
Uniqueness of NESS. $\xket{{\rm NESS}}$ is a unique stationary state of open quantum dynamics (\ref{eq:liouville}) if and only if all eigenvalues $\beta_j$ of $\mm{X}$ (\ref{eq:X}) lie away from the imaginary line ${\,{\rm Re}\,} \beta_j > 0$.
If this is not the case, then:
\begin{enumerate}
\item
For each zero rapidity $\beta_j=0$,
\begin{equation}
\ket{{\rm NESS};j,k} :=   {\hat b} '_{j,k,1} \xket{{\rm NESS}}
\end{equation}
we also have the stationarity ${\hat {\cal L}} \ket{{\rm NESS};j,k} = 0$.
\item
For each imaginary rapidity $\beta_j= {\rm i}  b$, $b \in \mathbb{R}\setminus\{0\}$, we have a corresponding negative rapidity $\beta_{j'}=- {\rm i}  b$, and
\begin{equation}
\ket{{\rm NESS};j,j',k,k'} :=  {\hat b} '_{j,k,1}  {\hat b} '_{j',k',1} \xket{{\rm NESS}},
\end{equation} which satisfies stationarity   ${\hat {\cal L}} \ket{{\rm NESS};j,j',k,k'} = 0$.
\end{enumerate}
\end{theorem}
\begin{proof}
Uniqueness of $\xket{{\rm NESS}}$ in case of strict positivity of the spectral gap of ${\hat {\cal L}}$, $\Delta = 2\min_j {\,{\rm Re}\,}\beta_j$ (theorem \ref{spctheorem}) is obvious. In the other cases, (i), the stationarity ${\hat {\cal L}} \ket{{\rm NESS};j,k} = 0$ follows from the facts that the term $j,k$ is absent in the
normal form of the Liouvillean (\ref{eq:dLL}), since $\beta_{j,k} = 0$ and $\ell_{j,k}=1$ (lemma \ref{lemma:stability}), whereas
$ {\hat b} '_{j,k,1}$ anticommutes with all the other terms,. So the annihilation maps $\un{ {\hat b} }$ can be commuted to the right and then (\ref{eq:vacua}) is used.
Similarly, in case (ii), applying ${\hat {\cal L}}$ (\ref{eq:dLL}) to  $\ket{{\rm NESS};j,j',k,k'}$ and commuting annihilation maps $\un{ {\hat b} }$ to
the right results in two more terms w.r.t. case (i), which however cancel each other since $\beta_j+\beta_{j'}=0$. \qed
\end{proof}
Note that (\ref{eq:V}) implies
\begin{equation}
\un{ {\hat b} }' = \mm{P}^{-1} \un{ {\hat c} }^\dagger, \quad \un{ {\hat c} }^\dagger = \frac{1}{\sqrt{2}}(\un{ {\hat a} }_1 +  {\rm i}  \un{ {\hat a} }_2)
\end{equation}
where $\un{ {\hat a} }_1,\un{ {\hat a} }_2$ represent, respectively, the first and the last $2n$ components of a $4n$ vector $\un{ {\hat a} }$.
Therefore, since $\un{ {\hat c} }^\dagger_j$ can be interpreted as {\em creation} maps of {\em Hermitian} fermionic Majorana operators $w_j$ \cite{njp}, we have the following obvious statements: in case (i), the row $(j,k,1)$ of $\mm{P}^{-1}$ which is the {\em left} eigenvector of a real matrix $\mm{X}$ of eigenvalue $0$ can be always chosen {\em real}, and hence $\ket{{\rm NESS};j,k}$ represents a {\em Hermitian} operator
, and similarly in case (ii), the rows $(j,k,1)$ and $(j',k,1)$ of $\mm{P}^{-1}$ can be chosen to be mutually complex conjugate (as they correspond to left eigenvectors w.r.t. a pair of mutually conjugated eigenvalues) and hence $\ket{{\rm NESS};j,j',k,k}$ represents a {\em Hermitian} operator.
In other words, $ {\hat b} '_{j,k,1}$ (i) and $ {\hat b} '_{j,k,1} {\hat b} '_{j',k,1}$ (ii) are, under the above conditions on the eigenvectors of $\mm{P}$,
Hermiticity preserving maps. Furthermore, using similar arguments one can see that the {\em non-diagonal} maps
$ {\hat b} '_{j,k,1} {\hat b} '_{j',k',1} +  {\hat b} '_{j,k',1} {\hat b} '_{j',k,1}$ and $ {\rm i} ( {\hat b} '_{j,k,1} {\hat b} '_{j',k',1} -  {\hat b} '_{j,k',1} {\hat b} '_{j',k,1})$ are Hermiticity preserving, so
\begin{eqnarray}
\ket{{\rm NESS};j,j',k,k',+}:=\ket{{\rm NESS};j,j',k,k'} &+& \ket{{\rm NESS};j,j',k',k} \nonumber \\
\ket{{\rm NESS};j,j',k,k',-} := {\rm i} \ket{{\rm NESS};j,j',k,k'} &-&  {\rm i} \ket{{\rm NESS};j,j',k',k}
\end{eqnarray}
correspond to Hermitian operators.

Note as well that all these elements of ${\cal K}$ correspond to {\em trace-zero} operators as we have trivially
${\,{\rm tr}\,} \rho_{{\rm NESS};j,k} = \braket{1}{{\rm NESS};j,k} = \bra{1} {\hat b} '_{j,k,1}\xket{{\rm NESS}} = 0$,
${\,{\rm tr}\,} \rho_{{\rm NESS};j,j',k,k'} = \braket{1}{{\rm NESS};j,j',k,k'} = \bra{1} {\hat b} '_{j,k,1} {\hat b} '_{j',k',1}\xket{{\rm NESS}} = 0$, due to (\ref{eq:vacua}).
Thus, in general a {\em convex set} of {\em real} coefficients should exist $\{ \alpha_{j,k},\alpha_{j,j',k,k',\pm}\in\mathbb{R}\}$, such that
\begin{equation}
\xket{{\rm NESS}} + \sum_{j,k}^{\beta_j = 0} \alpha_{j,k} \ket{{\rm NESS};j,k} + \!\!\!\!\!\sum_{j,j',k,k',\tau=\pm}^{\beta_j+\beta_{j'}=0}\!\!\!\!\!\!\!\alpha_{j,j',k,k',\tau} \ket{{\rm NESS};j,j',k,k',\tau}
\end{equation}
represent valid positive trace-one density operators.

Geometric characterization of this set of degenerate non-equilibrium steady states poses an interesting open problem.

\subsection{Physical observables in NESS}

\begin{theorem}
\label{observable}
Suppose the spectrum of $\mm{X}$ (\ref{eq:X}) is strictly away from the imaginary axis ${\,{\rm Re}\,} \beta_j > 0$. Then
the complete set of quadratic correlators in NESS is given by the matrix $\mm{Z}$, a (unique) solution of the Lyapunov equation (\ref{eq:Lyap}),
\begin{equation}
{\,{\rm tr}\,} w_{j} w_{k} \rho_{\rm NESS} = \delta_{j,k} + 4  {\rm i}  Z_{j,k}.
\label{eq:corr}
\end{equation}
\end{theorem}
\begin{proof}
Inverting (\ref{eq:bfroma})
\begin{equation}
\un{ {\hat a} } = \mm{V}^{-1}(\un{ {\hat b} }\oplus \un{ {\hat b} }')=\mm{V}^T(\un{ {\hat b} }'\oplus \un{ {\hat b} }),
\end{equation}
we express the first first $2n$ components of the canonical Majorana maps explicitely
\begin{equation}
\un{ {\hat a} }_1 = \frac{1}{\sqrt{2}}\left\{\mm{P}^{-T}\un{ {\hat b} } + (\mathbbm{1}_{2n}+4 {\rm i}  \mm{Z})\mm{P}\un{ {\hat b} }'\right\}.
\end{equation}
Then we note that, since $\ket{w_j w_k} = 2  {\hat a} _{1,j}  {\hat a} _{1,k}\ket{1}$ and
$\bra{w_j w_k} = 2 \bra{1}  {\hat a} _{1,j}  {\hat a} _{1,k}$, we can express the correlation matrix
$C_{j,k} :={\,{\rm tr}\,} w_{j} w_{k} \rho_{\rm NESS}$
as
\begin{eqnarray}
\mm{C} &=& 2\bra{1} \un{ {\hat a} }_1\otimes \un{ {\hat a} }_1\xket{{\rm NESS}}  \nonumber \\
&=& \bra{1}\left\{\mm{P}^{-T}\un{ {\hat b} } + (\mathbbm{1}_{2n}+4 {\rm i}  \mm{Z})\mm{P}\un{ {\hat b} }'\right\}\otimes\left\{\mm{P}^{-T}\un{ {\hat b} } + (\mathbbm{1}_{2n}+4 {\rm i}  \mm{Z})\mm{P}\un{ {\hat b} }'\right\}\xket{{\rm NESS}} \nonumber \\
&=& \mm{P}^{-T} \bra{1}\un{ {\hat b} }\otimes \un{ {\hat b} }'\xket{{\rm NESS}}[(\mathbbm{1}_{2n}+4 {\rm i} \mm{Z})\mm{P}]^T = \mathbbm{1}_{2n}+4 {\rm i} \mm{Z}
\end{eqnarray}
where we used (\ref{eq:vacua}), and $\bra{1}\un{ {\hat b} }\otimes \un{ {\hat b} }'\xket{{\rm NESS}} = \mathbbm{1}_{2n}$ due to almost-CAR (\ref{eq:almostCAR}).
\qed
\end{proof}

\section{Examples}

We conclude this note by showing in two physical examples that the situations of non-diagonalizable Liouvilleans and subspaces of degenerate NESS are not as pathological or difficult to encounter as it might seem from the first sight.

\subsection{A single qubit with non-diagonalizable Liouvillean dynamics}

Take a single fermion (or qubit), described in terms of a pair of Hermitian anticomuting
operators $w_1=c+c^\dagger$ and $w_2= {\rm i} (c-c^\dagger)$.
Take Hamiltonian $H =  {\rm i}  h w_1 w_2$ and a single Lindblad operator
$L = \sqrt{\Gamma} (w_1 + e^{ {\rm i} \theta} w_2)$
Then, the key matrices read
\begin{equation}
\mm{X} = 2 \pmatrix{ \Gamma & \Gamma \cos\theta +h \cr \Gamma \cos\theta - h & \Gamma},\;\;
\mm{M}_{\rm i} = \pmatrix{ 0 & -\Gamma\sin\theta \cr \Gamma \sin\theta & 0}.
\end{equation}
For $h=\Gamma\cos\theta$ the matrix $\mm{X}$ becomes nondiagonalizable.

\subsection{A pair of spins with degenerate NESS}
Consider a pair  ($n=2$) of qubits, or spins $1/2$, described by Pauli operators $\un{\sigma}_{1,2}$,
coupled via Ising interaction
$H = \frac{1}{2}J \sigma^1_1 \sigma^1_2$
and attached to a Lindblad reservoir only through one of the qubits $L_1 = \Gamma_1 \sigma^+_1, L_2 = \Gamma_2 \sigma^-_1$.
This model is fermionized in terms of Majorana operators $w_{1,2,3,4}$ \cite{njp} using Jordan-Wigner transformation $H=-\frac{ {\rm i} }{2}  J w_2 w_3$ and the key matrices read
\begin{equation}
\mm{X} = \pmatrix{ \Gamma_+ & 0 & 0 & 0 \cr 0 & \Gamma_+ & -J & 0 \cr 0 & J & 0 & 0 \cr 0 & 0 & 0 & 0},\;\;
\mm{M}_{\rm i} = \frac{\Gamma_-}{4}\pmatrix{ 0 & 1 & 0 & 0 \cr -1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0}
\end{equation}
where $\Gamma_{\pm} = \Gamma_2 \pm \Gamma_1$.
The rapidity spectrum is
\begin{equation}
\beta_1 = \Gamma_+, \quad \beta_{2,3} = \frac{\Gamma_+}{2} \pm \sqrt{\left(\frac{\Gamma_+}{2}\right)^2 - J^2}, \quad \beta_4=0.
\end{equation}
The Lyapunov equation (\ref{eq:Lyap}) for the steady state 2-point correlator (\ref{eq:corr}) has a unique solution, in spite of zero eigenvalue (since it has multiplicity $1$, see the proof of lemma \ref{lemma:similar}), namely
\begin{equation}
\mm{Z} = \frac{2\Gamma_-}{2\Gamma_+^2+ J^2}
\pmatrix{ 0 & \Gamma_+ & J & 0  \cr
- \Gamma_+ & 0 & 0 &  0 \cr
-J & 0 & 0 & 0 \cr
0 & 0 & 0 & 0}.
\end{equation}
However, the non-uniqueness of $\xket{{\rm NESS}}$ is manifested in non-vanishing and non-unique $1-$point functions, namely since the non-unique NESS is
parametrized as $\ket{{\rm NESS};\alpha} = \xket{{\rm NESS}} + \alpha \ket{{\rm NESS};4,1} $, where $\ket{{\rm NESS};4,1} = \hat{b}'_4 \xket{{\rm NESS}} = \hat{c}^\dagger_4 \xket{{\rm NESS}}$, we have
$\ave{w_j} = \delta_{j,4} \alpha$.  Using the (generalized) Wick theorem, we see that even order monomials are insensitive to non-uniqueness parameter
$\alpha$, whereas odd order monomials depend on $\alpha$.
The range of admissible values of $\alpha$ is obtained for example by explicit construction of the spectrum of the NESS density matrix
$\rho_{\rm NESS} = 4\sum_{\gamma_j\in\{0,1\}} \bra{1}\hat{a}_{1,1}^{\gamma_1}\hat{a}_{1,2}^{\gamma_2}\hat{a}_{1,3}^{\gamma_3}\hat{a}_{1,4}^{\gamma_4}\ket{{\rm NESS};\alpha}
w^{\gamma_1}_1 w^{\gamma_2}_2 w^{\gamma_3}_3 w^{\gamma_4}_4$
and require it to lie in the interval $[0,1]$.

Richer structure of degenerate NESS is obtained by adding further spins $1/2$ to such a one-sided open Ising chain, namely each new spin coupled to the rest of the chain
via Ising interaction adds a new pair of completely imaginary structure matrix eigenvalues $\pm i b$ (plus a pair of fully complex eigenvalues).

\section*{Acknowledgements}

The author is grateful to Iztok Pi\v zorn for providing some valuable information on the structure of the spectrum and eigenvectors of the Liouvillean structure matrix which motivated this work, and to Peter \v Semrl for extensive discussions and providing detailed arguments for the conjecture \ref{conjecture} underlying the Jordan canonical form structure of the
many-body nilpotent map. The work is supported by the Programme P1-0044 and the Grant J1-2208 of the Slovenian Research Agency.
Finally, hospitality of University of Potsdam
in the final stage of preparing this publication is warmly acknowledged.

\section*{References}

redpajama_set_name: RedPajamaArXiv


Mesquite MX, Mesquite, NV (10/8/2018) – Absolute Motorsports' Jacob Peter scored a third-place finish during the WORCS Series SXS Youth 250 Production race at Mesquite MX in Arizona. It marked the young racer's sixth podium of the season in the competitive class.
The weekend marked the first SXS race at the motocross facility and the surrounding desert. Featuring some of the biggest air on the tour, Peter took a conservative approach to the jumps early in the event before going all out at the end.
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Peter also competed on his quad and motorcycle during the weekend. The young iron man finished seventh in his motorcycle class. Going into the final race, he hopes to finish off the year with another victory.
Look for Peter to net his second victory of the season in the SXS class as the WORCS series finishes out its 2018 November 2-4 at Buffalo Bill's Casino in Primm, NV.


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Patience Jane Wheatcroft, baronne Wheatcroft (née le ) est une journaliste britannique et pair à vie.

Elle est rédactrice en chef du Wall Street Journal Europe . Elle quitte ce poste en devenant pair. Elle est auparavant rédactrice en chef du journal The Sunday Telegraph. Elle démissionne de ce poste en septembre 2007 après dix-huit mois de travail et quitte temporairement le journalisme.

Carrière
Wheatcroft fait ses études à la Wolverhampton Girls' High School et à l'Université de Birmingham (LLB, Law, 1972) . Elle et son mari, Tony Salter, lancent le magazine spécialisé Retail Week en 1988, et Wheatcroft en est consultante éditoriale jusqu'en 1992.

Elle travaille pour plusieurs journaux nationaux, dont le Daily Mail, le Sunday Times et le Daily Telegraph. Après avoir été rédactrice en chef adjointe de The Mail on Sunday, elle est nommée rédactrice en chef des affaires et de la city du Times en 1997, puis rédactrice en chef du Sunday Telegraph en mars 2006.

Wheatcroft remporte le Wincott Senior Journalist of the Year Award en 2001 et en 2003 est nommée London Press Club Business Journalist of the Year.

Elle démissionne de son poste de rédactrice en chef du Sunday Telegraph le 4 septembre 2007, étant remplacée par Ian MacGregor, qui était jusqu'alors rédacteur en chef adjoint du Daily Telegraph .

Postes d'administratrice
Du  janvier 2008 à 2009, Wheatcroft est administratrice non exécutive de Barclays plc. Du 27 février 2008 à 2009, elle est administratrice non-exécutive de Shaftesbury plc, une société d'investissement immobilier britannique avec des actifs dans le centre de Londres.

Le 8 mai 2008, elle est nommée à la tête du Panel d'audit judiciaire nouvellement créé par le nouveau maire de Londres, Boris Johnson. Le panel est chargé de surveiller et d'enquêter sur la gestion financière de la London Development Agency et de l'Autorité du Grand Londres .

Le 30 juillet 2010, le Premier ministre David Cameron nomme Wheatcroft au conseil d'administration du British Museum .

Le 22 décembre 2010, Wheatcroft est créée pair à vie en tant que baronne Wheatcroft, de Blackheath dans le quartier londonien de Greenwich. Elle siège à la Chambre des lordscomme pair conservateur, après avoir été une militante de longue date du parti, jusqu'en novembre 2019, mais siège maintenant en tant que pair non affilié .

Vie privée
Wheatcroft est mariée et mère de trois enfants. Son mari éditeur, Tony, est un militant du Parti conservateur œuvrant dans les quartiers londoniens de Greenwich et Lewisham .

Références

Liens externes

Étudiant de l'université de Birmingham
Journaliste britannique
Pair à vie
Naissance en septembre 1951

redpajama_set_name: RedPajamaWikipedia


On April 12, the far-right Jobbik candidate won a by-election in the individual constituency of Tapolca. The election took place due to the death of former Fidesz MP Jenő Lasztovicza in early January 2015.
The electorate dissatisfied with the performance of the government turns to the political side it considers capable of defeating Fidesz. On February 22, in the Veszprém by-election this meant giving the mandate to the individual candidate supported by left-wing parties, Zoltán Kész, and in Tapolca backing Jobbik candidate, Lajos Rig.
With this the governing party's "central power field" strategy that worked well in 2014 has suffered a setback, just as the myth that Fidesz can stop the rise of the far right.
Jobbik can consolidate its position as the challenger of Fidesz, and could even go first in the polls.
Read the complete Flash Report here.


yO7nn9 Im obliged for the article. Much obliged.
The MLS Board in our area will fine agents if photos are not uploaded within a certain time period. I think it is 2-3 days.It seems pretty sad that an agent wouldn't just automatically post photos with the listing.Even if the house is a fixer, the potential buyers would want to know at least what style the home is. Or, in the case of land, an idea of the setting where the site is located.
There are so many people out of jobs and are not reflected in the unemployment number because they did not apply for unemployment benefits. The inflation is so high and I think it will be a long way to economic recovery.
Is your faith in what God is doing or who God is?Most definitely who God is. . .and because I am confident in who God is, I think I'm able to see more clearly what he is doing in my life. More often I see what he's done–tracing those fingerprints back through times when I didn't understand his plan.Are you hoping in God's doing or God's being?God's being–but God being is also God doing–God is a verb–constant compassion, love, patience, wisdom, benevolence, etc.Do you trust an I DO, or an I AM?I AM. . .God is. . .God was. . .God will be–constant, eternal, faithful. Overwhelming.


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Stories: Notable New Yorkers
Happy Birthday, Leonard Bernstein
Tuesday, August 21, 2018 by Morgen Stevens-Garmon
August 25 marks the 100th birthday of Leonard Bernstein. We're getting in on the world-wide celebration of the conductor, composer, and New Yorker with a few highlights from the Museum's collection. Happy birthday, Lenny!
Stanley Kubrick for Look Magazine. [Leonard Bernstein at a piano.] 1949. Courtesy MCNY and SK Film Archives. X2011.4.12304.141
Bernstein was born in Lawrence, Massachusetts. He started taking piano lessons at the New England Conservatory of Music at age 13, and graduated with honors from Harvard University two months before his 21st birthday. After Harvard, Bernstein studied conducting at the Curtis Institute of Music in Philadelphia, and during the summer break, he made his symphonic conducting debut at the newly founded Tanglewood Music Festival. Though often in New England for gigs, Bernstein shared an apartment in New York City with his good friend Adolph Green. Green, his writing partner Betty Comden, and a few friends formed an act called The Revuers. When they performed in Greenwich Village, Bernstein would often serve as their accompanist.
Stanley Kubrick for Look Magazine. [Bernstein playing piano.] 1949. Courtesy MCNY and SK Film Archives. X2011.4.12304.102E.
Bernstein met the dancer Jerome Robbins in 1943. Robbins was looking for a composer for his ballet about three sailors on leave in New York. The resulting "Fancy Free" premiered in 1944 with Bernstein conducting. In December of that year, On the Town opened on Broadway. An expansion of "Fancy Free," the musical marked Bernstein's Broadway debut as well as the first Broadway appearance of Robbins's choreography and book and lyric authors Comden and Green.
Sheet music for "Lonely Town" from On the Town. 1944. Museum of the City of New York. 70.22.31.
Bernstein went on to compose five more full-length musicals including Wonderful Town and West Side Story. He also contributed original compositions to plays.
Leonard Bernstein (1918-1990). Fragment of score for Peter Pan. 1950. Museum of the City of New York. 51.73.
Around the time Bernstein met Robbins, the composer was hired as assistant conductor with the New York Philharmonic. When a guest conductor suddenly came down sick, Bernstein had to step in without much preparation. The Carnegie Hall concert was broadcast nationally, and it turned Bernstein into an instant sensation. From 1958 to 1969, Bernstein served as music director of the New York Philharmonic.
William Auerbach-Levy (1889-1964). [Leonard Bernstein.] 1940-1964. Museum of the City of New York. 64.100.180.
Find out where to join in the birthday celebrations here. And stop by the Museum to see more images of Leonard Bernstein as part of Through a Different Lens: Stanley Kubrick Photographs.
Stanley Kubrick for Look Magazine. [Leonard Bernstein reading.] 1949. Courtesy of Museum of the City of New York and SK Film Archives. X2011.4.12304.104E.
Stanley Kubrick for Look Magazine. [Leonard Bernstein with a dog.] 1949. Courtesy of Museum of the City of New York and SK Film Archives. X2011.4.12304.115D.
By Morgen Stevens-Garmon, Associate Curator, Theater Collection
Related Exhibition - May 3, 2018 - January 6, 2019
Through a Different Lens
Stanley Kubrick Photographs
City Arts Tuesday, September 26, 2017
When Love Comes so Strong
If you know the story of Romeo and Juliet, you know the story of West Side Story. Read about the creative forces behind the musical and its lasting legacy.
City Arts Thursday, October 26, 2017
Salsa on Stage
Rhythm & Power: Salsa in New York digs into salsa's history as an art form and social movement. Our Theater Collection curator takes a look at how that movement translated and transferred on stage.
City Arts Tuesday, April 3, 2018
Happy 75th Birthday, Oklahoma!
Oh, what a beautiful opening night for Oklahoma!, the first musical from composer-lyrist team Rodgers & Hammerstein. To celebrate its 75th birthday, we delve into the history of the legendary musical with objects from our Theater Collection.


The algorithm correctly predicted 8 of the 10 stocks forecasted to increase during the predicted time horizon of 14 days in the Pharma package forecast from 12/13/2015. The most impressive prediction was for RPTP, which offered a return of 17.84% during the predicted time horizon to investors who bought this stock. Other impressive returns were CTIC and SCMP which increased by 17.27% and 14.78%, respectively. The average return from this forecast for investors was 6.60%, offering a premium to investors over the S&P 500 return of 2.42% during the predicted time horizon.
Raptor Pharmaceutical Corp. (RPTP) is a biopharmaceutical company. The Company is focused on developing and commercializing transformative treatments for people affected by rare and debilitating diseases. The mean short-term price target for RPTP has been set at $13 a share.


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Compelling hoarding has recently been sent into the spotlight cheers to several fact television series of which showcase your disorder. Even though more people find out about this sickness than ever before, its presence on tv hasn't got an entirely positive effect on sufferers and those who enjoy them, seeing that hoarding is frequently treated as being a spectacle instead of as the problem that it is. Frequently it's easy to forget that those that are affected by this condition face psychological and functional challenges even after the digital cameras have been deterred.
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Measles. Whopping Cough. Diphtheria. Rubella. Polio. Small Pox. The list may be unfamiliar to a younger generation, but for anyone born before 1960 it can conjure up images of quarantines, "iron lungs" and even premature death. These diseases were once considered the scourge of society and responsible for high childhood mortality rates. Any visit to an old cemetery from the early part of the 20th century would reveal rows of headstones of children under five who died from "natural causes." Thanks to advances in science, medicine and public health, however, these childhood diseases have now been eliminated or significantly isolated. Key in this success has been the practice of vaccination. In 2000, the Center for Disease Control reported that all of the aforementioned diseases had been eliminated in the U.S. largely due to mandatory childhood immunizations.
Vaccinations have become one of the most effective tools in addressing infectious diseases and responsible for dramatically increasing life expectancies and decreasing under age 5 mortality rates across the globe. Vaccines have been found to be overwhelmingly safe, effective and cheap with new ones being formulated annually, as evidenced by recent prototypes for HIV/AIDS, Ebola and, the world's biggest killer, malaria. The U.S., combining proactive and visionary research with generous government and private endowments from groups such as the Gates Foundation, has spearheaded this effort. The result has been millions of lives saved.
While vaccination rates in the U.S. have consistently topped over 95% over the past 50 years, in some areas the rate is falling well below acceptable thresholds that guarantee universal protection. Our current childhood vaccination rates have now fallen to an average 91% for mumps, measles & rubella and only 83.4% for diphtheria, tetanus and pertussis. By comparison, the World Health Organization puts Albania, Nicaragua and Morocco at 99% in both categories. Public health professionals in the U.S. have recently noticed a new and troubling rise in many of these diseases with measles taking a conspicuous lead. Cases of the disease in the first three months of 2019 have already far exceeded all of 2018 with regional outbreaks occurring in 26 states and spreading. So why the backsliding?
The rising trend of "scientific skepticism" as manifested by the "anti-vax" movement largely lies at root of the current outbreaks. Working on half-truths, pseudoscience and a healthy dose of fear-based conjecture (typically propagated on social media), non-scientists and individuals with little to no medical training have championed the right to "opt out" of vaccinations for their children and even themselves. The reasons include a fear of vaccine-related complications (such as the debunked link to autism), freedom of choice and religious exemptions.
Faulty science and irrational thinking aside, the trend has nonetheless gained ground. It poses such a serious a problem that the World Health Organization has added the anti-vax movement as one of the top 10 threats to global health, alongside pandemic flu, non-communicable diseases and anti-microbial resistance. While vaccines can cause side effects, these are minimal with benefits to public health that far outweigh their potential risk. The statistics on their safety are among the highest for any medical intervention.
Our public health system now faces yet another challenge which is largely self-inflicted. Although "opting out" of shots is appropriate for some who have preexisting conditions, most people have no logical reason for not being immunized. In a world that is both smaller and less isolated, epidemics are now an airline flight or non-immunized child away. In this high-risk environment, taking sound precautions keeps in mind that we're all in this together and cannot carry on is if we're the only ones walking this earth. It's a question of recognizing common sense and serving the common good.
Mike Radoiu of Staunton is an optometrist who practices in Harrisonburg and is a retired member of the U.S. Navy Medical Service Corps. Write him at mradoiu@salus.edu.