TITLE: Showing that if a subgroup is normal, it's surjective homomorphic image is normal
QUESTION [6 upvotes]: $\alpha : G \to H$ is a surjective homomorphism. And $U \subset G$ is a subgroup of $G$. Verfiy the claim -
The image of $U$, ie $\alpha(U)$, is a subgroup of $H$, and if $U$ is normal in $G$, then $\alpha(U)$ is normal in $H$.
Answer:
Firstly, do I have to show $\alpha(U)$, is a subgroup of $H$ or is that statement just a statement of fact as part of the question?
Anyway here is what I have done..taking it as a given that $\alpha(U)$, is a subgroup of $H$ -
As $U$ is normal we have
$U = gUg^{-1}$
$\alpha(U) = \alpha(gUg^1) =$ {applying homomorphic mapping into H} = $\alpha(g)\alpha(U)\alpha(g^{-1})$
Is that correct? I have a feeling I should take the $-1$ exponent outside the bracket as an extra final step or is that superfluous?

REPLY [2 votes]: To show that $\alpha(U)$ is a normal subgroup, you need to prove that $\alpha(U) = x \alpha(U) x^{-1}$ for all $x \in H$.  But any $x \in H$ can be written in the form $\alpha(g)$ for $g \in G$ since $\alpha$ is surjective.  Thus, you need only prove that $\alpha(U) = \alpha(g) \alpha(U) \alpha(g)^{-1}$ for all $g \in G$.  Note the $-1$ exponent is "on the outside", so you really do need to take that last step as you suspected.
And yes, as written it is meant that you should prove that $\alpha(U)$ is a subgroup.  The proof requires you to carefully work through the definition/criterion for being a subgroup, but nothing beyond that.
Also, +1 for showing your reasoning and clearly indicating what you are unsure of.

REPLY [1 votes]: You have to prove that $\alpha(U)$ is normal subgroup in $H$, so for all $x\in H$, one has $x\cdot \alpha(U)\cdot x^{-1} \subseteq \alpha(U)$.


text:
\begin{document}

\pagestyle{plain}

\frontmatter

\maketitle

\begin{abstract}

The proposal of increasingly complex and innovative space endeavours poses growing demands for mission designers. In order to meet the established requirements and constraints while maintaining a low fuel cost, the use of low-energy trajectories is particularly interesting. These paths in space allow spacecraft to change orbits and move with little to no fuel, but they are computed using motion models of a higher fidelity than the commonly used two-body problem. For this purpose, perturbation methods that explore the third-body effect are especially attractive, since they can accurately convey the system dynamics of a three-body configuration with a lower computational cost, by employing mapping techniques or exploring analytical approximations.

The focus of this work is to broaden the knowledge of low-energy trajectories by developing new mathematical tools to assist in mission design applications. In particular, novel models of motion based on the third-body effect are conceived and classified by the forces they account for (conservative or non-conservative). The necessary numerical tools to complement the trajectory design are developed: this includes differential correction methods and targeting schemes, which take advantage of the Jacobian matrices derived from the presented models to generate full low-thrust control laws.

One application of this analysis focuses on the trajectory design for missions to near-Earth asteroids. Two different projects are explored: one is based on the preliminary design of separate rendezvous and capture missions to the invariant manifolds of libration point $L_2$. This is achieved by studying two specific, recently discovered bodies and determining dates, fuel cost and final control history for each trajectory. The other covers a larger study on asteroid capture missions, where several asteroids are regarded as potential targets. The candidates are considered using a multi-fidelity design framework. Its purpose is to filter through the trajectory options using models of motion of increasing accuracy, so that a final refined, low-thrust solution is obtained. The trajectory design hinges on harnessing Earth's gravity by exploiting encounters outside its sphere of influence, the named \textit{Earth-resonant encounters}.

An additional application explored in this investigation is the search and computation of periodic orbits for different planetary systems, following the current interest for missions involving distant retrograde and prograde orbits.

\textcolor{External}{In summary, this thesis presents four novel methods to model the third-body perturbation, distinct in their suitability for applications from real-time computations to long-term orbital predictions. These, together with the additionally developed tools for trajectory design, are applied in two asteroid mission cases. The developed Earth-resonant encounters allow for a very large increase in retrievable mass with respect to the state-of-the-art, namely for the cases of six near-Earth asteroids presented.}

\end{abstract}

\sstableofcontents

\sslistoffigures

\sslistoftables

\begin{listofabbreviations}
    \abbrev{2BP}{Two-Body Problem}
    \abbrev{ARRM}{Asteroid Redirect Robotic Mission}
    \abbrev{AU}{Astronomical Unit}
    \abbrev{CR3BP}{Circular-Restricted Three-Body Problem}
	\abbrev{CR4BP}{Circular-Restricted Four-Body Problem}
    \abbrev{DPO}{Distant Prograde Orbit}
	\abbrev{DRO}{Distant Retrograde Orbit}
	\abbrev{E-KM}{Euler-Keplerian Map}
	\abbrev{ESA}{European Space Agency}
	\abbrev{GVE}{Gauss' Variational Equations}
	\abbrev{GVE-3B}{Gauss' Variational Equations for the Third-Body Effect}
	\abbrev{JPL}{Jet Propulsion Laboratory}
	\abbrev{K3BP}{Keplerian Third-Body Potential}
    \abbrev{KM}{Keplerian Map}
    \abbrev{LCLM}{Low-Cost Likelihood Map}
    \abbrev{LPE}{Lagrange Planetary Equations}
    \abbrev{LPO}{Libration Point Orbit}
    \abbrev{LVLH}{Local Vertical, Local Horizontal}
    \abbrev{MPC}{Minor Planet Centre}
    \abbrev{NASA}{National Aeronautics and Space Administration}
    \abbrev{NEA}{Near-Earth Asteroid}
    \abbrev{NLP}{Non-Linear Programming}
    \abbrev{P3DRO}{Period-3 DRO}
    \abbrev{PAP-KM}{Periapsis-Apoapsis-Periapsis Keplerian Map}
    \abbrev{QPDRO}{Quasi-Periodic DRO}
    \abbrev{SOA}{State-of-the-Art}
	\abbrev{STM}{State Transition Matrix}
    \abbrev{VOP}{Variation of Parameters}
    \abbrev{TOF}{Time of Flight}
\end{listofabbreviations}

\chapter*{List of Symbols}
\addcontentsline{toc}{chapter}{List of Symbols}

\abbrev{$a$}{Semi-major axis}
\abbrev{$b$}{Semi-minor axis}
\abbrev{$C$}{Jacobi constant}
\abbrev{$E$}{Eccentric anomaly}
\abbrev{$e$}{Eccentricity}
\abbrev{$FM$}{Figure of merit}
\abbrev{$G$}{Gravitational constant}
\abbrev{$g_0$}{Gravitational acceleration}
\abbrev{$I_{SP}$}{Specific impulse}
\abbrev{$i$}{Inclination}
\abbrev{$J$}{Cost function}
\abbrev{$\text{\textbf{J}}$}{Jacobian matrix}
\abbrev{$L$}{Angular momentum}
\abbrev{$l$}{True longitude}
\abbrev{$\bm{l}$}{Eigenvector}
\abbrev{$M$}{Mean anomaly}
\abbrev{$m$}{Mass}
\abbrev{$n$}{Mean motion}
\abbrev{$O_{I xyz}$}{\textcolor{Both}{Inertial reference frame}}
\abbrev{$O_{R xyz}$}{\textcolor{Both}{Synodic reference frame}}
\abbrev{$O_{\Earth xyz}$}{\textcolor{Both}{Earth-pointing reference frame}}
\abbrev{$O_{r \theta h}$}{\textcolor{Both}{LVLH frame}}
\abbrev{$O_{eph}$}{\textcolor{Both}{Orbital plane frame}}
\abbrev{$o$}{\textcolor{Both}{Generalized momenta}}
\abbrev{$P$}{\textcolor{Both}{Number of periods}}
\abbrev{$p$}{\textcolor{Both}{Semi-latus rectum}}
\abbrev{$\mathpzc{R}$}{Disturbing function}
\abbrev{$\bm{r}$}{Position vector}
\abbrev{$\bm{s}$}{State vector}
\abbrev{$T$}{Orbital period}
\abbrev{$\mathpzc{T}$}{Kinetic energy}
\abbrev{$\mathpzc{U}$}{Potential energy}
\abbrev{$\bm{u}$}{Control vector}
\abbrev{$\bm{v}$}{Velocity vector}
\abbrev{$v_e$}{Exhaust velocity}
\abbrev{$_{\Earth}$}{Earth subscript}
\abbrev{$_{\Sun}$}{Sun subscript}
\begin{center}
	$\ast$
\end{center}
\abbrev{$\alpha$}{Phasing of the massless particle with the disturbing body}
\abbrev{$\Delta v$}{Velocity change}
\abbrev{$\theta$}{Angle formed by the line connecting primaries with the massless particle}
\abbrev{$\bm{\kappa}$}{Vector of orbital elements}
\abbrev{$\lambda$}{Eigenvalue}
\abbrev{$\mu$}{Gravitational parameter}
\abbrev{$\nu$}{True anomaly}
\abbrev{$\upsilon$}{Argument of latitude}
\abbrev{$\bm{\Phi}$}{State transition matrix}
\abbrev{$\Omega$}{Longitude of the ascending node}
\abbrev{$\omega$}{Argument of the periapsis}
\begin{center}
	$\ast$~$\ast$~$\ast$
\end{center}
\abbrev{$\text{\textbf{A}}$}{A matrix}
\abbrev{$\mathpzc{K}$}{A Keplerian element}
\abbrev{$\bm{z}$}{A vector}
\abbrev{$\norm{\bm{z}}$}{Euclidean norm of a vector}
\abbrev{$\bm{Z}$}{A free variables vector}

\chapter{Acknowledgements}

When I started this PhD journey, I had no idea what to expect of the three years ahead. I know now that it would have been impossible to get to the end without the help of all the people I'm naming here. I hope I have thanked all of you properly, and these words are just one more way to do so.

First of all, to my supervisor, Pau: my PhD experience wouldn't have been as good without your guidance. Thank you for always having time (and patience) to discuss things with me, for giving me freedom to pursue my own side projects, for challenging me with new opportunities, for editing so much of my written work and, mainly, for always encouraging me to do better.

To the friends who shared the PhD experience with me: Tiago, Eddie, Rob, Melissa, thank you for the coffee breaks, the times to complain and the unspoken understandings. To Cat, who has been supporting me since we were 6 and always found time for a chat and a musical in London. To my friends that encouraged me from afar, wherever you were: Ru, Spitafires, Francisco, In\^es, Sofia.

\vspace{5mm}

Para a minha fam\'ilia, por me obrigarem a explicar a minha pesquisa em almo\c{c}os de domingo e pela preocupa\c{c}\~ao que sempre tiveram comigo. M\~ae e Pai, obrigada por me aturarem, por virem visitar-me onde quer que esteja e por sempre me terem dado as condi\c{c}\~oes, os conselhos e a confian\c{c}a para arriscar. Isabel, por estarmos ligadas. N\~ao teria conseguido isto sem voc\^es.

Duarte, obrigada por me levares a passear quando est\'a bom tempo, por estares comigo todos os dias e saberes sempre o que preciso.
\mainmatter

\chapter{Introduction}
\label{chap:intro}
\input{intro.tex}

\chapter{Dynamical Models and Methods}
\label{chap:dyn}
\input{dyn.tex}

\chapter{The Keplerian Third-Body Potential}
\label{chap:k3bp}
\input{keplerian.tex}

\chapter{Application: Gauss' Variational Equations for Low-Thrust Design}
\label{chap:ast1}
\input{asteroid_gve.tex}

\chapter{Application: Trajectory Design Framework for Earth-Resonant Trajectories}
\label{chap:ast2}
\input{asteroid_km.tex}

\chapter{Application: Computation of Periodic Orbits}
\label{chap:dro}
\input{periodic.tex}

\chapter{Concluding Remarks}
\input{conclusions.tex}

\bibliographystyle{unsrtnat}
\bibliography{biblio}

\end{document}

config: arxiv


TITLE: Electron at rest
QUESTION [0 upvotes]: David Griffiths suggested a website in his book where I got this paper
http://www.hep.princeton.edu/~mcdonald/examples/electronatrest.pdf
Here the author says classically a particle at rest(in some frame) when its velocity and momentum vanish. In quantum mechanics simply requiring the expectation value of momentum to be zero implies only particle's average velocity is zero.
Something closer to the classical meaning of rest is achieved only when one requires that the expectation value  of the square of velocity be vanishingly small as well. As indicated in the statement of the question, the uncertainly principle then tell us that this condition can only be achieved if the particle is in an arbitrarily large box.
As I am new in quantum mechanics I don't understand why he said expectation value of the square of velocity? And also Griffiths in the 1st chapter calculated expectation of value of position and here took the time derivative and said this is not velocity of the particle ,this is the velocity of expectation value of the position.
Wait for 3rd chapter to know what exactly velocity of particle mean? Now I haven't read the 3rd chapter yet. I have seen the older questions, I think this is not duplicate of them.

REPLY [2 votes]: This has to do with what we really mean when we ask if something is "at rest." Specifically, we usually mean something like this, "Is the particle's momentum in our frame of reference exactly zero?"
As it turns out, we can't actually know a particle's momentum exactly in quantum mechanics because of the Uncertainty Principle. Supposing that we could would require $\sigma_p$ (or, if you like, $\Delta p$) to be zero which would in turn require $\sigma_x$ to be infinity to satisfy the Uncertainty Principle (this is why he says the particle can only achieve "rest" in an "arbitrarily large box").
But $\sigma_p$ is a Standard Deviation defined as $\sqrt{\langle p^2\rangle - \langle p\rangle^2}$. As you can see, the expectation value of $p^2$ needs to be zero as well if the whole expression is to be zero (which it must be if we are to know the particle's momentum exactly).


TITLE: Infinite sum convergence test.
QUESTION [0 upvotes]: I have to tell if the sum $$\sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n + (-1)^{n+1}}$$
Converges or not. Can i use a comprasion test here? To assume that my sum is bigger than $\sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n + 1}$? I dont think that's OK, but that's the only way I can think of right now.

REPLY [2 votes]: How about this? Write out the series:
$$-1(\frac{1}{1 + 1}) + 1(\frac{1}{2-1})-1(\frac{1}{3 + 1})+1(\frac{1}{4-1})-1(\frac{1}{5+1})....$$
$$=\frac{-1}{2} + 1 -\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\frac{1}{5}-\frac{1}{8}....$$
Then reorder terms...
$$1 - \frac{1}{2} + \frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}...$$
And it was the alternating harmonic series all along! Which has been proven to equal
$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}=ln(2)$$


TITLE: Term for theory that can be defined within another theory
QUESTION [8 upvotes]: Suppose we have two formal axiomatic theories of mathematics and the objects and operators in one theory can be defined in terms of the objects and operators in another theory such that the axioms of the former are theorems in the latter.  Is there a term for this type of relationship? Where would I go to learn more about questions about this type of relationship? My intuition is that goal of something like ZFC is to be a theory that's as simple and general as possible such that any other mathematics we were interested in could be reframed in terms of ZFC, something one might call a universal theory although I don't know what the actual terminology is. Maybe there are widely different types of universal theories.  Perhaps ZFC or even specific universes within ZFC can be constructed from ZFC itself.

REPLY [12 votes]: You are searching for the concept known as interpretation.

A model $N$ is interpretable in another model $M$ (perhaps a totally different language signature) if inside $M$ we can define a domain (perhaps a quotient by a definable equivalence relation on $k$-tuples) and define the fundamental relations and structure of $N$ on that domain, in such a way that the defined structure is isomorphic to $N$.

For example, the complex number field $\langle \mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field, by representing $a+bi$ with pairs of real numbers $(a,b)$. However, it turns out that the real number field is not interpretable in the complex field.

A theory $T$ is interpreted in another theory $S$, if we have a uniform manner of defining a model of $T$ inside any model of $S$.

Two models (or theories) are mutually interpretable, if each is interpreted in the other.

A strictly stronger notion is bi-interpretation. Two models are bi-interpretable, if they are mutually interpretable, in such a way that the models can define the isomorphism of themselves with their copy inside (the copy of) the other structure. By iterating the interpretations, one obtains a fractal like situation:

My recent paper

Alfredo Freire, Joel David Hamkins, "Bi-interpretation in weak set theories," https://arxiv.org/abs/2001.05262.

contains an elementary introduction of interpretation of models of set theory, including the proof that no two extensions of ZF set theory are bi-interpretable.
Here is a talk I gave for the Oxford set theory seminar on the topic: "Bi-interpretation of weak set theories"
https://youtu.be/8KxlpMOXoP4


text:
\section{Discussion}
\label{sec:discussion}

This work develops a sharper finite-sample analysis of the classical asynchronous Q-learning algorithm, highlighting and refining its dependency on intrinsic features of the Markovian trajectory induced by the behavior policy. Our sample complexity bound strengthens the state-of-the-art result by an order of at least $|\cS||\cA|$. A variance-reduced variant of asynchronous Q-learning is also analyzed, exhibiting improved scaling with the discount complexity $\frac{1}{1-\gamma}$.

Our findings and the analysis framework developed herein suggest a couple of directions for future investigation. For instance, our  improved sample complexity of asynchronous Q-learning has a dependence of $\frac{1}{(1-\gamma)^5}$ on the discount complexity, which is inferior to its model-based counterpart.  In the synchronous setting, \cite{wainwright2019stochastic} demonstrated an empirical lower bound $\frac{1}{(1-\gamma)^4}$ for Q-learning. It would be important to determine the exact scaling in this regard. In addition, it would be interesting to see whether the techniques developed herein can be exploited towards understanding model-free algorithms with more sophisticated exploration schemes \cite{dann2015sample}. Finally, asynchronous Q-learning on a single Markovian trajectory is closely related to coordinate descent with coordinates selected according to a Markov chain; one would naturally ask whether our analysis framework can yield improved convergence guarantees for general Markov-chain-based optimization algorithms \cite{sun2020markov,doan2020convergence}.


TITLE: Expectation of normal distribution as expectation of chi-squared distribution
QUESTION [0 upvotes]: This is Lemma 11.3 from the book the theory of linear models and multivariate analysis , which says Let $U\sim N_p(\theta,I)$ and K have a Poission distribution with mean $||\theta||^2/2$, then $E\frac{1}{||U||^2}=E\frac{1}{p-2+2K}$.
The proof is like: Let $V|K\sim \chi^2_{p+2K}(0) $, then by Lemma 1.8 of the book, $V\sim \chi_p^2(||\theta||).$ Also, by the defintion of a $\chi^2$ distribution, $||U||\sim \chi_p^2(||\theta||^2).$ Thereore $$E\frac{1}{||U||^2}=E\frac{1}{V}=E(E\frac{1}{V}|K)=E\frac{1}{p-2+2K}$$
However, I couldn't figure out why $E(\frac{1}{V}|K)=\frac{1}{p-2+2K}$, Can someone give me some help?

REPLY [0 votes]: The question is to understand why, if the distribution of $V_n$ is central chi-squared with $n\gt2$ degrees of freedom, then $E[V_n^{-1}]=(n-2)^{-1}$.
One can go back to the density $f_n$ of $V_n$, which is defined by $f_n(x)=c_n^{-1}x^{(n/2)-1}\mathrm e^{-x/2}$ with $c_n=2^{n/2}\Gamma(n/2)$ and note that
$$
E[V_n^{-1}]=\int_0^\infty x^{-1}f_n(x)\mathrm dx=\int_0^\infty c_n^{-1}x^{(n/2)-2}\mathrm e^{-x/2}\mathrm dx.
$$
One recognizes in the last integrand a multiple of $f_{n-2}(x)$, hence
$$
E[V_n^{-1}]=c_n^{-1}c_{n-2}=\frac{\Gamma((n/2)-1)}{2\Gamma(n/2)}=\frac1{n-2}.
$$
Another approach is to rely on the MGF of $V_n$, which is such that
$$
E[\mathrm e^{-tV_n}]=(1+2t)^{-n/2},
$$
at least for every $t\gt0$, and to use the identity
$$
E[V_n^{-1}]=\int_0^\infty E[\mathrm e^{-tV_n}]\mathrm dt,
$$
hence
$$
E[V_n^{-1}]=\int_0^\infty \frac{\mathrm dt}{(1+2t)^{n/2}}=\left.\frac{-1}{(n-2)(1+2t)^{n/2-1}}\right|_0^\infty=\frac1{n-2}.
$$


TITLE: What do nearby/vanishing cycles have to do with Fourier transforms?
QUESTION [6 upvotes]: Let $E$ be a vector bundle on some smooth algebraic variety and $E^*$ its dual. Suppose $A$ is a sheaf (constructible or a $D$-module) on $E$. Given a linear function $f$ on $E$, we may compute the stalk at $f$ of the Fourier transform of $A$.
Now I’ve heard a slogan along the lines of “the stalk of the Fourier transform of $A$ is the vanishing cycles of $A$ along $f$.” Obviously this doesn’t really make sense; the stalk is a sheaf on the base and the vanishing cycles a sheaf on $f^{-1}(0)$. Nevertheless, my guess is that somehow this idea can be made precise.
So, is there a precise statement relating nearby and vanishing cycles to Fourier transforms of sheaves?

REPLY [4 votes]: You probably want to say $A$ is a constructible sheaf / $D$-module with regular singularities. $D$-modules with irregular singularities (for example, those created by the Fourier transform) will behave very differently, even if $E$ is a vector bundle of rank $1$ on a point.

I'm going to pretend you asked the question for $\ell$-adic sheaves and describe the relevant phenomena there. I think the situation for $D$-modules will be analogous.
Let's start with the Fourier transform on $\mathbb A^1$. For an $\ell$-adic sheaf $A$ with tame ramification (analogue of regular singularities), the correct statement is:
The nearby cycles of the Fourier transform of $A$ at $\infty$ is isomorphic to a sum over points of $\mathbb A^1$ of the vanishing cycles of $A$ at that point. Furthermore, the Fourier transform is locally constant away from $0$, so its stalk at any point is non-canonically isomorphic to the nearby cycles of the Fourier transform at $\infty$ and thus non-canonically isomorphic to this sum.
Now let's consider the Fourier transform on $\mathbb A^n$, keeping the base still a point. The stalk of the Fourier transform at $\lambda f$ for a scalar $\lambda$ is going to be the stalk of the Fourier transform of $R f_* A$ at $\lambda$. Thus it will be non-canonically isomorphic to the sum of the stalks of the vanishing cycles of $R f_* A$ at each point of $\mathbb A^1$.
If $f$ were proper, it would then by isomorphic to the sum over $a$ of the cohomology of $f^{-1}(a)$ with coefficients in the vanishing cycles complex of $A$ along $f$. Maybe we can say that this is morally true in general. So we need to take $f^{-1}(a)$ for all $a$, not just $f^{-1}(0)$, and take the cohomology, to get the desired isomorphism. If the vanishing cycles are supported at finitely many points, this would just be the sum of the vanishing cycles at those points. Maybe that's what people who told you this were thinking of?
However, while this may be morally true, it will not be literally true in every case. The simplest counterexample I can think of is on $\mathbb A^n$, take $A$ to be the constant sheaf on the curve $xy=1$ and $f$ to be $y$. Then $A$ has no vanishing cycles on any point of $\mathbb A^2$ (none on $f^{-1}(0)$ since $A$ vanishes in a neighborhood of $f^{-1}(0)$, and none elsewhere because $A$ in each other fiber looks identical to the generic fiber) but its Fourier transform has a stalk of rank one at $y$.

Over a nontrivial base, I guess you want to take the vanishing cycles sheaf of $A$ along $f$ on $f^{-1}(a)$, push forward to the base, and then sum over $a$. This will be even more morally and less factually correct, but maybe there is some genericity condition under which it is literally true.


TITLE: $K(x,t)=2\pi\sum_{n=1}^\infty n\sin(n\pi x)e^{-n^2\pi^2t}$ satisfies some "pseudo-good-kernel" properties
QUESTION [7 upvotes]: Let $f\in C([0,\infty))$. Define
$$K(x,t)=2\pi\sum_{n=1}^\infty n\sin(n\pi x)e^{-n^2\pi^2t},\qquad x\in[0,1],\  t>0.$$
Show that
$$\lim_{x\to0^+}\int_0^tK(x,t-\tau)f(\tau)\,d\tau=f(t), \qquad t>0.$$

I can show the result by assuming that
$$\lim_{x\to0^+}\int_0^tK(x,t-\tau)\,d\tau=1,\tag{1}$$
and
$$\int_0^1 |K(x,t)|\,dt<\infty \text{ for all } x \text{ close to }0.\tag{2}$$
For any $\varepsilon>0$, we can find $\delta>0$ so that $|f(\tau)-f(t)|<\varepsilon$ for all $\tau\in(t-\delta,\delta)$. In $[\delta,t)$, the series defining $K$ is convergent absolutely and we can change the order of limitation and integration to get
$$\lim_{x\to0^+}\int_0^{t-\delta}K(x,t-\tau)f(\tau)\,d\tau=0.$$
As for the integration in $(t-\delta,t)$, we have
$$\int_{t-\delta}^t |K(x,t-\tau)||f(\tau)-f(t)|\,d\tau\leq \varepsilon\int_{t-\delta}^t |K(x,t-\tau)|\,d\tau\lesssim \varepsilon,$$
by $(2)$. And now the result follows from $(1)$.
I have no idea how to show $(1)$ and $(2)$. It seems that we need to carefully use the definition of $K$. For example, if we simply apply Fubini to $(2)$, we will get a divergent integral.
Any help would be appreciated!

REPLY [2 votes]: The alternative representation $$K(x,t)=\frac1{t\sqrt{\pi t}}\sum_{n=-\infty}^\infty\left(n+\frac x2\right)\exp\left[-\frac1t\left(n+\frac x2\right)^2\right]\tag{$\ast$}\label{altrep}$$ makes all the claims easy to obtain. Say, for $0<x<2$ we get $$\newcommand{\erfc}{\operatorname{erfc}}\int_0^t K(x,\tau)\,d\tau=\erfc\frac{x}{2\sqrt{t}}+\sum_{n=1}^\infty\left[\erfc\frac1{\sqrt{t}}\left(n+\frac x2\right)-\erfc\frac1{\sqrt{t}}\left(n-\frac x2\right)\right].$$ This tends to $1$ as $x\to 0^+$, which shows $(1)$. If we replace $[\ldots-\ldots]$ with $[\ldots+\ldots]$ above, we get an upper bound for $\int_0^t|K(x,\tau)|\,d\tau$, sufficient to see why $(2)$ holds.
To prove \eqref{altrep}, apply the Poisson summation formula $$\sum_{n=-\infty}^\infty g(n)=\sum_{n=-\infty}^\infty\hat{g}(n),\qquad\hat{g}(\eta)=\int_{-\infty}^\infty g(\xi)e^{-2\pi i\eta\xi}\,d\xi$$ to $g(\xi)=\pi\xi\sin x\pi\xi\ e^{-(\pi\xi)^2t}$, and obtain $\hat{g}(\eta)$ using known integrals $$\int_{-\infty}^\infty e^{-a^2 x^2}\cos 2bx\,dx=\frac{\sqrt\pi}{a}e^{-(b/a)^2}\underset{\partial/\partial b}{\implies}\int_{-\infty}^\infty xe^{-a^2 x^2}\sin 2bx\,dx=\frac{b\sqrt\pi}{a^3}e^{-(b/a)^2}.$$


TITLE: Prove $a^{1/k}+b^{1/k}\gt (a+b)^{1/k}$
QUESTION [1 upvotes]: Prove the statement(#):
(#) Suppose $k$ is an integer greater than $1$. Suppose $a,b$ are positive real numbers. Then $a^{1/k}+b^{1/k}\gt (a+b)^{1/k}$.
How should I start the proof? Any inequalities I can apply? Which topic is related to this proof? Plz give me some hints:( Thanks.

REPLY [2 votes]: Firstly whenever proving any sort of algebraic (in)equality I would try to isolate a useful property to simplify the equation. In this case,
$$
(a^{\frac{1}{k}}+b^{\frac{1}{k}})^{k} > [(a+b)^{\frac{1}{k}}]^{k} \\
(a^{\frac{1}{k}}+b^{\frac{1}{k}})^{k} > a+b
$$
as both sides have an exponent $1/k$.
Secondly, I would try to identify any useful properties that I know something about. So in this case, $(p+q)^{k}$ is a very common binomial form.
Then from this point onwards the proof is pretty straightforward. I would recommend always starting with this sort of approach. Sometimes you'll get stuck because novel ideas are required but it'll at least help you get started with most proofs of this form.


text:
\begin{document}
\maketitle

\begin{abstract}
 The purpose of this paper is to study the problem of generalizing the Belavkin-Kalman filter to the case where the classical measurement signal is replaced by a fully quantum non-commutative output signal. We formulate a least mean squares estimation problem that involves a non-commutative system as the filter processing the non-commutative output signal. We solve this estimation problem within the framework of non-commutative probability. Also, we find the necessary and sufficient conditions which make these non-commutative estimators physically realizable. These conditions are restrictive in practice.
\end{abstract}

\bf Keyword: \rm Linear quantum stochastic differential equations (QSDEs), quantum noises, Kalman filtering, physical realizability, linear least mean squares estimators, non-commutative outputs, coherent observers.

\section{Introduction}
Quantum filtering theory as a fundamental theory in quantum optics, which was implicit in the
   work of Davies in the 1960s~\cite{davies1969quantum,davies1976quantum} concerning open quantum systems and generalized measurement theory, and culminating in the general theory developed and initiated  by Belavkin during the 1980s \cite{Belavkinone,belavkin2004towards,belavkin1992quantum}.
   The quantum filter is a stochastic differential equation for the conditional state, from which the best estimates of the system observables may be obtained. In related work by Carmichael, the quantum filter is referred to as the stochastic master equation~\cite{HC93,WM10}.

One application of the quantum filter, or variants of it,  is in measurement feedback control~\cite{VPB83,EB05,J05,WM10,JK10,sayrin2011real,amini2013feedback}. As in classical control theory, optimal measurement feedback control strategies may be expressed as functions of information states, of which the conditional state is a particular case~\cite{J05,MJ12}. However, feedback control of quantum systems need not involve measurements, and indeed the topic of coherent quantum  feedback is evolving~\cite{NJP09,mabuchi2008coherent,wiseman1994all,james2008h,yanagisawa2003transfer,lloyd2000coherent,MJ12}, though a general theory of optimal design of coherent quantum feedback systems is at present unavailable.
In coherent quantum feedback control, the controller is also a quantum system, and information flowing in the feedback loop is also quantum (e.g. via a quantum field).
This type of feedback has recently led to new proposals for quantum memories, quantum error
correction, and ultra-low power classical photonic signal processing~\cite{kerckhoff2010designing,mabuchi2011nonlinear,mabuchi2012qubit,mabuchi2011coherent}.

 The purpose of this paper is to contribute to the knowledge of coherent quantum estimation and control by developing further a non-commutative formulation of the quantum filter given by Belavkin in 1980s~\cite{Belavkinone}. While the main results obtained by Belavkin apply only to the commutative measurement  case, the problem formulation he used was more general. Belavkin's theory of quantum filtering concerns the estimation of the variables of quantum systems conditioned on classical (commutative) measurement records. For linear quantum stochastic systems, Belavkin's filter has the same form as the classical Kalman filter. The Belavkin-Kalman filter is a classical system that processes the incoming measurements to produce the desired estimates. The estimates may be used for monitoring and/or feedback control of the quantum system.

 In our study, we formulate and solve a problem of optimal estimation of a linear quantum system variables  given the non-commutative outputs within the framework of non-commutative probability theory. In particular, we derive a system of non-commutative stochastic differential equations (the non-commutative Belavkin-Kalman filter) that minimizes a least squares error criterion. Such non-commutative filtering equations are well defined mathematically, even if they do not correspond to a physical system. However, if we wish to implement the non-commutative Belavkin-Kalman filter within the class of physically realizable linear quantum stochastic systems (such as linear quantum optical systems), then we find that the conditions for physical realizability impose strong restrictions.  In this paper,
 we find physical realizability conditions for general case and also for some particular cases. These strong physical realizability conditions are a key contribution of this paper.

 We remark  that our contribution here is different from the problem studied in~\cite{vladimirov2013coherent}. Since, in~\cite{vladimirov2013coherent}, the authors propose another physically realizable quantum system considered as a filter, connected to the output of the plant whose dynamics can be determined by minimizing the mean square discrepancy between the plant's state and the output of the filter. Also, they suppose an additional vacuum noise other than the plant's noises in the form of filter's dynamics.
However, in this paper, we focus firstly on finding the form of linear least mean squares estimators for the non-commutative outputs by temporarily excluding the physical realizability constraints. To do this, we proceed as classical Kalman filtering and Belavkin-Kalman filtering by supposing that the mean squares estimator should satisfy a linear dynamics of innovation processes and we don't suppose any additional vacuum noises other than $dw$ which is the input process of the plant. As such, we obtain the form of least mean squares estimators for non commutative outputs. Then, we seek necessary and sufficient conditions which make such linear least mean squares estimators automatically physically realizable. As we can observe in examples, for some particular forms of plants, we are obliged to add additional vacuum noises to the least mean squares estimators to ensure physical realizability. These estimators which track asymptotically the plant's state and are physically realizable are called coherent observers \cite{Miao2012}. Roughly speaking, coherent least mean squares estimators and observers are another physical systems connected to the main system in cascade~\cite{vuglar2014,Miao2012,NJP09}. We remark that coherent linear least squares estimators and observers  could in principle be used   for coherent feedback control, although this matter is outside the scope of the present paper.

   \medskip

This paper is organized as follows. In Section~\ref{sec:three}, we  present general quantum linear stochastic dynamics. In Section~\ref{sec:four}, we obtain non-commutative linear least mean squares estimators for the general linear quantum stochastic dynamics, expressed in Theorem~\ref{thm:main}. In Section~\ref{sec:phys}, we study the physical realizability of such linear least mean squares estimators. The main results of this section are expressed in Theorem~\ref{thm:phys} and Corollaries~\ref{coro:phys1}-\ref{coro:case2}. Moreover, some illustrative examples are provided. Finally, the conclusion is given in Section~\ref{sec:six}.

\section{Quantum linear stochastic dynamics}~\label{sec:three}
Consider linear quantum possibly non-commutative stochastic systems of the form~\cite{james2008h}
\begin{eqnarray}~\label{eq:sdq}
&&dx\left(t\right)=Ax\left(t\right)dt+Bdw\left(t\right),\nonumber\\
&&dy\left(t\right)=Cx\left(t\right)dt+Ddw\left(t\right).
\end{eqnarray}
Here, $A,$ $B,$ $C$ and $D$ are real matrices in $\mathbb R^{n\times n},$ $\mathbb R^{n\times n_w},$ $\mathbb R^{n_y\times n},$ and $\mathbb R^{n_y\times n_w},$ where $n,n_w,n_y$ are positive integers with $n_w\geq n_y$. Also,

\noindent $x(t)=[x_1(t),\ldots,x_n(t)]^T$ is a vector of self-adjoint possibly non-commutative system variables defined on a Hilbert space $\mathcal H$. The initial state $x(0)$ is Gaussian with state $\rho_0$ and satisfies  the commutation relations~\footnote{The notation $[A,B]$ corresponds to $AB-BA.$}
\begin{equation*}
[x_j(0),x_k(0)]=2i\Theta_{jk},\quad j,k=1,2,\ldots,n
\end{equation*}
where $\Theta$ is a real antisymmetric matrix with components $\Theta_{jk}$ and $i=\sqrt{-1}.$ We assume that the matrix $\Theta$ can take one of the two following forms:
\begin{itemize}
\item Canonical if $\Theta=\textrm{diag}_{\frac{n}{2}}(J),$ with $n$ even or
\item Degenerate canonical if  $\Theta=\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(J)\right),$ with

$0< n'\leq n$ and $n-n'$ even.~\footnote{Here $0_{m\times n}$ corresponds to $m\times n$ zero matrix.}
\end{itemize}
Here $J$ corresponds to the real skew-symmetric $2\times 2$ matrix
$$J=\left[\begin{array}{cc}0 & 1\\-1 & 0\end{array}\right],$$
and the "diag" notation corresponds to a block diagonal matrix. Also, $\textrm{diag}_m(J)$ denotes a $m\times m$ block diagonal
matrix with $m$ matrices $J$ on the diagonal.

The noise $dw(t)$ is a vector of self-adjoint quantum noises with It\={o} table
\begin{equation}~\label{eq:fwone}
dw\left(t\right)dw\left(t\right)^{T}=F_wdt,
\end{equation}
where $F_w$ is a non-negative Hermitian matrix~\footnote{The notation $X^T$ corresponds to the transpose of the matrix $X.$} (see e.g.,~\cite{parthasarathy1992introduction,belavkin1991continuous}).
Indeed, the special case $F_w=I_{n_w\times n_w}$ describes a  classical noise vector $dw$. However, the more general case
 $$
 F_w=I_{n_w\times n_w}+i\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n_w-n'}{2}}(J)\right)
 $$
 presents $n'$ classical noises and $n_w-n'$ conjugate quantum noises.~\footnote{Here $I_{n\times n}$ is the $n\times n$ identity matrix.}
 Here, the self-adjoint entries of the vector $w(t)$ which act on the
Boson Fock space $\mathcal{F}$
are the quantum noises driving the system and they correspond to boson quadratures (see e.g.,~\cite{HP84,belavkin1992quantum,parthasarathy1992introduction}). This determines the following
commutation relations for the noise components
 \begin{equation}~\label{eq:comw}
 [dw(t),dw^T(t)]=2T_w\,dt,
 \end{equation}
 with $T_w=\frac{1}{2}(F_w-F_w^T).$~\footnote{If X and Y are column vectors of operators, the commutator is defined by $[X,Y^T]=XY^T-(YX^T)^T.$}

Similarly, the process $y$ has the following It\={o} table
\begin{equation*}
dy\left(t\right)dy\left(t\right)^{T}=F_ydt,
\end{equation*}
where $F_y$ is a non-negative Hermitian matrix.
Indeed, the special case $F_y=I_{n_y\times n_y}$ describes a  classical output vector $dy$. However, the more general case
 $$
 F_y=I_{n_y\times n_y}+i\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n_y-n'}{2}}(J)\right)
 $$
 presents $n'$ classical outputs and $n_y-n'$ conjugate quantum outputs. The commutation relations for the processes $dy$ is determined by the matrix $T_y$ given by the following
 \begin{equation*}
 [dy(t),dy^T(t)]=2T_y\,dt,
 \end{equation*}
 with $T_y=\frac{1}{2} (F_y-F_y^T).$ Note that we have the following relations
 \begin{equation*}
 F_y=DF_w D^T,\quad\textrm{and}\quad T_y=DT_w D^T.
 \end{equation*}
As discussed in~\cite{james2008h}, we can always set up the following conventions by appropriately enlarging $dw$, $dy$, $B$ and $C$: i)  $n_y$ is even ii) $F_w$ has the following form
\begin{equation}~\label{eq:noisec}
F_w=I_{n_w\times n_w}+i\textrm{diag}_{\frac{n_w}{2}}(J),
\end{equation}
hence $n_w$ should be even.

\subsection*{Physical realizability of linear QSDEs}
Not all QSDEs of the form~\eqref{eq:sdq} represent the dynamics of physically meaningful open quantum systems. In the case that $\Theta$ is canonical, the system is physically realizable if it presents an open quantum harmonic oscillator. Now we give the formal definition of physical realizability (see e.g.,~\cite[Definition 3.3]{james2008h}).
\begin{Def}
\rm The system~\eqref{eq:sdq} is said to be physically realizable if one of the following holds:
\begin{itemize}
\item $\Theta$ is canonical and Equation~\eqref{eq:sdq} represents the dynamics of an open quantum harmonic oscillator.
\item $\Theta$ is degenerate canonical and there exists an augmentation of Equation~\eqref{eq:sdq} (see more details in~\cite{james2008h}) such that the new QSDEs represent the dynamics of an open quantum harmonic oscillator.
\end{itemize}
The system~(\ref{eq:sdq}) describes an open quantum harmonic oscillator if there exists a quadratic Hamiltonian $H = \frac{1}{2} x(0)^T R x(0)$,
with a real, symmetric, $n \times n$ matrix $R,$ and a  coupling operator $L = \Lambda x(0)$, with a complex-valued $\frac{1}{2} n_w \times n$ coupling matrix $\Lambda$
	such that
	\begin{eqnarray}
	x_k(t) &=& U(t)^\dagger (x_k(0) \otimes 1) U(t),\quad k=1,\cdots,n \nonumber \\
	y_l(t) &=& U(t)^\dagger (1 \otimes w_l(t)) U(t), \quad l=1,\cdots,n_y \label{eqn:unitary}
\end{eqnarray}
where $\{U(t),\quad t\geq 0\}$ is an adapted process of unitary operators satisfying the following QSDE~\cite{HP84}
	\begin{equation*}
	dU(t)=\left(-iH\,dt-\frac{1}{2}L^\dag L\,dt+[-L^\dag\,\, L^T]\Gamma dw(t)\right)U(t),\quad U(0)=I.
	\end{equation*}
In this case, the matrices $A$, $B$, $C$ and $D$ are given by
	\begin{eqnarray*}
		A = 2 \Theta \left(R + \mathfrak{Im}\left(
			\Lambda^{\dagger}\Lambda \right) \right),
			\label{eqn:a2} \\
		B = 2i \Theta \left[ \begin{array}{cc}
			-\Lambda^{\dagger} & \Lambda^T \end{array} \right] \Gamma,
			\label{eqn:b2} \\
		C = \mathcal P^T \left[ \begin{array}{cc}
				\Sigma & 0 \\ 0 & \Sigma \end{array} \right]
			\left[ \begin{array}{cc} \Lambda + \Lambda^\# \\
				-i\Lambda + i\Lambda^\# \end{array} \right] ,
			\label{eqn:c2} \\
		D = \left[\begin{array}{cc}
				I_{n_y \times n_y} &
				0_{n_y \times (n_w - n_y)}
			\end{array} \right].
	\end{eqnarray*}
	Here,
	$\Gamma$ is a
	$n_w \times n_w$ matrix and
	\begin{eqnarray*}
	\Gamma &=&  \mathcal P \textrm{diag} (M), \\
	M &=&  \frac{1}{2}
		\left[ \begin{array}{cc} 1 & i \\ 1 & -i \end{array} \right], \\
	\Sigma &=&  \left[ \begin{array}{cc}
				I_{\frac{1}{2}n_y \times
				\frac{1}{2}n_y} &
				0_{\frac{1}{2}n_y \times \frac{1}{2}\left(
			n_w - n_y \right) } \end{array} \right].
	\end{eqnarray*}
\rm $\mathcal P$ is the appropriately dimensioned square permutation
	matrix such that
	$$ \mathcal P \left[ \begin{array}{cccc}a_1 & a_2 & \cdots & a_{2m} \end{array} \right]
		= \left[ \begin{array}{cccccccc} a_1 & a_3 & \cdots & a_{2m-1} &
		a_2 & a_4 & \cdots & a_{2m} \end{array} \right].$$
	Also, note that $\mathfrak{Im}\left(.\right)$
	denotes the imaginary part of a matrix, ${X}^\dagger$ denotes the adjoint of an operator $X,$ and
	${X}^\#$ denotes the complex conjugate of a matrix $X.$
	\end{Def}

The following theorem borrowed from~\cite{james2008h} provides necessary and sufficient conditions for physical realizability of Equation~\eqref{eq:sdq} for any $\Theta$ (canonical or degenerate canonical).
\begin{thm}[\cite{james2008h}]~\label{thm:prp}
\rm The system~\eqref{eq:sdq} is physically realizable if and only if
\begin{eqnarray}~\label{eq:pr}
&&iA\Theta+i\Theta A^{T}+BT_wB^{T}=0,\nonumber\\
&&BD^T=\Theta C^T\textrm{diag}_{\frac{n_y}{2}}(J).
\end{eqnarray}
Here, $D=[I_{n_y\times n_y} \quad 0_{n_y\times (n_w-n_y)}].$
Moreover, for canonical $\Theta,$ the Hamiltonian and coupling matrices have explicit expressions as follows. The Hamiltonian matrix $R$ is uniquely given by $R=\frac{1}{4}(-\Theta A+A^T\Theta),$ and the coupling matrix $\Lambda$ is given uniquely by
\begin{equation*}
\Lambda=-\frac{1}{2}i\,[0_{\frac{n_w}{2}\times \frac{n_w}{2}}\quad I_{\frac{n_w}{2}\times \frac{n_w}{2}}](\Gamma^{-1})^TB^T\Theta.
\end{equation*}
In the case that $\Theta$ is degenerate canonical, a physically realizable augmentation of the system can be constructed to determine the associated Hamiltonian and coupling operators using the above explicit formulas.
\end{thm}
In the following lemma, we prove that the non-demolition property holds for non-commutative outputs if system~\eqref{eq:sdq} is physically realizable.
\begin{lemma}
\rm If the system~\eqref{eq:sdq} is physically realizable, then  the non-demolition property holds, i.e.,
$$[x(t),y(s)^T]=0,\quad\textrm{for any}\quad t\geq s.$$
\end{lemma}
\begin{proof}
\rm In~\cite[Lemma 4]{wang2014network}, it was shown that the condition $[x(t),y(s)^T]=0,$ for any $t\geq s$ is equivalent  to the following
\begin{equation}~\label{eq:ndem}
\Theta C^T+BT_wD^T=0.
\end{equation}
Now we show that if the plant is supposed physically realizable, i.e., if condition~\eqref{eq:pr} holds, then, the above equality holds too. Since, by condition~\eqref{eq:pr}, we have
\begin{equation*}
BD^T=\Theta C^T\textrm{diag}_{\frac{n_y}{2}}(J),
\end{equation*}
which is equivalent to $\Theta C^T=-BD^T\textrm{diag}_{\frac{n_y}{2}}(J),$ as $(\textrm{diag}_{\frac{n_y}{2}}(J))^2=-I_{n_y\times n_y}.$ Now it is easy to verify that condition~\eqref{eq:ndem} is satisfied, because we have
$$D^T\textrm{diag}_{\frac{n_y}{2}}(J)=T_w D^T,$$ therefore
$$BD^T\textrm{diag}_{\frac{n_y}{2}}(J)=BT_w D^T,$$ which is exactly condition~\eqref{eq:ndem}.
\end{proof}
In the following lemma, we show that the non-commutative outputs don't have self-non-demolition property.
\begin{lemma}
\rm The non-commutative output processes $y$ have the following commutation relations
\begin{equation*}
[y(t),y(s)^T]=2DT_wD^Ts,\quad\textrm{for all}\quad t\geq s.
\end{equation*}
\end{lemma}
\begin{proof}
First note that we have the following property
\begin{equation*}
[y(t),y(s)^T]=[y(s)+\int_s^t dy(s'),y(s)^T]=[y(s),y(s)^T],
\end{equation*}
since
\begin{equation*}
[\int_s^t dy(s'),y(s)^T]=[\int_s^t C x(s')+Ddw(s'),y(s)^T]=0,
\end{equation*}
as $[x(s'),y(s)^T]=0$ for any $s'\geq s,$ by previous lemma and $[dw(s'),y(s)^T]=0,$ for any $s'\geq s.$
Now it is sufficient to prove the lemma for $[y(s),y(s)^T].$ By taking the differentiation of this commutator, by It\={o} rule, we find
\begin{align*}
d[y(s),y(s)^T]&=[dy(s),y(s)^T]+[y(s),dy(s)^T]+[dy(s),dy(s)^T]\\
&=[Cx(s)ds+Ddw(s),y(s)^T]+[y(s),x(s)^TC^Tds+dw(s)^TD^T]\\
&+[Cx(s)ds+Ddw(s),x(s)^TC^Tds+dw(s)^TD^T]\\
&=D[dw(s),dw(s)^T]D^T=2DT_w D^Tds,
\end{align*}
where we have used the following facts: $[x(s),y(s)^T]=0,$ $[y(s),x(s)^T]=0,$ $[dw(s),y(s)^T]=0,$ $[y(s),dw(s)^T]=0,$ $(ds)^2=0,$ $dw(s)ds=0,$ and $ds dw(s)^T=0.$ Also, for the last equality, we have used the commutation relations for the processes $dw$ given in~\eqref{eq:comw}.

Finally, we get the following
$$[y(s),y(s)^T]=2DT_wD^Ts,$$
since $[y(0),y(0)^T]=0_{n_y\times n_y}.$
\end{proof}
\begin{rem}
\rm We recall that when $y$ is commutative, we have $$T_y=DT_wD^T=0_{n_y\times n_y},$$ which implies that the process $y$ is self-commuting.
\end{rem}
Before starting the next section, let us present the following definition.
\begin{Def}
\rm For any vector of zero mean self-adjoint operators $\zeta,$ the symmetric covariance is defined by
\begin{eqnarray}~\label{eq:symcov}
C_\zeta = \frac{1}{2} \mathbb{E}[ \zeta \zeta^T  + (  \zeta \zeta^T   )^T ] .
\end{eqnarray}
The matrix $C_\zeta$ is non-negative,  real and symmetric.  If $\zeta$ does not have zero mean, the covariance is defined by subtracting the mean.
\end{Def}
\section{Linear least mean squares estimation}
\label{sec:four}
In this section, we formulate a linear least squares estimation problem for the non-commutative linear system (\ref{eq:sdq}), with non-commutative output process  $y(t)$.   The problem concerns finding an operator $\hat x(t)$, called an estimator,  such that the dynamical evolution of $\hat x(t)$ depends causally on the output process $y(t)$ and the length of the error
\begin{eqnarray}~\label{eq:error}
e(t) = x(t) - \hat x(t)
\end{eqnarray}
is minimized.  The idea is that $\hat x(t)$ \lq\lq{tracks}\rq\rq  \ the plant operator $x(t)$.
The vector $\hat x(t)$ has self-adjoint operator components defined on a generally  larger space than the system (\ref{eq:sdq}). More precisely, the vector $\hat x$ consists of entries which are self-adjoint operators acting on the tensor product Hilbert Space $\mathcal H\bigotimes \mathcal F\bigotimes \mathcal H_1,$ where $\mathcal H_1$ is the initial Hilbert space of the least squares estimator $\hat x,$ which is a copy of the system space and independent of the system (\ref{eq:sdq}).

Take $\mathcal Y(t)$ for the von Neumann algebra generated by the output process $y(s)$ for $0\leq s \leq t.$  When $\mathcal Y(t)$ is commutative, i.e., $y(t)$ is a classical measurement process (by the Spectral Theorem~\cite[Theorem 3.3]{BHJ07}), the optimal filter in the least
squares sense is obtained by computing the conditional expectation onto $\mathcal Y(t)$~\cite{belavkin1992quantum,BHJ07}. The non-demolition property $([x(t), y(s)^T] = 0,$ for any $t \geq s$) is sufficient to conclude the existence of the commutative conditional expectation~\cite{belavkin1991continuous}.

In contrast to commutative
output, it is not shown whether the least mean squares estimator that we define in Definition~\ref{def:lms}, is equivalent to conditional expectation. This problem
is related to the existence of a non-commutative conditional expectation which is not always guaranteed, and we do not consider this matter in this paper  (see more details in~\cite{takesaki1972conditional}).

Firstly, we define the class $\xi$ of linear estimators of the form,
\begin{equation}~\label{eq:qkalman}
d\check x(t)=A\check{x}(t)dt+K (t)( dy(t) - C\check{x}(t)dt), \ \ \check{x}(0)=\hat x_0,
\end{equation}
where $y$ is the adapted process defined in Equation~\eqref{eqn:unitary} (see~\cite{BHJ07,parthasarathy1992introduction,HP84} for a discussion of adapted quantum processes).
Equation (\ref{eq:qkalman}) has the standard form of an observer or Kalman filter.  The real gain matrix $K(t)$ is to be determined.

The initial condition  $\hat x(0)$ satisfies the commutation relations
$$
\hat x(0) \hat x(0)^T-(\hat x(0) \hat x(0)^T)^T=2i\Theta .
$$
The state of $\hat x(0)$ is taken to be $\hat \rho_0$.    Consequently, the initial state of the composite system  (\ref{eq:sdq}) and  (\ref{eq:qkalman}) is the Gaussian state $\rho= \rho_0 \otimes \hat \rho_0$.
\begin{Def}~\label{def:lms}
\rm A linear least mean squares estimator $\hat x $ for the non-commutative linear system (\ref{eq:sdq}) has the following properties,
\begin{itemize}
 \item it is defined on the class $\xi,$ i.e., it is a linear system of the form (\ref{eq:qkalman}), and
 \item  the real matrix $K(t)$ is chosen to minimize the symmetrized mean squares error defined as follows
\begin{equation}
J(K(t)):=\mathrm{Tr}[P(t)],
\label{eq:mmse}
\end{equation}
where $P(t)$ is the symmetric error covariance matrix defined by
\begin{equation*}~\label{eq:ricsym}
P(t):=C_{e(t)}=\frac{1}{2} \mathbb{E}_\rho[ e(t) e(t)^T+(e(t)e(t)^T)^T].
\end{equation*}
\end{itemize}
\end{Def}
Prior to state our main theorem, we need the following equations
\begin{align}~\label{eq:P-ric}
\dot{P}\left(t\right)&=\left(A-K(t)C\right)P(t)+P(t)\left(A-K(t)C\right)^{T}
+(B-K(t)D)(B-K(t)D)^T
\nonumber\\
P(0)&=C_{e(0)}.
\end{align}
Here, $C_{e(0)}$ is the initial symmetric error covariance.
\begin{thm}~\label{thm:main}
\rm Suppose that the plant~\eqref{eq:sdq} is physically realizable. Then, the linear system (\ref{eq:qkalman})  is a linear least mean squares estimator for the system (\ref{eq:sdq})  if and only if the gain $K(t)$ is given by
\begin{eqnarray}
K(t)=BD^T+P(t)C^T,
\label{eq:K-optimal}
\end{eqnarray}
where $P(t)$ is the symmetric positive definite solution to the Riccati equation (\ref{eq:P-ric}).
Furthermore, the innovations process
\begin{equation}~\label{eq:inn}
dr(t):=dy(t)-C\hat{x}(t)dt=Ce(t)dt+Ddw(t), \ \ r(0)=0,
\end{equation}
is a quantum Wiener process with symmetrized covariance $t$ and    It\={o} table
\begin{eqnarray}
dr(t) dr^T(t) = D F_w D^T dt .
\end{eqnarray}
\end{thm}
\begin{proof}
The proof is a modification of the well known methods documented in \cite{kwakernaak1972linear}. It follows from \cite[Lemma 3.1]{kwakernaak1972linear}  that $J(K)$ is minimized, if and only if
$$K(t)=(BD^T+P(t)C^T)(DD^T)^{-1},$$ where $P(t)$ is the solution to (\ref{eq:P-ric}). Below, we show that the symmetrized error covariance matrix satisfies the Riccati equation~\eqref{eq:P-ric}.

The error as defined in Equation~\eqref{eq:error} satisfies  the following dynamics
\begin{equation}~\label{eq:sde}
de(t)=\Big(A-K(t)C\Big)e(t)\,dt+\Big(B-K(t)D\Big)\,dw(t).
\end{equation}
Fix any real matrix $K(t)$ and let $\hat x(t)$ be the solution of Equation (\ref{eq:qkalman}). Let $e(t)=x(t)-\hat x(t)$ be the associated error, and
consider the real symmetric error covariance
\begin{eqnarray}
P(t) = C_{e(t)}=\frac{1}{2} \mathbb{E}_\rho[ e(t) e(t)^T+(e(t)e(t)^T)^T] .
\end{eqnarray}
Take the derivation of the above equation, by the quantum It\={o} rule, we have
\begin{align*}
dP(t)&=\frac{1}{2}\Big(\mathbb{E}_\rho\Big[ de(t) e(t)^T+e(t)de(t)^T+de(t)de(t)^T\nonumber\\
&+\left(de(t) e(t)^T+e(t)de(t)^T+de(t)de(t)^T\right)^T\Big] \Big)
\end{align*}
Now, it is sufficient to replace the expression of $de$ given in Equation~\eqref{eq:sde} in above, we get
\begin{align}~\label{eq:ricito}
dP(t)&=\frac{1}{2}\Big((A-KC)\mathbb{E}_\rho[e(t)e(t)^T]\,dt+\mathbb{E}_\rho[e(t)e(t)^T](A-KC)^T\,dt\nonumber\\
&+(B-KD)F_w(B-KD)^T\,dt
+\mathbb{E}_\rho[(e(t)e(t)^T)^T](A-KC)^T\,dt\nonumber\\
&+(A-KC)\mathbb{E}_\rho[(e(t)e(t)^T)^T]\,dt+(B-KD)F_w^T(B-KD)^T\,dt\Big),
\end{align}
where we have used the followings: $\mathbb E_\rho[dw(t)e(t)^T]=0,$ $\mathbb E_\rho[e(t)dw(t)^T]=0,$ $\mathbb E_\rho[(dt)^2]=0,$ $\mathbb E_\rho[dtdw^T]=0,$ $\mathbb E_\rho[dwdt]=0,$ and Equation~\eqref{eq:fwone}.

From Equation~\eqref{eq:ricito}, we can write the following
\begin{align}~\label{eq:P-new}
\dot{P}\left(t\right)&=\left(A-K(t)C\right)P(t)+P(t)\left(A-K(t)C\right)^{T}\nonumber\\
&+(B-K(t)D)    (B-K(t)D)^T\nonumber\\
P(0)&=C_{e(0)}
\end{align}
Since by Equation~\eqref{eq:noisec}, we know $\frac{F_w+F_w^T}{2}=I_{n_w\times n_w}.$

Now the mean squares error can be expressed in terms of $P(t)$ as follows
\begin{eqnarray}
J(K(t)) = \mathrm{Tr}[ P(t) ].
\end{eqnarray}
As Equation~\eqref{eq:P-new} has the same form as the standard Riccati equation considered in~\cite{kwakernaak1972linear}, we can apply~\cite[Lemma 3.1]{kwakernaak1972linear} to find the the minimum of $J(K),$ we get
\begin{equation}~\label{eq:kgain}
K(t)=(BD^T+P(t)C^T)(DD^T)^{-1}.
\end{equation}
The gain given in above can be further simplified as
$$K(t)=BD^T+P(t)C^T,$$ since $DD^T=I_{n_y\times n_y}$ by physical realizability of the plant. This finishes the proof of the first part of Theorem~\ref{thm:main}.

\medskip

Next, following \cite[Sec. 4.3.6]{kwakernaak1972linear}, let
\begin{eqnarray}
\Gamma(t) = \left[
\begin{array}{cc}
\Gamma_{11}(t)  & \Gamma_{12}(t)
\\
\Gamma_{12}^T(t) & \Gamma_{22}(t)
\end{array}
\right]
\end{eqnarray}
be the (symmetrized) covariance matrix for the vector
$
 \left[
 \begin{array}{c}
r(t)
\\
e(t)
\end{array}
\right].
$
By the It\={o} rule and taking expectations, we find that
{\small\begin{eqnarray*}
\dot \Gamma_{11}(t)   &= & C\Gamma_{12}^T(t) + \Gamma_{12}(t) C^T + I , \ \ \Gamma_{11}(0)=0_{n_y\times n_y},
\\
 \dot \Gamma_{12}(t)   &= &   C \Gamma_{22}(t) + \Gamma_{12}(t) (A-K(t)C)^T + D(B - K(t)D)^T, \ \ \Gamma_{12}(0)=0_{n_y\times n},
 \\
 \dot \Gamma_{22}(t)   &= &  (A-K(t)C) \Gamma_{22}(t)  + \Gamma_{22}(t) (A-K(t)C)^T
 \nonumber \\
 &&  + (B-K(t)D) (B-K(t)D)^T, \ \ \Gamma_{22}(0)=C_{e(0)} .
\end{eqnarray*}}
Comparing with Equation (\ref{eq:P-ric}), we find that $\Gamma_{22}(t)=P(t)$, and from (\ref{eq:K-optimal}), we have
\begin{eqnarray}
 \dot \Gamma_{12}(t)   =   \Gamma_{12}(t) (A-KC)^T, \ \ \Gamma_{12}(0)=0_{n_y\times n},
\end{eqnarray}
which implies $\Gamma_{12}(t)=0_{n_y\times n}$ for all $t \geq 0$. From this, we see that
\begin{eqnarray}
\dot \Gamma_{11}(t) =I_{n_y\times n_y},\ \ \Gamma_{11}(0)=0_{n_y\times n_y},
\end{eqnarray}
and hence $\Gamma_{11}(t) =tI_{n_y\times n_y}.$ Also, it is obvious that we have the following relations for the innovation processes $dr,$
$$dr(t) dr^T(t) = D F_w D^T dt,$$ since $(dt)^2=0,$ $dw(t)dt=0,$ and $dtdw(t)^T=0.$
\end{proof}
\section{Results on physical realizability}~\label{sec:phys}
In this section, we will study the physical realizability of the least mean squares estimators announced in Theorem~\ref{thm:main}.
In Theorem \ref{thm:main}, we don't assume the linear least mean squares estimator (\ref{eq:qkalman}) to be physically realizable.

Let us write $B$ as
\begin{equation}~\label{eq:Bsubm}
B=[B'_{n\times n_y}\quad B''_{n\times n_w-n_y}].
\end{equation}
In the following, we will announce a general theorem which gives necessary and sufficient conditions ensuring  physical realizability of the least mean squares estimators given in Theorem~\ref{thm:main}.
 \begin{thm}~\label{thm:phys}
\rm Assume that the plant~\eqref{eq:sdq} satisfies the physical realizability conditions announced in Theorem~\ref{thm:prp}. Then, the linear least mean squares estimator announced in Theorem~\ref{thm:main} is a physical realizable estimator if  and only if
\begin{align}~\label{eq:prc}
&-B\textrm{diag}_{\frac{n_w}{2}}(J)B^T+3B'\textrm{diag}_{\frac{n_y}{2}}(J)B'^T+2PC^T\textrm{diag}_{\frac{n_y}{2}}(J)B'^T\nonumber\\
&+2B'\textrm{diag}_{\frac{n_y}{2}}(J)CP+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)CP=0,
\end{align}
with $P$ satisfying the following Riccati equation
\begin{align}~\label{eq:ric}
\dot{P}(t)&=(A-B'C)P(t)+P(t)(A-B'C)^T-P(t)(C^TC)P(t)+B''B''^T,\nonumber\\
P(0)&=C_{e(0)}
\end{align}
\end{thm}
\begin{proof}
The estimator of the form~\eqref{eq:qkalman} can be rewritten as the following form
\begin{equation*}
d\hat{x}(t)=(A-KC)\hat{x}(t)dt+K (t) dy(t) , \ \ \hat{x}(0)=\hat x_0.
\end{equation*}
If we impose the physical realizability constraints on the estimator
of the form given in above, we get the following condition
\begin{equation*}
i(A-KC)\Theta+i\Theta (A-KC)^T+KD T_w D^T K^T=0.
\end{equation*}
Now it is sufficient to replace $K$ by its value determined by Theorem~\ref{thm:main} (Equation~\eqref{eq:K-optimal}). We find
\begin{align}~\label{eq:prg}
&A\Theta+\Theta A^T-BD^T C\Theta-\Theta C^T DB^T-PC^TC\Theta-\Theta C^T C P\nonumber\\
&+BD^T\textrm{diag}_{\frac{n_y}{2}}(J)DB^T+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)CP\nonumber\\
&+BD^T\textrm{diag}_{\frac{n_y}{2}}(J)CP+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)DB^T=0,
\end{align}
where we have used $D\textrm{diag}_{\frac{n_w}{2}}(J)D^T=\textrm{diag}_{\frac{n_y}{2}}(J).$ Now use the following facts
$C\Theta=-\textrm{diag}_{\frac{n_y}{2}}(J)DB^T$ and $A\Theta+\Theta A^T=-B\textrm{diag}_{\frac{n_w}{2}}(J)B^T,$ which are derived from the physical realizability of the plant, i.e., Equation~\eqref{eq:pr}. Also, note that $BD^T=B'.$ From these equalities, Equation~\eqref{eq:prc} can be derived from Equation~\eqref{eq:prg}.

Moreover, Equation~\eqref{eq:ric} is derived from the Riccati equation~\eqref{eq:P-new} by replacing $K$ by its value given in Equation~\eqref{eq:K-optimal} and using the physical realizability of the plant.
\end{proof}
\subsection{Some special cases}
In the following, first, we study the physical realizability of the least mean squares estimator announced in Theorem~\ref{thm:main} for the case where

\noindent $B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0,$ with $B'$ defined in Equation~\eqref{eq:Bsubm}. Second, we study the case where $B'=0.$
\subsubsection*{Case $\bf 1:$ $\bf B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0$}
 As it is demonstrated in the following corollary, the physical realizability constraint announced in~\eqref{eq:prc} can be simplified.
 \begin{coro}~\label{coro:phys1}
\rm If $B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0.$ Then, the estimator of the form~\eqref{eq:qkalman} is a physical realizable least mean squares estimator if and only if the following constraints are satisfied.
\begin{itemize}
\item [(i)] $K=B'+PC^T.$
\item[(ii)] For $\Theta$ canonical,
\begin{align}~\label{eq:prc1m}
B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T+2P\Theta B'B'^T+2B'B'^T\Theta P=0,
\end{align}
with $P$ the unique symmetric positive definite solution of the following Riccati equation,
\begin{align}~\label{eq:rc}
\dot{P}(t)&=AP(t)+P(t)A^T+P\Theta B'B'^T\Theta P(t)+B''B''^T,\nonumber\\
P(0)&=C_{e(0)}.
\end{align}
\item [(iii)] For $\Theta$ degenerate canonical,
\begin{itemize}
\item [(i)] if $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T=C^T$, then
\begin{align*}
B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T+2P\Theta B'B'^T+2B'B'^T\Theta P=0,
\end{align*}
with $P$ satisfying dynamics~\eqref{eq:rc}.

\item [(ii)] But if $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T\neq C^T$ holds, then,
\begin{align}~\label{eq:prc2m}
&-B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T+2PC^T\textrm{diag}_{\frac{n_y}{2}}(J)B'^T\nonumber\\
&+2B'\textrm{diag}_{\frac{n_y}{2}}(J)CP+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)CP=0,
\end{align}
with $P$ satisfying Equation~\eqref{eq:ric}.
\end{itemize}
\end{itemize}
\end{coro}
\begin{proof}
By Theorem~\ref{thm:phys}, we know that the least mean squares estimator~\eqref{eq:qkalman} is physically realizable if and only if the condition~\eqref{eq:prc} is satisfied. By the assumption $B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0,$ we get
\begin{align*}
B\textrm{diag}_{\frac{n_w}{2}}(J)B^T=B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T.
\end{align*}
Then, the condition~\eqref{eq:prc} becomes
\begin{align}~\label{eq:prcc}
-B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T&+2PC^T\textrm{diag}_{\frac{n_y}{2}}(J)B'^T\nonumber\\
&+2B'\textrm{diag}_{\frac{n_y}{2}}(J)CP+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)CP=0.
\end{align}
We know by the physical realizability of the plant $C\Theta=-\textrm{diag}_{\frac{n_y}{2}}(J)DB^T$
(and similarly, $\Theta C^T=-BD^T\textrm{diag}_{\frac{n_y}{2}}(J)$).
Now suppose that $\Theta$ is canonical, then by multiplying the above conditions by $\Theta,$ we find
$C=\textrm{diag}_{\frac{n_y}{2}}(J)DB^T\Theta$ (and similarly, $C^T=\Theta BD^T\textrm{diag}_{\frac{n_y}{2}}(J)$). Finally, by replacing the values of $C$ and $C^T$ found as such, we get the constraint~\eqref{eq:prc1m} given in the second part of the corollary.

Now we consider the case $\Theta$ degenerate canonical, in this case if we multiply the expression of $C\Theta$ (and similarly, $\Theta C^T$) given in above by $\Theta$, we get
$C\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)=\textrm{diag}_{\frac{n_y}{2}}(J)DB^T\Theta$
(and similarly,

\noindent $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T=\Theta BD^T\textrm{diag}_{\frac{n_y}{2}}(J)$). Now it is clear that if the condition $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T=C^T$ holds, then we get exactly the constraint given in the first part. However, if this condition does not hold, from~\eqref{eq:prc}, we get the constraint~\eqref{eq:prcc} which is exactly the condition~\eqref{eq:prc2m}.
\end{proof}
\begin{rem}
\rm We remark that the condition $B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0$ was considered in order to simplify the physical realizability constraints in Equation~\eqref{eq:prc}. As the corollary in above shows, in most of the times, this case is equivalent to eliminating the quadratic terms in Equation~\eqref{eq:prc}. Also, note that if $n_y=n=2,$ the condition $B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0$ is equivalent to the condition $\det(B')=0$, i.e., the quadratures are linearly dependent.
\end{rem}
\paragraph{Particular case: $n_y=n_w.$} Consider the case $n_y=n_w.$ Then, a physical realizable plant should satisfy $D=I_{n_y\times n_y}.$ As a result, the plant given in~\eqref{eq:sdq} takes the following form
\begin{eqnarray}~\label{eq:sdqone}
&&dx\left(t\right)=Ax\left(t\right)dt+B\,dw\left(t\right),\nonumber\\
&&dy\left(t\right)=Cx\left(t\right)dt+dw\left(t\right).
\end{eqnarray}
Now let us state the following corollary as analogue of Corollary~\ref{coro:phys1}.
\begin{coro}~\label{coro:phystwo}
\rm The estimator of the form~\eqref{eq:qkalman} associated to the plant's dynamics~\eqref{eq:sdqone} is a physical realizable least mean squares estimator if and only if the following constraints are satisfied
\begin{itemize}
\item [(i)] $K=B+PC^T.$
\item [(ii)] For $\Theta$ canonical,
\begin{align}~\label{eq:prcone}
2P\Theta BB^T+2BB^T\Theta P=0,
\end{align}
with $P$ satisfying the following Riccati equation
\begin{align}~\label{eq:rcone}
\dot{P}(t)&=AP(t)+P(t)A^T+P(t)\Theta BB^T\Theta P(t),\nonumber\\
P(0)&=C_{e(0)}.
\end{align}
\item [(iii)] For $\Theta$ degenerate canonical,
\begin{itemize}
\item[(i)] if $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T=C^T$, then
\begin{align*}
2P\Theta BB^T+2BB^T\Theta P=0,
\end{align*}
with $P$ satisfying dynamics~\eqref{eq:rcone}.

\item [(ii)] But if $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T\neq C^T$ holds, then,
\begin{align}~\label{eq:cons3}
2PC^T\textrm{diag}_{\frac{n_y}{2}}(J)B^T
&+2B\textrm{diag}_{\frac{n_y}{2}}(J)CP\nonumber\\
&+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)CP=0,
\end{align}
with $P$ satisfying the following dynamics
\begin{align}~\label{eq:rics}
\dot{P}(t)&=(A-BC)P(t)+P(t)(A-BC)^T-P(t)(C^TC)P(t)\nonumber\\
P(0)&=C_{e(0)}
\end{align}
\end{itemize}
\end{itemize}
\end{coro}
\begin{proof}
The proof of this corollary can be done by the same arguments provided for Corollary~\ref{coro:phys1}. Since, if $n_y=n_w,$ the condition $B'\textrm{diag}_{\frac{n_y}{2}}(J) B'^T=0$ is equivalent to $B\textrm{diag}_{\frac{n_w}{2}}(J) B^T=0.$
\end{proof}
\paragraph{Particular case: $n=2,$ $n_y=2,$ $n_w=4,$ and $\Theta=J.$} Consider the simple case $n=2,$ $n_y=2,$ $n_w=4,$ and $\Theta=J.$ Take $A=\begin{pmatrix}a_1&a_2\\
a_3&a_4\end{pmatrix},$ $P=\begin{pmatrix}p_1&p_2\\
p_2&p_4\end{pmatrix},$ $B'=\begin{pmatrix}b_1&b_2\\
b_3&b_4\end{pmatrix},$ and $B''=\begin{pmatrix}d_1&d_2\\d_3&d_4\end{pmatrix}.$
In the following, we find the constraints which guarantee the physical realizability of the least mean squares estimator announced in Theorem~\ref{thm:main}.
\begin{coro}~\label{coro:spp}
\rm Take $\Theta=J$. If $B'J B'^T=0,$ then, the least mean squares estimator given in Theorem~\ref{thm:main} is physically realizable if and only if  \begin{equation}~\label{eq:phntwo}
2p_1(-b_4^2-b_3^2)+2p_2(2b_1b_3+2b_2b_4)+2p_4(-b_1^2-b_2^2)-\det(B'')=0,\end{equation} with $P$ satisfying the Riccati equation~\eqref{eq:rc}.
\end{coro}
\begin{proof}
The proof can be directly derived from Equation~\eqref{eq:prc1m}.
\end{proof}
Now, we can conclude the following corollary.
\begin{coro}~\label{coro:spp1}
\rm Suppose $b_1=b_3,$ $b_2=b_4,$ and  $\det(B'')=0.$ Then, the linear least mean squares estimator announced in Theorem~\ref{thm:main}, is physically realizable if and only if
$p_1+p_4=2p_2.$
\end{coro}
The following corollary shows the difficulty of finding a physical realizable least mean squares estimator for some particular forms of $P,$ $B'$ and $B''$.
\begin{coro}~\label{coro:spp2}
\rm Suppose $b_1=b_3,$ $b_2=b_4,$ $d_1=d_3,$ and $d_2=d_4.$ Then, it is impossible to realize physically a linear least mean squares estimator of the form given in~\eqref{eq:qkalman} such that $p_1=p_2=p_4$.
\end{coro}
\begin{proof}
By Corollary~\ref{coro:spp1}, we know that if $b_1=b_3,$ $b_2=b_4,$ and $\det(B'')=0,$ then the physical realizability condition~\eqref{eq:phntwo} implies that $p_1+p_4=2p_2.$ Thus, when $p_1=p_2=p_4,$ this condition is satisfied.

However, note that a least mean squares estimator should satisfy Equation~\eqref{eq:rc}. Also, we should take into account the facts that $b_1=b_3,$ $b_2=b_4,$ $d_1=d_3,$ and $d_2=d_4.$ Therefore, the steady state solution $P=\lim_{t\to\infty} C_{e(t)}$ of the Riccati equation~\eqref{eq:rc}, if there exists, should satisfy the following
\begin{align}~\label{eq:consric}
2(a_1p_1+a_2p_2)-(b_1^2+b_2^2)(p_1-p_2)^2+d_1^2+d_2^2&=0\nonumber\\
a_2p_4+a_3p_1+(p_1-p_2)(p_4-p_2)+d_1^2+d_2^2&=0\nonumber\\
2(a_3p_2-a_1p_4)-(b_1^2+b_2^2)(p_2-p_1)^2+d_1^2+d_2^2&=0,
\end{align}
where we have used $a_4=-a_1,$ since by physical realizability of the plant, $A$ should satisfy the following
$$AJ+JA^T=0.$$
We can observe that if $p_1=p_2=p_4,$ the matrix $A$ should have the following form
\begin{equation}~\label{eq:forma}
A=\begin{pmatrix}
a_1&a_2\\
a_3&-a_1
\end{pmatrix},\quad \textrm{with}\quad a_3-a_2=2a_1.
\end{equation}

 Note that in this case $K=B',$ since $PC^T=0.$ Also, we know that $A-KC=A-B'C$ should be a Hurwitz matrix (see e.g.,~\cite{kwakernaak1972linear,speyer2008stochastic}). However, we have  $A-B'C=A,$ since $B'C=B'JB'^TJ=0.$ Now, it remains to show that $A$ with its particular form given in~\eqref{eq:forma} could not be a Hurwitz matrix. It is sufficient to find the eigenvalues of $A.$ We have
$$\det(A-\lambda I_{2\times 2})=-a_1^2+\lambda^2-a_3 a_2=0,$$ with $a_3-a_2=2a_1.$ This implies that
$\lambda^2=(a_1+a_2)^2.$
Now, it is clear that $A$ could not be a Hurwitz matrix, i.e., all of its eigenvalues have negative real parts.
\end{proof}
This result shows that in order to obtain conditions on $B,$ which make the linear least mean squares estimator given in Theorem~\ref{thm:main} physically realizable (for e.g., see Equation~\eqref{eq:phntwo}), we need to suppose some constraints on $P.$ This demonstrates the difficulty to find an appropriate plant whose least mean squares estimator is physically realizable.
\subsubsection*{Case $\bf 2:$ $\bf B'=0$}
Let us announce the following corollary for this special case.
\begin{coro}~\label{coro:case2}
\rm If $B'=0.$ Then, the estimator of the form~\eqref{eq:qkalman} is a physical realizable least mean squares estimator if and only if
\begin{itemize}
\item[(I)] For canonical $\Theta,$ we have

$K=0,$ and
$B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T=0;$

\item [(II)] For degenerate canonical $\Theta=\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(J)\right),$ we have
\begin{itemize}
\item [(i)] If $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T=C^T$, then

$K=0,$ and
$B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T=0.$
\item [(ii)]
If $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T\neq C^T,$ then
$K=PC^T,$ and

$-B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T+PC^T\textrm{diag}_{\frac{n_y}{2}}(J)CP=0,$ with $P$ satisfying
\begin{align}~\label{eq:riccase3}
\dot{P}(t)&=AP(t)+P(t)A^T-P(t)(C^TC)P(t)+B''B''^T,\nonumber\\
P(0)&=C_{e(0)}.
\end{align}
\item [(iii)] Let us write $C=[C'_{n_y\times n'} \quad C''_{n_y\times (n-n')}].$ Then,
if $$\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T\neq C^T$$ and $C'^T\textrm{diag}_{\frac{n_y}{2}}(J)C'=0,$ then
$K=PC^T,$ and

$B''\textrm{diag}_{\frac{n_w-n_y}{2}}(J)B''^T=0,$ with $P$ satisfying the Riccati Equation~\eqref{eq:riccase3}.
\end{itemize}
\end{itemize}
\end{coro}
\begin{proof}
If $\Theta$ is canonical, then we find $K=0,$ since $B'=0$ implies $C=0$ by physical realizability conditions given in Theorem~\ref{thm:prp} (Equation~\eqref{eq:pr}). Then, we can use the results of Corollary~\ref{coro:phys1}, where by replacing $B'=0$ in Equations~\eqref{eq:prc1m} and~\eqref{eq:rc}, we find the conditions given in part (I).

However, if $\Theta$ is degenerate canonical, $C$ is not necessarily zero if $C\Theta=0.$ In this case, we find $K=PC^T.$
If $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T= C^T,$ then $C=0.$ This proves the results given in (i) of part (II).

But if $\textrm{diag}\left(0_{n'\times n'},\textrm{diag}_{\frac{n-n'}{2}}(I)\right)C^T\neq C^T,$ then we have to replace $B'=0$ in Equations~\eqref{eq:prc2m} and~\eqref{eq:ric}, but $C$ is not necessarily zero.
This proves the conditions $(ii)$ of part (II). The condition $(iii)$ in part (II) can be derived from condition $(ii).$ Also, by noting that $C\Theta=0,$ implies $C''=0.$
\end{proof}
Note that $B'=0$ implies $C\Theta=0.$ Roughly speaking, when $C\Theta=0,$ the non-commutative filter obtained in the above theorem, could also be realized with Homodyne  or Hetrodyne detection. Since, no quantum information is transferred from the plant to the filter in this case. This is like the classical filtering cases of Homodyne or Heterodyne detection, where one always ends up taking a single quadrature of the field.
\subsection{Consistency with standard results}
In this subsection, we recall the standard results, i.e., Belavkin-Kalman and classical Kalman filtering. They can be respectively considered as special cases when the output is commutative but the plant's dynamics is non-commutative and when the output and the plant's dynamics are both commutative.
\paragraph{Non-commutative dynamics, commutative (classical) outputs.} It can easily be shown that the least mean squares estimators found in Theorem~\ref{thm:main} are reduced to Belavkin-Kalman filters~\cite{EB05} under the assumptions that Belavkin used, i.e., the commutativity of the outputs and the non-demolition property.

Take $\mathcal{Y}_t$ to be commutative, that is $y(t)$ is self-adjoint for each $t$ and $[y(t), y(s)^T]=0$ for all $s, t$.
By the Spectral Theorem, \cite[Theorem 3.3]{BHJ07}), $y(t)$ corresponds to a classical stochastic process, the measurement process. Such continuous measurement signal arise in Homodyne detection~\cite{WM10}.

For the commutative outputs, we have the following correlation for the observation process $dy$
\begin{equation*}
dy(t)dy^T(t)=F_y\,dt,
\end{equation*}
with $F_y=I_{n_y\times n_y}.$ Note that we have the following relation between $F_w$ and $F_y,$
$$DF_w D^T=F_y,$$ with $D=[I_{n_y\times n_y} \quad 0_{n_y\times (n_w-n_y)}].$

\medskip

Therefore, $F_w$ takes the following form
\begin{align}~\label{eq:fw}
F_w=I_{n_w\times n_w}+i\textrm{diag}\left(0_{n_y\times n_y},\textrm{diag}_{\frac{n_w-n_y}{2}}(J)\right).
\end{align}

It is well established that the optimal filter satisfies the following dynamics
\begin{equation}~\label{eq:fopt}
d\hat{x}(t)=A\hat{x}(t)dt+(BD^T+P(t)C^T)( dy(t) - C\hat{x}(t)dt),
\end{equation}
with $\widehat x(0)=\mathbb E[x(0)]=\bar x(0),$ and $$\Sigma_0=\frac{1}{2}\mathbb E\left[(x(0)-\bar x(0))(x(0)-\bar x(0))^T+\Big((x(0)-\bar x(0))(x(0)-\bar x(0))^T\Big)^T\right].$$

Remark that the variable $\widehat x$ has zero commutation relation for any $t\geq 0$, i.e.,
\begin{equation}~\label{eq:com}
\widehat x(t)\widehat x(t)^T-(\widehat x(t)\widehat x(t)^T)^T=0,
\end{equation}
which means that $\widehat x$ is a classical variable.

The real symmetric matrix $P(t):=\frac{1}{2}\mathbb{E}\left[e\left(t\right)e\left(t\right)^{T}+(e\left(t\right)e\left(t\right)^{T})^T\right]$ satisfies the following Riccati equation
\begin{align*}~\label{eq:P}
\dot{P}\left(t\right)&=\left(A-K(t)C\right)P(t)+P(t)\left(A-K(t)C\right)^{T}
+(B-K(t)D)(B-K(t)D)^T\nonumber\\
P(0)&=\Sigma_0.
\end{align*}
\paragraph{Classical Kalman filtering.} The classical linear stochastic dynamics is described by classical variables as follows
\begin{eqnarray}~\label{eq:sd}
dx(t)&=A x(t)\,dt+B dw\nonumber\\
dy(t)&=C x(t)\,dt+D\, dw,
\end{eqnarray}
where $A,$ $B,$ $C$ and $D$ are real matrices in $\mathbb R^{n\times n},$ $\mathbb R^{n\times n_w}$ and $\mathbb R^{n_y\times n}$ and $\mathbb R^{n_y\times n_w},$ and $w$ is a vector of classical Wiener processes, with $dw(t)dw(t)^T=I_{n_w\times n_w}\,dt.$  Take $\mathcal Y(t)$ as the algebra generated by the observation processes previous to time $t,$ defined by $\mathcal Y(t):=\textrm{span}\{(y(s))_{0\leq s\leq t}\}.$

\medskip

It is well known that the classical Kalman filter~\cite{speyer2008stochastic,kwakernaak1972linear} satisfies the following dynamics
\begin{equation*}
d\widehat x(t)=A\widehat x(t)\,dt+(BD^T+P(t)C^T)(DD^T)^{-1}\left(dy(t)-C\widehat x(t)\,dt\right),
\end{equation*}
with $\widehat x(0)=\mathbb E[x(0)]=\bar x(0)$ and $\mathbb E[(x(0)-\bar x(0))(x(0)-\bar x(0))^T]=\Sigma_0.$

\medskip

The covariance of the error $P(t)=\EE{e(t)e^T(t)}$  satisfies the following Riccati equation
\begin{align}~\label{eq:ricatti}
\dot{P}\left(t\right)&=(A-KC)\,P(t)+P(t)\,(A-KC)^T
+\big(B(t)-K(t)D\big)\big(B(t)-K(t)D\big)^T,\nonumber\\
P(0)&=\Sigma_0.
\end{align}
Note that  for the classical Kalman filter, we have $$F_w=I_{n_w\times n_w},\quad F_y=I_{n_y\times n_y}\quad\textrm{and} \quad\Theta=0_{n\times n}.$$

\subsection{Construction of coherent observers with least mean squares estimators}

Suppose that the linear least mean squares estimator~\eqref{eq:qkalman} does not satisfy the constraints of physical realizability given in Theorem~\ref{thm:phys}. Then, we can allow additional vacuum noise inputs to the least mean squares estimator~\eqref{eq:qkalman} such that the
resulting system is physically realizable. Suppose that the following estimator is physically realizable
\begin{equation*}
d\tilde x=(A-KC)\tilde x+Kdy+bdv,
\end{equation*}
with $dvdv^T=F_v dt,$ where $
F_v=I_{n_v\times n_v}+i\textrm{diag}_{\frac{n_v}{2}}(J),$ and with $n_v$ positive even integer. Also, we suppose that $dv$ is independent of $dw.$ This estimator is called a coherent observer \cite{Miao2012}, since it tracks in average the plant dynamics~\eqref{eq:sdq} when $A-KC$ is Hurwitz, and is physically realizable. The error covariance matrix is defined by the following
\begin{equation*}
\tilde P(t):=C_{\tilde e}(t),
\end{equation*}
where $C_{\tilde e}$ is defined in Equation~\eqref{eq:symcov} and $\tilde e=x-\tilde x$. It is not difficult to show that the error covariance matrix satisfies the following Riccati equation
\begin{align}~\label{eq:rictilde}
\dot{\tilde P}(t)&=(A-B'C)\tilde P(t)+\tilde P(t)(A-B'C)^T-\tilde P(t)(C^TC)\tilde P(t)+B''B''^T+bb^T,\nonumber\\
\tilde P(0)&=C_{\tilde e(0)}.
\end{align}
The steady state solution of the above Riccati equation, if there exists, is given by $\tilde P=\lim_{t\to\infty} C_{\tilde e(t)}.$ Then, the performance can be defined by the following
$$\tilde J=\tr{\tilde P}.$$
In the following, we give some examples. However, in this paper, we don't discuss different algorithms that can be considered to make least mean squares estimators physically realizable. We choose a matrix $b$ that can make the least mean squares estimator physically realizable and we compare the performance of the estimator $\tilde x$ with the least mean squares estimator $\hat x$ (see e.g.,~\cite{vuglar2014,Miao2012} for more details on different algorithms to design coherent observers).
 \subsection{Examples}
 In the following,  we give some examples from the literature to illustrate the results of this section. Also, these examples show the difficulty to find an example where construction of a physically realizable least mean squares estimator is feasible.
 \paragraph{Example 1.}
Consider an optical cavity of the form
\begin{align*}
dx(t)&=-\kappa/2\,x(t)\,dt-\sqrt{\kappa}dw(t)\\
dy(t)&=\sqrt{\kappa}\,x(t)\,dt+dw(t),
\end{align*}
where $dw(t)dw(t)^T=(I_{2\times 2}+iJ)\,dt$ and $\Theta=J.$ Therefore, we have $[dy(t),dy^T(t)]=2J\,dt,$ i.e., the output processes are non-commutative.

Take $P=\begin{pmatrix}p_1&p_2\\
p_2&p_4\end{pmatrix}.$ Note that for $\kappa> 0$ arbitrary,
we get the following Riccati equation
\begin{align*}
\kappa p_1-\kappa p_1^2-\kappa p_2^2&=0,\\
\kappa p_2-\kappa p_1 p_2-\kappa p_2 p_4&=0\\
-\kappa p_2^2+\kappa p_4-\kappa p_4^2&=0.
\end{align*}
This implies that $P=I_{2\times 2}$ and $K=0_{2\times 2}.$ Therefore, we get the following estimator
\begin{align}~\label{eq:llmc}
d\hat x(t)&=-\kappa/2\,\hat x(t)\,dt.
\end{align}
This estimator seems trivial as there is no $dy$ term in the dynamics of the estimator. So no information from the system is used to compute the estimate. As a consequence, it does not matter if $y$ be a commutative or non-commutative process, since $K=0_{2\times 2},$ and $dy,$ it does not appeared in  the dynamics of the estimator. However, note that the estimator is physically realizable if and only if $\kappa=0,$ since $\hat x$ is a process with the commutation $\Theta=J.$ Also, remark that $\kappa=0$ means that the system would be decoupled from the field. Particularly, the estimator~\eqref{eq:llmc} is not useful in practice, as there is no $dy$ term. Hence, there is no interest to make it physically realizable (when $\kappa\neq 0$) by adding some vacuum noises.
\paragraph{Example 2.}
Now consider a dynamic squeezer. This is an optical cavity with a non-linear element inside. After appropriate linearizations, an optical squeezer can be described by the
following QSDE (see e.g.,~\cite{gardiner2004quantum,roy2015coherent}) if we assume that $\chi=\chi_r+i\chi_i,$ and $\chi_r=0$,
\begin{align*}
dx&=\begin{pmatrix}
-\frac{1}{2}(\kappa_1+\kappa_2)&-\chi_i\\
-\chi_i&-\frac{1}{2}(\kappa_1+\kappa_2)
\end{pmatrix}xdt-\sqrt{\kappa_1}dw_1-\sqrt{\kappa_2}dw_2\\
dy&=\sqrt{\kappa_1}xdt+dw_1,
\end{align*}
where $dw_1(t)dw_1(t)^T=(I_{2\times 2}+iJ)\,dt,$ $dw_2(t)dw_2(t)^T=(I_{2\times 2}+iJ)\,dt,$ and $\Theta=J.$ We  have the following commutation relations for the output processes, $[dy(t),dy^T(t)]=2J\,dt.$

For any arbitrary parameters $\kappa_1\geq 0,$ $\kappa_2\geq 0,$ and $\chi_i,$ reals, the physical realizability constraint~\eqref{eq:prc} is satisfied if and only if
\begin{equation}~\label{eq:ex2p}
2 \kappa_1 - \kappa_2 - 2 \kappa_1 p_1 - \kappa_1 p_2^2 - 2 \kappa_1 p_4 + \kappa_1 p_1 p_4=0,
\end{equation}
with $P=\begin{pmatrix}
p_1&p_2\\
p_2&p_4
\end{pmatrix}.$ The matrix $P$ should satisfy the Riccati Equation~\eqref{eq:ric}, which becomes
\begin{align}~\label{eq:ex2ric}
\kappa_2 + (\kappa_1 - \kappa_2) p_1 - \kappa_1 p_1^2 - \kappa_1 p_2^2 - 2 p_2 \chi_i&=0\nonumber\\
(\kappa_1 - \kappa_2) p_2 - \kappa_1 p_1 p_2 - \kappa_1 p_2 p_4 - p_1 \chi_i - p_4 \chi_i&=0\nonumber\\
\kappa_2 - \kappa_1 p_2^2 + (\kappa_1 -  \kappa_2) p_4 - \kappa_1 p_4^2 - 2 p_2 \chi_i&=0.
\end{align}

 If we take $B'=0$ (with previous notation), i.e., $\kappa_1=0,$ the physically realizable constraint~\eqref{eq:ex2p} is satisfied if and only if $\kappa_2=0.$ This means that both field channels are decoupled from the system. Moreover, if we take $\kappa_1=\kappa_2=0,$ the Riccati Equation~\eqref{eq:ex2ric} has not a unique solution. Also, for $\kappa_2\geq 0$ and $\kappa_1>0,$ the physical realizability condition given in~\eqref{eq:ex2p} imposes a constraint on the form of $P.$ This shows the restrictiveness of physical realizability constraints.

Now take $\kappa_1=0.1,$ $\kappa_2=0.2,$ and $\chi_i=0.01.$ In this case, we find $P=\begin{pmatrix}
1.0030  & -0.0667\\
   -0.0667  &  1.0030
\end{pmatrix},$ and $K=\begin{pmatrix}
0.0009 &  -0.0211\\
   -0.0211   & 0.0009
\end{pmatrix}.$
Therefore, we get the following least mean squares estimator,
\begin{equation}~\label{eq:ex2}
d\hat x=\begin{pmatrix}
-0.1503 &  -0.0033\\
   -0.0033 &  -0.1503
\end{pmatrix}\hat x dt+\begin{pmatrix}
0.0009  & -0.0211\\
   -0.0211  &  0.0009
\end{pmatrix}dy
\end{equation}
which is not physically realizable.  We have $J(K)=\mathrm {Tr}(P)=2.0060.$

Obviously, we can make the least mean squares estimator~\eqref{eq:ex2} physically realizable by adding a vacuum noise as follows
\begin{equation*}
d\tilde x=\begin{pmatrix}
-0.1503 &  -0.0033\\
   -0.0033 &  -0.1503
\end{pmatrix}\tilde x dt+\begin{pmatrix}
0.0009  & -0.0211\\
   -0.0211  &  0.0009
\end{pmatrix}dy+\begin{pmatrix}
0.5486    &     0\\
         0  &  0.5486
         \end{pmatrix}dv,
\end{equation*}
with $dv(t) dv(t)^T=(I_{2\times 2}+iJ)dt.$ We find $\tilde P=\begin{pmatrix}
1.7955   &-0.0782\\
   -0.0782  &  1.7955
\end{pmatrix}.$ Therefore, $\tilde J(K)=\mathrm {Tr}(\tilde P)=3.5910.$ Remark that the form of the estimator in above is not unique. We recall that the study of different algorithms to design coherent observers is beyond the scope of this paper.
 \paragraph{Example 3.}
    Consider a degenerate parametric amplifier (DPA) described as follows (in the quadrature representation)
    \begin{align*}
dx&=\begin{pmatrix}
-\frac{1}{2}\kappa+\epsilon_r&\epsilon_i\\
\epsilon_i&-\frac{1}{2}\kappa-\epsilon_r
\end{pmatrix}xdt-\sqrt{\kappa}dw\\
dy&=\sqrt{\kappa}xdt+dw.
\end{align*}
Here, $\Theta=J,$ $dw(t)dw^T(t)=(I_{2\times 2}+iJ)dt,$ then, $[dy(t),dy^T(t)]=2J\,dt.$ Suppose $\kappa\geq 0,$ $\epsilon_r,$ and $\epsilon_i$ are reals. Also, take $P=\begin{pmatrix}
p_1&p_2\\
p_2&p_4
\end{pmatrix}.$ Then, $P$ satisfies the following Riccati equation obtained from Equation~\eqref{eq:ric},
\begin{align}~\label{eq:ex3r}
(2\epsilon_r + \kappa)p_1 - \kappa p_1^2 + 2\epsilon_i p_2 - \kappa p_2^2&=0,\nonumber\\
\epsilon_i p_1 +\kappa p_2  -
 \kappa p_1 p_2 + \epsilon_i p_4 -\kappa p_2 p_4&=0\nonumber\\
2\epsilon_i p_2 - \kappa p_2^2 + (\kappa - 2\epsilon_r ) p_4 - \kappa p_4^2&=0.
\end{align}
The physical realizability of the least mean squares estimator is satisfied if $\kappa=0,$ since the constraint~\eqref{eq:prc} becomes
$$2 \kappa - 2 \kappa p_1 - \kappa p_2^2 - 2 \kappa p_4 + \kappa p_1 p_4=0.$$
Once again, if $\kappa=0,$ the system would be decoupled from the field. Moreover, in this case, the Riccati equation~\eqref{eq:ex3r} has not a unique solution. Moreover, if $\kappa>0,$ the physical realizability condition in above imposes a constraint on the form of $P.$ This example also, illustrates the difficulty to find the least mean squares estimator which is physically realizable.

Now take the following parameters: $\kappa=0.1,$ $\epsilon_r=0.01,$ and $\epsilon_i=0.01.$ In this case, we get $P=\begin{pmatrix}
1.2000  &  0.2000\\
    0.2000 &   0.8000\end{pmatrix}$ and $K=\begin{pmatrix}
 0.0632&    0.0632\\
    0.0632&   -0.0632
\end{pmatrix}.$ The performance of the least mean squares estimator is given by $J(K)=\mathrm{Tr}(P)=2.$

The linear least mean squares estimator has the following form
\begin{equation*}
d\hat x=\begin{pmatrix}
-0.0600  & -0.0100\\
   -0.0100  & -0.0400
\end{pmatrix}\hat x dt+\begin{pmatrix}
 0.0632&    0.0632\\
    0.0632&   -0.0632\end{pmatrix}dy,
\end{equation*}
which is not  physically realizable. We can make this estimator physically realizable as follows
\begin{equation*}
d\tilde x=\begin{pmatrix}
-0.0600  & -0.0100\\
   -0.0100  & -0.0400
\end{pmatrix}\tilde x dt+\begin{pmatrix}
 0.0632&    0.0632\\
    0.0632&   -0.0632\end{pmatrix}dy+\begin{pmatrix}
   0.3286   &      0\\
         0  &  0.3286
\end{pmatrix}dv,
\end{equation*}
with $dv(t)dv(t)^T=(I_{2\times 2}+iJ)dt.$ We find $\tilde P=\begin{pmatrix}
1.8034  &  0.1431\\
    0.1431 &   1.5172\end{pmatrix}.$ Therefore $\tilde J(K)=\mathrm{Tr}(\tilde P)=3.3206.$
\paragraph{Example 4.} Consider the following plant
\begin{align*}
dx&=\begin{pmatrix}
0&\Delta\\
-\Delta&0
\end{pmatrix}x\,dt+\begin{pmatrix}
0&0&0&0\\
0&-2\sqrt{\kappa_2}&0&-2\sqrt{\kappa_3}
\end{pmatrix}\begin{pmatrix}
dw_1\\
dw_2
\end{pmatrix},\\
dy&=\begin{pmatrix}
2\sqrt{\kappa_2}&0\\
0&0
\end{pmatrix}x\,dt+dw_1,
\end{align*}
with $\Theta=J,$ $F_w=I_{4\times 4}+i\textrm{diag}_{2}(J),$ and $F_y=I_{2\times 2}+iJ.$ Then, $[dy(t),dy^T(t)]=2J\,dt,$ which means that the output processes are non-commutative. This plant may
be thought of as representing the scenario of an atom trapped
between two mirrors of a three mirror cavity in the strong coupling
limit in which the cavity dynamics are adiabatically eliminated (see more details in~\cite{doherty1999feedback,gough2007singular}).

Here, $B'=\begin{pmatrix}0&0\\
0&-2\sqrt{\kappa_2}\end{pmatrix}.$ It is trivial that the condition $B'JB'^T=0$ is satisfied. We can easily show that the physical realizability constraint (condition~\eqref{eq:prc}) is reduced to \begin{equation}~\label{eq:ex4pr}
8\kappa_2 p_1=0,
\end{equation} with $P=\begin{pmatrix}p_1&p_2\\
p_2&p_4\end{pmatrix}$ satisfying the Riccati Equation~\eqref{eq:ric}, which takes the following form
\begin{align}~\label{eq:ex4ric}
\kappa_2 - 4 \kappa_2 p_1^2 + 2 \Delta p_2&=0\nonumber\\
-\Delta p_1 - 4 \kappa_2 p_1 p_2 + \Delta p_4&=0\nonumber\\
\kappa_2 - 2 \Delta p_2 - 4 \kappa_2 p_2^2+4\kappa_3&=0.
\end{align}

 So if $\kappa_2=0,$ the physical realizability condition~\eqref{eq:ex4pr} is satisfied. This means that the system should be decoupled from the field  channel $dw_1$. However, if $\kappa_2=0,$ for $\Delta\neq 0,$ the Riccati equation~\eqref{eq:ex4ric} has no solution if $\kappa_3\neq 0.$ Moreover, the Riccati equation~\eqref{eq:ex4ric} has not a unique solution if $\kappa_2=\kappa_3=0.$ Also, if $p_1=0,$ the condition~\eqref{eq:ex4pr} is satisfied. However, it is not difficult to show that there is no positive definite solution to  the Riccati equation above in this case. This illustrates once again the restrictive nature of the physical realizability conditions.

Now take $\kappa_2=\kappa_3=0.1$ and $\Delta=0.01.$ The linear least mean squares estimator takes the following form
\begin{equation*}
d\hat x=\begin{pmatrix} -0.0883 &0.0100\\
 -0.4001&0
\end{pmatrix}\hat x dt+\begin{pmatrix}0.1397 &0\\0.6168  &-0.6325\end{pmatrix}dy.
\end{equation*}
This estimator is not physically realizable which is also conform with Corollary~\ref{coro:phys1}, since the condition~\eqref{eq:prc1m} is not satisfied in this example.

We can certainly add vacuum noise term $bdv$ (with $dv(t) dv(t)^T=(I_{2\times 2}+iJ)dt$) which is independent of $dw_1$  to the estimator above to make it physically realizable. Therefore, take the following estimator
\begin{equation*}
d\tilde x=\begin{pmatrix} -0.0883 &0.0100\\
 -0.4001&0
\end{pmatrix}\tilde x dt+\begin{pmatrix}0.1397 &0\\0.6168  &-0.6325\end{pmatrix}dy+\begin{pmatrix}
0.4204    &     0\\
         0  &  0.4204
\end{pmatrix}dv.
\end{equation*}
 Let us write the performance for estimators $\hat x$ and $\tilde x$ respectively as follows,
$P=\begin{pmatrix}
0.2208 &  0.9753\\
   0.9753  &  8.8359
\end{pmatrix},$ therefore, $J(K)=9.0567.$

For $\tilde x,$ we find $\tilde P=\begin{pmatrix}
0.7075&    1.1760\\
    1.1760 &  33.9878
    \end{pmatrix},$ then $\tilde J(K)=\tr{\tilde P}=34.6953.$
     \paragraph{Example 5.} Consider the following example which is borrowed from~\cite{wiseman1994all,gough2009series}
\begin{align*}
dx&=\gamma\begin{pmatrix}
-1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&-1-\cos(\theta)
\end{pmatrix}xdt+\sqrt{\gamma}\begin{pmatrix}
-1-\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&-1-\cos(\theta)
\end{pmatrix}dw\\
dy&=\sqrt{\gamma}\begin{pmatrix}
1+\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&1+\cos(\theta)
\end{pmatrix}xdt+dw,
\end{align*}
where $dw(t)dw^T(t)=(I_{2\times 2}+iJ)dt,$ $\Theta=J,$ and $[dy(t),dy^T(t)]=2J\,dt.$
This is a simple example of all-optical feedback scheme where the light from one end of a cavity is taken and reflect
it back into the other. For simplicity, it is assumed a bath in the vacuum state and a cavity with equal transmitivities at both end-mirrors.

Our aim is to see whether appropriate parameters $\theta$ and $\gamma$ exist such that the linear least mean squares estimator becomes automatically physically realizable. The  matrix $P=\begin{pmatrix}
p_1&p_2\\
p_2&p_4
\end{pmatrix}$ should satisfy Riccati Equation~\eqref{eq:ric}.
After some calculations, we get
\begin{align*}
2 (1 + \cos(\theta)) p_1+ 2 \sin(\theta) p_2 - (2 + 2 \cos(\theta)) (p_1^2 + p_2^2)& = 0, \\
  2 (1 + \cos(\theta)) p_2 + \sin(\theta) p_4 -p_1 \sin(\theta) - (2 + 2 \cos(\theta)) (p_1 p_2 + p_2 p_4) &= 0, \\
  2 (1 + \cos(\theta)) p_4 - 2 \sin(\theta) p_2 - (2 + 2 \cos(\theta)) (p_4^2 + p_2^2) &=0.
  \end{align*}
  Thus, we find $P=I_{2\times 2}$ and $K=0_{2\times 2}.$ Therefore, the linear least mean squares estimator has the following form
\begin{equation*}
 d\hat x=\gamma\begin{pmatrix}
-1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&-1-\cos(\theta)
\end{pmatrix}\hat xdt,
 \end{equation*}
which is not an interesting estimator in practice as there is no $dy$ term, similar to Example $1.$

Now we find $\theta$ and $\gamma$ such that the least mean squares estimator proposed by Theorem~\ref{thm:main} be physically realizable. To do so, we should solve the following equation which comes from the physical realizability condition given in Equation~\eqref{eq:prc}
\begin{equation*}
\gamma\begin{pmatrix}
0&-2-2\cos(\theta)\\
2+2\cos(\theta)&0
\end{pmatrix}=0_{2\times 2}.
\end{equation*}
This equation is satisfied if $\gamma=0$ or (and)  $\theta=k\pi$ (with $k$ an odd number).  When $\gamma=0,$ this means that the coupling to the field is zero. Also, $\theta$ corresponds to the phase of the vacuum light which is picked up when reflected by the cavity mirror. So, when $\theta=k\pi,$ with $k$ an odd number, this means that the damping through the mirrors can be completely eliminated (see more details in~\cite{wiseman1994all}). Obviously, for these cases, we have $A=B=C=0,$ (with previous notations)  which is meaningless.

  \medskip

We have observed in the examples above, constructing physically realizable least mean squares estimators was impossible when $B'\neq 0$ or we should consider some constraints on the matrix $P$ which makes the problem hard and sometimes impossible to solve. This shows the restrictiveness of the physical realizability constraints. Also, when $B'=0,$ the physically realizable least mean squares estimators are not well defined. Supported by these examples and some others which are not given in this paper, we conclude that maybe it is impossible to find examples which could result in physically realizable least mean squares estimators without any additional quantum noises when $B'\neq 0$. (Note that the case $B'=0$ is not an interesting case, since it could also be realized with Homodyne or Hetrodyne detection, as mentioned before, below Corollary 6.) However, we couldn't show this in general case, maybe it is wrong. Also, note that finding examples is a hard problem since we should solve the quadratic equations in $P$ (Equation~\eqref{eq:ric}) where we obtain $P$ as a function of free parameters of the matrix $A,$ and $B.$ Then, these free parameters could  be determined by replacing $P$ in the physical realizability constraints (Equation~\eqref{eq:prc}).
\section{Conclusion}~\label{sec:six}
We have obtained non-commutative linear least mean squares estimators for linear QSDEs by extending Belavkin-Kalman filters to the case where the output processes are non-commutative.  We have assumed that these least mean squares estimators are given as a linear combination of innovation processes. Furthermore, we studied the physical realizability of such estimators for the general case and some special cases.

We have observed that when $B'=0,$ it is more simple to construct a physically realizable least mean squares estimator, specially  for $\Theta$ degenerate canonical and when $C^T\textrm{diag}_{\frac{n_y}{2}}(J)C=0$. Since, in this case, the physical realizability condition does not depend on the form of $P$ (see more details in Corollary~\ref{coro:case2}). However, roughly speaking, for this case, the non-commutative filter could also be realized by Homodyne or Hetrodyne detection as $C\Theta=0.$ In general, finding examples which satisfy physical realizability conditions, it is difficult without any assumptions on $P.$ These assumptions create constraints on their associated Riccati equations (see e.g., Theorem~\ref{thm:phys} and Corollaries~\ref{coro:phys1}-~\ref{coro:case2}). Moreover, based on our observations, we can conclude that maybe, the construction of a physically realizable least mean squares estimator without any additional quantum noises is impossible when $B'\neq 0$. Generally speaking, the results presented here show the restrictive nature of physical realizability conditions.

Indeed, this work does not show that the best estimate based on the knowledge of the non-commutative output processes, and under the constraints of the physical realizability, has the form of the proposed linear estimator~\eqref{eq:qkalman}. Further research is required to solve the optimal filtering problem under the non-convex constraints imposed by physical realizability conditions. Furthermore, the optimal filtering problem when the coherent controllers are added into the plant's dynamics (see e.g.,~\cite{NJP09,hamerly2013coherent}) can be considered as a future research plan.

\vspace{0.4cm}

\noindent\bf{Acknowledgments.} \rm The authors gratefully acknowledge Professor T. Duncan for helpful discussions. This work was supported by the Australian Research Council Centre of Excellence for Quantum Computation and Communication Technology (Project No. CE110001027) and Air Force Office of Scientific Research (AFOSR) under Grant No. FA2386-12-1-4075.
The first author would like to thank Ryan Hamerly and Nikolas Tezak for valuable discussions. Nina H. Amini acknowledges the support of Math+X Postdoctoral Fellowship from the
Simons Foundation.

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TITLE: Convergence of the recursive sequence $x_n = x_{n-1} - \frac12\lambda(x_{n-1}-1)^{-3/2}$
QUESTION [0 upvotes]: Let $\lambda \geq 0$ and define the recursive sequence
$$
x_n = x_{n-1} - \frac{\lambda}{2(x_{n-1}-1)^{\frac{3}{2}}}
$$
I know that this sequence is bounded below by 1 and I also need to determine a starting point $x_0$ such that the serie converges (maybe expressed as a function of $\lambda$).
I start setting up the problem, by using the inequality
$$
1 \leq x - \frac{\lambda}{2(x-1)^{\frac{3}{2}}} \leq x
$$
However, I have absolutely no idea about how to handle such inequality.
Thanks to anybody willing to help! :)

REPLY [1 votes]: Hint
as we have
$$x_{n}=x_{n-1}-\frac{\lambda}{2(x_{n-1}-1)^\frac32}$$
if $(x_n)$ converges to $L$ then
we will have
$$L=L-\frac{\lambda}{2(L-1)^\frac32}$$
impossible if $\lambda>0$ and $(x_n)$ diverges.
if $\; \lambda=0\;  \;x_n=x_0=C^t$


TITLE: will spin increase the forward momentum of an object being hit in the centre of a spinning object?
QUESTION [1 upvotes]: if you have scenario
A)a flat square or flat circular object, and it moves in a straight line, towards an object e.g. a ball.
vs
B)A flat square or flat circular object, moves in a straight line towards an object e.g. a ball (so same so far), but also, the flat object is spinning in the Y axis too, and the centre of the spinning flat object hits the ball. So the ball still goes straight.
Will the ball go faster or slower or same speed, in A, or B?
here is a 2min video that might help demonstrate what I mean
https://www.youtube.com/watch?v=aOykv3GKLSM

REPLY [1 votes]: The ball will rebound with the same speed in both cases. The relative velocity between the ball and plate at the point of contact is the same. Unlike the situation in which the ball is struck by the advancing or receding side of the spinning plate. Then the relative velocity is different.
Another way of looking at this is from the reference frame in which the plate is still spinning but the centre of the plate (or more generally, the point at which contact is made) is stationary. It is now the ball which moves towards the centre of the plate. Provided that the ball hits the centre of the plate - which is stationary in both cases - it will rebound with the same speed whether the plate is stationary or spinning.
If the collision takes place when the plate is tilted, the ball will not rebound in the direction from which it came, but the speed will still be the same.
I am assuming that the collision time is very small compared with the time it takes for the plate to make 1 revolution.


TITLE: Lipschitz choice of "norm attaining maps'
QUESTION [3 upvotes]: I am not very sure if the following problem has been treated in the literature and if so, whether it always holds:

A Banach space $X$ is isomorphic to a Hilbert space if the 'norm attaining' map $F$ from $S_{X^*}$ into $S_X$ satisfying $x^*(F(x^*)) = 1$ is a bilipschitz equivalence. Here $S_Z$ denotes the unit sphere of the Banach space $Z$.

REPLY [1 votes]: I think this will be correct, if I interpret your assumption correctly.
The assumption implies that the duality mapping on $X$ with gauge $t\mapsto t^2$, $$j_X(x) := \bigl\{ x^* \in X^* \colon x^*(x) = \|x^*\|\,\|x\|,~\|x^*\| = \|x\|\bigr\},$$ is bijective and Lipschitz continuous via $j_X(x) = \|x\|\, F^{-1}(x/\|x)$. Hence $X$ is $2$-smooth [Remark 5, XR]), so of type $2$ [Prop. IV.5.10, DGZ]. In particular, it is reflexive and $F$ induces the duality mapping $j_{X^*}$ on $X^*$ in the same way as $F^{-1}$ did on $X$. Analogously, $X^*$ is $2$-smooth, and this implies that $X$ is of cotype $2$ [Prop. IV.5.12, DGZ]. Finally, a Banach space which is both of type and cotype $2$ is isomorphic to a Hilbert space by Kwapien's theorem.
[XR] Xu, Zong-Ben; Roach, G. F., Characteristic inequalities of uniformly convex and uniformly smooth Banach spaces, J. Math. Anal. Appl. 157, No. 1, 189-210 (1991)
[DGZ] Deville, Robert; Godefroy, Gilles; Zizler, Václav, Smoothness and renormings in Banach spaces, Pitman Monographs and Surveys in Pure and Applied Mathematics. 64. Harlow: Longman Scientific & Technical. New York: John Wiley & Sons, Inc.. 376 p. (1993).