text:
\section{Introduction}
Our main goal is to prove the following theorem
\begin{theorem}[Main result]\label{maint}\ref{theorem1}
There exists a commutative ring $R$ and a thick subcategory $\mathsf{C}$ of $\mathsf{Perf}(R)$ such that the space $\mathrm{K}(\mathsf{C})$ of Waldhausen $K$-theory is equivalent to an Eilenberg-MacLane space.
\end{theorem}
In our opinion this theorem seems counterintuitive at the first glance. There is very few examples of rings for which the algebraic $K$-theory groups were computed in all degrees (e.g. the $K$-theory of finite fields computed by Quillen). Another source for such computation is the Farrell-Jones conjecture. We will compute explicitly the $K$-groups for some particular (commutative) group ring \ref{F-J}.
\begin{conjecture}[Classical Farrell-Jones \cite{luck2010k}]\label{1}
For any regular ring $k$ and any torsionfree group $G$, the assembly map
$$\mathrm{H}_{n}(\mathrm{B}G; \mathbf{K}(k))\longrightarrow \mathrm{K}_{n}(k[G]) $$
is an isomorphism for any $n\in \mathds{Z}$.
\end{conjecture}
We refer to \cite{waldhausen1985algebraic} for the definition of the $K$-theory spectrum $\mathbf{K}(k)$ of a ring $k.$ We recall that $\mathrm{B}G$ is the classifying space of the group $G$ and that $k[G]$ is the associated group ring with a natural augmentation $k[G]\rightarrow k$.
We recall also that $\mathrm{H}_{n}(\mathrm{B}G; \mathbf{K}(k))$ is the same thing as the $n$-th stable homotopy group of the spectrum $\mathrm{B}G_{+}\wedge \mathbf{K}(k)$. More precisely the assembly map is induced by the following map of spectra
$$\mathrm{B}G_{+}\wedge \mathbf{K}(k)\rightarrow \mathbf{K}(k[G]). $$
The conjecture \ref{1} admits a positive answer in the case where $k$ is regular ring and $G$ is a torsionfree abelian group, it is a particular case of the main result of \cite{wegner2015farrell}.

\section{Fibre sequence for Waldhausen K-theory }
\begin{notation}\label{notaion} We fix the following notations:
\begin{enumerate}
\item Let $\mathcal{E}$ be any (differential graded) ring. Let $ \mathsf{Mod}_{\mathcal{E}}$ denotes the (differential graded) model category of $\mathcal{E}$-complexes \cite{hovey2007model}. And $\mathsf{Perf}(\mathcal{E})$ denotes the (differential graded) category of perfect (i.e. compact) $\mathcal{E}$-complexes.
\item For any (differential graded) ring map $\mathcal{E}\rightarrow \mathcal{A}$, $\mathsf{Perf}(\mathcal{E},\mathcal{A})$ denotes the thick subcategory of $\mathsf{Perf}(\mathcal{E})$ such that $M\in \mathsf{Perf}(\mathcal{E},\mathcal{A})$ if and only if $M\otimes^{\mathbb{L}}_{\mathcal{E}}\mathcal{A}\simeq 0$ i.e.  $M\otimes^{\mathbb{L}}_{\mathcal{E}}\mathcal{A}$ is quasi-isomorphic to $0$. By the symbol $\otimes^{\mathbb{L}}_{\mathcal{E}}$ we do mean the derived tensor product over $\mathcal{E}$.
\end{enumerate}
\end{notation}
\begin{lemma}\label{main}
Let $\mathcal{E}\rightarrow \mathcal{A}$ be a morphism of (differential graded) rings such that $\mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}}\mathcal{A}\simeq \mathcal{A}$, then
$$ \mathrm{K}(\mathcal{E},\mathcal{A})\rightarrow \mathrm{K}(\mathcal{E})\rightarrow  \mathrm{K}(\mathcal{A})$$
is a fibre sequence of (infinite loop) spaces where $\mathrm{K}(\mathcal{E},\mathcal{A}):= \mathrm{K}(\mathsf{Perf}(\mathcal{E},\mathcal{A}))$.
\end{lemma}
\begin{proof}
Let $\mathbf{w}$ be the class of equivalences in $ \mathsf{Mod}_{\mathcal{E}}$ defined as follows:  a map $P\rightarrow P^{'} $ is $\mathbf{w}$-equivalence if and only if $\mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}}P\rightarrow \mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}}P^{'} $ is a quasi-isomorphism ($\mathbf{q.i.}$).

The left Bousfield localization \cite{hirschhorn2009model} of the model category $\mathsf{Mod}_{\mathcal{E}}$ with respect to the class $\mathbf{w}$ exists and it is denoted by $\mathrm{L}_{\mathbf{w}} \mathsf{Mod}_{\mathcal{E}}$. Since $\mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}}\mathcal{A}\simeq \mathcal{A}$ we obtain a Quillen equivalence
$$\xymatrix{\mathrm{L}_{\mathbf{w}}\mathsf{Mod}_{\mathcal{E}}\ar@<2pt>[r]^-{\mathcal{A}\otimes_{\mathcal{E}}-} & \mathsf{Mod}_{\mathcal{A}}\ar@<2pt>[l]^-{U}} $$
More precisely, for any $M\in \mathsf{Mod}_{\mathcal{A}}$ the (derived) counit map $ \mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}} U(M)\rightarrow M $
is a quasi-isomorphism (because it is a quasi-isomorphism for $\mathcal{A}=M$, the functor $\mathcal{A}\otimes_{\mathcal{E}}^{\mathbb{L}}-$ commutes with homotopy colimits and $\mathcal{A}$ is a generator for the homotopy category of $\mathsf{Mod}_{\mathcal{A}}$). On another hand, the derived unit map $P\rightarrow \mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}} U(P)$ is an equivalence in $\mathrm{L}_{\mathbf{w}}\mathsf{Mod}_{\mathcal{E}}$ for any $P\in \mathsf{Mod}_{\mathcal{E}}$ by definition. In particular the subcategory of compact objects in $\mathrm{L}_{\mathbf{w}}\mathsf{Mod}_{\mathcal{E}}$ is equivalent to $\mathsf{Perf}(\mathcal{A}$). Thus, by \cite[theorem 3.3]{sagave2004algebraic}, we have an equivalence of the $K$-theory spaces
$$\mathrm{K}((\mathsf{Perf}(\mathcal{E}),\mathbf{w}))\simeq \mathrm{K}((\mathsf{Perf}(\mathcal{A}),\mathbf{q.i.})):= \mathrm{K}(\mathcal{A}).$$
By Waldhausen fundamental theorem \cite[Theorem 1.6.4]{waldhausen1985algebraic}, the sequence of Waldhausen categories
$$ (\mathsf{Perf}(\mathcal{E})^{\mathbf{w}},\mathbf{q.i.})\rightarrow (\mathsf{Perf}(\mathcal{E}),\mathbf{q.i.})\rightarrow (\mathsf{Perf}(\mathcal{E}),\mathbf{w}) $$
induces a fibre sequence of $K$-theory spaces
$$\mathrm{K}((\mathsf{Perf}(\mathcal{E})^{\mathbf{w}},\mathbf{q.i.}))\rightarrow \mathrm{K}(\mathcal{E})\rightarrow  \mathrm{K} (\mathcal{A})$$
 where $\mathsf{Perf}(\mathcal{E})^{\mathbf{w}}$ is the full subcategory of $\mathsf{Perf}(\mathcal{E})$ such that $E\in \mathsf{Perf}(\mathcal{E})^{\mathbf{w}}$ if and only if $\mathcal{A}\otimes^{\mathbb{L}}_{\mathcal{E}}E\simeq 0$. It is obvious by definition that  $\mathsf{Perf}(\mathcal{E})^{\mathbf{w}}=\mathsf{Perf}(\mathcal{E},\mathcal{A})$.\\
Hence $$\mathrm{K}(\mathcal{E},\mathcal{A})\rightarrow \mathrm{K}(\mathcal{E})\rightarrow \mathrm{K}(\mathcal{A})$$ is a homotopy fibre sequence of spaces.
\end{proof}
A similar result can be found in \cite[Theorem 0.5]{neeman2004noncommutative} and in \cite[Lemma 5.1]{chen2012recollements}.

\section{Farrell-Jones conjecture}

\begin{notation} We fix the following notations:
\begin{enumerate}
\item $k=\mathds{F}_{2}$ is the finite field with two elements.
\item $R$ is the group algebra $k[\mathds{Q}]$, where $\mathds{Q}$ is the additive abelian group of rational numbers.
\end{enumerate}
\end{notation}
\begin{proposition}\label{prop}
If $\mathds{V}$ is a rational vector space and $A$ is a finite abelian group then
\[\mathrm{H}_{\ast}(\mathrm{B}\mathds{V}; \mathds{Z})=
\begin{cases}
\mathds{Z}& \quad \textrm{ if } n= 0\\
\mathds{V} & \quad \textrm{ if } n= 1\\
0 & \quad \textrm{ else } \\
\end{cases}\]
and
\[\mathrm{H}_{\ast}(\mathrm{B}\mathds{V}; A)=
\begin{cases}
A & \quad \textrm{ if } n= 0\\
0 & \quad \textrm{ else } \\
\end{cases}\]

\end{proposition}
\begin{lemma}\label{F-J}

\[\pi_{n}\mathrm{K}(R):=\mathrm{K}_{n}(R)=
\begin{cases}
\mathrm{K}_{n}(k) & \quad \textrm{ if } n\neq 1\\
 \mathds{Q} & \quad \textrm{ if } n=1 \\
\end{cases}\]
\end{lemma}
\begin{proof}
By Quillen theorem \cite{quillen1972cohomology}, the algebraic $K$-theory of the finite field $k$ is given by
\[\mathrm{K}_{n}(k)=
\begin{cases}
\mathds{Z} & \quad \text{ if } n=0 \\
0 & \quad \text{ if } n \text{ even } >0 \\
 \mathds{Z}/(2^{j}-1) & \quad \text{ if } n=2j-1 \text{ and } j>0 \\
\end{cases}
\]
Since $\mathds{Q}$ is a rational vector space and $\mathrm{K}_{n}(k)$ are finite abelian groups (for $n>0$) then by proposition \ref{prop} we have that
\[\mathrm{H}_{p}(\mathrm{B}\mathds{Q}; \mathrm{K}_{q}(k))=
\begin{cases}
\mathds{Q} & \quad \text{ if } p=1 \text{ and } q=0\\
\mathrm{K}_{q}(k) & \quad \text{ if } p=0 \text{ and } q\geq 0 \\
0 & \quad \text{ else }\\
\end{cases}
\]
The second page $E^{2}_{p,q}=\mathrm{H}_{p}(\mathrm{B}\mathds{Q}; \mathrm{K}_{q}(k))$ of the converging Atiyah-Hirzebruch spectral sequence \cite{luck2010k}
$$\mathrm{H}_{p}(\mathrm{B}\mathds{Q}; \mathrm{K}_{q}(k))  \Longrightarrow\mathrm{H}_{p+q}(\mathrm{B}\mathds{Q}; \mathbf{K}(k))$$
has graphically the following shape:\\
\begin{center}
\begin{tikzpicture}
  \matrix (m) [matrix of math nodes,
    nodes in empty cells,nodes={minimum width=5ex,
    minimum height=5ex,outer sep=-5pt},
    column sep=1ex,row sep=1ex]{
    \vdots     &  \vdots  &      0        &   0   &   0         &\dots &  0  &\dots\\
    q     &  |[draw,red,circle]|\mathrm{K}_{q}(k)  & 0             &   0   &   0         &\dots &  0  &\dots\\
          \vdots     &  \vdots  & \vdots             & \vdots  & \vdots              &  \cdots & \vdots &\dots  \\
          5     &  |[draw,red,circle]|\mathds{Z}/(7) &  0          &   0 &   0   & \dots & 0  &\dots\\
          4     &  0 &  0          &   0 &   0   & \dots & 0  &\dots\\
          3     &  |[draw,red,circle]|\mathds{Z}/(3) &  0          &   0 &   0   & \dots & 0  &\dots\\
          2     &  0 &  0          &  |[draw,circle]|0  &   0   & \dots &0  &\dots\\
          1     &  0 &  0         &   0   &   0 & \dots &0  &\dots\\
          0     & |[draw,red,circle]| \mathds{Z} & |[draw,red,circle]|\mathds{Q} &  0   &   0 & \dots &0  &\dots\\
    \quad\strut &   0  &  1  &  2  & 3 &    \dots & p &\dots\strut\\ };
\draw[ultra thick] (m-1-1.north east) -- (m-10-1.east) ;
\draw[ultra thick] (m-10-1.north) -- (m-10-8.north) ;
\draw[blue,ultra thick,->] (m-7-4.south) -- (m-6-2.north);
\end{tikzpicture}\\
\end{center}
where the differentials $d^{2}: E_{p,q}^{2}\rightarrow E_{p-2,q+1}^{2}$ are obviously identical to $0$. It means that the spectral sequence collapses, hence in our particular case it implies that
$$ \mathrm{H}_{p}(\mathrm{B}\mathds{Q}; \mathrm{K}_{q}(k))= \mathrm{H}_{p+q}(\mathrm{B}\mathds{Q}; \mathbf{K}(k)).$$
Since the Farrell-Jones conjecture is true in the case of torsionfree abelian groups \cite{wegner2015farrell}, we obtain that
\[\mathrm{K}_{n}(R)\cong\mathrm{H}_{n}(\mathrm{B}\mathds{Q}; \mathbf{K}(k))=
\begin{cases}
    \mathrm{K}_{n}(k)  & \quad \text{if } n\neq 1 \\
    \mathds{Q} & \quad \text{ if } n=1\\
   \end{cases}
\]
\end{proof}

\begin{lemma}\label{lemma}
There is a fibre sequence of Waldhausen $K$-theory spaces given by
$$ \mathrm{K}(R,k)\rightarrow \mathrm{K}(R)\rightarrow  \mathrm{K}(k)$$
\end{lemma}
\begin{proof}
Since $k$ is a finite field (in particular a finite abelian group) and $\mathds{Q}$ is a rational vector space, it follows by \ref{prop} that

\[\mathrm{H}_{n}(\mathrm{B}\mathds{Q};k)=\mathrm{Tor}^{R}_{n}(k,k)=\begin{cases}
    k & \quad \text{if } n=0 \\
    0 & \quad \text{ else } \\
   \end{cases}
\]
therefore $k\otimes_{R}^{\mathbb{L}}k\simeq k$. The conclusion follows from lemma \ref{main} when $k=\mathcal{A}$ and $R=\mathcal{E}$.
\end{proof}

\begin{theorem}\label{theorem1}
With the same notation, the $K$-theory space of the thick subcategory $\mathsf{Perf}(R,k)$ is equivalent to the Eilenberg-MacLane space $\mathrm{B}\mathds{Q}$.
\end{theorem}
\begin{proof}

Since the Farrell-Jones conjecture is true for $G=\mathds{Q}$. Combining lemma \ref{lemma} and lemma \ref{F-J}, we have by Serre's long exact sequence that the homotopy groups of the homotopy fibre $\mathrm{K}(R,k)$ of
$\mathrm{K}(R)\rightarrow \mathrm{K}(k)$ are given by
 \[\mathrm{K}_{n}(R,k)=\begin{cases}
    \mathds{Q} & \quad \text{if } n=1 \\
    0 & \quad \text{ else } \\
   \end{cases}
\]
and by definition $\mathrm{K}(R,k):=\mathrm{K}(\mathsf{Perf}(R,k))$, hence we have proved the main theorem \ref{maint}.
\end{proof}

 \bibliographystyle{plain}


text:
\section{Introduction}
\vspace{-5pt}
Recently, hybrid beamfomring (HB) technique has demonstrated its great potentials in improving the secrecy performance of mmWave systems, where an eavesdropper (Eve) attempts to obtain the confidential information from legitimate transmitter (Alice) to legitimate receiver (Bob) \cite{ChenGLOBECOM2019,YangCL2021,WangAccess2019}.
To protect the legitimate transmission, Alice employs HB to send both the information signal (IS) and artificial noise (AN), in order to enhance the reception of Bob and suppress the reception of Eve simultaneously by utilizing the directionality and antenna gain offered by the HB.
However, such an idea cannot work properly in all possible scenarios, especially when the legitimate and eavesdropping links are highly correlated.

To overcome this issue, intelligent reflecting surface (IRS) has been introduced in the secure beamforming design of mmWave systems \cite{XiuCL2021,LuWCL2020,XiuAccess2020,QiaoWCL2020}, where the passive beamforming (PB) is employed by IRS to provide additional spatial degree-of-freedom and reconstruct wireless environment.
By integrating the full-digital beamforming (FDB) with the PB of IRS, several alternating optimization algorithms were proposed in \cite{XiuAccess2020,LuWCL2020,XiuCL2021} to maximize the secrecy capacity of mmWave systems. Since each antenna requires a dedicated radio frequency (RF) chain in FDB, the hardware complexity may be unacceptable if massive antennas are employed by Alice.
To reduce the hardware complexity of Alice, the work in \cite{QiaoWCL2020} proposed two secrecy capacity oriented alternating algorithms for downlink mmWave systems, where Bob employs only a single antenna.

In mmWave systems, the millimeter-level signal wavelength facilitates deploying multiple antennas at both Alice and Bob. Thus, it is practically meaningful to jointly design the HB and the PB for enhancing the security of IRS-assisted multiple-input-multiple-output (MIMO) mmWave systems.
However, such an issue remains unexplored so far, to the best of authors' knowledge. On the other hand, the work in \cite{QiaoWCL2020}  cannot be extended to IRS-assisted MIMO mmWave systems, since its proposed algorithms rely on the single antenna assumption for Bob.

Motivated by the above facts, we investigate the secure beamforming design for an IRS-assisted mmWave MIMO system.
To maximize the secrecy capacity, the joint optimization of HB and PB is formulated as a non-convex problem with constant-modulus constraints. To efficiently solve such a challenging problem, the original problem is decomposed into an orthogonal forcing passive beamforming subproblem and a null-space jamming hybrid beamforming subproblem. Then these two subproblems are sequentially solved by the proposed convex approximation assisted alternating direction method of multipliers (CA-ADMM) algorithm and block coordinate descent aided orthogonal matching pursuit (BCD-OMP) algorithm. Simulation results demonstrate the superior performance of proposed algorithms.

\section{System Description}\label{Sec:system}

In the considered secure mmWave system, an IRS is deployed to help the legitimate transmission from Alice to Bob under the eavesdropping of Eve. The IRS consists of $N_{\rm I}$ low-cost reflecting elements, while Alice, Bob and Eve are equipped with $N_{\rm A}$, $N_{\rm B}$ and $N_{\rm E}$ antennas, respectively. Due to the use of mmWave signals, we consider that Alice employs $N_{\rm RF}(\ll N_{\rm A})$ RF chains to realize an HB architecture. To suppress the eavesdropping of Eve, Alice sends both IS ${\bf s}\in\mathcal{C}^{L_s\times 1}$ and AN ${\bf z}\in\mathcal{C}^{L_z\times 1}$ simultaneously  with using HB, where $L_{s}$ and $L_{z}$ represent the length of data streams for the IS and the AN, respectively. Therefore, the transmitted signal of Alice can be expressed as
\begin{align}\label{eqn:tran_A}
{\bf x}_{\rm A}&={\bf F}({\bf W}_{s}{\bf s}+{\bf W}_{z}{\bf z}),
\end{align}
where $\mathbb{E}[{\bf ss}^H]={\bf I}_{L_s}$, $\mathbb{E}[{\bf zz}^H]={\bf I}_{L_z}$, ${\bf F}$ is the analog beamformer, ${\bf W}_{s}$ and ${\bf W}_z$ represent the digital beamformers for the IS and the AN, respectively.

Let us denote the channel matrix between nodes $i$ and $j$ as ${\bf H}_{ij}$, where $i,j\in\{\rm{Alice(A)},\rm{Bob(B)},\rm{Eve(E)},\rm{IRS(I)}\}$ and $i\neq j$. Thus, the equivalent channel from Alice to node $i(\in\{\mathrm{B},\mathrm{E}\})$ via the IRS can be defined as
${\bf H}_{{\rm eq},i}({\bf \Theta})\triangleq{\bf H}_{{\rm A}i}+{\bf H}_{{\rm I}i}{\bf \Theta}{\bf H}_{\rm{AI}}$, where ${\bf\Theta}={\rm diag}(e^{j\theta_1},e^{j\theta_2},\cdots,e^{j\theta_{N_{\rm I}}})$ is the reflection matrix of the IRS.
Accordingly, the received signals of node $i(\in\{\mathrm{B},\mathrm{E}\})$ can be written as
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{equation}
{\bf y}_{i
={\bf H}_{\mathrm{eq},i}({\bf \Theta}){\bf x}_{\rm A}+{\bf n}_{i},\label{eqn:rec_B}
\end{equation}}
where ${\bf n}_{i}\sim \mathcal{CN}({\bf{0}}, \sigma^2{\bf I})$ represent the vector of zero-mean additive white Gaussian noise (AWGN) with variance $\sigma^2$.
Then, by the received signal shown in \eqref{eqn:rec_B}, the secrecy capacity can be expressed as
\begin{align}\label{eqn:secap}
&C_{\rm{sec}}({\bf W}_s, {\bf W}_z,{\bf F},{\bf \Theta})=\big\{\log{\rm det} \left[{\bf I}_{N_{\rm B}}+{\bf S}_{\rm B}({\bf \Theta}){\bf C}_{\rm B}^{-1}({\bf \Theta}) \right]\nonumber\\
&\hspace{20mm}-\log{\rm det} \left[{\bf I}_{N_{\rm E}}+{\bf S}_{\rm E}({\bf \Theta}){\bf C}_{\rm E}^{-1}({\bf \Theta}) \right]\big\}^+,
\end{align}
where ${\bf S}_i({\bf \Theta})\triangleq{\bf H}_{{\rm eq},i}({\bf \Theta}){\bf F}{\bf W}_{s}{\bf W}_{s}^H{\bf F}^H{\bf H}_{{\rm eq},i}^H({\bf \Theta})$, ${\bf C}_i({\bf \Theta})\triangleq{\bf H}_{{\rm eq},i}({\bf \Theta}){\bf F}{\bf W}_{z}{\bf W}_{z}^H{\bf F}^H{\bf H}_{{\rm eq},i}^H({\bf \Theta})+\sigma^2{\bf I}_{N_i}$, for $i(\in\{\mathrm{B},\mathrm{E}\})$.

\section{Secrecy Capacity Maximization Problem}\label{Sec:problem}

To maximize the secrecy capacity defined in \eqref{eqn:secap}, the joint optimization of PB and HB can be formulated as follows
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{subequations}\label{eqn:know_problem}
\begin{align}
(\mathcal{P}_0):\hspace{1mm}\max_{{\bf W}_{s},{\bf W}_{z},{\bf F},{\bf \Theta}}&\hspace{1mm}C_{\rm{sec}}({\bf W}_{s},{\bf W}_{z},{\bf F},{\bf \Theta})
\label{eqn:objective_function}\\
{\rm s.t.}\hspace{6.5mm}
&\hspace{1mm}\|{\bf F}{\bf W}_{s}\|_{\rm F}^2+\|{\bf F}{\bf W}_{z}\|_{\rm F}^2\leq P_{\max},\label{eqn:powcons}\\
&\hspace{1mm}|{\bf F}|=\frac{1}{\sqrt{N_{\rm A}}}{\bf 1},\label{eqn:analogcon}\\
&\hspace{1mm}{\bf\Theta}={\rm diag}(e^{j\theta_1},e^{j\theta_2},\cdots,e^{j\theta_{N_{\rm I}}}),\label{eqn:IRScm1}\\
&\hspace{1mm}\theta_n\in [0, 2\pi),\hspace{2mm} n=1,\cdots,N_{\rm I}.\label{eqn:IRScm2}
\end{align}
\end{subequations}}
In the above problem, constraints \eqref{eqn:powcons}$\sim$\eqref{eqn:IRScm2} are explained as follows: \eqref{eqn:powcons} represents the transmit power constraint, where $P_{\max}$ is the maximum transmit power; \eqref{eqn:analogcon} means the constant-modulus constraint for analog beamformer at Alice; \eqref{eqn:IRScm1} and \eqref{eqn:IRScm2} represent the continuous diagonal constant-modules constraint for PB. Note that the PB and HB are highly coupled in objective function \eqref{eqn:objective_function}.
To decouple the formulated problem, we have the assumption that the optimal IS hybrid beamforming ${\bf W}^{\star}({\bf \Theta}){\bf F}^{\star}({\bf \Theta})\approx a_1{\bf V}_{\rm B}({\bf \Theta})+a_2{\bf V}_{\rm B}^{\perp}({\bf \Theta})$, where ${\bf V}_{\rm B}$ is the matrix of the strongest $L_s$ right singular vectors of equivalent channel ${\bf H}_{\rm{eq,B}}({\bf \Theta})$ and ${\bf V}_{\rm B}^{\perp}(\bf \Theta)$ belongs to the null space of ${\bf H}_{\rm{eq,B}}({\bf \Theta})$, and $a_1+a_2=\frac{P_{\max}}{L_s}$. With the above assumption, we have the following lemma.
\begin{lemm}\label{lem:lowboundCsec}
The optimal objective of problem $({\mathcal{P}}_0)$ is no less than the optimal objective of the following lower-bound problem at high signal-to-noise ratio (SNR):
\begin{align}
({\mathcal P}_1):\hspace{4mm}&\max_{\bf \Theta}\hspace{1mm}{\hat C}_{\rm{sec}}({\bf \Theta})\hspace{3mm}{\rm s.t.}\hspace{3mm} \eqref{eqn:IRScm1}, \eqref{eqn:IRScm2}.\nonumber
\end{align}
where $\hat{C}_{\rm{sec}}({\bf \Theta})\triangleq\log(1+\frac{P_{\max}}{\sigma^2L_s}\|{\bf H}_{\rm{eq,B}}({\bf \Theta})\|_{\rm F}^2)-\log{\rm det}$ $({\bf I}+\frac{P_{\max}}{\sigma^2L_s}{\bf H}_{\rm{eq,B}}({\bf \Theta}){\bf H}_{\rm{eq,E}}^H({\bf \Theta}){\bf H}_{\rm{eq,E}}({\bf \Theta}){\bf H}_{\rm{eq,B}}^H({\bf \Theta}))$.
\end{lemm}
{\bf Proof:} Here, an sub-optimal HB for IS can be expressed as
\begin{equation}
{\bf F}^{\dagger}({\bf \Theta}){\bf W}_{s}^{\dagger}({\bf \Theta})\approx \sqrt{P_{\max}\psi^{\star}({\bf\Theta})/L_s}{\bf V}_{\rm B}({\bf \Theta}).
\end{equation}
Replacing ${\bf F}^{\star}({\bf \Theta}){\bf W}_{s}^{\star}({\bf \Theta})$ by ${\bf F}^{\dagger}({\bf \Theta}){\bf W}_{s}^{\dagger}({\bf \Theta})$, we have
\begin{IEEEeqnarray}{rcl}\label{eqn:lowbound2}
C_{\rm{sec}}({\bf \Theta})&\geq& \log{\rm det} [{\bf I}_{L_s}+P_{\max}\psi^{\star}({\bf\Theta}){\bf \Sigma}_{\rm B}^2({\bf \Theta})/(\sigma_{\rm B}^2L_s)]\nonumber\\
&&-\log{\rm det}[{\bf I}_{L_s}+P_{\max}\psi^{\star}({\bf\Theta}){\bf V}_{\rm B}^H({\bf \Theta}){\bf H}_{\rm{eq,E}}^H({\bf \Theta})\nonumber\\
&&\times {\bf H}_{\rm{eq,E}}({\bf \Theta}){\bf V}_{\rm B}({\bf \Theta})/(L_s(\sigma_{\rm{E}}^2{\bf I}+{\bf A}_z(\bf\Theta)))]\nonumber\\
&{\geq}& \log(1+\rho_1^{\rm LB}\|{\bf H}_{\rm{eq,B}}({\bf\Theta})\|_{\rm F}^2)-\log{\rm det}\nonumber\\
&&[{\bf I}_{L_s}+\rho_2^{\rm LB}{\bf V}_{\rm B}^H({\bf \Theta}){\bf H}_{\rm{eq,E}}^H({\bf \Theta}){\bf H}_{\rm{eq,E}}({\bf \Theta}){\bf V}_{\rm B}({\bf \Theta})]\nonumber\\
&=& \hat{ C}_{\rm{sec}}({\bf \Theta})
\end{IEEEeqnarray}
where $\rho_1^{\rm LB}=P_{\max}\psi^{\star}({\bf\Theta})/(L_s\sigma_{\rm B}^2)$ and $\rho_2^{\rm LB}=P_{\max}/(L_s\sigma_{\rm E}^2)$. The proof is completed.

Based on the above lemma, we can find that problem $(\mathcal{P}_1)$ is only with respect to ${\bf \Theta}$. Therefore, the original problem can be decomposed into the following two subproblems:

First, when the second term of $\hat{C}_{\rm{sec}}({\bf \Theta})$ is sufficient small, the maximization problem $(\mathcal{P}_1)$ approximates to minimizing the following \textbf{Orthogonal-Forcing Passive Beamforming (OF-PB)} subproblem
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align} \label{eqn:PBsub}
&\min_{{\bf\Theta}}\|{\bf H}_{\rm{eq,B}}({\bf \Theta}){\bf H}_{\rm{eq,E}}^H({\bf \Theta})\|_{\rm F}^2-\alpha_B\|{\bf H}_{\rm{eq,B}}({\bf \Theta})\|_{\rm F}^2,\\
&\hspace{2mm}{\rm s.t.}\hspace{2mm}\eqref{eqn:IRScm1}, \eqref{eqn:IRScm2}.\nonumber
\end{align}}
where $\alpha_B$ is a given parameter.

Applying the optimal solution of OF-PB subproblem, denoted by ${\bf \Theta}^\star$, into original problem $(\mathcal{P}_0)$, we further obtain the following  \textbf{Null-Space Jamming Hybrid Beamforming (NSJ-HB)} subproblem:
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align}
\label{eqn:p2reformu}
&\max_{{\bf W}_{s},{\bf W}_{z},{\bf F}}\hspace{3mm}C_{\rm{sec}}({\bf W}_{s},{\bf W}_{z},{\bf F},{\bf \Theta}^{\star})\hspace{3mm}{\rm s.t.}\hspace{3mm}
\eqref{eqn:powcons},\eqref{eqn:analogcon}.
\end{align}}
With the above decomposition, the suboptimal solution of original problem $({\mathcal{P}}_0)$ can be obtained by sequentially solving OF-PB subproblem and NSJ-HB subproblem by algorithms that will be proposed in the next section.
\vspace{-12pt}
\section{Proposed algorithms for OF-PB and NSJ-HB subproblems}\label{sec:algorithm}
\vspace{-10pt}
\subsection{CA-ADMM algorithm for solving OF-PB subproblem}
\vspace{-5pt}
By introducing a variable ${\bf x}\in{\mathcal{C}}^{N_{\rm I}}$, which denotes the vector of main diagonal elements of ${\bf \Theta}$, the objective function of OF-PB subproblem can be reformulated as
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align}\label{eqn:obj2}
&\|\tilde{\bf h}_{a}+{\bf H}_{PB,1}{\bf x}^*+{\bf H}_{PB,2}{\bf x}+ {\bf H}_{PB,3}{\rm vec}({\bf x}{\bf x}^H)\|_{\rm 2}^2\nonumber\\
&-\alpha_B\|{\bf h}_{\text {AB}}+{\bf H}_{PB,4}{\bf x}\|_2^2,
\end{align}}
where $\tilde{\bf h}_a={\rm vec}\big({\bf H}_{\rm {AB}}{\bf H}_{\rm {AE}}^H\big)$, ${\bf H}_{PB,1}={\bf H}_{\rm{IE}}^*\odot({\bf H}_{\rm {AB}}{\bf H}_{\rm{AI}}^H)$, ${\bf H}_{PB,2}=({\bf H}_{\rm{AI}}{\bf H}_{\rm {AE}}^H)^T\odot{\bf H}_{\rm{IB}}$, ${\bf H}_{PB,4}={\bf H}_{\rm{AI}}^T\odot{\bf H}_{\rm{IB}}$, ${\bf H}_{PB,3}=({\bf H}_{\rm{IE}}^*\otimes{\bf H}_{\rm{IB}}){\rm diag}\big({\rm vec}({\bf H}_{\rm{AI}}{\bf H}_{\rm{AI}}^H)\big)$ and ${\bf h}_{\rm {AB}}={\rm vec}({\bf H}_{\rm {AB}})$.

Then, to deal with the constant-modulus constraint and fourth-order objective function \eqref{eqn:obj2}, we introduce variable ${\bf y}_1$ to instead ${\bf x}$ in the fourth-order term of \eqref{eqn:obj2} and introduce variable ${\bf y}_2$ to instead ${\bf x}$ in the quadratic term of \eqref{eqn:obj2}. In this way, the objective function of problem \eqref{eqn:obj2} is reformulated as
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align}
&f({\bf x},{\bf y}_1,{\bf y}_2)\\
&\triangleq\|\tilde{\bf h}_{a}+{\bf H}_{PB,1}{\bf x}^*+{\bf H}_{PB,2}{\bf x}+ {\bf H}_{PB,3}{\rm vec}({\bf y}_1{\bf y}_1^H)\|_{\rm 2}^2\nonumber\\
&-\alpha_B\|{\bf h}_{\text {AB}}+{\bf H}_{PB,4}{\bf x}\|_2^2-{\bf x}^H{\bf H}_{PB,5}{\bf x}\nonumber\\
&-2{\rm Re}({\bf x}^T{\bf H}_{PB,6}{\bf x})+{\bf y}_2^H{\bf H}_{PB,5}{\bf y}_2+2{\rm Re}({\bf y}_2^T{\bf H}_{PB,6}{\bf y}_2),\nonumber
\end{align}}
where ${\bf H}_{PB,6}={\bf H}_{PB,1}^H{\bf H}_{PB,2}$ and ${\bf H}_{PB,5}=({\bf H}_{PB,1}^H{\bf H}_{PB,1})^*$ $+{\bf H}_{PB,2}^H{\bf H}_{PB,2}-\alpha_B{\bf H}_{PB,4}^H{\bf H}_{PB,4}$.

Therefore, problem \eqref{eqn:obj2} is equivalent to be
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align} \label{eqn:pb2}
&\min_{|{\bf x}|={\bf 1},{\bf y}_1,{\bf y}_2}f({\bf x},{\bf y}_1,{\bf y}_2)\hspace{4mm}{\rm s.t.}\hspace{2mm} {\bf y}_1={\bf x},\hspace{2mm}{\bf y}_2={\bf x}.
\end{align}}
Then, the augmented Lagrangian function of problem \eqref{eqn:pb2} is given by
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align}\label{eqn:lagra}
&{\mathcal L}({\bf x},{\bf y}_1,{\bf y}_2,{\boldsymbol \lambda}_1,{\boldsymbol \lambda}_2)\nonumber\\
&=f({\bf x},{\bf y}_1,{\bf y}_2)+{\rm Re}({\boldsymbol \lambda}_1^H({\bf y}_1-{\bf x})+{\boldsymbol \lambda}_2^H({\bf y}_2-{\bf x}))\nonumber\\
&\hspace{5mm}+\frac{\rho_1}{2}\|{\bf y}_1-{\bf x}\|_2^2+\frac{\rho_2}{2}\|{\bf y}_2-{\bf x}\|_2^2
\end{align}}
Note that the constant-modulus constraint is only imposed to variable ${\bf x}$ and are not related to variables ${\bf y}_1$ and ${\bf y}_2$. Therefore, denoting $k$ as the iteration number, the ADMM algorithm framework for solving problem \eqref{eqn:lagra} is given by
\vspace{-2mm}
\begin{subequations}\label{eqn:admm}
\begin{align}
&{\bf x}^{k+1} = \underset{|{\bf x}|={\bf 1}}{{\rm arg}\min}\hspace{2mm}{\mathcal L}({\bf x},{\bf y}_1^k,{\bf y}_2^k,{\boldsymbol \lambda}_1^k,{\boldsymbol \lambda}_2^k)\label{eqn:admmx}\\
&{\bf y}_1^{k+1} = \underset{{\bf y}_1}{{\rm arg} \min} \hspace{2mm}{\mathcal L}({\bf x}^{k+1},{\bf y}_1,{\bf y}_2^k,{\boldsymbol \lambda}_1^k,{\boldsymbol \lambda}_2^k)\label{eqn:admmy}\\
&{\bf y}_2^{k+1} = \underset{{\bf y}_2}{{\rm arg} \min} \hspace{2mm}{\mathcal L}({\bf x}^{k+1},{\bf y}_1^{k+1},{\bf y}_2,{\boldsymbol \lambda}_1^k,{\boldsymbol \lambda}_2^k)\label{eqn:admmz}\\
&{\boldsymbol \lambda}_1^{k+1}={\boldsymbol \lambda}_1^{k}+\rho_1({\bf y}_1^{k+1}-{\bf x}^{k+1})\label{eqn:admmlambda1}\\
&{\boldsymbol \lambda}_2^{k+1}={\boldsymbol \lambda}_2^{k}+\rho_2({\bf y}_2^{k+1}-{\bf x}^{k+1})\label{eqn:admmlambda2}
\end{align}
\end{subequations}

As observed from the ADMM algorithm framework \eqref{eqn:admm}, the closed-form solutions of the constant-modulus constrained linear programming problem \eqref{eqn:admmx} and the non-constrained quadratic programming problem \eqref{eqn:admmz} can be derived following \cite{Liang2016TSP}. The main challenge of utilizing the ADMM algorithm lies on solving problem \eqref{eqn:admmy}.
To solve the fourth-order non-convex problem \eqref{eqn:admmy}, we first introduce a variable $\hat{\bf y}_1=[{\rm Re}({\bf y}_1);{\rm Im}({\bf y}_1)]$, and convert ${\mathcal L}({\bf x}^{k+1},{\bf y}_1,{\bf y}_2^k,{\boldsymbol \lambda}_1^k,{\boldsymbol \lambda}_2^k)$ to be an equivalent function with real-value variable as follows:
{\small\begin{equation}\label{eqn:y_Lagreal}
\hat{\mathcal{L}}_y^k(\hat{\bf y}_1)=f_y^k(\hat{\bf y}_1)+{\hat{\boldsymbol \lambda}}_1^{k^T}({\hat{\bf y}_1}-{\hat{\bf x}}^{k+1}))+\frac{\rho_1}{2}\|{\hat{\bf y}_1}-{\hat{\bf x}}^{k+1}\|_2^2,
\end{equation}}
where $f_y^k(\hat{\bf y}_1)=\|{\hat{\bf H}}_{PB,3}[({\bf K}_1{\hat{\bf y}_1\otimes{\bf K}_2}+{\bf K}_3{\hat{\bf y}_1\otimes{\bf K}_4}){\hat{\bf y}_1}]+{\hat{\bf a}}_{y}^k\|_2^2$, with $\hat{\bf H}_{PB,3}=\begin{bmatrix}{\rm Re}({\bf H}_{PB,3}), &-{\rm Im}({\bf H}_{PB,3})\\{\rm Im}({\bf H}_{PB,3}), &{\rm Re}({\bf H}_{PB,3})\end{bmatrix}, \hat{\bf a}_y^k=\begin{bmatrix}{\rm Re}({\bf a}_y^k)\\{\rm Im}({\bf a}_y^k)\end{bmatrix},$ ${\bf a}_{y}^{k}=\tilde{\bf h}_{a}+{\bf H}_{PB,1}{\bf x}^{k+1^*}+{\bf H}_{PB,2}{\bf x}^{k+1}$, ${\hat{\boldsymbol \lambda}}_1^k=\begin{bmatrix}{\rm Re}({\boldsymbol \lambda}_1^k)\\{\rm Im}({\boldsymbol \lambda}_1^k)\end{bmatrix}, {\hat{\bf x}}^{k+1}=\begin{bmatrix}{\rm Re}({\bf x}^{k+1})\\{\rm Im}({\bf x}^{k+1})\end{bmatrix}, {\bf K}_1=\begin{bmatrix}{\bf I}_N, &{\bf 0}\\{\bf 0}, &-{\bf I}_N\end{bmatrix},$ ${\bf K}_2=\begin{bmatrix}{\bf I}_N,{\bf 0}_{N\times N}\end{bmatrix}, {\bf K}_3=\begin{bmatrix}{\bf 0},&{\bf I}_N\\{\bf I}_N, &{\bf 0}\end{bmatrix},{\bf K}_4=\begin{bmatrix}{\bf 0}_{N\times N}, {\bf I}_N\end{bmatrix}.$

To solve this fourth-order problem, we can find that $\hat{\mathcal{L}}_y^k(\hat{\bf y}_1)$ is Lipschitz continuous when $L_y>8c_2^2|\sum_i\sum_j\tilde{\bf H}_{3}$ $(i,j)|+4c_2^2|\sum_i\sum_j\tilde{\bf H}_{3}(i,j)|+4\|\tilde{\bf H}_{PB,3}^T{\tilde{\bf a}}_{y}\|_1$,
with $c_2=\max\{c,1\}$, $\tilde{\bf H}_{3}=\hat{\bf H}_{PB,3}^T\hat{\bf H}_{PB,3}$, $\tilde{\bf H}_{PB,3}(i,j)=|\hat{\bf H}_{PB,3}(i,j)|$, and ${\tilde{\bf a}}_{y}(i)=|\tilde{\bf h}_{a}(i)+\sum_i{\bf H}_{PB,1}(i,:)+\sum_i{\bf H}_{PB,2}(i,:)|$. Therefore, the upper-bound quadratic function of $\hat{\mathcal{L}}_y^k(\hat{\bf y}_1)$ can be formulated as
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align}\label{eqn:uyk}
\mathcal{U}_y^k(\hat{\bf y}_1)\triangleq& f_y^k(\hat{\bf x}^{k+1})+(\nabla f_y^k(\hat{\bf x}^{k+1})+{\hat{\boldsymbol \lambda}}_1^{k})^T({\hat{\bf y}_1}-{\hat{\bf x}}^{k+1})\nonumber\\
&+\frac{\rho_1+L_y}{2}\|{\hat{\bf y}_1}-{\hat{\bf x}}^{k+1}\|_2^2
\end{align}}
Applying the convex approximation (CA) technique, we customize the ADMM algorithm by minimizing $\mathcal{U}_y^k(\hat{\bf y}_1)$ instead of \eqref{eqn:admmy}. The CA-ADMM algorithm is summarized in Algorithm \ref{Alg:cADMM_IRS}.
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{algorithm}[t]
{\baselineskip 12pt
\caption{The CA-ADMM Algorithm}
\label{Alg:cADMM_IRS}
\begin{algorithmic}[1]
\STATE \textbf{Initialize}: Set Lipschitz constant $L_y$ and iteration index $k=1$, and initialize $\{{\bf x}^1, {\bf y}_1^1, {\bf y}_2^1, {\boldsymbol \lambda}_1^{1}, {\boldsymbol \lambda}_2^{1}\}$ randomly.
\WHILE {$\|{\bf x}^k-{\bf y}_1^k\|_2^2+\|{\bf x}^k-{\bf y}_2^k\|_2^2\geq \epsilon_1$}
\STATE Compute ${\bf x}^{k+1}$ via \eqref{eqn:admmx};
\STATE Compute ${\bf y}_1^{k+1}$ by minimizing $\mathcal{U}_y^k(\hat{\bf y}_1)$ in \eqref{eqn:uyk} and compute ${\bf y}_2^{k+1}$ via \eqref{eqn:admmz};
\STATE Compute ${\boldsymbol \lambda}_1^{k+1}$ and ${\boldsymbol \lambda}_2^{k+1}$ in parallel via \eqref{eqn:admmlambda1} and \eqref{eqn:admmlambda2}, respectively.
\ENDWHILE
\RETURN{${\bf \Theta}^{\star}={\rm {diag}}({\bf x}^{k+1})$.}
\end{algorithmic}
}
\end{algorithm}}

\subsection{BCD-OMP algorithm for solving NSJ-HB subproblem}\label{sec:solveHP}

To solve NSJ-HB subproblem \eqref{eqn:p2reformu}, we first denote IS full-digital beamforming (ISFDB) as $\tilde{\bf W}_{s}\triangleq{\bf F}{\bf W}_{s}$ and AN full-digital beamforming (ANFDB) as $\tilde{\bf W}_{z}\triangleq{\bf F}{\bf W}_{z}$. Replacing the variables of NSJ-HB subproblem \eqref{eqn:p2reformu} as the ISFDB $\tilde{\bf W}_{s}$ and the ANFDB $\tilde{\bf W}_{z}$, we will get the full-digital beamforming (FDB) problem, which can be solved by applying the Block Coordinate Descent (BCD) algorithm proposed in \cite{Pan2020TCOM}.

Second, the hybrid beamforming ${\bf W}_{s}$ and ${\bf F}$ can be solved by minimizing the Euclidean distance from the optimal ISFDB $\tilde{\bf W}_{s}^{\star}$, leading to the following problem
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{align} \label{eqn:hpproblem}
&\min_{{\bf F},{\bf W}_{s}}\hspace{3mm} \|\tilde{\bf W}_{s}^{\star}-{\bf F}{\bf W}_{s}\|_{\rm F}^2, \hspace{6mm}{\rm s.t.}
\hspace{2mm}\eqref{eqn:analogcon}.
\end{align}}
The above problem \eqref{eqn:hpproblem} can be solved by using Orthogonal Matching Pursuit (OMP) algorithm proposed in \cite{AyachTcom2014}.

Finally, the optimal digital artificial noise beamforming ${\bf W}_{z}^{\star}$ can be obtained by letting ${\bf W}_{z}^{\star}$ lie on the null space of the effective full-digital channel ${\bf H}_{\rm{eq,B}}({\bf \Theta}^{\star}){\bf F}^{\star}$.
The BCD-OMP algorithm is summarized in \textbf{Algorithm} \ref{Alg:BCD-OMP}.
{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{algorithm}[t]
{\baselineskip 12pt
\caption{The BCD-OMP algorithm}
\label{Alg:BCD-OMP}
\begin{algorithmic}[1]
\STATE \textbf{Input}: ${\bf \Theta}^{\star}$ solved by CA-ADMM algorithm;
\STATE Compute the optimal FDB $\tilde{\bf W}_{s}^{\star}$ by applying BCD algorithm in \cite{Pan2020TCOM};
\STATE Compute the optimal digital transmit beamforming ${\bf W}_{s}^{\star}$ and the optimal analog beamforming ${\bf F}^{\star}$ with optimized FDB $\tilde{\bf W}_{s}^{\star}$ via \eqref{eqn:hpproblem} by using OMP algorithm proposed in \cite{AyachTcom2014};
\STATE Compute the optimal digital artificial noise beamforming ${\bf W}_{z}^{\star}$ by letting ${\bf W}_{z}^{\star}$ lie on the null space of the effective full-digital channel ${\bf H}_{\rm{eq,B}}({\bf \Theta}^{\star}){\bf F}^{\star}$.
\RETURN{${\bf W}_{s}^{\star}$, ${\bf F}^{\star}$, and ${\bf W}_{z}^{\star}$.}
\end{algorithmic}
}
\end{algorithm}}

\section{Convergence and complexity analysis}\label{Sec:performance}

\subsection{Convergence Analyses of the CA-ADMM algorithm}
We have the following proposition to show that the proposed CA-ADMM algorithm converges to a stationary point under some conditions.
\vspace{-5pt}
\begin{Prop}\label{prop:optimal}
If penalty parameter $\rho_1,\rho_2$ and Lipschitz constant $L_y$ satisfy the inequalities $\frac{\rho_1+\rho_2}{2}-\frac{8L_y^2\rho_1+32L_y^3}{\rho_1^2}\geq 0$, $\frac{\rho_1-7L_y}{2}-\frac{2\rho_1+8L_y}{\rho_1^2}\geq 0$, and $\rho_1-5L_y\geq 0$, the proposed CA-ADMM algorithm converges to a stationary point of the equivalent OF-PB subproblem \eqref{eqn:pb2}.
\end{Prop}
{\bf Proof:} To prove this proposition, we first provide the following two lemmas to show the convergence of the proposed CA-ADMM algorithm.

\begin{lemm}\label{lem:suff_decrease}
For the proposed CA-ADMM algorithm, the augmented Lagrangian function (10) has the following inequality
\begin{IEEEeqnarray}{rcl}\label{eqn:descent}
&&\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})\nonumber\\
&\geq&\epsilon_x\|{\bf x}^{k+1}-{\bf x}^k\|_2^2+\sum_{i=1}^2\epsilon_{y_i}\|{\bf y}_i^{k+1}-{\bf y}_i^k\|_2^2,
\end{IEEEeqnarray}
where $\epsilon_x=\frac{\rho_1+\rho_2}{2}-\frac{8L_y^2\rho_1+32L_y^3}{\rho_1^2}$, $\epsilon_{y_1}=\frac{\rho_1-7L_y}{2}-\frac{2\rho_1+8L_y}{\rho_1^2}$, $\epsilon_{y_2}=\frac{L_{y_2}\rho_2}{2}-\frac{L_{y_2}}{\rho_2^2}$, with $L_{y_2}$ be the absolute value of maximum element in matrix $[{\bf H}_{PB,4}+{\bf H}_{PB,5}+{\bf H}_{PB,5}^*]$. If $\epsilon_x, \epsilon_{y_1}, \epsilon_{y_2}\geq 0$, $\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})$ decreases sufficiently.
\end{lemm}
\begin{IEEEproof}
Please refer to  Appendix \ref{app:lemma_3} for details.
\end{IEEEproof}
\begin{lemm}\label{lem:lowerbound}
The augmented Lagrangain function $\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})$ is lower bounded, i.e.,
\begin{equation}
\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})\geq 0.
\end{equation}
\end{lemm}
\begin{IEEEproof}
Please refer to  Appendix \ref{app:lemma_4} for details.
\end{IEEEproof}

Lemmas \ref{lem:suff_decrease} and \ref{lem:lowerbound}  show that the fourth-order augmented Lagrangian function ${\mathcal{L}}(\cdot)$ decreases sufficiently and has a lower bound, which indicates it is convergent.

Based on the above lemmas, we have the following proposition to present the proposed CA-ADMM algorithm converges to a stationary point of the OF-PB subproblem (5). Please refer to Appendix \ref{app:optimal} for details.

\subsection{Complexity Analysis}

The computational complexity of proposed CA-ADMM and BCD-OMP algorithms is analyzed as follows:

 \textbf{Complexity of CA-ADMM Algorithm}: In each iteration of the  CA-ADMM algorithm \ref{Alg:cADMM_IRS}, the complexity of computing ${\bf x}^{k+1}$, ${\boldsymbol \lambda}_1^{k+1}$ and ${\boldsymbol \lambda}_2^{k+1}$ is $\mathcal{O}(N)$; the complexity of computing ${\bf y}_2^{k+1}$ is $\mathcal{O}(4N^3)$; according to \eqref{eqn:admmy}$\sim$\eqref{eqn:uyk}, the complexity of computing ${\bf y}_1^{k+1}$ is $\mathcal{O}(\max\{N_{\text B}N_{\text E}N^2,4N^3\})$ . Therefore, the overall complexity for computing ${\bf \Theta}$ in CA-ADMM algorithm is $\mathcal{O}(T_1\max\{N_{\text B}N_{\text E}N^2,4N^3\})$, where $T_1$ is the iteration number in Algorithm \ref{Alg:cADMM_IRS}.

 \textbf{Complexity of BCD-OMP Algorithms}: The complexity of BCD-OMP Algorithms is dominated by two aspects: 1) computing the FDB $\tilde{\bf W}^{\star}$, 2) the complexity of solving problem \eqref{eqn:hpproblem}. For the first aspect, the complexity of computing the FDB $\tilde{\bf W}^{\star}$ is $\mathcal{O}(T_2\max\{2N_{\text A}^3,2N_{\text A}^2N_{\text E}\})$ in the BCD-OMP algorithm, where $T_2$ is the iteration number of the BCD algorithm. For the second aspect, the BCD-OMP algorithm utilizes the low-complexity OMP algorithm for solving problem \eqref{eqn:hpproblem}, indicating that the complexity of second aspect is reasonable to be accepted.

\vspace{-10pt}
\section{Numerical Simulations}\label{Sec:simulation}
\vspace{-5pt}
In our simulations, channels ${\bf H}_{ij}$ is modeled as sparse channel \cite{Alkhateeb2014JSTSP}, where the number of paths is set as 4, the small scale propagation fading gain obeys `Rician' fading with $\kappa=13.2$.
The path loss exponents of the direct links from Alice to Bob and from Alice to Eve are 4, and the reflected links from Alice to IRS, from IRS to Bob and Eve are 2. The AoDs/AoAs and distances $d_{ij}$ are obtained from the locations of the Alice, Bob, Eve, and IRS, where $A=[0,5]$, $B=[60,0]$, $E=[45,0]$, $I=[55,5]$. The Bob's and Eve's noise power are $\sigma^2=-59$dBm.
Unless stated otherwise, we set the numbers of antennas and RF chains are $N_{\rm{A}}=N_{\rm{I}}=32$, $N_{\rm{B}}=2$, $N_{\rm{E}}=2$, $N_{\rm{RF}}=4$, $L_s=L_z=2$. The parameters $\alpha_B$, $L_y$, $\rho_1$, $\rho_2$, and $\epsilon_1$ in the CA-ADMM Algorithm \ref{Alg:cADMM_IRS} are set as $\alpha_B=\sigma^2$, $L_y=8$, $\rho_1=\rho_2=16$, and $\epsilon_1=10^{-5}$.

{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
  \begin{figure}\label{Fig:NIRS}
    \centering
  \centerline{\includegraphics[width=0.4\textwidth]{NIRSNa}}
\caption{The SR comparisons among different algorithms versus (a) the number of IRS elements $N_{\rm I}$, (b) the number of antennas at Alice $N_{\rm{A}}$.}
\end{figure}}
Fig.1 depicts the SR of \emph{proposed CA-ADMM and BCD-OMP based HB strategy}, the \emph{BCD-MM based FDB strategy}, the \emph{random IRS strategy} and \emph{no IRS strategy}, versus the number of IRS elements $N_{\rm I}$ and the number of antennas $N_{\rm A}$, respectively.

As seen from the Fig.1(a), we have the following observations. First, when $N_{\rm I}=16$, the SR performance of the BCD-MM based FDB strategy is higher than the proposed HB strategy; while when $N_{\rm I}\geq32$, the proposed HB strategy outperforms the BCD-MM based FDB strategy. Moreover, the gaps of the SR performances achieved by the proposed HB strategy and the BCD-MM based FDB strategy increases as the increasing of $N_{\rm I}$. This observation demonstrates that the proposed HB strategy is more suitable for large-scale IRS scenarios. Second, the IRS assisted strategies outperform the no-IRS strategy. Thirdly, the performance gaps between the proposed HB strategy and random PB strategy are increasing as the increasing of the number of IRS elements $N_{\rm I}$.

As observed from Fig.1(b), under the varying number of antennas, the proposed HB strategy can achieve the similar performance to the BCD-MM based FDB strategy. Moreover, the gaps between the the proposed HB strategy and the random PB/without IRS strategies are increasing as the increasing of the number of antennas $N_{\rm A}$.

{\setlength\abovedisplayskip{1.5pt}
\setlength\belowdisplayskip{1.5pt}
\begin{figure}[htb]
  \centering
  \label{Fig:conver}
  \centerline{\includegraphics[width=0.35\textwidth]{conver.eps}}
\caption{The convergence performance of the proposed CA-ADMM algorithm.}
\end{figure}}

Fig.2 shows the convergence performance of the proposed CA-ADMM algorithm. As observed from this figure, the proposed CA-ADMM algorithm converges with different parameters $L_y$, $\rho_1$, and $\rho_2$ within 200 iterations to the same objective value of the OF-PB subproblem. Moreover, as the increasing of the parameter $L_y$, more iteration number is required to guarantee the convergence.

\begin{appendices}

\renewcommand{\theequation}{\thesection.\arabic{equation}}

\section{Proof of Lemma \ref{lem:suff_decrease}}
\label{app:lemma_3}

To facilitate the derivations, we define the following quantities
\begin{align}
&\Delta_{\bf x}^k=\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k},{\boldsymbol \lambda}_i^{k}\}_{i=1,2}),\nonumber\\
&\Delta_{{\bf y}_i}^k=\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k}\}_{i=1,2}),\nonumber\\
&\Delta_{{\boldsymbol \lambda}_i}^k=\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2}).\nonumber
\end{align}
Therefore, the left side of the inequality in Lemma \ref{lem:suff_decrease} can be rewritten as
\begin{align}\label{eqn:delta_all}
&\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})\Delta_{\bf x}^k+\sum_{i=1}^2(\Delta_{{\bf y}_i}^k+\Delta_{{\boldsymbol \lambda}_i}^k).
\end{align}
\subsubsection{For $\Delta_{\bf x}^k$}
According to ${\mathcal{L}}(\cdot)$'s strong convexity related to ${\bf x}$, we get
\begin{align}\label{eqn:deltaxk1}
\Delta_{\bf x}^k\geq&{\rm Re}(\nabla_{\bf x}{\mathcal{L}}\big({\bf x}^{k+1},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2} \big)({\bf x}^k-{\bf x}^{k+1}))+\sum_{i=1}^2\frac{\rho_i}{2}\|{\bf x}^{k+1}-{\bf x}^k\|_2^2.
\end{align}
Since ${\bf x}^{k+1}=\arg\min_{|{\bf x}|={\bf 1}}\hspace{2mm}{\mathcal{L}}\big({\bf x},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2} \big)$, we have
\begin{align}\label{eqn:x_inequality1}
{\rm Re}(\nabla_{\bf x}{\mathcal{L}}\big({\bf x}^{k+1},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2} \big)({\bf x}^k-{\bf x}^{k+1}))\geq 0.
\end{align}
Plugging \eqref{eqn:x_inequality1} into \eqref{eqn:deltaxk1}, we obtain
\begin{align}\label{eqn:deltaxk2}
\Delta_{\bf x}^k\geq \sum_{i=1}^2\frac{\rho_i}{2}\|{\bf x}^{k+1}-{\bf x}^k\|_2^2.
\end{align}

\subsubsection{For $\Delta_{{\bf y}_i}^k$}
\begin{align}\label{eqn:deltayk1}
\Delta_{{\bf y}_1}^k&\geq \mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{U}_y^k({\bf y}_1^{k+1})\nonumber\\
&=\underbrace{\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{U}_y^k({\bf y}_1^{k})}_{\Delta_{\mathcal{LU}}^k}+\underbrace{\mathcal{U}_y^k({\bf y}_1^{k})-\mathcal{U}_y^k({\bf y}_1^{k+1})}_{\Delta_{\mathcal{U}}^k}
\end{align}
\begin{align}\label{eqn:deltaulk}
\Delta_{\mathcal{LU}}^k&=f_y^k(\hat{\bf y}_1^{k})-f_y^k(\hat{\bf x}^{k+1})-\nabla f_y^k(\hat{\bf x}^{k+1})^T({\hat{\bf y}_1^k}-{\hat{\bf x}}^{k+1})-\frac{L_y}{2}\|{\hat{\bf y}_1^k}-{\hat{\bf x}}^{k+1}\|_2^2 -2L_y\|{\hat{\bf y}_1^k}-{\hat{\bf x}}^{k+1}\|_2^2\nonumber\\
&\geq -4L_y\|{\hat{\bf y}_1^{k+1}}-{\hat{\bf x}}^{k+1}\|_2^2-4L_y\|{\hat{\bf y}_1^{k+1}}-{\hat{\bf y}}_1^{k}\|_2^2\nonumber\\
&=-\frac{4L_y}{\rho_1^2}\|{\boldsymbol \lambda}_1^{k+1}-{\boldsymbol \lambda}_1^{k}\|_2^2-4L_y\|{{\bf y}_1^{k+1}}-{{\bf y}}_1^{k}\|_2^2
\end{align}
Moreover, since ${\mathcal{U}}(\cdot)$ is a quadratic strong convex problem, we have
\begin{equation}\label{eqn:deltauk}
\Delta_{\mathcal{U}}^k\geq (\nabla_{\hat{\bf y}_1}\mathcal{U}_y^k(\hat{\bf y}_1^{k+1})^T(\hat{\bf y}_1^{k+1}-\hat{\bf y}_1^{k})+\frac{\rho_1+L_y}{2}\|\hat{\bf y}_1^{k+1}-\hat{\bf y}_1^{k}\|_2^2.
\end{equation}
Since $\hat{\bf y}_1^{k+1}=\arg\min_{\hat{\bf y}_1}\mathcal{U}_y^k(\hat{\bf y}_1)$, we have
\begin{equation}
\nabla_{\hat{\bf y}_1}\mathcal{U}_y^k(\hat{\bf y}_1^{k+1})={\bf 0}.
\end{equation}
Substituting it into \eqref{eqn:deltauk}, we have
\begin{equation}\label{eqn:deltauk2}
\Delta_{\mathcal{U}}^k\geq \frac{\rho_1+L_y}{2}\|{\bf y}_1^{k+1}-{\bf y}_1^{k}\|_2^2
\end{equation}
Plugging \eqref{eqn:deltaulk} and \eqref{eqn:deltauk2} into \eqref{eqn:deltayk1}, we have
\begin{equation}\label{eqn:deltayk2}
\Delta_{{\bf y}_1}^k\geq -\frac{4L_y}{\rho_1^2}\|{\boldsymbol \lambda}_1^{k+1}-{\boldsymbol \lambda}_1^{k}\|_2^2+\frac{\rho_1-7L_y}{2}\|{{\bf y}_1^{k+1}}-{\bf y}_1^{k}\|_2^2\nonumber
\end{equation}
Similar to \eqref{eqn:deltauk}, since $\mathcal{L}({\bf x}^{k+1},{\bf y}_1^k,{\bf y}_2\{{\boldsymbol \lambda}_i^{k}\}_{i=1,2})$ is a quadratic strong convex problem related to ${\bf y}_2$, we have
\begin{equation}\label{eqn:deltay2}
\Delta_{{\bf y}_2}^k\geq \frac{\rho_2}{2}\|{{\bf y}_2^{k+1}}-{\bf y}_2^{k}\|_2^2
\end{equation}
\subsubsection{For $\Delta_{{\boldsymbol \lambda}_i}^{k}$}
\begin{equation}
\Delta_{{\boldsymbol \lambda}_i}^{k}=-\frac{1}{\rho_i}\|{\boldsymbol \lambda}_i^{k+1}-{\boldsymbol \lambda}_i^{k}\|_2^2.
\end{equation}
Let $\tilde{f}_1({\bf x},{\bf y}_1)=\|\tilde{\bf h}_{a}+{\bf H}_{PB,1}{\bf x}^*+{\bf H}_{PB,2}{\bf x}+{\bf H}_{PB,3}{\rm vec}({\bf y}_1{\bf y}_1^H)\|_{\rm 2}^2$ and $\tilde{f}_2({\bf y}_2)={\bf y}_2^H{\bf H}_{PB,4}{\bf y}_2+2{\rm Re}({\bf y}_2^T{\bf H}_{PB,5}{\bf y}_2)$. Then, solving $\nabla_{\hat{\bf y}_1}\mathcal{U}_y^k(\hat{\bf y}_1)=0$ and $\nabla_{{\bf y}_2}{\mathcal{L}}({\bf y}_2)=0$, we get
\begin{subequations}\label{eqn:lamk1}
\begin{align}
&{\boldsymbol \lambda}_1^{k+1}=-\nabla\tilde{f}_1({\bf x}^{k+1},{\bf x}^{k+1})-L_y({\bf y}_1^{k+1}-{\bf x}^{k+1}),\\
&{\boldsymbol \lambda}_2^{k+1}=-\nabla\tilde{f}_2({\bf y}_2^{k+1}).
\end{align}
\end{subequations}
With the derivations in \eqref{eqn:lamk1}, we will have
\begin{align}\label{eqn:deltalam1}
\|{\boldsymbol \lambda}_1^{k+1}-{\boldsymbol \lambda}_1^{k}\|_2^2&=\|-\nabla\tilde{f}_1({\bf x}^{k+1},{\bf x}^{k+1})+\nabla\tilde{f}_1({\bf x}^{k},{\bf x}^{k})+L_y({\bf x}^{k+1}-{\bf y}_1^{k+1}-{\bf x}^k+{\bf y}_1^k)\|_2^2\nonumber\\
&\leq L_y^2\big(8\|{\bf x}^{k+1}-{\bf x}^k\|_2^2+2\|{\bf y}_1^{k+1}-{\bf y}_1^k\|_2^2\big)
\end{align}
The inequality holds due to the fact that $\nabla\tilde{f}_1({\bf x}^{k+1},{\bf x}^{k})-\nabla\tilde{f}_1({\bf x}^{k},{\bf x}^{k})=0$. Moreover, we have
\begin{align}\label{eqn:deltalam2}
\|{\boldsymbol \lambda}_2^{k+1}-{\boldsymbol \lambda}_2^{k}\|_2^2&=\|-\nabla\tilde{f}_2({\bf y}_2^{k+1})+\nabla\tilde{f}_2({\bf y}_2^{k})\|_2^2\nonumber\\
&\leq L_{y_2}\|{\bf y}_2^{k+1}-{\bf y}_2^k\|_2^2
\end{align}
where $L_{y_2}$ is the absolute value of maximum element in matrix $[{\bf H}_{PB,4}+{\bf H}_{PB,5}+{\bf H}_{PB,5}^*]$.
Substituting \eqref{eqn:deltaxk2}, \eqref{eqn:deltay2}, \eqref{eqn:deltalam1}, and \eqref{eqn:deltalam2} into \eqref{eqn:delta_all}, we get
\begin{IEEEeqnarray}{rcl}
&&\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})\\
&&\geq \epsilon_x\|{\bf x}^{k+1}-{\bf x}^k\|_2^2+\epsilon_{y_1}\|{\bf y}_1^{k+1}-{\bf y}_1^k\|_2^2+\epsilon_{y_2}\|{\bf y}_2^{k+1}-{\bf y}_2^k\|_2^2,\nonumber
\end{IEEEeqnarray}
where $\epsilon_x=\frac{\rho_1+\rho_2}{2}-\frac{8L_y^2\rho_1+32L_y^3}{\rho_1^2}$, $\epsilon_{y_1}=\frac{\rho_1-7L_y}{2}-\frac{2\rho_1+8L_y}{\rho_1^2}$, $\epsilon_{y_2}=\frac{L_{y_2}\rho_2}{2}-\frac{L_{y_2}}{\rho_2^2}$.
If $\epsilon_x, \epsilon_{y_1}, \epsilon_{y_2}\geq 0$, $\mathcal{L}({\bf x}^k,\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})$ decreases sufficiently. This completes the proof.

\setcounter{equation}{0}
\section{Proof of Lemma \ref{lem:lowerbound}}
 \label{app:lemma_4}
Plugging \eqref{eqn:lamk1} into $\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})$, we get
\begin{align}
\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})=&\tilde{f}_1({\bf x}^{k+1},{\bf y}_1^{k+1})-\tilde{f}_2({\bf x}^{k+1})+\tilde{f}_2({\bf y}_2^{k+1})+\sum_{i=1}^2\frac{\rho_i}{2}\|{\bf y}_i^{k+1}-{\bf x}^{k+1}\|_2^2\nonumber\\
&+{\rm{Re}}[(-\nabla\tilde{f}_1^k({\bf x}^{k+1})-L_y({\bf y}_1^{k+1}-{\bf x}^{k+1}))^H({\bf y}_1^{k+1}-{\bf x}^{k+1})]\nonumber\\
&+{\rm{Re}}[(-\nabla\tilde{f}_2({\bf y}_2^{k+1}))^H({\bf y}_2^{k+1}-{\bf x}^{k+1})]\nonumber\\
\overset{(a)}{\geq}& \tilde{f}_1({\bf x}^{k+1},{\bf y}_1^{k+1})+{\rm{Re}}[-\nabla\tilde{f}_1^{k^H}({\bf x}^{k+1})({\bf y}_1^{k+1}-{\bf x}^{k+1})]\nonumber\\
&+(\frac{\rho_1}{2}-L_y)\|{\bf y}_1^{k+1}-{\bf x}^{k+1}\|_2^2+\rho_2\|{\bf y}_2^{k+1}-{\bf x}^{k+1}\|_2^2\nonumber\\
\overset{(b)}{\geq}& \tilde{f}_1({\bf x}^{k+1},{\bf y}_1^{k+1})+{\rm{Re}}[-\nabla\tilde{f}_1^{k^H}({\bf y}_1^{k+1})({\bf y}_1^{k+1}-{\bf x}^{k+1})]\nonumber\\
&+(\frac{\rho_1-5L_y}{2})\|{\bf y}_1^{k+1}-{\bf x}^{k+1}\|_2^2+\rho_2\|{\bf y}_2^{k+1}-{\bf x}^{k+1}\|_2^2\nonumber\\
\overset{(c)}{\geq}& \tilde{f}_1({\bf x}^{k+1},{\bf x}^{k+1})+(\frac{\rho_1-5L_y}{2})\|{\bf y}_1^{k+1}-{\bf x}^{k+1}\|_2^2+\rho_2\|{\bf y}_2^{k+1}-{\bf x}^{k+1}\|_2^2
\end{align}
where inequality (a) holds due to the strong convexity of function $\tilde{f}_2(\cdot)$, and  inequalities (b) and (c) comes from Decent Lemma \cite{1998Nonlinear}. Moreover, since $\tilde{f}_1({\bf x}^{k+1},{\bf x}^{k+1})\geq 0$, if $\rho_1-5L_y\geq 0$, $\mathcal{L}(\cdot)$ is lower bounded by 0. This completes the proof.

\setcounter{equation}{0}
\section{Proof of Proposition \ref{prop:optimal}}\label{app:optimal}

First of all, we set the parameters $\rho_1$ and $L_y$ satisfies the conditions in lemmas \ref{lem:suff_decrease} and \ref{lem:lowerbound}. For $k=1,2,\cdots,+\infty$, according to \eqref{eqn:descent} in Lemma \ref{lem:suff_decrease}, we have
\begin{align}
&\mathcal{L}({\bf x}^1,\{{\bf y}_i^1,{\boldsymbol \lambda}_i^{1}\}_{i=1,2})-\mathcal{L}({\bf x}^{k+1},\{{\bf y}_i^{k+1},{\boldsymbol \lambda}_i^{k+1}\}_{i=1,2})\nonumber\\
&\geq\sum_{k=1}^{+\infty}\epsilon_x\|{\bf x}^{k+1}-{\bf x}^k\|_2^2+\epsilon_{y_1}\|{\bf y}_1^{k+1}-{\bf y}_1^k\|_2^2+\epsilon_{y_2}\|{\bf y}_2^{k+1}-{\bf y}_2^k\|_2^2
\end{align}
Since $\epsilon_x,\epsilon_{y_1},\epsilon_{y_2}\geq 0$ and $\mathcal{L}({\bf x}^1,\{{\bf y}_i^1,{\boldsymbol \lambda}_i^{1}\}_{i=1,2})$ is finite, we conclude that
\begin{align}\label{eqn:finite1}
\lim_{k\rightarrow+\infty}{\bf x}^{k+1}-{\bf x}^k=0, \lim_{k\rightarrow+\infty}{\bf y}_i^{k+1}-{\bf y}_i^k=0.
\end{align}
Pugging (11d) into \eqref{eqn:deltalam1} and (11e) into \eqref{eqn:deltalam2}, we get
\begin{align}\label{eqn:finite2}
\lim_{k\rightarrow+\infty}{\boldsymbol \lambda}_i^{k+1}-{\boldsymbol \lambda}_i^k=0, \lim_{k\rightarrow+\infty}{\bf x}^{k+1}-{\bf y}_i^{k+1}=0.
\end{align}
Moreover, combining \eqref{eqn:finite1} and \eqref{eqn:finite2}, and letting ${\bf x}^{\star}$ be some point of the proposed ADMM algorithm, we have
\begin{align}\label{eqn:finite3}
\lim_{k\rightarrow+\infty}{\bf y}_i^{k+1}=\lim_{k\rightarrow+\infty}{\bf x}^{k+1}={\bf x}^{\star}.
\end{align}
Since ${\bf x}^{k+1}=\arg\min_{|{\bf x}|={\bf 1}}\mathcal{L}({\bf x},\{{\bf y}_i^k,{\boldsymbol \lambda}_i^{k}\}_{i=1,2})$ and $\mathcal{L}(\cdot)$ is a linear function related to ${\bf x}$, we have the inequality
\begin{align}\label{eqn:station}
&{\rm {Re}}[(\nabla\tilde{f}_1({\bf x}^{k+1},{\bf y}_1^{k})-\nabla\tilde{f}_2({\bf x}^{k+1})+\sum_{i=1}^2(-{\boldsymbol \lambda}_i^{k}+\rho_i({\bf y}_i^{k}-{\bf x}^{k+1})))^H({\bf x}-{\bf x}^{k+1})]\geq 0, \hspace{2mm}\forall |{\bf x}|={\bf 1},
\end{align}
When $k\rightarrow\infty$, substituting \eqref{eqn:lamk1} and \eqref{eqn:finite3} into \eqref{eqn:station}, we have
\begin{align}\label{eqn:station2}
\lim_{k\rightarrow\infty}{\rm {Re}}[\nabla\tilde{f}_1^H({\bf x}^{\star},{\bf x}^{\star})({\bf x}-{\bf x}^{\star})]\geq 0, \hspace{2mm}\forall |{\bf x}|={\bf 1},
\end{align}
Since $\tilde{f}_1({\bf x}^{\star},{\bf x}^{\star})=F({\bf x}^{\star})$, the proof is completed.

\end{appendices}

redpajama_set_name: RedPajamaArXiv


text:
\section{Prelude}

It is well known that for many physical quantum potentials, the spectrum of
the Hamiltonian is a discrete set (i.e. the set of energy
eigenvalues $E_n$ is a countable set) associated with a separable
Hilbert space of eigenstates $\psi _n$.\footnote{Over compact spaces the spectrum of the Hamiltonian is separable.} It is also well known that in Quantum Mechanics, $\left| \psi (x)\right| ^2$ defines a probability
distribution over coordinate space, in the observer world. Using the formula
for distance in noncommutative geometry \cite{con}, we were recently able to
show that one might then also associate the relevant metric with particle
energy levels.
In noncommutative geometry, one uses the spectral triple $(A,H,D)$ , where
A $ is the algebra of functions on the space (which according to the
Gelfand-Naimark theorem \cite{gn} is fully characteristic of the space), $H$
is a Hilbert space, and $D$ is a Dirac operator, which acts as a linear
operator on that Hilbert space. The Dirac operator is a self-adjoint
operator, with compact resolvent. In recent years,noncommutative
geometry \cite{models} has provided the means of further extending the
geometric approach to gauge theories with spontaneous symmetry
breakdown. In this study, we point out yet another possible
application, namely the definition of a metric associated with quantum
spectra. At this stage we limit our treatment to its evaluation and a
discussion of its consistency with the probabilistic intuitive picture.\\

The distance formula in noncommutative geometry is:
\begin{equation}
\label{dis}d(a,b)=\sup _f\left\{ \left| f(a)-f(b)\right| \;:\;f\in
A,\;\left\| \;\left[ \;D,f\;\right] \;\right\| \leq 1\right\}
\end{equation} \\
where $a,b\in X,\;f\in A,$\ $A$ is the algebra of functions on $X,$ and
the norm on the r.h.s is the norm of operators in $H$. Both the
noncommutative geometrical and the classical (i.e. the infimum among all the
paths which connect $a$ and $b$) definitions for distance give the same
result when the base space is a Riemannian manifold of genus zero. However
the n.c.g definition has the advantage of being applicable to discrete
spaces too, on which the concept of a ``path'' is not well defined.\\

The main advantage of the distance formula, eq.(\ref{dis}), is that it
involves not just the algebra of functions defined on the space, but
also the Hilbert
space, which is usually associated with Quantum Mechanics. Had
the geometry's defining triple not involved a Hilbert space of states, the
algebra of functions would have been applied to a fixed manifold (i.e. a
topological manifold, without any dynamics). However, once we have a
Hamiltonian $H$, its spectrum is associcated with a Hilbert space of
eigenstates. The eigenstates essentially fix the probability
distribution of the particle's position in space.\footnote{In Quantum
  Mechanics the probability distribution of the particle's position in
  space applies to a single particle, and not only as a statistical
  result due to large number of measurements. This is just as in the
  two slits experiment, where a single particle passes through both slits.} The base space can thus now be considered as a dynamical space, i.e., a space depending on the quantum state (and therefore also on the energy level). This can be seen in every scattering process, e.g. note how visible
light is scattered on drops forming a rainbow, due to spectral
decomposition, so that each frequency has it own geodesic line. One might
well claim that there is an interplay between the fixed base space, and the Hilbert space.\\

In the following section we will give a metric interpretation, as
``seen'' by the particle, in its own world, which is consistent with
the distributive aspects assigned to the wave function, in quantum
mechanics for the observable world.\\

\section{The Quantum Mechanical Distance Formula}
Let assume for simplicity that we are giving a one dimensional
Hamiltonian system which found in its lowest eigenstate $\psi$. While using eq.(\ref{dis}) over a smooth manifold, one can prove that:
\begin{equation}
\label{sup}\sup _f\left| f(a)-f(b)\right| =\sup _f\left| \int_a^b\left|
\nabla f\right| ds\right| =\left| \int_a^b\sup _f\left| \nabla f\right|
ds\right|
\end{equation}
(where a, b not necessarily ordered).\\
The norm  in eq.(\ref{dis}) usually defined as follows:
\begin{equation}
\label{normst}\left\| \;\left[ \;D,f\;\right] \;\right\| = \sup
\left\{\left\| \;\left[ \;D,f\;\right]\;\psi\;\right\| \;\;\;\;\;
  \mbox{where}\;\;\;\left\|\psi\right\|\le 1\right\}
\end{equation}
However, we suggest to modify the norm condition in eq.(\ref{dis}) to be
interpreted as a condition on the expectation value of
$\left|\left[\;D,f\;\right]\right|$ in the quantum state $\psi$ - that is:
\begin{equation}
\label{norm}1\ge \left\langle\;\left|\left[ \;D,f\;\right]\right|\;\right\rangle =
\left\langle \psi \left| \left| \left[ \;D,f\;\right] \right| \right| \psi
\right\rangle = \int_v\psi ^{*}(x)\left| \left[ \;D,f\;\right] \right| \psi (x)dx
\end{equation}
where $v$ is the entire base space, and we are assuming also that
$\psi$ is normalized:
\begin{equation}
\label{nirmul}\left\langle \psi |\psi \right\rangle =\int_v\psi ^{*}(x)\psi
(x)dx=1
\end{equation}
Therefore, in order to be consistent with both eq.(\ref {nirmul}) and
the norm condition in eq.(\ref{dis}) replaced by the new ``norm''
condition in eq.(\ref{norm}), it is sufficient to require the following
local condition:
\begin{equation}
\label{jakob}\left| \left[ \;D,f\;\right]
\right|\left.\right|_{x} =\left| \nabla f\right|\left.\right|_{x}\leq \frac 1{Vol(v)\psi ^{*}(x)\psi (x)}
\end{equation}
where $Vol(v)$ is the world volume that we assume to be finite. The
idea to replace the norm condition in eq.(\ref{dis}) by eq.(\ref{norm}),
follows from the fact that in Q.M. one considers only expectation values as
observables. This would enable a quantum state to affect the observed
metric as will be shown immediately.
By applying eq.(\ref{jakob}) to eq.(\ref{sup}), one should use the
equality in eq.(\ref{jakob}). So if $\forall x\in X,\exists \psi (x)\not =0$
(except on the boundaries), which is always the situation at the first
energy level, the distance formula eq.(\ref{dis}) becomes:
\begin{equation}
\label{no0}d(a,b)=\frac 1{Vol(v)}\left| \int_a^b\frac 1{\psi ^{*}(x)\psi
(x)}ds\right|
\end{equation}
where $ds$ is a path element on the straight line which connects the points
a$ and $b$.\\

Assume now that we are working in a $D$-dimensional base
space, such that the wave function of a particle which is found at some
energy level is given by: $\psi (x_1,...,x_D)=\prod_{i=1}^D\psi _{n_i}(x_i)
, where the $i$-index labels the coordinates, and the $n_i$-index indicates
in which eigenstate the particle is found (according to zero locus of
$\psi$), over the $i$-coordinate (i.e. we assume separability of the
variables). The base space is thus being
divided naturally by the distribution probability, defined on the base
space, into $N\equiv \prod_{i=1}^Dn_i$ parts (induced by the zero
locus of $\psi$). One should thus consider the
particle as being situated in any of the N parts, i.e. the location of the
particle is now being represented by an $N$-dimensional vector. Therefore,
one considers the particle as situated at one spatial point, given in $N
-dimensional space in its own world, but in $D$-dimensions in the outside
world.\\ \\
\vspace{10mm} \epsfxsize=8truecm \centerline{\epsfbox{rep2.ps}}
\vspace{5mm}
\centerline{\parbox{13truecm}{Figure 1. {\footnotesize Two points
      $a\;,\;b$ in the particles` world, while the particle is found in
      a two dimensional infinite square well potential at the energy
      level $n=2,\; m=3$. So $N=6$ (since there are 6 cells) and $D=2$ and therefore in this case its location will be defined by a $6\;\times\;2$ matrix}}}
\\ \\ The full representation of the location of the particle thus requires
an $N\times D$ - matrix, where each row represents a domain. The distance
formula will thus measure the distance between two points in $N$-dimensional
space, treated as a vector space, i.e., weighted in a Pythagorean way, and
inside each domain as in eq.(\ref{no0}). The quantum distance formula now
takes the following form:
\begin{equation}
\label{e0}d_N(a,b)=\sqrt{\sum_{i=1}^N\left( \frac 1{Vol(v_i)}\left|
\int_{a_i}^{b_i}\frac 1{\psi ^{*}(x)\psi (x)}ds_i\right| \right) ^2}
\end{equation}
where the $i$-index runs over the domains (i.e. over the rows), and by
$Vol(v_i)$ one means:
\begin{equation}\label{volvi}
Vol\left(v_i\right) = Vol(v)\int_{v_i}\psi ^{*}(x)\psi (x)dx
\end{equation}\\
In order that eq.(\ref{e0}) would be applied also to spaces with
infinite volume we suggest to replace the definition of $Vol(v)$. We
choose either to define $Vol(v)$ as the classical volume\footnote{For
  example: for harmonic oscillator one can take the volume as based on
  the amplitude of the corresponding energy level, or for the hydrogen
  atom one can take the appropriate Bohr radius (see examples 3.3, 3.4).} or to
define $\frac{1}{Vol(v)}$ as the square of the normalization
pre-factor appears in the eigenstate (up to an arbitrary scale).\\

In cases where the particle state is given by a superposition of
eigenstates, one has to find the $(D-1)\;-$ subspace on which the wave
function vanishes. The space is thereby naturally separated into
parts. As explained, the particle
is represented in each of these parts. The $D\;-$ dimensional volume of each
part should also be evaluated, using eq.(\ref{e0}). However, the main
difference is that in this case the metric evolves with time. Each
eigenstate in the superposition has its own time evolution, therefore the
probability distribution $\left| \psi \right| ^2$ also depends on time
(whereas there is no such time-dependence in the case in which the particle
is found in an eigenstate).\\

In the following section we apply the quantum
distance formula eq.(\ref{e0}) to several examples and discuss some of the
implications of the results.

\section{Examples}

\subsection{One Dimensional - Infinite Square Well Potential}

The potential is defined as:
\begin{equation}
\label{pot1}U\left( x\right) =\left\{
\begin{array}{cc}
0 & \;\;\;0\le x\le L \\
\infty & \;\;\;\mbox{elsewhere}
\end{array}
\right.
\end{equation}
\\ So, the eigenstates are: $\psi _n(x)=\sqrt{\frac 2L}\sin \left( \frac
n\pi x}L\right) $ and $Vol(v)=L$. The distances at the first energy
level are therefore,
\begin{equation}
d_1(a,b)=\frac 1L\left| \int_a^b\frac 1{\frac 2L\sin ^2\left( \frac{\pi x
L\right) }dx\right| =\frac L{2\pi }\left| \cot \left( \frac{\pi b}L\right)
-\cot \left( \frac{\pi a}L\right) \right|
\end{equation}
\\ \vspace{0mm} \epsfxsize=15truecm \centerline{\epsfbox{iswp.ps}}
\vspace{5mm}
\centerline{\parbox{13truecm}{Figure 2.{\footnotesize  The distances in
      an infinite square well potential, at the first energy level, as
      seen from the points a=0.5L and a=0.9L}}} \\ \\ The distance at the
n^{th}$- level then becomes:
\begin{equation}
d_n(a,b)=\frac {n}{L}\sqrt{\sum_{i=1}^n\left| \int_{a_i}^{b_i}\frac 1{\frac
2L\sin ^2\left( \frac{n\pi x}L\right) }dx\right| ^2}=\frac L{2\pi }\sqrt
\sum_{i=1}^n\left| \cot \left( \frac{n\pi b_i}L\right) -\cot \left( \frac
n\pi a_i}L\right) \right| ^2}
\end{equation}
where $a_i,b_i\;\in \;\left( \frac{(i-1)}nL\;,\;\frac inL\right) $ and
i=1,2,...,n$.\\

\subsection{Two Dimensional - Infinite Square Well Potential}

The potential is defined as follows:
\begin{equation}
\label{pot2}U\left( x,y\right) =\left\{
\begin{array}{ccc}
0 & \;\;\;0\le x\le L &  \\
0 & \;\;\;0\le y\le L &  \\
\infty & \;\;\;\mbox{else where} &
\end{array}
\right.
\end{equation}
\\ The eigenstates are then: $\psi _{n,m}(x)=\frac 2L\sin \left( \frac{n\pi
}L\right) \sin \left( \frac{m\pi y}L\right) $\\ The distance formula then
becomes:
\begin{equation}
\label{ddis}d_{\left( n,m\right) }\left( a,b\right) =\frac{nm}{L^2}\sqrt
\sum_{i=1}^{nm}\left| \int_{a_i}^{b_i}\frac 1{\frac 4{L^2}\sin ^2\left(
\frac{n\pi x}L\right) \sin ^2\left( \frac{m\pi y}L\right) }ds_i\right| ^2}
\end{equation}
where the $i-$index stands for the $i^{th}-$zone (where there are $nm$
zones), and $a_i\;,\;b_i$ are respectively the initial and final points in
the $i^{th}-$zone.\\

\subsection{One Dimensional Harmonic Oscillator}

The potential is:
\begin{equation}
\label{pot3}U\left( x\right) =\frac 12m\omega ^2x^2
\end{equation}
\\ The eigenfunctions are $Hermite\;polinomials$:
\begin{equation}
\label{herm}\phi _n(x)=\left[ \frac 1{2^nn!}\left( \frac \hbar {m\omega
}\right) ^n\right] ^{1/2}\left( \frac{m\omega }{\pi \hbar }\right)
^{1/4}\left[ \frac{m\omega }\hbar x-\frac d{dx}\right] ^ne^{-\frac 12\frac
m\omega }\hbar x^2}
\end{equation}
\\ so that the $0^{\;th}$ - eigenstate is:
\begin{equation}
\label{phi0}\phi _0(x)=\left( \frac{m\omega }{\pi \hbar }\right)
^{1/4}e^{-\frac 12\frac{m\omega }\hbar x^2}
\end{equation}
As the volume we take the ``classical'' amplitude $A_0$ at this energy
level which can be found by equating the energy of this quantum level with the potential energy - i.e.
\frac 12m\omega ^2A_0^2=\frac 12\hbar \omega $. Thus, the ``classical''
amplitude at this energy level is $A_0=\sqrt{\frac \hbar {m\omega}}$.
Therefore the distance at the $0^{\;th}\;energy-level$ is:
\begin{eqnarray}
\label{dis3}d_0\left(a,b\right)=\int_{a}^{b}\sqrt{\pi}e^{\frac
  {m\omega}{\hbar}x^2}dx = \left(\frac{\pi}{2}\sqrt{\frac
    {\hbar}{m\omega}}\right)\left|\left[{\mbox erf}
    \left(i\sqrt{\frac{m\omega}{\hbar}}b\right)-{\mbox erf}
    \left(i\sqrt{\frac{m\omega}{\hbar}}a\right)\right]\right|=\\
\frac{\pi}{2} A_0 \left|\left[{\mbox erf}\left(i \; \frac{b}{A_0}\right)-{\mbox erf}\left(i\; \frac a{A_0}\right)\right] \right| \nonumber
\end{eqnarray}\\
For higher energy levels one has to consider the
``classical'' amplitudes $A_n=A_0\sqrt{2n+1}$ as the total volume -
Vol\left( v\right) $. One has then to find the $n-$roots of the $n^{\;th}-
Hermite polynomial. Denote the $n-$roots by $\left( \alpha _1,...,\alpha
_n\right) $. The partial volumes $Vol\left( v_i\right) $ in eq.(\ref{e0})
become $Vol\left( v_i\right) =Vol\left( v\right) \int_{\alpha
_{i-1}}^{\alpha _i}\left| \phi _n\left( x\right) \right| ^2dx\;$ where
\;\alpha _0=-\infty \;$ and $\;\alpha _{n+1}=\infty $. One can now make use
of eq.(\ref{e0}) to find the distances at higher energy
levels. Evaluating the distance associated with the harmonic oscillator at higher dimensions is a straightforward calculation, making use of eq.(\ref{e0}).\\
\vspace{0mm} \epsfxsize=15truecm
\centerline{\epsfbox{oscil.ps}} \vspace{5mm}
\centerline{\parbox{13truecm}{Figure 3.{\footnotesize  The distances in
      a one dimensional harmonic oscillator, at the n=0 energy level, as seen from the points a=0[A], a=0.8[A] and a=1.2[A]}}}
\vspace{5mm}\\

\subsection{The Hydrogen Atom}

The potential is given by:
\begin{equation}
V(r)=-\frac{e^2}r
\end{equation}
For simplicity, we compute the distances, as seen in the electron's world,
in the $1s$ and the $2s$ levels and take only distances between two points
situated on the same radius vector.\\ The $1s$ and the $2s$ eigenstate are:
\begin{equation}
\begin{array}{l}
1s:\,\,\,\,\,\,\,\,\,\phi _{1s}=\frac 1{
\sqrt{\pi a_0^3}}\,e^{-r/a_0}\\
2s:\,\,\,\,\,\,\,\,\,\phi _{2s}=\frac 1{
\sqrt{8\pi a_0^3}}\,\left( 1-\frac r{2a_0}\right) \,e^{-r/(2a_0)}
\end{array}
\end{equation}
where $a_0$ is the Bohr radius. In the $1s$ case we define $Vol\left(
v\right) =\frac{4\pi }3a_0^3$\thinspace \thinspace \thinspace and the
distance between two points along the same radius vector is:
\begin{equation}
d_{1s}\left( r_1,r_2\right) =\frac 38\,\left| {\mbox e}^{2r_2/a_0}-{\mbox e
^{2r_1/a_0}\right| \,a_0
\end{equation}
In the $2s$ case one sees that the eigenstate vanishes at $r=2a_0$ and the
space is thereby naturally split into two domains: $r<2a_0$ and $r>2a_0$.
This means that according to our definition the electron's location is
represented in both sectors - \\ i.e. $\left\{ {}\right. \left(
x_1,x_2\right) \,\,\,\,\,$where$\,\,0\le x_1<2a_0\,\,\,\,\,$and
\,\,2a_0<x_2<\infty \left. {}\right\} $. The inner and outer volumes,
appearing in eq.(\ref{e0}), will be:
\begin{equation}
\begin{array}{l}
Vol\left( v_1\right) =
\frac{4\pi }3(R_2)^3\int_0^{2a_0}\left| \phi _{2s}\left( r\right) \right|
^24\pi r^2dr=\frac{256}3\pi a_0^3\;[1-\frac 7{{\mbox e}^2}] \\ Vol\left(
v_2\right) =\frac{4\pi }3(R_2)^3\int_{2a_0}^\infty \left| \phi _{2s}\left(
r\right) \right| ^24\pi r^2dr=\frac{256}3\pi a_0^3\;\frac 7{{\mbox e}^2}
\end{array}
\end{equation}
where we make use of the Bohr radius of the n$-th$ level, defined as:
R_n=n^2a_0$. One can now make use of eq.(\ref{e0}) and the distance between
the point $(x_1,x_2)$ and the point $(x_1^{\prime },x_2^{\prime })$ becomes:
\begin{equation}
d_{\left( x_1,x_2\right) }^{\left( x_1^{\prime },x_2^{\prime }\right) }
\sqrt{\left( \frac 1{Vol\left( v_1\right) }\left| \int_{x_1}^{x_1^{\prime
}}\left| \phi _{2s}\right| ^{-2}dr\right| \right) ^2+\left( \frac
1{Vol\left( v_2\right) }\left| \int_{x_2}^{x_2^{\prime }}\left| \phi
_{2s}\right| ^{-2}dr\right| \right) ^2}
\end{equation}
\section{Discussion and Conclusions}
It is worth reviewing
the distances evaluated in the infinite square well potential, as compared
to those reached in the harmonic oscillator case. While in an infinite
square well potential the distances at each energy level are independent of
both the mass of the particle and $\hbar $, in the harmonic oscillator the
distances depend on the mass of the particle and on the oscillator frequency
- and also on $\hbar $. The fact that $\hbar $ does not appear in the
distance formula in the infinite square well potential case might hint at
one of the following:\\ 1. Such potentials do not really exist in nature,
but can be constructed as a limiting case of a superposition of harmonic
oscillators.\footnote
There should be an infinite of them, since the energy spectrum of the
infinite square well potential is proportional to $n^2$ while for the
harmonic oscillator it is proportional to $n$.}
In that case the walls of the infinite square well
potential should be associated with the ``classical amplitudes'' of the
harmonic oscillator (i.e. $A_i=\sqrt{\frac \hbar {m_iw_i}}$ where the $i-$
index label the harmonic oscillators - see also eq.(\ref{dis3})\,).\\
2. On the
other hand, if infinite square well potentials do exist in nature, then a
spectrum of $mw$ should be induced, due to our former reasoning.\\

As one can see the distance between any point $x$
for which $\psi (x)\neq 0$ and a point $y$ for which $\psi (y)=0$ becomes
infinite. The less the probability distribution in the outside world, the
larger the distances in the particle world at that neighborhood, i.e. the
metric is `stretched'. In other words, while $\left| \psi (x)\right| ^2$ is
the probability distribution as defined over the space in the outside world,
$\frac 1{\left| \psi (x)\right| ^2}$ is the distribution of points as seen
within the particle world. This can conceptually be understood from the
similarity to the W.K.B. interpretation.\\ In the W.K.B approximation $\frac
1{\left| \psi \left( x\right) \right| ^2}$ is interpreted as the average
velocity of the particle in space. The velocity is however responsible for
the particle's translation in space. Thus, the velocity can be interpreted
as a particle's passage through a density of points in space. The larger the
velocity, the larger the spatial point-density. The differential element $ds$
in the quantum distance formula should therefore be associated with time,
for the quantum formula to have a metrical meaning. Thus, in the particle
world, ``farther'' means (according to the W.K.B approximation) a longer
time in the spectator's outside world. That is to say that the spectator
would have to wait a long time until he might find the particle at a place
with a low probability density.\footnote
Note that the interplay between time and distance elements already
occurs when, one compares the metric outside and inside a black
hole. It seems not to be an accident, since in both cases we have a compact space which is
disconnected from the observer.}\\

Another fact one should be aware of is
that the natural splitting induced by the probability distribution $\left|
\psi \left( x\right) \right| ^2$ over space, in the spectator world, is
being translated in the particle world into the fixing of the dimension of
the space. The above conclusions can be observed in all the examples
given. In this manner, the higher the energy level, the higher is the
space dimensionality in the particle world.\\

One might ask - is the metric (in the particle world) detectable?! We might
claim that any measuring process would insert a new interaction potential
into the Hamiltonian, and that this would modify the eigenstates and thereby
change the metric. Moreover, due to the effect of the turning on of that
interaction potential, the change in the metric would be time dependent.
This is just one more case in Quantum Mechanics, in which the measuring
process affects the system which is being measured. However the question is
how drastic is that effect. We claim that it might sometime be quite
drastic, if the new eigenstate of the system and of the measurement device
be changed in such a way that the quantum level is modified. This would
cause a topological change - i.e. a change in the dimension (due to the
change in the number of zones, as defined by the probability distribution)
\footnote
It is well known that in the two slits experiment, if one tries to detect
the ``geodesics (i.e the metric)'' a change in the topology is induced.}
However, if the energy level is slightly shifted - i.e. in such a way that
the subspace on which $\psi \left( x\right)=0 $ changes smoothly, then the change in the metric will be a minor
one.\footnote
The same effect exists classically - since in any process of measuring
a metric, at the end, the information should be sent back to the
observer. However, the information is being carried by some particle which
has some energy. Thus, $T_{00}$ is reduced and with it the metric is
changed. If the observer himself wants to be part of the metric, then by
that he is changing the $T_{00}$ and with it the endowed metric.}\\
\newpage
{\large Acknowledgment}\\
I'm most greatful to Prof. Y. Ne'eman and Prof. L. P. Horwitz for useful
discussions. I am greatful to Prof. A. Connes for inviting me to the IHES, where this work was partially done, and to the IHES for its warm hospitality.\\