# Stabilizing Preparation of Quantum Gaussian States via Continuous
Measurement

Liying Bao<EMAIL_ADDRESS>Bo Qi<EMAIL_ADDRESS>Daoyi Dong
<EMAIL_ADDRESS>Key Laboratory of Systems and Control, Academy of
Mathematics and Systems Science,
Chinese Academy of Sciences, Beijing 100190, China  University of Chinese
Academy of Sciences, Beijing 100049, China  School of Engineering and
Information Technology, University of New South Wales, Canberra ACT 2600,
Australia

###### Abstract

This paper provides a stabilizing preparation method for quantum Gaussian
states by utilizing continuous measurement. The stochastic evolution of the
open quantum system is described in terms of the quantum stochastic master
equation. We present necessary and sufficient conditions for the system to
have a unique stabilizing steady Gaussian state. The conditions are much
weaker than those existing results presented in the approach of preparing
Gaussian states through environment engineering. Parametric conditions of how
to prepare an arbitrary pure Gaussian state are provided. This approach
provides more degrees of freedom to choose the system Hamiltonian and the
system-environment coupling operators, as compared with the case where
dissipation-induced approach is employed. The stabilizing conditions for the
case of imperfect measurement efficiency are also presented. These results may
benefit practical experimental implementation in preparing quantum Gaussian
states.

###### keywords:

Quantum Gaussian state; Continuous measurement; Riccati equation; Linear
quadratic Gaussian system.

††thanks: The material in this paper was not presented at any conference.

, , [cor1]Corresponding author.

##  1 Introduction

Continuous variable systems, which are quantum systems with infinite
dimensional Hilbert spaces, have been an important platform for quantum
cryptography, quantum information and quantum computation [Edwards & Belavkin,
2005, van Handel, Stockton, & Mabuchi, 2005, James, Nurdin, & Petersen, 2008,
Wiseman & Milburn, 2010, Nurdin, Petersen, & James, 2012, Pan, Zhang, & James,
2016, Ma, Woolley, Petersen, & Yamamoto, 2018, Gao, Zhang, & Petersen, 2020,
Gao, Dong, Petersen, & Ding, 2020, Ghalaii, Ottaviani, Kumar, Pirandola, &
Razavi, 2021, Zhang & Pan, 2020, Zhang & Petersen, 2020, Gao, Dong, Petersen,
& Ding, 2021, Guo, Peng, Liao, & Wang, 2021]. Gaussian states, which include a
wide and important class of quantum states such as vacuum states, squeezed
vacuum, squeezed coherent states, quasi-free states and ground states of some
free Hamiltonians, are the basis for various continuous variable quantum
information processing [De Palma, Trevisan, & Giovannetti, 2017, Holevo, 2020,
Srikara, Thapliyal, & Pathak, 2020]. Since Gaussian states can be completely
characterized by their first and second moments and the Gaussian character can
be preserved under typical quantum operations, they are relatively easy to be
studied analytically and are often used as theoretical testing ground
[Bondurant, Kumar, Shapiro, & Maeda, 1984, Erkmen & Shapiro, 2008, Yokoyama et
al., 2013, Chabaud, 2021]. Besides, protocols based on Gaussian states and
Gaussian measurements (e.g., homodyne detection) are relatively experimentally
friendly as compared to operations of general quantum states [Pinel, Jian,
Treps, Fabre, & Braun, 2013, Banchi, Braunstein, & Pirandola, 2015, McDonald &
Clerk, 2020, Bao, Qi, Dong, & Nori, 2021]. Thus Gaussian states are of great
importance in both theoretical studies and experimental implementations.

The preparation of a desired Gaussian state is clearly a pivotal task in
continuous variable quantum information processing [Diehl et al., 2008,
Verstraete, Wolf, & Ignacio, 2009, Krauter et al., 2011, Koga & Yamamoto,
2012, Yamamoto, 2012, Ma, Woolley, Petersen, & Yamamoto, 2014, Ma, Woolley,
Jia, & Zhang, 2019, Häffner, Zanin, Gomes, Céleri, & Souto Ribeiro, 2020,
Giovanni, Brunelli, & Genoni, 2021]. There have been some results proposed on
the basis of the dissipation-induced approach or the environment engineering
approach [Diehl et al., 2008, Verstraete, Wolf, & Ignacio, 2009, Krauter et
al., 2011, Koga & Yamamoto, 2012, Yamamoto, 2012, Ma, Woolley, Petersen, &
Yamamoto, 2014]. The basic idea behind those results is that one can utilize
the dissipative environment by engineering a proper Hamiltonian and
appropriately designed dissipative channels such that the corresponding stable
state is a desired useful Gaussian state. In [Koga & Yamamoto, 2012],
necessary and sufficient conditions for a Gaussian master equation to have a
unique steady pure state were provided and then based on those conditions a
systematic procedure to prepare a desired state via dissipation was proposed.
Based on the quantum stochastic differential equation, [Yamamoto, 2012]
further investigated the pure Gaussian state generation and clarified a
physical meaning that the nullifier dynamics of any Gaussian system generating
a unique steady pure state is passive. An alternative method of preparing pure
Gaussian state was presented in [Ma, Woolley, Petersen, & Yamamoto, 2014],
where it was shown that a desired pure Gaussian state can be prepared by a
cascade of one-dimensional open quantum harmonic oscillators, without any
direct interaction Hamiltonians between these oscillators.

As mentioned, Gaussian states can be completely characterized by their first
and second moments. In practice, most of the useful properties of Gaussian
states such as the purity [Paris, Illuminati, Serafini, & De Siena, 2003],
entanglement [Marian & Marian, 2008] and squeezing [Petersen, Madsen, &
Mølmer, 2005] are only related with the second moment. In previous results,
the covariance matrix which depicts the second moment of the Gaussian states
is described in terms of the quantum master equation (QME) in essence [Koga &
Yamamoto, 2012, Yamamoto, 2012, Ma, Woolley, Petersen, & Yamamoto, 2014]. In
this paper, we provide an alternative way in preparing quantum Gaussian states
by utilizing continuous measurement. To account for the randomness of quantum
measurement, the evolution of the open quantum system is described in terms of
the quantum stochastic master equation (QSME). The basic idea is that the
covariance matrix determined by the QSME is smaller (in the sense of matrix
partial order) than that determined by the QME if all the parameterizations
are the same. Thus, by utilizing continuous measurement, the useful properties
of Gaussian states can be improved. We first present necessary and sufficient
conditions for the system to have a unique stabilizing steady pure Gaussian
state. The conditions are much weaker than those presented in the dissipation-
induced approach. Then general parametric conditions of generating an
arbitrarily desired pure Gaussian state are provided. They provide more
degrees of freedom to choose the system Hamiltonian and coupling channels as
compared with previous results. For completeness and practical applications,
we also consider the case where the detection is imperfect.

The paper is organized as follows. In Section II, we first give some
preliminaries on Gaussian states and the QSME, and then set up the linear
quadratic Gaussian system model to be utilized. The main results focusing on
pure Gaussian states are presented in Section III. Section IV discusses the
case where the detection efficiency is less than 1. Section V concludes the
paper.

##  2 Gaussian states, QSME & linear quadratic Gaussian system

This section briefly introduces the Wigner representation of Gaussian states,
the framework of continuous measurement and the setting of linear quadratic
Gaussian system. After an appropriate setting is given, the preparation
problem of Gaussian states is restated as a stabilizing steady state problem
of a Riccati equation.

###  2.1 Wigner representation of Gaussian states

We start from reviewing the Wigner phase-space representation [Paris,
Illuminati, Serafini, & De Siena, 2003, Wiseman & Milburn, 2010], which is one
of the most commonly used representations for quantum Gaussian states.

Denote $\hat{a}_{k}$ and $\hat{a}^{\dagger}_{k}$, $k=1,...,m$, as the
canonical bosonic annihilation and creation operators of the $k$-th mode,
respectively. The corresponding canonical quadrature operators $\hat{q}_{k}$
and $\hat{p}_{k}$ are defined via

$\hat{q}_{k}=\frac{\hat{a}_{k}+\hat{a}_{k}^{\dagger}}{2}~{}~{}\text{and}~{}~{}\hat{p}_{k}=i\frac{\hat{a}_{k}^{\dagger}-\hat{a}_{k}}{2},$

which satisfy the canonical commutation relation

$[\hat{q}_{k},\hat{p}_{j}]=i\delta_{kj},$

where $i=\sqrt{-1}$ and $\delta_{kj}$ is the Kronecker delta function. Denote
the vector of quadrature operators for the $m$ mode system as

$\hat{\textbf{X}}=(\hat{q}_{1},\cdots,\hat{q}_{m},\hat{p}_{1},\cdots,\hat{p}_{m})^{\top},$

where $\top$ is the matrix transpose. The operator vector $\hat{\textbf{X}}$
satisfies the canonical commutation relations [Koga & Yamamoto, 2012,
Yamamoto, 2012, Ma, Woolley, Petersen, & Yamamoto, 2014]

$\left[\hat{\textbf{X}},\hat{\textbf{X}}^{\top}\right]=\hat{\textbf{X}}\hat{\textbf{X}}^{\top}-(\hat{\textbf{X}}\hat{\textbf{X}}^{\top})^{\top}=iJ,$

where $J=\begin{pmatrix}0&I_{m}\\\ -I_{m}&0\end{pmatrix}$, and $I_{m}$ is the
$m\times m$ identity matrix.

The joint displacement operator can be described in terms of
$\hat{\textbf{X}}$ and $J$ as

$\hat{D}(\bm{\alpha})=\textrm{exp}(i~{}\hat{\textbf{X}}^{\top}J\bm{\alpha}),$

where $\bm{\alpha}\in\mathbb{R}^{2m}$ [Paris, Illuminati, Serafini, & De
Siena, 2003, Wiseman & Milburn, 2010]. On the basis of $\hat{D}(\bm{\alpha})$,
for an arbitrary quantum state operator $\hat{\rho}$, its Wigner
characteristic function is defined as

$\chi_{\hat{\rho}}(\bm{\alpha})=\textrm{Tr}[\hat{\rho}\hat{D}(\bm{\alpha})].$

Now the Wigner function $W_{\hat{\rho}}(X)$ of the state $\hat{\rho}$ can be
described as

$W_{\hat{\rho}}(X)=\int_{\mathbb{R}^{2m}}\frac{d\bm{\alpha}}{(2\pi)^{2m}}\textrm{exp}(-iX^{\top}J\bm{\alpha})\chi_{\hat{\rho}}(\bm{\alpha}),$

where $X=(q_{1},...,q_{m},p_{1},\cdots,p_{m})^{\top}$ with $q_{i}$ and $p_{j}$
being real numbers [Paris, Illuminati, Serafini, & De Siena, 2003, Wiseman &
Milburn, 2010].

A state is called Gaussian if its Wigner function is in the form of

$W(X)=\frac{\textrm{exp}[-\frac{1}{2}(X-\bar{X})^{\top}V^{-1}(X-\bar{X})]}{(2\pi)^{m}\sqrt{\textrm{det}[V]}},$
(1)

where
$\bar{X}=\langle\hat{\textbf{X}}\rangle=\textrm{Tr}(\hat{\textbf{X}}\hat{\rho})$
is the mean value vector whose element
$\bar{X}_{i}=\langle\hat{\textbf{X}}_{i}\rangle=\textrm{Tr}(\hat{\textbf{X}}_{i}\hat{\rho})$,
$V$ is the covariance matrix whose element
$V_{ij}=\frac{1}{2}\langle\Delta\hat{\textbf{X}}_{i}\Delta\hat{\textbf{X}}_{j}+\Delta\hat{\textbf{X}}_{j}\Delta\hat{\textbf{X}}_{i}\rangle$
with
$\Delta\hat{\textbf{X}}_{i}=\hat{\textbf{X}}_{i}-\langle\hat{\textbf{X}}_{i}\rangle$,
and $\textrm{det}[V]$ denotes the determinant of matrix $V$ [Edwards &
Belavkin, 2005, Wiseman & Milburn, 2010]. In addition to the positive
semidefinite condition $V\geq 0$, $V$ should satisfy the Heisenberg
uncertainty principle [Koga & Yamamoto, 2012, Yamamoto, 2012, Ma, Woolley,
Petersen, & Yamamoto, 2014]

$V\geq\pm\frac{i}{2}J.$

It is straightforward to see that a Gaussian state is determined by the mean
vector $\bar{X}$ and the covariance matrix $V$.

In this paper, we only focus on the covariance matrix $V$. This is because, on
the one hand, that one can always adjust the mean of a Gaussian state to any
target mean by a suitable Weyl operator [Parthasarathy, 2010]. On the other
hand it is noted that most of the useful properties of Gaussian states such as
the purity, entanglement and squeezing are only related with the covariance
matrix in practice. To be specific, the purity of a $m$ mode Gaussian state
depends only on the covariance matrix as
$\textrm{Tr}[\hat{\rho}^{2}]=\frac{1}{2^{m}\sqrt{\textrm{det}[V]}}$ [Paris,
Illuminati, Serafini, & De Siena, 2003]. Moreover, for a two-mode pure
Gaussian state, its $4\times 4$ covariance matrix contains all the necessary
information to determine its entanglement properties for both entanglement
criteria and entanglement measures [Rendell & Rajagopal, 2005, Marian &
Marian, 2008]. In addition, squeezed coherent states have been extensively
studied in quantum information, such as quantum teleportation and ghost
imaging. The squeezing properties are also determined by the covariance matrix
[Petersen, Madsen, & Mølmer, 2005].

###  2.2 Quantum stochastic master equation

In previous results, the commonly used model in preparing a target Gaussian
state is to engineer a dissipative system described by the quantum master
equation [Edwards & Belavkin, 2005, Wiseman & Milburn, 2010, Koga & Yamamoto,
2012, Yamamoto, 2012]

$\frac{d\hat{\rho}_{t}}{dt}=-i[\hat{H},\hat{\rho}_{t}]+\sum_{i=1}^{m}(\hat{L}_{i}\hat{\rho}_{t}\hat{L}_{i}^{\dagger}-\frac{1}{2}\hat{L}_{i}^{\dagger}\hat{L}_{i}\hat{\rho}_{t}-\frac{1}{2}\hat{\rho}_{t}\hat{L}_{i}^{\dagger}\hat{L}_{i}),$
(2)

where $\hat{\rho}_{t}$ is the density operator, $\hat{H}$ is the effective
Hamiltonian of the system and $\hat{L}_{i},~{}i=1,\ldots,m$, are the
dissipative channels which describe the interaction between the system and the
environment. The dissipative QME (2) can be viewed as an average
(unconditional) evolution of the quantum state under the assumption that all
the measurement results are discarded. Intuitively if one can utilize the
measurement information, the corresponding conditional covariance matrix may
be reduced, and thus the useful properties of the Gaussian states can be
improved. In this paper, we present how to prepare a desired quantum Gaussian
state by utilizing continuous measurement. To account for the randomness of
quantum measurement, the evolution of the open quantum system is described in
terms of the quantum stochastic master equation.

To proceed, we specify the following notation for clarity. For a matrix
$Q=(q_{ij})$, let $Q^{\dagger}$, $Q^{\top}$ and $Q^{*}$ represent the
conjugate transpose, transpose and conjugate for all the elements of $Q$,
respectively. For example, $Q^{\dagger}=(q^{*}_{ji}),~{}Q^{\top}=(q_{ji}),$
and $Q^{*}=(q^{*}_{ij})=(Q^{\dagger})^{\top}$.

The dynamics of the open system coupled with $m$ independent measurement
channels $\\{\hat{L}_{i}\\}^{m}_{i=1}$ can be described by a family of unitary
operators $\\{\hat{U}_{t}\\}_{t\in\mathbb{R_{+}}}$, which satisfy

$\displaystyle d$
$\displaystyle\hat{U}_{t}=-\hat{K}\hat{U}_{t}dt+\sum^{m}_{i=1}(\hat{L}_{i}d\hat{A}^{\dagger}_{i,t}-\hat{L}_{i}^{\dagger}d\hat{A}_{i,t})\hat{U}_{t},$
(3) $\displaystyle\hat{U}_{0}=I,$

where
$\hat{K}=i\hat{H}+\frac{1}{2}\sum^{m}_{i=1}\hat{L}^{\dagger}_{i}\hat{L}_{i}$
[Edwards & Belavkin, 2005]. Here we have assumed $\hbar=1$. The differential
of the annihilation operators of the bath $d\hat{A}_{i,t}$ obeys the quantum
It$\hat{\textrm{o}}$ rules:

$d\hat{A}_{i,t}d\hat{A}_{j,t}^{\dagger}=\delta_{i,j}dt,~{}d\hat{A}_{i,t}d\hat{A}_{j,t}=0,$
$d\hat{A}^{\dagger}_{i,t}d\hat{A}^{\dagger}_{j,t}=0,~{}d\hat{A}^{\dagger}_{i,t}d\hat{A}_{j,t}=0.$

For any system operator $\hat{X}$, its time evolution in the Heisenberg
picture is represented by

$j_{t}(\hat{X})=\hat{U}_{t}^{\dagger}(\hat{X}\otimes I)\hat{U}_{t}.$

From (3) and the quantum It$\hat{\textrm{o}}$ rules, we have

$\displaystyle dj_{t}(\hat{X})=j_{t}(\mathcal{L}[\hat{X}])dt+$
$\displaystyle\sum^{m}_{i=1}\Big{(}j_{t}([\hat{L}^{\dagger}_{i},\hat{X}])dA_{i,t}$
(4) $\displaystyle+j_{t}([\hat{X},\hat{L_{i}}])dA^{\dagger}_{i,t}\Big{)},$

where

$\mathcal{{L}}[\hat{X}]=i[\hat{H},\hat{X}]+\sum_{i=1}^{m}(\hat{L}_{i}^{\dagger}\hat{X}\hat{L}_{i}-\frac{1}{2}\hat{L}_{i}^{\dagger}\hat{L}_{i}\hat{X}-\frac{1}{2}\hat{X}\hat{L}_{i}^{\dagger}\hat{L}_{i}).$

Consider $m$ family of measurement operators
$\\{Y_{i,t}\\}_{t\in\mathbb{R_{+}}}$, where

$\hat{Y}_{i,t}=\hat{U}^{\dagger}_{t}(\hat{A}_{i,t}+\hat{A}^{\dagger}_{i,t})\hat{U}_{t}$

for $i=1,\cdots,m$. From (3) and the quantum It$\hat{\textrm{o}}$ rules, we
have

$d\hat{Y}_{i,t}=j_{t}(\hat{L}_{i}+\hat{L}_{i}^{\dagger})dt+d\hat{A}_{i,t}+d\hat{A}^{\dagger}_{i,t}.$
(5)

Once
$\hat{Y}_{i,t}=\hat{U}^{\dagger}_{t}(\hat{A}_{i,t}+\hat{A}^{\dagger}_{i,t})\hat{U}_{t}$,
$i=1,\cdots,m$, are observed, one can associate a conditional expectation to
each observable $\hat{X}$ of the system. The conditional expectation gives the
least squares estimator of $j_{t}(\hat{X})$ as
$\mathcal{E}[j_{t}(\hat{X})|\hat{Y}^{t}_{0}]$, where $\hat{Y}^{t}_{0}$ is the
algebra generated by $\hat{Y}_{i,s\leq t}$, $i=1,\cdots,m$. Since
$\hat{Y}^{t}_{0}$ is nondemolition, the observation process is in essence
equivalent to $m$ classical stochastic processes

$\\{y_{i,s\leq t},i=1,\cdots,m\\}.$

Thus $\mathcal{E}[j_{t}(\hat{X})|\hat{Y}^{t}_{0}]$ actually represents the
expectation of $\hat{X}$ conditioned on the classical stochastic observations
$\\{y_{i,s\leq t}$, $i=1,\cdots,m\\}$. This conditional expectation can be
conveniently written in the Schrödinger picture as

$\mathcal{E}[j_{t}(\hat{X})|\hat{Y}^{t}_{0}]=\text{Tr}(\hat{\rho}^{c}_{t}\hat{X})$

with $\hat{\rho}^{c}_{t}$ denoting the conditional system state at time $t$.
The stochastic master equation of $\rho^{c}_{t}$ is described as [van Handel,
Stockton, & Mabuchi, 2005, Edwards & Belavkin, 2005, Wiseman & Milburn, 2010]

$\displaystyle
d\hat{\rho}_{t}^{c}=\mathcal{L}^{\dagger}[\hat{\rho}^{c}_{t}]dt+$
$\displaystyle\sum_{i=1}^{m}\Big{(}\hat{L}_{i}\hat{\rho}^{c}_{t}+\hat{\rho}^{c}_{t}\hat{L}^{\dagger}_{i}$
(6)
$\displaystyle-\text{Tr}[\hat{\rho}^{c}_{t}(\hat{L}_{i}+\hat{L}_{i}^{\dagger})]\hat{\rho}^{c}_{t}\Big{)}dw_{i,t},$

where the innovations

$dw_{i,t}=d{y}_{i,t}-\langle\hat{\rho}^{c}_{t},\hat{L}_{i}+\hat{L}_{i}^{\dagger}\rangle
dt$

are Gaussian, and

$\mathcal{{L}}^{\dagger}[\hat{\rho}_{t}^{c}]=-i[\hat{H},\hat{\rho}_{t}^{c}]+\sum_{i=1}^{m}(\hat{L}_{i}\hat{\rho}_{t}^{c}\hat{L}_{i}^{\dagger}-\frac{1}{2}\hat{L}_{i}^{\dagger}\hat{L}_{i}\hat{\rho}_{t}^{c}-\frac{1}{2}\hat{\rho}_{t}^{c}\hat{L}_{i}^{\dagger}\hat{L}_{i}).$

It is clear that by taking the expectation of (6), we can obtain the quantum
master equation (2).

###  2.3 Quantum linear quadratic Gaussian system

We are interested in systems of $m$ degrees of freedom with the phase space
operator vector

$\hat{\textbf{X}}=(\hat{q}_{1},\cdots,\hat{q}_{m},\hat{p}_{1},\cdots,\hat{p}_{m})^{\top}$

as defined in subsection 2.1. Recall that the operator vector
$\hat{\textbf{X}}$ satisfies the canonical commutation relation

$\displaystyle\left[\hat{\textbf{X}},\hat{\textbf{X}}^{\top}\right]$
$\displaystyle=\hat{\textbf{X}}\hat{\textbf{X}}^{\top}-(\hat{\textbf{X}}\hat{\textbf{X}}^{\top})^{\top}=iJ.$

The system is coupled to $m$ measurement channels via the operator vector
$\hat{\textbf{L}}=\Lambda\hat{\textbf{X}}$, where
$\Lambda\in\mathbb{C}^{m\times 2m}$. The Hamiltonian which is quadratic in
$\hat{\textbf{X}}$ is represented as

$\hat{H}(u_{t})=\frac{1}{2}\hat{\textbf{X}}^{\top}G~{}{\hat{\textbf{X}}}+\hat{\textbf{X}}^{\top}Ku_{t}+u_{t}^{\top}K^{\dagger}\hat{\textbf{X}},$

where $G=G^{\top}\in\mathbb{R}^{2m\times 2m}$, $K\in\mathbb{C}^{2m\times m}$
and $u_{t}\in\mathbb{R}^{m}$ is the control. For such a system, from (3)-(5),
we have the quantum linear equations in the Heisenberg picture as

$\displaystyle
d\hat{\textbf{X}}_{t}=(A\hat{\textbf{X}}_{t}+Bu_{t})dt+d\hat{\textbf{V}}_{t},$
$\displaystyle
d\hat{\textbf{Y}}_{t}=C\hat{\textbf{X}}_{t}dt+d\hat{\textbf{W}}_{t},$

where

$\displaystyle A$
$\displaystyle=J\Big{(}G+\frac{1}{2i}(\Lambda^{\dagger}\Lambda-\Lambda^{\top}\Lambda^{*})\Big{)}\in\mathbb{R}^{2m\times
2m},$ $\displaystyle B$ $\displaystyle=J(K+K^{*})\in\mathbb{R}^{2m\times m},$
$\displaystyle C$ $\displaystyle=\Lambda+\Lambda^{*}\in\mathbb{R}^{m\times
2m},$

and the quantum noise increments are given by

$\displaystyle
d\hat{\textbf{V}}_{t}=iJ(\Lambda^{\top}d\hat{\textbf{A}}_{t}^{\dagger}-\Lambda^{\dagger}d\hat{\textbf{A}}_{t}),$
$\displaystyle
d\hat{\textbf{W}}_{t}=d\hat{\textbf{A}}_{t}+d\hat{\textbf{A}}_{t}^{\dagger}.$

For the quantum linear quadratic system, if the initial state $\hat{\rho}$ of
the system is Gaussian, then under the nondemolition measurement of the output
operators $\hat{\textbf{Y}}_{t}$, the conditional dynamics (6) is also
Gaussian. Thus for the linear quadratic Gaussian system, its dynamics can be
sufficiently described by the time flow of the conditional mean vector
$\bar{X}_{t}=\textrm{Tr}(\hat{\textbf{X}}\hat{\rho}^{c}_{t})$ and the
covariance matrix $V_{t}$, whose element

$V_{t}{{}_{,ij}}=\frac{1}{2}\langle\Delta_{c}\hat{\textbf{X}}_{i}\Delta_{c}\hat{\textbf{X}}_{j}+\Delta_{c}\hat{\textbf{X}}_{j}\Delta_{c}\hat{\textbf{X}}_{i}\rangle_{c}$

with
$\Delta_{c}\hat{\textbf{X}}_{i}=\hat{\textbf{X}}_{i}-\langle\hat{\textbf{X}}_{i}\rangle_{c}$
and
$\langle\hat{\textbf{X}}_{i}\rangle_{c}=\text{Tr}(\hat{\textbf{X}}_{i}\rho^{c}_{t})$.
Using (6), we have the conditional expectation $\bar{X}_{t}$ of
$\hat{\textbf{X}}$ under the measurement of $\hat{\textbf{Y}}_{t}$ as

$\displaystyle d\bar{X}_{t}$
$\displaystyle=(A\bar{X}_{t}+Bu_{t})dt+(V_{t}C^{\top}+M)d\tilde{y}_{t},$ (7)
$\displaystyle\bar{X}_{0}$ $\displaystyle=\bar{X},$

where

$M=\frac{i}{2}J(\Lambda^{\top}-\Lambda^{\dagger})\in\mathbb{R}^{2m\times m},$

and

$d\tilde{y}_{t}=dy_{t}-C\bar{X}_{t}dt$

is the Gaussian innovation which describes the information gain from the the
measurement. The conditional covariance matrix satisfies the matrix Riccati
equation

$\displaystyle\frac{d}{dt}V_{t}$
$\displaystyle=AV_{t}+V_{t}A^{\top}+N-(V_{t}C^{\top}+M)(V_{t}C^{\top}+M)^{\top},$
(8) $\displaystyle V_{0}$ $\displaystyle=V,$

where

$N=\frac{1}{2}J(\Lambda^{\dagger}\Lambda+\Lambda^{\top}\Lambda^{*})J^{\top}\in\mathbb{R}^{2m\times
2m}.$

It is worth noting that the equation for $V_{t}$ is deterministic and does not
depend on the stochastic measurement results.

Taking expectation values of (7) and (8), we have the moment equations for the
unconditional approach as

$\displaystyle d\bar{X}_{t}=(A\bar{X}_{t}+Bu_{t})dt,$ (9)
$\displaystyle\frac{d}{dt}V_{t}=AV_{t}+V_{t}A^{\top}+N,$

which correspond to the quantum master equation (2). To prepare a desired
Gaussian state, most of the previous results are on the basis of the moment
equations (9). Note that the final term in (8) causes a reduction in the
conditional covariance matrix as compared with that in (9). Recall that most
of the useful properties of the Gaussian states are determined by the
covariance matrix. Thus intuitively by utilizing the conditional covariance
matrix equation (8), one may provide weaker conditions in preparing a desired
Gaussian state.

Let us first look at the equation of the conditional expectation (7). We can
choose $B$ such that its column space is the same as that of $VC^{\top}+M$,
where $V$ is the steady solution of (8). Then we choose $F$ such that it
satisfies

$BF=-VC^{\top}-M.$

Thus, for the Markovian feedback (or direct feedback) $u_{t}dt=Fdy_{t}$
[Wiseman & Milburn, 2010], the equation (7) of the first moment becomes
deterministic in the limit $t\rightarrow\infty$ as

$\frac{d}{dt}\bar{X}_{t}=(A-MC-VC^{\top}C)\bar{X}_{t}.$ (10)

Therefore, if $A-MC-VC^{\top}C$ is stable, then the solution of (10) will
approach to 0 approximately. In the following we will give necessary and
sufficient conditions for the conditional covariance matrix equation (8) to
have a steady solution and meanwhile $A-MC-VC^{\top}C$ is stable. From now on
we focus on the conditional covariance matrix equation (8).

##  3 Main results

In subsection 3.1, we first present necessary and sufficient condition for the
Riccati equation (8) to have a unique stabilizing steady solution, which
corresponds to a conditional covariance matrix of some Gaussian state. In
subsection 3.2, we further prove that the corresponding Gassian state is pure.
Then in subsection 3.3, we provide algebraic characterization of how to
prepare a pure Gaussian state with the desired covariance matrix.

###  3.1 Stabilizing steady solution of the Riccati equation

Above all, we present the following lemma (Theorem 3.4 in [Zhou, Doyle, &
Glover, 1995]) concerning the properties of detectability, which is to be
utilized in the main results.

######  Lemma 1.

The following are equivalent:
(i) $[C,~{}A]$ is detectable;
(ii) The matrix $\begin{bmatrix}A-\lambda I\\\ C\end{bmatrix}$ has full column
rank for all $\textrm{Re}(\lambda)\geq 0$;
(iii) For all $\lambda$ and $x\neq 0$ such that $Ax=\lambda x$ and
$\textrm{Re}(\lambda)\geq 0$, $Cx\neq 0$;
(iv) There exists a matrix $F$ such that $A+FC$ is stable, i.e., all the
eigenvalues of $A+FC$ have negative real parts;
(v) $(A^{\top},~{}C^{\top})$ is stabilizable.

The definition of stabilizing steady solution is defined as the following.

######  Definition 1.

A matrix $V$ is called a stabilizing steady solution of (8), if

(i) it satisfies the algebraic Riccati equation

$AV+VA^{\top}+N-(VC^{\top}+M)(VC^{\top}+M)^{\top}=0,$

or, equivalently,

$(A-MC)V+V(A-MC)^{\top}-VC^{\top}CV+\frac{1}{4}JC^{\top}CJ^{\top}=0,$ (11)

where

$A-MC=J\Big{(}G+\frac{1}{2i}(\Lambda^{\top}\Lambda-\Lambda^{\dagger}\Lambda^{*})\Big{)}\in\mathbb{R}^{2m\times
2m}.$

(ii) $A-MC-VC^{\top}C$ is stable, i.e., all the eigenvalues of $A-MC-
VC^{\top}C$ sit in the left half plane.

Now we provide necessary and sufficient conditions for the Riccati equation
(8) to have a unique stabilizing steady solution, which corresponds to a
quantum Gaussian state.

######  Theorem 1.

The Riccati equation (8) has a unique stabilizing steady solution $V$ which is
positive semidefinite if and only if $[C,~{}A]$ is detectable.

Proof: Sufficiency: Firstly, according to Theorem 13.7 in [Zhou, Doyle, &
Glover, 1995], if

  1. 1.

$[C,~{}A-MC]$ is detectable, and

  2. 2.

$[CJ^{\top},~{}(A-MC)^{\top}]$ has no unobservable modes on the imaginary
axis,

then the Riccati equation (11) has a unique stabilizing solution $V$, i.e.,
$A-MC-VC^{\top}C$ is stable. Moreover, $V$ is positive semidefinite.

Secondly, let us prove that $[CJ^{\top},~{}(A-MC)^{\top}]$ has no unobservable
modes on the imaginary axis if $[C,~{}A-MC]$ is detectable. Suppose, on the
contrary, that $[CJ^{\top},~{}(A-MC)^{\top}]$ has an unobservable mode on the
imaginary axis. From Lemma 1, there exists $w\in\mathbb{R}$ and a
corresponding $x\neq 0$, such that

$\displaystyle(A-MC)^{\top}x$ $\displaystyle=iwx,$ $\displaystyle CJ^{\top}x$
$\displaystyle=0.$

Thus, we have

$\displaystyle(A-MC)J^{\top}x$
$\displaystyle=J(G+\frac{1}{2i}(\Lambda^{\top}\Lambda-\Lambda^{\dagger}\Lambda^{*}))J^{\top}x$
$\displaystyle=J(A-MC)^{\top}x$ $\displaystyle=Jiwx$
$\displaystyle=-iwJ^{\top}x$

Combining this conclusion with $CJ^{\top}x=0$ contradicts with the condition
that $[C,~{}A-MC]$ is detectable. Thus, $[CJ^{\top},~{}(A-MC)^{\top}]$ has no
unobservable modes on the imaginary axis if $[C,A-MC]$ is detectable.
Moreover, it is straightforward to verify that $[C,~{}A-MC]$ being detectable
is equivalent to $[C,~{}A]$ being detectable.

Necessity: Since the unique stabilizing solution of the Riccati equation (11)
can be represented as [Zhou, Doyle, & Glover, 1995]

$\displaystyle V=$
$\displaystyle\int_{0}^{\infty}\textrm{e}^{\big{(}(A-MC)^{\top}-C^{\top}CV\big{)}^{\top}t}~{}\big{(}VC^{\top}CV$
$\displaystyle+\frac{1}{4}JC^{\top}CJ^{\top}\big{)}~{}\textrm{e}^{\big{(}(A-MC)^{\top}-C^{\top}CV\big{)}t}dt,$

the matrix $(A-MC)^{\top}-C^{\top}CV$ should be stable. This implies that
$[(A-MC)^{\top},~{}C^{\top}]$ is stabilizable, and $[C,~{}A]$ is detectable
accordingly. $\blacksquare$

######  Remark 3.1.

A similar result was also given in [Wiseman & Milburn, 2010], while our proof
is different from that there. In existing results, to ensure the system
described by (9) to have a unique steady state, the system matrix $A$ is
usually supposed to be stable. However, under the continuous measurement and
quantum filtering framework, Theorem 1 provides a much weaker condition for
the Riccati equation concerning the conditional covariance matrix to have a
unique stabilizing steady solution.

It can be seen that if $[C,~{}A]$ is detectable, $A-MC-VC^{\top}C$ is stable,
and the solution of (10) will approximately approach to 0.

From Theorem 1, if $[C,~{}A]$ is detectable, then there is a unique
stabilizing solution of the Riccati equation (11) satisfying $V\geq 0$. But as
a covariance matrix depicting quantum Gaussian state, $V$ must satisfy the
Heisenberg uncertainty principle,

$V\geq\pm\frac{i}{2}J.$

In fact, $V\geq\pm\frac{i}{2}J$ essentially implies $V>0$ as proved in
[Pirandola, Serafini, & Lloyd, 2009].

A necessary and sufficient condition for the solution $V$ of the Riccati
equation (11) to satisfy the Heisenberg uncertainty principle is given in
Theorem 2.

######  Theorem 2.

The Riccati equation (8) has a unique stabilizing steady solution $V$
satisfying $V\geq\pm\frac{i}{2}J$ if and only if $[C,~{}A]$ is detectable.

Since the proof of Theorem 2 involves a Riccati equation in the complex field.
We introduce two lemmas for Riccati equations in the complex field.

######  Lemma 3.2.

Suppose $F\in\mathbb{C}^{n\times n},~{}O\in\mathbb{C}^{n\times m},$ and
$Z\in\mathbb{C}^{k\times n}$. If $[O^{\dagger},~{}F^{\dagger}]$ is detectable
and $[Z,F]$ has no unobservable modes on the imaginary axis, then the Riccati
equation for $X$

$F^{\dagger}X+XF-XOO^{\dagger}X+Z^{\dagger}Z=0$

has a unique stabilizing solution which is positive semidefinite.

The proof of Lemma 3.2 is presented in Appendix B by analyzing the Riccati
equation in the complex domain (see Appendix A).

######  Lemma 3.3.

$[CJ^{\top},~{}(A-MC)^{\top}]$ has no unobservable modes on the imaginary axis
if and only if $A-MC-\frac{i}{2}JC^{T}C$ has no eigenvalues on the imaginary
axis.

Proof: Sufficiency: Suppose, on the contrary, that
$[CJ^{\top},~{}(A-MC)^{\top}]$ has an unobservable mode on the imaginary axis.
Then there exists $\lambda\in\mathbb{R}$ and a corresponding vector $x\neq 0$,
such that

$\displaystyle(A-MC)^{\top}x$ $\displaystyle=i\lambda x,$ $\displaystyle
CJ^{\top}x$ $\displaystyle=0.$

Thus, we have

$\displaystyle(A-MC-\frac{i}{2}JC^{\top}C)^{\top}x$ $\displaystyle=$
$\displaystyle(A-MC)^{\top}x-\frac{i}{2}C^{\top}CJ^{\top}x$ $\displaystyle=$
$\displaystyle i\lambda x,$

which contradicts with the condition that $A-MC-\frac{i}{2}JC^{\top}C$ has no
eigenvalues on the imaginary axis.

Necessity: Suppose $A-MC-\frac{i}{2}JC^{\top}C$ has an eigenvalue on the
imaginary axis. Then there exists $\lambda\in\mathbb{R}$ and a corresponding
vector $x\neq 0$, such that

$(A-MC-\frac{i}{2}JC^{\top}C)x=i\lambda x.$ (12)

Multiplying the above equation from left by $x^{\dagger}J^{\top}$ yields

$\displaystyle x^{\dagger}J^{\top}(A-MC-\frac{i}{2}JC^{\top}C)x$
$\displaystyle=$ $\displaystyle
x^{\dagger}\big{(}G+\frac{1}{2i}(\Lambda^{\top}\Lambda-\Lambda^{\dagger}\Lambda^{*})\big{)}x-\frac{i}{2}x^{\dagger}C^{\top}Cx,$
$\displaystyle=$ $\displaystyle i\lambda x^{\dagger}J^{\top}x.$

Since $(x^{\dagger}Jx)^{\dagger}=-x^{\dagger}Jx$ is a pure imaginary number
and $G+\frac{1}{2i}(\Lambda^{\top}\Lambda-\Lambda^{\dagger}\Lambda^{*})$ as
well as $C^{\top}C$ are real symmetric matrices, we have $Cx=0$, and
$(A-MC)x=i\lambda x$ from (12), accordingly. Then, since
$(A-MC)J^{\top}=J(A-MC)^{\top}$, we have

$\displaystyle(A-MC)^{\top}Jx$ $\displaystyle=J^{\top}(A-MC)J^{\top}Jx$
$\displaystyle=J^{\top}i\lambda x$ $\displaystyle=-i\lambda Jx,$
$\displaystyle CJ^{\top}Jx$ $\displaystyle=0,$

which contradicts with the condition that $[CJ^{\top},~{}(A-MC)^{\top}]$ has
no unobservable modes on the imaginary axis.

Thus, $[CJ^{T},~{}(A-MC)^{T}]$ has no unobservable modes on the imaginary axis
if and only if $A-MC-\frac{i}{2}JC^{T}C$ has no eigenvalues on the imaginary
axis. $\blacksquare$

Now we give the proof of Theorem 2.

Proof: Sufficiency: From Theorem 1, if $[C,~{}A]$ is detectable, then the
Riccati equation $(\ref{conditioncovariance})$ has a unique stabilizing steady
solution $V$ which is positive semidefinite. Since $V$ is real and symmetric,
and $J$ is antisymmetric, we just need to prove $V\geq\frac{i}{2}J$.

Suppose $Y=V-\frac{i}{2}J$. From $(\ref{riccati})$ we have the Riccati
equation for $Y$ as

$\displaystyle(A-MC-$
$\displaystyle\frac{i}{2}JC^{\top}C)Y+Y(A-MC-\frac{i}{2}JC^{\top}C)^{\dagger}$
(13) $\displaystyle-YC^{\top}CY=0.$

From the proof of Theorem 1, if $[C,~{}A-MC]$ is detectable, then
$[CJ^{\top},~{}(A-MC)^{\top}]$ has no unobservable modes on the imaginary
axis. This conclusion combined with Lemma 3.3 implies that
$A-MC-\frac{i}{2}JC^{\top}C$ has no eigenvalues on the imaginary axis. Thus,
$[0,(A-MC-\frac{i}{2}JC^{\top}C)^{\dagger}]$ has no unobservable modes on the
imaginary axis. Moreover, $[C,~{}A-MC-\frac{i}{2}JC^{\top}C]$ is detectable if
$[C,~{}A]$ is detectable. According to Lemma 3.2, the Riccati equation (13)
for $Y$ has a unique stabilizing solution which is positive semidefinite,
i.e., $V\geq\frac{i}{2}J$, if $[C,~{}A]$ is detectable.

Necessity: It can be obtained from the necessity part of Theorem 1.
$\blacksquare$

###  3.2 Stabilizing steady pure Gaussian state

From Theorem 2 we can see that if $[C,~{}A]$ is detectable, then the
stabilizing steady state solution of (8) corresponds to a conditional
covariance matrix of some Gaussian state. Since pure Gaussian states of
continuous variable systems are the key ingredients of secure optical
communications and Heisenberg limited interferometry, the characterization of
pure Gaussian states has attracted much attention.

Note that the purity of a $m$ mode Gaussian state is simply

$\textrm{Tr}[\hat{\rho}^{2}]=\frac{1}{2^{m}\sqrt{\textrm{det}[V]}}.$

In [Koga & Yamamoto, 2012], by utilizing a dissipation-induced approach on the
basis of the model (9), an explicit algebraic characterization of the purity
of Gaussian state was given. For the model (9), to have a unique stabilizing
steady pure state with covariance matrix $V_{s}$, besides $A$ being stable,
$V_{s}$ must satisfy the following matrix equations:

$(V_{s}+\frac{i}{2}J)\Lambda^{\top}=0,~{}JGV_{s}+V_{s}GJ^{\top}=0.$ (14)

However, these matrix equations are not easy to verify without knowing the
solution $V_{s}$.

A natural question is whether the unique stabilizing steady solution $V$ of
the Riccati equation (8) satisfying $V\geq\pm\frac{i}{2}J$ corresponds to some
pure Gaussian state. The following Theorem 3 gives an affirmative answer.

######  Theorem 3.

If $[C,~{}A]$ is detectable, then the Riccati equation (8) has a unique
stabilizing steady solution $V$, which can be considered as a conditional
covariance matrix of some pure Gaussian state.

Proof: To prove the steady state being pure, we resort to the correspondence
of Gaussian states between the Schr$\ddot{\textrm{o}}$dinger picture and the
Heisenberg picture.

Consider the following stochastic master equation of $\hat{\rho}_{t}$ starting
from an initial pure Gaussian state $\hat{\rho}_{0}$,

$d\hat{\rho}_{t}=\mathcal{L}^{\dagger}[\hat{\rho}_{t}]dt+\sum_{i=1}^{m}\Big{(}\hat{L}_{i}\hat{\rho}_{t}+\hat{\rho}_{t}\hat{L}^{\dagger}_{i}-\text{Tr}[\hat{\rho}_{t}(\hat{L}_{i}+\hat{L}_{i}^{\dagger})]\hat{\rho}_{t}\Big{)}dw_{i,t},$
(15)

where the innovations

$dw_{i,t}=d{y}_{i,t}-\langle\hat{\rho}_{t},\hat{L}_{i}+\hat{L}_{i}^{\dagger}\rangle
dt$

are Gaussian, and

$\mathcal{{L}}^{\dagger}[\hat{\rho}_{t}]=-i[\hat{H},\hat{\rho}_{t}]+\sum_{i=1}^{m}(\hat{L}_{i}\hat{\rho}_{t}\hat{L}_{i}^{\dagger}-\frac{1}{2}\hat{L}_{i}^{\dagger}\hat{L}_{i}\hat{\rho}_{t}-\frac{1}{2}\hat{\rho}_{t}\hat{L}_{i}^{\dagger}\hat{L}_{i}).$

It is straightforward to verify that

$d~{}\text{Tr}(\hat{\rho}^{2}_{t})=0.$

This combined with the initial state $\hat{\rho}_{0}$ being pure implies that
the state $\hat{\rho}_{t}$ is pure for all time $t$. Physically this means
that if the initial state is pure and there is no information loss during the
measurement process, then the state will always be pure.

Moreover, the evolution (15) preserves the Gaussian property. Thus (15) can be
fully depicted by the equations of the first- and second-moment in the
Heisenberg picture whose dynamics are the same as (7) and (8), while the
initial values should correspond to the pure state $\hat{\rho}_{0}$,
respectively. Recall that the purity of the Gaussian state only depends on the
second moment. Hence, from the correspondence between the
Schr$\ddot{\textrm{o}}$dinger picture and the Heisenberg picture for Gaussian
states, if (8) has a unique stabilizing steady solution, it must correspond to
a pure Gaussian state. Thus, from Theorem 2, if $[C,~{}A]$ is detectable, then
the Riccati equation (8) has a unique stabilizing steady solution $V$, which
can be considered as a conditional covariance matrix of some pure Gaussian
state. $\blacksquare$

To prepare a Gaussian state, the approach via continuous measurement has the
following advantages:

(i) From Theorem 3, to prepare a pure Gaussian state, the conditions that need
to be satisfied via the continuous measurement are much weaker as compared
with conditions given by the dissipation-induced approach.

(ii) For the unconditional approach, if $V_{s}$ is a unique steady pure state
of (9), then it must also be a unique steady pure state of the conditional
covariance matrix equation (8). This is because that $V_{s}$ being a unique
steady pure state of (9) is equivalent to that (14) holds. This further yields

$V_{s}C^{\top}+M=0.$

Thus, (8) and (9) have the same steady pure state.

(iii) Under the same system Hamiltonian and coupling operators, the
conditional covariance matrix $V_{\text{c}}$ obtained via continuous
measurement approach evolving under (8) takes the form [Giovanni, Brunelli, &
Genoni, 2021]

$V_{\text{unc}}=V_{c}+\Sigma,$

where $V_{\text{unc}}$ obeys (9), and

$\Sigma=\textrm{E}[\bar{X}_{c}\bar{X}_{c}^{\top}+\bar{X}_{c}^{\top}\bar{X}_{c}]-\Big{(}\textrm{E}[\bar{X}_{c}]\textrm{E}[\bar{X}_{c}^{\top}]+\textrm{E}[\bar{X}_{c}^{\top}]\textrm{E}[\bar{X}_{c}]\Big{)},$

with $\bar{X}_{c}$ obeying (7). Note that the Gaussian state which corresponds
to $V_{\text{unc}}$ may be mixed. Since most of the useful properties of a
Gaussian state are determined by the covariance matrix, generally the smaller
the covariance matrix (in the matrix partial order) is, the more useful the
Gaussian state is. Thus, continuous measurement approach can prepare more
useful Gaussian states under the same system parameters.

To better illustrate the advantage of the conditional method in preparing pure
Gaussian states, we give a single mode example.

Example 1. Consider the Hamiltonian as

$\hat{H}=\kappa\frac{\hat{a}^{2}+\hat{a}^{\dagger
2}+\hat{a}^{\dagger}\hat{a}+\hat{a}\hat{a}^{\dagger}}{2}=\kappa\hat{q}^{2},$

with $\kappa$ the damping rate of the cavity. The corresponding $G$ matrix is
$G=\begin{pmatrix}2\kappa&0\\\ 0&0\end{pmatrix}$. It can be verified that
there does not exist a positive definite matrix $V$ such that it satisfies the
second equation of (14). In other words, if the system Hamiltonian is
$\hat{H}=\kappa\hat{q}^{2}$, no matter how to choose the coupling operator
$\hat{L}$, no pure Gaussian states can be prepared by utilizing the
dissipation-induced (unconditional) approach.

Now we employ the conditional approach, and verify that pure Gaussian states
can be prepared via continuous measurement. We take
$V=\frac{1}{2}\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ as an example. It is
clear that $V$ corresponds to a pure Gaussian state. To prepare it, select the
measurement operator as

$\hat{L}=\sqrt{\kappa}\Big{(}(\sqrt{2}-\frac{\sqrt{2}}{2}i)\hat{a}-\frac{\sqrt{2}}{2}i\hat{a}^{\dagger}\Big{)}=\sqrt{\kappa}\Big{(}(1-i)\hat{q}+i\hat{p}\Big{)}.$

The corresponding parameters are

$\Lambda=\sqrt{\kappa}(1-i,~{}i),~{}C=\sqrt{\kappa}(2,0),~{}A-MC=\begin{pmatrix}\kappa&0\\\
0&-\kappa\end{pmatrix}.$

It can be verified that $[C,~{}A-MC]$ is detectable, and
$V=\frac{1}{2}\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ is a solution of (11).

###  3.3 Prepare a Gaussian state with desired covariance matrix

In this subsection, we give an algebraic characterization of how to prepare a
Gaussian state with the desired covariance matrix. Let the complex matrix
$\Lambda$ be decomposed into

$\Lambda=\textrm{R}+i\textrm{Im}$

with its real part $\textrm{R}\in\mathbb{R}^{m\times 2m}$ and imaginary part
$\textrm{Im}\in\mathbb{R}^{m\times 2m}$, respectively.

######  Theorem 4.

Let $V_{s}$ be a covariance matrix corresponding to a pure Gaussian state.
Then, this is a unique stabilizing steady solution of (8) if the following two
conditions hold:
(i) $\text{Rank}\begin{pmatrix}\textrm{R}V_{s}J\\\ \textrm{
R}\end{pmatrix}=2m$;
(ii) $G=-\textrm{R}^{\top}\textrm{Im
}-\textrm{Im}^{\top}\textrm{R}+2J^{\top}V_{s}\textrm{R}^{\top}\textrm{R}+2\textrm{R}^{\top}\textrm{R}V_{s}J$.

Proof: We first prove that $V_{s}$ is a solution of (11) if condition (ii) is
satisfied. By substituting $\Lambda=\textrm{R}+i\textrm{Im}$ into (11), this
is equivalent to prove that $V_{s}$ satisfies

$\displaystyle
J(G+\textrm{R}^{\top}\textrm{Im}+\textrm{Im}^{\top}\textrm{R})V_{s}+V_{s}(G+\textrm{R}^{\top}\textrm{Im}+\textrm{Im}^{\top}\textrm{R})J^{\top}$
$\displaystyle-4V_{s}\textrm{R}^{\top}\textrm{R}V_{s}+J\textrm{R}^{\top}\textrm{R}J^{\top}=0.$

Multiplying the above equation by $J^{\top}$ from left and by $J$ from right,
we have the following equivalent equation

$\displaystyle(G+\textrm{R}^{\top}\textrm{
Im}+\textrm{Im}^{\top}\textrm{R})V_{s}J+J^{\top}V_{s}(G+\textrm{R}^{\top}\textrm{Im}+\textrm{Im}^{\top}\textrm{R})$
(16)
$\displaystyle-4J^{\top}V_{s}\textrm{R}^{\top}\textrm{R}V_{s}J+\textrm{R}^{\top}\textrm{R}=0.$

Since $V_{s}$ corresponds to a pure Gaussian state, we have [Wolf, Giedke,
Kr$\ddot{\textrm{u}}$ger, Werner, & Cirac, 2004]

$JV_{s}JV_{s}=-\frac{I}{4}.$

Then $\textrm{R}^{\top}\textrm{R}$ can be expressed as

$\textrm{R}^{\top}\textrm{R}=-2\textrm{R}^{\top}\textrm{R}V_{s}JV_{s}J-2J^{\top}V_{s}J^{\top}V_{s}\textrm{R}^{\top}\textrm{R}.$

Substituting this expression of $\textrm{R}^{\top}\textrm{R}$ into the left
hand side of (16), we have

$\displaystyle~{}~{}~{}(G+\textrm{R}^{\top}\textrm{
Im}+\textrm{Im}^{\top}\textrm{R}-2J^{\top}V_{s}\textrm{R}^{\top}\textrm{R}-2\textrm{R}^{\top}\textrm{R}V_{s}J)V_{s}J$
$\displaystyle+J^{\top}V_{s}(G+\textrm{R}^{\top}\textrm{Im
}+\textrm{Im}^{\top}\textrm{R}-2J^{\top}V_{s}\textrm{R}^{\top}\textrm{R}-2\textrm{R}^{\top}\textrm{R}V_{s}J)$
$\displaystyle=0,$

if condition (ii) is met. Thus $V_{s}$ is a solution of (11).

Now we prove that $V_{s}$ is a unique stabilizing steady solution of (8).
According to Theorem 2, we just need to prove that conditions (i) and (ii)
imply that $[C,~{}A-MC]$ is detectable. On the contrary, there exists
$\lambda$ with $\textrm{Re}(\lambda)\geq 0$, and a corresponding eigenvector
$x$, such that

$\displaystyle(A-MC)x$ $\displaystyle=\lambda x,$
$\displaystyle\textrm{R}~{}x$ $\displaystyle=0.$

By utilizing condition (ii) and after some calculations, we have

$\displaystyle 2\textrm{R}^{\top}\textrm{R}V_{s}Jx=-\lambda Jx.$

Multiplying this equation by $x^{\dagger}J^{\top}V_{s}$ from left yields

$\displaystyle 2x^{\dagger}J^{\top}V_{s}\textrm{R}^{\top}\textrm{R
}V_{s}Jx=-\lambda x^{\dagger}J^{\top}V_{s}Jx.$ (17)

Note that condition (i) $\text{Rank}\begin{pmatrix}\textrm{R}V_{s}J\\\
\textrm{R}\end{pmatrix}=2m$ and $\textrm{R}x=0$ imply that
$\textrm{R}V_{s}Jx\neq 0$. Thus,
$x^{\dagger}J^{\top}V_{s}\textrm{R}^{\top}\textrm{R}V_{s}Jx>0$. In addition,
since $V_{s}$ corresponds to a Gaussian state, this yields $V_{s}>0$, and
$x^{\dagger}J^{\top}V_{s}Jx>0$ accordingly. Then from (17), $\lambda<0$. But
this contradicts with $\textrm{Re}(\lambda)\geq 0$. Therefore, $[C,~{}A-MC]$
is detectable. Hence, $V_{s}$ is a unique stabilizing steady solution of (8)
under conditions (i) and (ii). $\blacksquare$

######  Remark 3.4.

Under condition (ii) in Theorem 4, condition (i) holds if and only if
$[C,~{}A-MC]$ is detectable. The necessity part has been proven in the proof
of Theorem 4. To prove the sufficiency, we first note that under condition
(ii), $[C,~{}A-MC]$ being detectable is equivalent to
$[\textrm{R},~{}J\textrm{R}^{\top}\textrm{R}V_{s}J]$ being detectable.
Further, if condition (i) does not hold, then there exists $x\neq 0$, such
that $\textrm{R}V_{s}Jx=0$ and $\textrm{R}x=0$. This clearly contradicts with
$[\textrm{R},~{}J\textrm{R}^{\top}\textrm{R}V_{s}J]$ being detectable.

######  Remark 3.5.

To make condition (i) of Theorem 4 hold, there are a lot of degrees of freedom
to choose R. Here we give a simple choice such that
$\textrm{R}=\begin{pmatrix}I&0\end{pmatrix}$, where $I$ is the $m\times m$
identity matrix and $0$ is the $m\times m$ null matrix. Now let us check
conditon (i). Given a target covariance matrix $V_{s}$ described by

$V_{s}=\begin{pmatrix}V_{11}&V_{12}\\\ V_{12}^{\top}&V_{22}\end{pmatrix},$

since it corresponds to a pure Gaussian state,

$V_{11}=V_{11}^{\top}>0~{}and~{}V_{22}=V_{22}^{\top}>0.$

Then we have $\textrm{R}V_{s}J=\begin{pmatrix}-V_{12}&V_{11}\end{pmatrix}.$ It
is clear that

$\text{Rank}\begin{pmatrix}\textrm{R}V_{s}J\\\ \textrm{
R}\end{pmatrix}=\textrm{Rank}\begin{pmatrix}-V_{12}&V_{11}\\\
I&0\end{pmatrix}=2m.$

######  Remark 3.6.

Note that in Theorem 4, to prepare a desired pure Gaussian state with
covariance matrix $V_{s}$, there is no restriction on the imaginary part of
the coupling matrix $\Lambda$, which is helpful to relax the restrictions of
experimental realizations.

##  4 Imperfect detection efficiency

In practical applications, we may not be able to achieve perfect detection
efficiency and the imperfect efficiency $\eta<1$ may have significant impact
on the performance in preparing Gaussian states. In this section we consider
the case where the detection efficiency $\eta<1$. Here we assume that all the
detection efficiencies are the same for simplicity. In this case the dynamics
of the conditional state $\hat{\rho}^{c}_{t}$ is described by

$\displaystyle d\hat{\rho}_{t}^{c}=$
$\displaystyle\mathcal{L}^{\dagger}[\hat{\rho}^{c}_{t}]dt$ (18)
$\displaystyle+$
$\displaystyle\sqrt{\eta}\sum_{i=1}^{m}\Big{(}\hat{L}_{i}\hat{\rho}^{c}_{t}+\hat{\rho}^{c}_{t}\hat{L}^{\dagger}_{i}-\text{Tr}[\hat{\rho}^{c}_{t}(\hat{L}_{i}+\hat{L}_{i}^{\dagger})]\hat{\rho}^{c}_{t}\Big{)}dw_{i,t},$

where the innovations

$dw_{i,t}=d{y}_{i,t}-\sqrt{\eta}\langle\hat{\rho}^{c}_{t},\hat{L}_{i}+\hat{L}_{i}^{\dagger}\rangle
dt$

are Gaussian, and

$\mathcal{{L}}^{\dagger}[\hat{\rho}_{t}^{c}]=-i[\hat{H},\hat{\rho}_{t}^{c}]+\sum_{i=1}^{m}(\hat{L}_{i}\hat{\rho}_{t}^{c}\hat{L}_{i}^{\dagger}-\frac{1}{2}\hat{L}_{i}^{\dagger}\hat{L}_{i}\hat{\rho}_{t}^{c}-\frac{1}{2}\hat{\rho}_{t}^{c}\hat{L}_{i}^{\dagger}\hat{L}_{i}).$

If the initial state is pure Gaussian, (18) still preserves the Gaussian
nature. However, due to the imperfect detection, there is information loss
during the measurement. This leads to that the Gaussian state may become mixed
as the state evolving.

For the $m$ mode system considered in subsection 2.3, the first- and second-
moment equations corresponding to (7) and (8) become

$\displaystyle d\bar{X}_{t}$
$\displaystyle=(A\bar{X}_{t}+Bu_{t})dt+\sqrt{\eta}(V_{t}C^{\top}+M)d\tilde{y}_{t},$
(19) $\displaystyle\frac{d}{dt}V_{t}$
$\displaystyle=AV_{t}+V_{t}A^{\top}+N-\eta(V_{t}C^{\top}+M)(V_{t}C^{\top}+M)^{\top}.$

If $V$ is a stabilizing steady solution of (19), then $V$ needs to satisfy the
following Riccati equation

$AV+VA^{\top}+N-\eta(VC^{\top}+M)(VC^{\top}+M)^{\top}=0,$

which is equivalent to

$\displaystyle(A-\eta MC)V$ $\displaystyle+V(A-\eta MC)^{\top}$ (20)
$\displaystyle-\eta VC^{\top}CV+N-\eta MM^{\top}=0.$

Similar to Theorem 2 where the detection efficiency $\eta=1$, we have the
following conclusion.

######  Theorem 5.

The Riccati equation (19) has a unique stabilizing steady solution satisfying
$V\geq\pm\frac{i}{2}J$ if and only if $[C,~{}A-MC]$ is detectable.

To prove Theorem 5, we first rearrange (19) into the following form

$(A-\eta MC)V+V(A-\eta MC)^{\top}-\eta VC^{\top}CV+Q_{V}^{\top}Q_{V}=0,$

where $Q_{V}=\begin{bmatrix}\textrm{R}\\\
\sqrt{1-\eta}~{}\textrm{Im}\end{bmatrix}J^{\top}$. We have the following
lemma.

######  Lemma 4.7.

$[Q_{V},~{}(A-\eta MC)^{\top}]$ has no unobservable modes on the imaginary
axis if $[C,~{}A-MC]$ is detectable.

Proof: Suppose, on the contrary, that there exists $\lambda\in\mathbb{R}$ and
a corresponding vector $x\neq 0$, such that

$\displaystyle(A-\eta MC)^{\top}x=i\lambda x,$ $\displaystyle
Q_{V}x=\begin{bmatrix}\textrm{R}\\\
\sqrt{1-\eta}\textrm{Im}\end{bmatrix}J^{\top}x=0.$

We have $CJ^{\top}x=2RJ^{\top}x=0$. Moreover, since

$(A-\eta MC)J^{\top}=J(A-\eta MC)^{\top},$

we have

$(A-\eta MC)J^{\top}x=J(A-\eta MC)^{\top}x=i\lambda Jx,$

which yields

$(A-\eta MC)J^{\top}x=-i\lambda J^{\top}x.$

Combining this conclusion with $CJ^{\top}x=0$ contradicts with the condition
that $[C,~{}A-MC]$ is detectable. Thus, $[Q_{V},~{}(A-\eta MC)^{\top}]$ has no
unobservable modes on the imaginary axis if $[C,~{}A-MC]$ is detectable.
$\blacksquare$

Secondly, similar to (13), denoting $Y=V-\frac{i}{2}J$, we have the Riccati
equation about $Y$ as

$\displaystyle(A-\eta MC-\frac{i}{2}\eta JC^{\top}C)Y+Y(A-\eta
MC-\frac{i}{2}\eta JC^{\top}C)^{\top}$ $\displaystyle-\eta
YC^{\top}CY+Q_{Y}^{\top}Q_{Y}=0,$

where $Q_{Y}=\sqrt{1-\eta}\begin{bmatrix}\textrm{R}\\\
\textrm{Im}\end{bmatrix}J^{\top}$. Then we have the following lemma.

######  Lemma 4.8.

$[Q_{V},~{}(A-\eta MC)^{\top}]$ has no unobservable modes on the imaginary
axis if and only if $[Q_{Y},~{}(A-\eta MC-\frac{i}{2}\eta JC^{\top}C)^{\top}]$
has no unobservable modes on the imaginary axis.

Proof: Sufficiency: Suppose, on the contrary, that there exists
$\lambda\in\mathbb{R}$ and a corresponding vector $x\neq 0$, such that

$\displaystyle(A-\eta MC)^{\top}x=i\lambda x,$ $\displaystyle
Q_{V}x=\begin{bmatrix}\textrm{R}\\\
\sqrt{1-\eta}\textrm{Im}\end{bmatrix}J^{\top}x=0.$

Then we have $CJ^{\top}x=2\textrm{R}J^{\top}x=0$. Therefore,

$\displaystyle(A-\eta MC-\frac{i}{2}\eta JC^{\top}C)^{\top}x$
$\displaystyle=(A-\eta MC)^{\top}x=i\lambda x,$ $\displaystyle Q_{Y}x$
$\displaystyle=0,$

which contradicts with the condition that $[Q_{Y},(A-\eta MC-\frac{i}{2}\eta
JC^{\top}C)^{\top}]$ has no unobservable modes on the imaginary axis.

Necessity: Suppose there exists $\lambda\in\mathbb{R}$ and a corresponding
vector $x\neq 0$, such that

$\displaystyle(A-\eta MC-\frac{i}{2}\eta JC^{\top}C)^{\top}x=i\lambda x,$
$\displaystyle Q_{Y}x=\sqrt{1-\eta}\begin{bmatrix}\textrm{R}\\\ \textrm{
Im}\end{bmatrix}J^{\top}x=0.$

Then we have $CJ^{\top}x=0$. Therefore,

$\displaystyle(A-\eta MC)^{\top}x$ $\displaystyle=(A-\eta MC-\frac{i}{2}\eta
JC^{\top}C)^{\top}x=i\lambda x,$ $\displaystyle Q_{V}x$ $\displaystyle=0,$

which contradicts with the condition that $[Q_{V},~{}(A-\eta MC)^{\top}]$ has
no unobservable modes on the imaginary axis.

Thus, $[Q_{V},~{}(A-\eta MC)^{\top}]$ has no unobservable modes on the
imaginary axis if and only if $[Q_{Y},~{}(A-\eta MC-\frac{i}{2}\eta
JC^{\top}C)^{\top}]$ has no unobservable modes on the imaginary axis.
$\blacksquare$

Now we give the proof of Theorem 5.

Proof: Sufficiency: If $[C,~{}A-MC]$ is detectable, then from Lemma 4.7,
$[Q_{V},~{}(A-\eta MC)^{\top}]$ has no unobservable modes on the imaginary
axis. Thus according to Theorem 13.7 in [Zhou, Doyle, & Glover, 1995], the
Riccati equation (20) about $V$ has a unique stabilizing solution which is
positive semidefinite.

To prove $V\geq\frac{i}{2}J$, from Lemma 3.2, we just need to verify that
$[C,~{}A-MC-\frac{i}{2}\eta JC^{\top}C]$ is detectable and $[Q_{Y},(A-\eta
MC-\frac{i}{2}\eta JC^{\top}C)^{\top}]$ has no unobservable modes on the
imaginary axis. In fact, as $[C,~{}A-MC]$ is detectable, so is
$[C,~{}A-MC-\frac{i}{2}\eta JC^{\top}C]$. Moreover, from Lemmas 4.7 and 4.8,
$[Q_{Y},(A-\eta MC-\frac{i}{2}\eta JC^{\top}C)^{\top}]$ has no unobservable
modes on the imaginary axis. Thus, the Riccati equation (19) has a unique
stabilizing steady solution satisfying $V\geq\pm\frac{i}{2}J$ if $[C,~{}A-MC]$
is detectable.

Necessity: According to [Zhou, Doyle, & Glover, 1995], the solution of (20)
has the following form

$\displaystyle Y=\int_{0}^{\infty}$ $\displaystyle\textrm{e}^{\big{(}(A-\eta
MC)^{\top}-\eta C^{\top}CV\big{)}^{\top}t}\big{(}\eta VC^{\top}CV$
$\displaystyle+N-\eta MM^{\top}\big{)}\textrm{e}^{\big{(}(A-\eta
MC)^{\top}-\eta C^{\top}CV\big{)}t}.$

It is clear that $(A-\eta MC)^{\top}-\eta C^{\top}CV$ is stable. This yields
$[C,~{}A-MC]$ being detectable. $\blacksquare$

Although the Gaussian state may be not pure if the detection is imperfect, the
steady solution $V_{\text{c}}$ of the conditional covariance matrix equation
(19) is still smaller than the steady solution $V_{\text{unc}}$ obtained via
the unconditional dissipation-induced approach (9).

##  5 Conclusion

In this paper we consider the problem of how to prepare Gaussian states via
continuous measurement. We present some necessary and sufficient conditions
for the conditional dynamics of the covariance matrix to have a unique
stabilizing steady solution. This conditional method is much superior to the
unconditional approaches in the following two aspects. On one hand, the
conditions given by our method is much weaker than those given by the
unconditional methods, which may be beneficial to experimental realizations.
On the other hand, under the same system parameters, the Gaussian states
prepared via the continuous measurement may be more useful in quantum
information processing.

This work was supported by the National Natural Science Foundation of China
(Nos. 11688101, 61773370, 61833010 and 61621003) and the Australian Research
Councils Discovery Projects funding scheme under Project DP190101566.

## Appendix

##  Appendix A Riccati equation in the complex domain

In this Appendix, we present some results concerning a matrix Ricaati equation
in the complex domain, which are similar to those in Section 13.2 of [Zhou,
Doyle, & Glover, 1995], where the domain considered is real.

Let $F\in\mathbb{C}^{n\times n},~{}P=P^{\dagger}\in\mathbb{C}^{n\times n}$ and
$K=K^{\dagger}\in\mathbb{C}^{n\times n}$. The matrix Riccati equation to be
considered is

$F^{\dagger}X+XF+XPX+K=0.$ (21)

We associate a $2n\times 2n$ matrix in the complex domain with the Riccati
equation (21) as

$H=\begin{bmatrix}F&P\\\ -K&-F^{\dagger}\end{bmatrix}.$

Noting that the $2n\times 2n$ matrix

$J=\begin{bmatrix}0&I\\\ -I&0\end{bmatrix}$

has the property $J^{2}=-I$. Then we have

$J^{-1}HJ=-JHJ=-H^{\dagger}.$

Thus, $H$ and $-H^{\dagger}$ are similar. This further implies that $\lambda$
is an eigenvalue of $H$ if and only if $-{\lambda}^{*}$ is, where
$\lambda^{*}$ denotes the conjugate of $\lambda$.

Assume that $H$ has no eigenvalues on the imaginary axis. Then it must have
$n$ eigenvalues in the left half plane and $n$ eigenvalues in the right half
plane. Denote the corresponding two $n$-dimensional spectral subspaces as
$\mathcal{X}_{-}(H)$ and $\mathcal{X}_{+}(H)$, respectively. To be specific,
the former is the invariant subspace corresponding to eigenvalues in the left
half plane and the latter corresponds to eigenvalues in the right half plane.
By finding a basis of $\mathcal{X}_{-}(H)$, stacking the basis vectors up to
form a matrix, and partitioning the matrix, we have

$\mathcal{X}_{-}(H)=\textit{Span}\begin{bmatrix}X_{1}\\\ X_{2}\end{bmatrix},$

where $X_{1},~{}X_{2}\in\mathbb{C}^{n\times n}$. If $X_{1}$ is nonsingular, we
can define

$X=X_{2}X_{1}^{-1}.$

It can be verified that $X$ is independent of a specific choice of basis of
$\mathcal{X}_{-}(H)$, and is uniquely determined by $H$, which can be
described by the map

$\textit{Ric}:H\rightarrow X.$

We will take the domain of Ric, denoted by $\textrm{dom}(\textit{Ric})$, to
consist of matrices $H$ with the following two properties:

  * •

Stability property: $H$ has no eigenvalues on the imaginary axis;

  * •

Complementarity: the two subspaces $\mathcal{X}_{-}(H)$ and
$\textit{span}\begin{bmatrix}0\\\ I\end{bmatrix}$ are complementary, which is
equivalent to $X_{1}$ being nonsingular.

Now we have Theorem 6 concerning the properties of $X$.

######  Theorem 6.

Suppose $H\in\textrm{dom}(\textit{Ric})$ and $X=\textit{Ric}(H)$. Then

(i) $X=X^{\dagger}$;

(ii) $X$ satisfies Riccati equation (21), i.e.,

$F^{\dagger}X+XF+XPX+K=0,$

(iii) $F+PX$ is stable.

Proof: (i) Since $H\in\textrm{dom}(\textit{Ric})$, there exists a stable
matrix denoted by $H_{-}\in\mathbb{C}^{n\times n}$, which is a matrix
representation of $H|_{\mathcal{X}_{-}(H)}$, such that

$H\begin{bmatrix}X_{1}\\\ X_{2}\end{bmatrix}=\begin{bmatrix}X_{1}\\\
X_{2}\end{bmatrix}H_{-}.$ (22)

Multiplying this equation from left by $\begin{bmatrix}X_{1}\\\
X_{2}\end{bmatrix}^{\dagger}J$ yields

$\begin{bmatrix}X_{1}\\\ X_{2}\end{bmatrix}^{\dagger}JH\begin{bmatrix}X_{1}\\\
X_{2}\end{bmatrix}=\begin{bmatrix}X_{1}\\\
X_{2}\end{bmatrix}^{\dagger}J\begin{bmatrix}X_{1}\\\ X_{2}\end{bmatrix}H_{-}.$

Noting that $JH=(JH)^{\dagger}$, the left hand side of the above equation is
Hermitian, and so is the right hand side. Thus, we have

$\displaystyle(-X_{1}^{\dagger}X_{2}+X_{2}^{\dagger}X_{1})H_{-}$
$\displaystyle=H_{-}^{\dagger}(-X_{1}^{\dagger}X_{2}+X_{2}^{\dagger}X_{1})^{\dagger}$
$\displaystyle=-H_{-}^{\dagger}(-X_{1}^{\dagger}X_{2}+X_{2}^{\dagger}X_{1}).$

This is a Lyapunov equation. Since $H_{-}$ is stable, the unique solution is

$-X_{1}^{\dagger}X_{2}+X_{2}^{\dagger}X_{1}=0.$

Since $X_{1}$ is nonsingular, $X$ can be represented as

$X=(X_{1}^{\dagger})^{-1}X_{1}^{\dagger}X_{2}X_{1}^{-1}.$

This combined with $X_{1}^{\dagger}X_{2}=X_{2}^{\dagger}X_{1}$ yields
$X=X^{\dagger}$.

(ii) Start with the equation

$H\begin{bmatrix}X_{1}\\\ X_{2}\end{bmatrix}=\begin{bmatrix}X_{1}\\\
X_{2}\end{bmatrix}H_{-}.$

Multiplying the above equation from right by $X_{1}^{-1}$ yields

$H\begin{bmatrix}I\\\ X\end{bmatrix}=\begin{bmatrix}I\\\
X\end{bmatrix}X_{1}H_{-}X_{1}^{-1}.$ (23)

Now multiplying from left by $\begin{bmatrix}X&-I\end{bmatrix}$, we have

$\begin{bmatrix}X&-I\end{bmatrix}H\begin{bmatrix}I\\\ X\end{bmatrix}=0.$

This is precisely the Riccati equation (21).

(iii) Multiplying (23) from left by $\begin{bmatrix}I&0\end{bmatrix}$, we have

$F+PX=X_{1}H_{-}X_{1}^{-1}.$

Thus $F+PX$ is stable because $H_{-}$ is stable. $\blacksquare$

A solution $X$ of (21) is called stabilizing if $F+PX$ is stable. A necessary
and sufficient condition for the existence of a unique stabilizing solution of
(21) is stated in the following theorem.

######  Theorem 7.

Suppose $H$ has no imaginary eigenvalues and $P$ is either positive semi-
definite or negative semi-definite. Then $H\in\textrm{dom}(\textit{Ric})$ if
and only if $[P,~{}F^{\dagger}]$ is detectable.

Proof: Sufficiency: To prove that $H\in\textrm{dom}(\textit{Ric})$, we just
need to show that $X_{1}$ is nonsingular, i.e., $\textrm{Ker}(X_{1})=0$.

First, let us demonstrate that $\textrm{Ker}(X_{1})$ is $H_{-}$-invariant,
where $H_{-}$ is given in (22). To prove this, let $x\in\textrm{Ker}(X_{1})$.
Multiplying (22) from left by $\begin{bmatrix}I&0\end{bmatrix}$ yields

$FX_{1}+PX_{2}=X_{1}H_{-}.$ (24)

Multiplying the above equation from left by $x^{\dagger}X_{2}^{\dagger}$, and
by $x$ from right, and recalling the fact that
$X_{2}^{\dagger}X_{1}=(X_{2}^{\dagger}X_{1})^{\dagger}$, we have

$x^{\dagger}X_{2}^{\dagger}PX_{2}x=0.$

Since $P$ is semi-definite, this implies that $PX_{2}x=0$. Then, from (24), we
have $X_{1}H_{-}x=0$, i.e., $H_{-}x\in\textrm{Ker}(X_{1})$, which proves that
$\textrm{Ker}(X_{1})$ is $H_{-}$-invariant.

Now to prove that $X_{1}$ is nonsingular, suppose, on the contrary, that
$\textrm{Ker}(X_{1})\neq 0$. Then $H_{-}|_{\textrm{Ker}(X_{1})}$ has an
eigenvalue $\lambda$ with $\textrm{Re}(\lambda)<0$, and a corresponding
eigenvector $x\neq 0$, i.e.,

$\displaystyle H_{-}x=\lambda x,~{}\textrm{Re}(\lambda)<0,$ $\displaystyle
x\in\textrm{Ker}(X_{1}),~{}x\neq 0.$

Multiplying (22) from left by $\begin{bmatrix}0&I\end{bmatrix}$ yields

$-KX_{1}-F^{\dagger}X_{2}=X_{2}H_{-}.$

Multiplying the above equation by $x$ from right, we have

$(F^{\dagger}+\lambda I)X_{2}x=0.$

Since $PX_{2}x=0$, we have

$\begin{bmatrix}F^{\dagger}+\lambda I\\\ P\end{bmatrix}X_{2}x=0.$

Then the detectability of $[P,~{}F^{\dagger}]$ implies $X_{2}x=0$, which
contradicts with the fact that $\begin{bmatrix}X_{1}\\\ X_{2}\end{bmatrix}$
has full column rank.

Necessity: From Theorem 6, $H\in\textrm{dom}(\textit{Ric})$ implies that $X$
is a stabilizing solution such that $F+PX$ is stable. From (v) of Lemma 1,
this further implies that $[P,~{}F^{\dagger}]$ is detectable. $\blacksquare$

######  Theorem 8.

Suppose $H$ has the form

$H=\begin{bmatrix}F&-OO^{\dagger}\\\ -Z^{\dagger}Z&-F^{\dagger}\end{bmatrix}.$

Then,

(i) $H\in\textrm{dom}(\textit{Ric})$ if and only if
$[O^{\dagger},~{}F^{\dagger}]$ is detectable and $[Z,~{}F]$ has no
unobservable modes on the imaginary axis;

(ii) $X=\textit{Ric}(H)\geq 0$ if $H\in\textrm{dom}(\textit{Ric})$;

(iii) $H\in\textrm{dom}(\textit{Ric})$, then $\textrm{Ker}(X)=0$ if and only
if $[Z,~{}F]$ has no stable unobservable modes.

Proof: To prove (i), noting that from Theorem 7, the detectability of
$[O^{\dagger},~{}F^{\dagger}]$ is necessary. Hence, from Theorem 7, we only
need to show that under the condition $[O^{\dagger},~{}F^{\dagger}]$ being
detectable, $H$ has no imaginary eigenvalues if and only if $[Z,~{}F]$ has no
unobservable modes on the imaginary axis.

Suppose that there exists a real $\omega$ such that $i\omega$ is an eigenvalue
of $H$, and $\begin{bmatrix}x\\\ z\end{bmatrix}\neq 0$ is the corresponding
eigenvector. Then, we have

$\displaystyle Fx-OO^{\dagger}z=i\omega x,$
$\displaystyle-Z^{\dagger}Zx-F^{\dagger}z=i\omega z.$

Re-arranging the above equation yields

$\displaystyle(F-i\omega I)x=OO^{\dagger}z,$ (25) $\displaystyle-(F-i\omega
I)^{\dagger}z=Z^{\dagger}Zx.$

Thus

$\displaystyle z^{\dagger}(F-i\omega
I)x=z^{\dagger}OO^{\dagger}z=||O^{\dagger}z||^{2},$
$\displaystyle-x^{\dagger}(F-i\omega
I)^{\dagger}z=x^{\dagger}Z^{\dagger}Zx=||Zx||^{2}.$

Hence, $\ x^{\dagger}(F-i\omega I)^{\dagger}z$ is real and

$-||Zx||^{2}=x^{\dagger}(F-i\omega I)^{\dagger}z=z^{\dagger}(F-i\omega
I)x=||O^{\dagger}z||^{2}.$

Therefore, $Zx=0$ and $O^{\dagger}z=0$. From (25),

$\displaystyle(F-i\omega I)x=0,$ $\displaystyle-(F-i\omega I)^{\dagger}z=0.$

Combining this with $Zx=0$ and $O^{\dagger}z=0$, we have

$\displaystyle\begin{bmatrix}F-i\omega I\\\ Z\end{bmatrix}x=0,$
$\displaystyle\begin{bmatrix}F^{\dagger}+i\omega I\\\
O^{\dagger}\end{bmatrix}z=0.$

Under the condition that $[O^{\dagger},~{}F^{\dagger}]$ is detectable, we have
$z=0$, and $x\neq 0$. Now it is straightforward to see that $i\omega$ is an
eigenvalue of $H$ if and only if $i\omega$ is an unobservable mode of
$[Z,~{}F]$.

To prove (ii), let $X=\textit{Ric}(H)$. The Riccati equation is

$F^{\dagger}X+XF-XOO^{\dagger}X+Z^{\dagger}Z=0,$

or equivalently,

$(F-OO^{\dagger}X)^{\dagger}X+X(F-OO^{\dagger}X)+XOO^{\dagger}X+Z^{\dagger}Z=0.$
(26)

Noting that $F-OO^{\dagger}X$ is stable by (iii) of Theorem 6, we have

$\displaystyle X=$
$\displaystyle\int_{0}^{\infty}\textrm{e}^{(F-OO^{\dagger}X)^{\dagger}t}(XOO^{\dagger}X+Z^{\dagger}Z)\textrm{e}^{(F-OO^{\dagger}X)t}dt.$

Since $XOO^{\dagger}X+Z^{\dagger}Z$ is positive semi-definite, so is $X$.

Finally, we prove (iii).

Sufficiency: Let us first show that $\textrm{Ker}(X)$ is an $F$-invariant
subspace. Suppose $x\in\textrm{Ker}(X)$, then $Xx=0$. Multiplying (26) from
left by $x^{\dagger}$ and right by $x$ yields

$Zx=0.$

Now multiply (26) from right by $x$ to get

$XFx=0.$

Thus, $\textrm{Ker}(X)$ is $F$-invariant.

Now if $\textrm{Ker}(X)\neq 0$, then there exists a nonzero
$x\in\textrm{Ker}(X)$ and a corresponding $\lambda$ such that

$\displaystyle Fx$ $\displaystyle=(F-OO^{\dagger}X)x=\lambda x,$
$\displaystyle Zx$ $\displaystyle=0.$

From (iii) of Theorem 6, $F-OO^{\dagger}X$ is stable, so
$\textrm{Re}(\lambda)<0$. Thus, $\lambda$ is a stable unobservable mode, which
is a contradiction.

Necessity: Suppose, on the contrary, that $[Z,~{}F]$ has an unobservable
stable mode $\lambda$, i.e., there is $x\neq 0$ such that $Fx=\lambda
x,~{}\textrm{Re}(\lambda)<0,$ and $Zx=0$. By multiplying the Riccati equation
(26) from left by $x^{\dagger}$ and right by $x$, we have

$2\textrm{Re}(\lambda)x^{\dagger}Xx-x^{\dagger}XOO^{\dagger}Xx=0.$

Since $X$ is positive semi-definite, i.e., $x^{\dagger}Xx\geq 0$, and
$x^{\dagger}XOO^{\dagger}Xx\geq 0$, we have $x^{\dagger}Xx=0$, i.e., $X$ is
singular, which is a contradiction. $\blacksquare$

##  Appendix B Proof of Lemma 3.2

It is straightforward to see that Lemma 3.2 can be obtained from (i), (ii) of
Theorem 8 and (iii) of Theorem 6.

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