\documentclass[11pt]{article}
\usepackage{latexsym}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{hyperref}
%\usepackage{epsfig}
%\usepackage{psfig}

\newcommand{\handout}[5]{
  \noindent
  \begin{center}
  \framebox{
    \vbox{
      \hbox to 5.78in { {\bf PHYS 7895: Gaussian Quantum Information } \hfill #2 }
      \vspace{4mm}
      \hbox to 5.78in { {\Large \hfill #5  \hfill} }
      \vspace{2mm}
      \hbox to 5.78in { {\em #3 \hfill #4} }
    }
  }
  \end{center}
  \vspace*{4mm}
}

\newcommand{\lecture}[4]{\handout{#1}{#2}{#3}{Scribe: #4}{Lecture #1}}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{observation}[theorem]{Observation}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{assumption}[theorem]{Assumption}

% 1-inch margins, from fullpage.sty by H.Partl, Version 2, Dec. 15, 1988.
\topmargin 0pt
\advance \topmargin by -\headheight
\advance \topmargin by -\headsep
\textheight 8.9in
\oddsidemargin 0pt
\evensidemargin \oddsidemargin
\marginparwidth 0.5in
\textwidth 6.5in

\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\bra}[1]{\langle #1 |}
\newcommand{\tr}{\operatorname{Tr}}

\DeclareMathOperator{\arccoth}{arccoth} 

\parindent 0in
\parskip 1.5ex
%\renewcommand{\baselinestretch}{1.25}

\begin{document}

\lecture{10}{}{Lecturer: Mark M.\ Wilde}{Kunal Sharma}

This document is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Unported License.


%\tableofcontents
\section{Overview}

In the last lecture, we represented faithful Gaussian states as thermal states of quadratic Hamiltonians and discussed the Williamson theorem. 

In this lecture, we first review a method to find the symplectic eigenvalues of a positive definite matrix. We then derive a relation between the Hamiltonian matrix and the covariance matrix corresponding to a faithful Gaussian state. Finally,	 we determine formulas for the purity and von Neumann entropy of a Gaussian state. We point readers to \cite{AS17} for background on some of the topics covered in this lecture. 

	\section{Symplectic eigenvalues of a positive definite matrix} 

In this section, we discuss a method to find the symplectic eigenvalues of a positive definite matrix~$M$. 


Let $\Omega$ denote the real, canonical, anti-symmetric form defined as
\begin{align} \label{eq:omega}
\Omega = I_n \otimes \Omega_1,
\end{align}
where \begin{align}\Omega_1  = \begin{bmatrix}
0 & 1 \\
- 1 & 0 
\end{bmatrix} , 
\end{align}
which encodes the canonical commutation relations of the quadrature operators.
Note that $\Omega\Omega^T = -\Omega^2 = I$.

Let $S$ denote a sympletic matrix such that $S \Omega S^T = \Omega$. It follows that such a matrix $S$ is invertible with inverse given by $S^{-1} = \Omega S^T \Omega^T$. We begin by showing that $\Omega S = S^{-T} \Omega$. This is a direct consequence of the fact that $S^T$ is symplectic, which can be seen from the following steps:
\begin{align}
S \Omega S^T &= \Omega, \\
\Rightarrow S \Omega S^T \Omega &= -I, \\
\Rightarrow S \Omega S^T \Omega S &= - S,\\
\Rightarrow S^{-1}S \Omega S^T \Omega S&= - S^{-1}S,\\
\Rightarrow \Omega S^T \Omega S& = -I, \\
\Rightarrow S^T \Omega S &= \Omega~.
\end{align} 
It then follows that 
\begin{align}\label{eq:sym-mat-rel1}
\Omega S = S^{-T}\Omega. 
\end{align}

 
 
As discussed in the  previous lecture, a positive definite $2n \times 2n$ matrix $M$ has the following symplectic decomposition: 
\begin{align}\label{eq:sym-decomp}
M = S D S^T, 
\end{align}
where $d_j>0$ for all $j\in\{1,\ldots, n\}$,
\begin{align}\label{eq:symp-eigval}
D = \bigoplus_{j=1}^{n} d_j \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} = D_n \otimes I_2, 
\end{align}
and 
\begin{align}
D_n = \text{diag}(d_1, d_2, \dots, d_n ).
\end{align}

We now establish a connection between the symplectic eigenvalues of a positive definite matrix $M$ and the eigenvalues of the matrix $i\Omega M$. Consider the following chain of equalities: 
\begin{align}
i \Omega M & = i \Omega S D S^T \\
&  = i \Omega S (D_n \otimes I_2) S^T\\
& = S^{-T} (i \Omega)(D_n \otimes I_2) S^T \\
& = S^{-T} (I_n \otimes i \Omega_1)(D_n \otimes I_2) S^T\\
& = S^{-T} (D_n \otimes i \Omega_1) S^T\\
& = S^{-T}(D_n \otimes -\sigma_Y) S^T\\
& = \underbrace{S^{-T} (I_n \otimes U_2)}_{B} (D_n \otimes -\sigma_Z) \underbrace{(I_n \otimes U_2^{\dagger}) S^T}_{B^{-1}}. \label{eq:eigen-decomp-iomegaM}
\end{align}
The first equality follows from \eqref{eq:sym-decomp}. The second equality follows from \eqref{eq:symp-eigval}. The third equality follows from \eqref{eq:sym-mat-rel1}. The fourth equality follows from the definition of $\Omega$ as defined in \eqref{eq:omega}. The last two equalities follow from the fact that 
\begin{align}\label{eq:omega1-sigmay}
i \Omega_1 = -\sigma_Y = U_2 (- \sigma_Z)U^{\dagger}_2,
\end{align}
where 
\begin{align}
U_2 = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\
i & - i 
\end{bmatrix}. 
\end{align}


From \eqref{eq:eigen-decomp-iomegaM} and from the fact that 
$D_n \otimes -\sigma_Z  = \text{diag}(-d_1, d_1, -d_2, d_2, \dots, -d_n, d_n)$, it follows that the usual eigendecomposition of $i \Omega M$ is given by $B (D_n \otimes -\sigma_Z) B^{-1}$, where 
\begin{align}\label{eq:eigenvec-mat}
B = S^{-T} (I_n \otimes U_2)
\end{align}
is the matrix of eigenvectors. We note that $S^{-T}$ can be expressed in terms of $\Omega$ and $S$. Since $S \Omega S^T  = \Omega$, it follows that $S \Omega S^T \Omega^T = \Omega \Omega^T$. Since $\Omega \Omega^T =I$, $S^{-1} = \Omega S^T \Omega^T$. Therefore, $S^{-T} = \Omega S \Omega^T$. 

Therefore, a method to find the symplectic eigenvalues of a positive definite matrix $M$ is as follows. We first find the usual eigendecomposition  of the matrix $i \Omega M$, and the corresponding eigenvalues  provide the information of symplectic eigenvalues of the matrix $M$. Moreover, the symplectic matrix $S$ corresponding to the transformation $M = S D S^T$, can be found from the eigenvector matrix $B = S^{-T} (I_n \otimes U_2) = \Omega S \Omega^T (I_n \otimes U_2) $ as defined in \eqref{eq:eigenvec-mat}, i.e., $S = B^{-T}( I_n \otimes U_2^T  ) = \Omega^T B ( I_n \otimes U_2^\dag  ) \Omega$. 


\section{Relationship between the Hamiltonian matrix and the covariance matrix for a faithful Gaussian state}

In this section, we derive the following relations between the Hamiltonian matrix and the covariance matrix corresponding to a faithful Gaussian state:
\begin{align}
\sigma &= \coth\bigg(\frac{i \Omega H }{2}\bigg) i \Omega, \\
H & = 2 \arccoth( i \Omega \sigma) i \Omega~. \label{eq:H-sigma}
\end{align}

As discussed in the previous lecture,  a positive definite matrix $H$ can be represented in the following symplectic diagonalized form: 
\begin{align}
H = S^T \bigoplus_{j =1}^n \lambda_j \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} S~,\label{eq:decomp-of-H}
\end{align}
where $\lambda_j > 0, \forall j \in \{1, \dots, n  \}$. 

Moreover, the corresponding covariance matrix $\sigma$ can be written as
\begin{align}\label{eq:simga-symp-rep}
\sigma = S^{-1} \bigoplus_{j=1}^n \coth\bigg( \frac{\lambda_j}{2} \bigg) \begin{bmatrix}
1 & 0 \\
0 & 1 
\end{bmatrix} S^{-T}~,
\end{align}
where $\nu_j \equiv\coth( \lambda_j/2 )  $ for $j\in \{1,\ldots,n\}$ are the symplectic eigenvalues of $\sigma$. 


From  \eqref{eq:eigen-decomp-iomegaM} and \eqref{eq:decomp-of-H}, it follows that
\begin{align}\label{eq:eigen-decomp-iomegaH}
\frac{1}{2} i  \Omega H = \frac{1}{2}S^{-1} ( I_n \otimes U_2 ) (D_n \otimes - \sigma_Z ) (I_n \otimes U^{\dagger}_2) S,
\end{align}
where $D_n  = \text{diag}(\lambda_1, \lambda_2, \dots, \lambda_n )$. 


Consider the following chain of equalities:
\begin{align}
\coth\bigg(\frac{i \Omega H }{2}\bigg) &= S^{-1} (I_n \otimes U_2) \coth\bigg(\frac{D_n \otimes - \sigma_Z}{2}\bigg) (I_n \otimes U^{\dagger}_2) S\\
& = S^{-1} (I_n \otimes U_2)\big(\coth(D_n/2) \otimes - \sigma_Z\big) (I_n \otimes U^{\dagger}_2) S\\
& = S^{-1} (\coth(D_n/2)\otimes i \Omega_1)S\\
& = S^{-1}(\coth(D_n/2)\otimes I_2) (I_n \otimes i \Omega_1) S\\
& = S^{-1} (\coth(D_n/2) \otimes I_2) i \Omega S\\
&  = S^{-1} (\coth(D_n/2) \otimes I_2) S^{-T} i \Omega \\
& = \sigma i \Omega~.
\end{align}
The first equality follows from \eqref{eq:eigen-decomp-iomegaH}. The second equality follows from the fact that $\coth(\cdot)$ is an odd function. The third equality follows from \eqref{eq:omega1-sigmay}. The fifth equality follows from \eqref{eq:omega}. The sixth equality follows from \eqref{eq:sym-mat-rel1}. The last equality follows from \eqref{eq:simga-symp-rep}. 

Therefore, we get 
\begin{align}
\coth\bigg(\frac{i \Omega H}{2}\bigg) i \Omega & = \sigma (i \Omega) (i \Omega)\\
& = \sigma. 
\end{align}

Similarly, the relation in \eqref{eq:H-sigma} can be derived. 


\section{Uncertainty relation and symplectic eigenvalues of a covariance matrix}

Previously, we proved that the  following uncertainty relation holds for any $n$-mode quantum state that has a finite covariance matrix $\sigma$:
\begin{align} \label{eq:uncertainty-relation}
\sigma + i \Omega \geq 0~. 
\end{align}

 We now discuss the restriction imposed by the uncertainty relation in \eqref{eq:uncertainty-relation} on the symplectic eigenvalues of $\sigma$. Let $S$ be the symplectic matrix diagonalizing $\sigma$ as 
 \begin{align}
 S \sigma S^T = D = \bigoplus_{j=1}^n d_j \begin{bmatrix}
 1 & 0 \\
 0 & 1
 \end{bmatrix}. 
 \end{align}
  
 We now prove that \eqref{eq:uncertainty-relation} implies  $d_j \geq 1, \forall j$.
 Consider the following chain of inequalities:
 \begin{align}
 \sigma + i \Omega &\geq 0 \\
 \Rightarrow S( \sigma + i \Omega) S^T & \geq 0 \\
\Rightarrow S \sigma S^T + i S \Omega S^T &\geq 0 \\
\Rightarrow D + i \Omega &\geq 0 \\
\Rightarrow \bigoplus_{j=1}^{n} d_j \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} + i \begin{bmatrix}
0 & 1 \\
-1 & 0 
\end{bmatrix} &\geq 0\\
\Rightarrow \bigoplus_{j=1}^{n} \begin{bmatrix}
d_j & i \\
-i & d_j 
\end{bmatrix} &\geq 0\\
\Rightarrow \begin{bmatrix}
d_j & i \\
-i & d_j 
\end{bmatrix} &\geq 0, \forall j. \label{eq:sym-eigval-pos}
 \end{align}
Since the eigenvalues of $\begin{bmatrix}
d_j & i \\
-i & d_j 
\end{bmatrix}$ are $d_j+1$ and $d_j - 1$, it follows from \eqref{eq:sym-eigval-pos} that $d_j \geq 1, \forall j$. 

Thus, any quantum covariance matrix $\sigma$ (i.e., obeying \eqref{eq:uncertainty-relation}) has all of its symplectic eigenvalues greater than or equal to one.

\section{Purification of a Gaussian state}

In this section, we study Gaussian purifications of Gaussian states. We begin by determining the mean vector and covariance matrix for a  tensor product of two Gaussian states. 


\subsection{Tensor product of two Gaussian states}

Let $\bar{r}_A$  denote the mean vector and $\sigma_A$ denote the covariance matrix of a Gaussian state $\rho_A$. 
Let $\bar{r}_B$ denote the mean vector and $\sigma_B$ denote the covariance matrix of a Gaussian state $\rho_B$. Then the mean vector of the tensor product state $\rho_A \otimes \rho_B$ is given by
 \begin{align}
\bar{r}_{AB} \equiv
\begin{bmatrix}
\bar{r}_A \\
\bar{r}_B
\end{bmatrix}.
\end{align} 
Moreover, the covariance matrix of $\rho_A \otimes \rho_B$ is given by
\begin{align}
\sigma_{AB} \equiv \sigma_A \oplus \sigma_B  = \begin{bmatrix}
\sigma_A & 0 \\
0 & \sigma_B
\end{bmatrix}. 
\end{align}

Similarly, if the  mean vector of a Gaussian state is $\begin{bmatrix}
\bar{r}_A \\
\bar{r}_B
\end{bmatrix}$ and the covariance matrix is $\begin{bmatrix}
\sigma_A & 0 \\
0 & \sigma_B
\end{bmatrix}$, then the Gaussian state is a tensor product of two Gaussian states. 

\subsection{Gaussian purifications of Gaussian states} 



A thermal state with mean number of photons $\bar{n} \geq 0$ can be expressed in the photon-number basis as follows. 
\begin{align}\label{eq:thermal-state}
\theta(\bar{n}) = \frac{1}{\bar{n}+1} \sum_{n=0}^{\infty} \bigg(\frac{\bar{n}}{\bar{n}+1}\bigg)^n  \ket{n}\bra{n}~. 
\end{align}
Alternatively, 
\begin{align}
\theta(\lambda) = \frac{1}{z(\lambda)} \sum_{n=0}^{\infty} \exp(-\lambda (n + 1/2)) \ket{n}\bra{n}, 
\end{align}
where $z(\lambda) = (e^{\lambda/2} - e^{-\lambda/2})^{-1}$ for $\lambda >0$   (note that $\lambda=\ln(1+ 1/\bar{n})$). 

A purification of the thermal state  $\theta_A(\bar{n})$ is given by the following two-mode squeezed vacuum (TMS) state:
\begin{align}
\ket{\psi_{\operatorname{TMS}}(\bar{n})}_{AR} = \frac{1}{\sqrt{ \bar{n}+1 }} \sum_{n=0}^{\infty} \sqrt{\bigg(\frac{\bar{n}}{\bar{n}+1}\bigg)^n} \ket{n}_A \ket{n}_R,
\end{align}
where $R$ is a reference system. 
 
The covariance matrix of the two-mode squeezed vacuum state $\ket{\psi_{\operatorname{TMS}}(\bar{n})}_{AR} $ is given by
\begin{align}
\begin{bmatrix}
2 \bar{n} + 1 & 0 & 2 \sqrt{ \bar{n} (\bar{n} + 1)  } & 0 \\
0 & 2 \bar{n} + 1 & 0 & - 2 \sqrt{ \bar{n} (\bar{n} + 1)  }\\
2 \sqrt{ \bar{n} (\bar{n} + 1)  }& 0 & 2\bar{n}+1 & 0 \\
0 & - 2 \sqrt{ \bar{n} (\bar{n} + 1)  }& 0 & 2 \bar{n}+1
\end{bmatrix},
\end{align}
which can be written in the following compact form:
\begin{align}
\begin{bmatrix}
(2 \bar{n}+1) I & 2 \sqrt{ \bar{n} (\bar{n} + 1)  }\sigma_Z \\
2 \sqrt{ \bar{n} (\bar{n} + 1)  } \sigma_Z & (2\bar{n}+1) I 
\end{bmatrix}. 
\end{align}

By the Williamson theorem, any $n$-mode Gaussian state $\rho$ can be written as 
\begin{align}\label{eq:Will-decomp-nmodeGstate}
\rho = \hat{D}_{-\bar{r}} \hat{S} \left[  \bigotimes_{j = 1}^n \theta_{A_j}(\bar{n}_j)  \right] \hat{S}^{\dagger} \hat{D}_{\bar{r}},
\end{align}
where $\hat{S}$ is a unitary generated by a quadratic Hamiltonian. Then a Gaussian purification of $\rho$ is given by 
\begin{align}
\left[ \hat{D}_{-\bar{r}} \hat{S}   \right]_{A^n} \bigotimes_{j=1}^n \ket{\psi_{\operatorname{TMS}}(\bar{n}_j)}_{A_j R_j}~. 
\end{align}

The mean vector of this purification is $\begin{bmatrix}
\bar{r}\\
0 
\end{bmatrix}$. Moreover, the covariance matrix of this purification is 
\begin{align}\begin{bmatrix}
	\sigma &   S\bigoplus_{j=1}^n 2 \sqrt{  \bar{n}_j (\bar{n}_j +1) }\sigma_Z   \\
	\left(\bigoplus_{j=1}^n 2 \sqrt{  \bar{n}_j (\bar{n}_j +1) }\sigma_Z \right)S^T & \bigoplus_{j=1}^n (2 \bar{n}_j +1) I_2 
\end{bmatrix}~. 
	\end{align}

One can arrive at this conclusion from the fact that 
\begin{align}
\sigma = S \left(\bigoplus_{j=1}^n (2 \bar{n}_j + 1)I_2\right)S^T
\end{align}
and the covariance matrix  for $\bigotimes_{j=1}^n \ket{ \psi_{\operatorname{TMS}}(\bar{n}_j)  }_{A_j R_j}$ is 
\begin{align}
\begin{bmatrix}
\bigoplus_{j=1}^n (2\bar{n}_j + 1) I_2 & \bigoplus_{j=1}^n 2 \sqrt{\bar{n}_j (\bar{n}_j+1)} \sigma_Z \\
\bigoplus_{j=1}^n 2 \sqrt{\bar{n}_j (\bar{n}_j+1)} \sigma_Z & \bigoplus_{j=1}^n(2\bar{n}_j + 1) I_2 
\end{bmatrix}. 
\end{align}

We note that the symplectic matrix for the unitary evolution $\hat{S}_{A^n} \otimes I_{R^n}$ is given by
\begin{align}
\begin{bmatrix}
S & 0 \\
0 & I
\end{bmatrix}. 
\end{align}


\section{Purity of a quantum state}
The purity of a quantum state $\rho$ is defined as $\tr\{ \rho^2   \}$. We now show that $\tr\{ \rho^2 \} \leq 1$. Consider the following spectral decomposition of the state $\rho$:
\begin{align}
\rho = \sum_x \lambda_x \ket{\phi_x} \bra{\phi_x}. 
\end{align}
Then
\begin{align}
\tr\{ \rho^2 \} = \sum_x \lambda^2_x~. 
\end{align}
Since $\lambda_x \leq 1 \Rightarrow \lambda_x^2 \leq 1$ and since $\sum_x \lambda_x = 1 \Rightarrow \sum_x \lambda_x^2 \leq 1$. Therefore, if a state is pure, then $\tr\{ \rho^2  \} =1$.


We now show that if $\tr\{  \rho^2  \} = 1$, then the state is pure. Consider that 
\begin{align}
1 &= \tr\{ \rho^2  \}\\
& = \sum_x \lambda_x^2~.  
\end{align}

Moreover, $\tr\{  \rho\} = \sum_x \lambda_x =1 \Rightarrow \tr\{  \rho \}^2 = \sum_{x, y} \lambda_x \lambda_y = 1$. 

Consider the following chain of inequalities:
\begin{align}
\Rightarrow 0 & =\tr\{  \rho^2 \}  - \tr\{ \rho \}^2 \\
& = \sum_x \lambda_x^2 - \bigg[  \sum_{x, y} \lambda_x \lambda_y  \bigg]\\
& = \sum_x \lambda_x^2 - \bigg[ \sum_x \lambda_x^2 + \sum_{x \neq y } \lambda_x \lambda_y  \bigg]\\
& =\sum_{x \neq y} \lambda_x \lambda_y~. \label{eq:lambdaxy}
\end{align}
Since $\lambda_x, \lambda_y\geq0$, the only possibility to satisfy \eqref{eq:lambdaxy} is that $\lambda_x = 1$ and $\lambda_y = 0, \forall y \neq x$. Thus, $\tr\{  \rho^2 \} = 1$ implies that $\rho$ is a pure state.

\subsection{Purity of a Gaussian state}
In this section, we calculate the purity for Gaussian states. From the Williamson decomposition of an $n$-mode Gaussian state as defined in \eqref{eq:Will-decomp-nmodeGstate} and from the fact that the purity is invariant under  unitary transformations, we get
\begin{align}
\tr\{ \rho^2  \} = \prod_{j=1}^n \tr\{   \theta^2(\bar{n}_j) \}~. 
\end{align} 

Consider the following chain of equalities:
\begin{align}
\tr\{ \theta^2(\bar{n}_j)  \} & = \frac{1}{( \bar{n}_j +1  )^2} \sum_{n = 0}^{\infty}   \bigg(  \frac{\bar{n}_j }{\bar{n}_j +1} \bigg)^{2n}\\
& = \frac{1}{( \bar{n}_j +1  )^2}  \frac{1}{ 1 - \big(\bar{n}_j/(\bar{n}_j +1
	)\big)^2   } \\
& =  \frac{1}{(\bar{n}_j +1)^2 - \bar{n}_j^2 }\\
& = \frac{1}{2 \bar{n}_j + 1 }\\
& = \frac{1}{\nu_j}~,
\end{align}
where $\nu_j$ denotes the symplectic eigenvalue of $\theta(\bar{n}_j)$. The first equality follows from the definition of a thermal state as defined in \eqref{eq:thermal-state}. The second equality follows from the sum of an infinite geometric series. 

Therefore, 
\begin{align}
\tr\{  \rho^2 \} &= \prod_{j =1}^n \frac{1}{\nu_j}\\
& = \sqrt{   \prod_{j =1}^n \frac{1}{\nu_j^2}   }\\
& = \frac{1}{\sqrt{  \prod_{j =1}^n \nu_j^2   }}\\
& = \frac{1}{ \text{Det}(\sigma)  }~.
\end{align}
The last equality follows from \eqref{eq:simga-symp-rep} and from the fact that for any symplectic matrix $S$, $\text{Det}(S) =1$. 

Therefore, the purity of a Gaussian state is 
\begin{align}
\tr\{\rho^2  \} = \frac{1}{\sqrt{ \text{Det}(\sigma)   }}~,
\end{align}
which implies that a Gaussian state is pure if and only if $\text{Det}(\sigma) = 1$. Since $\nu_j \geq 1$, an equivalent condition for the purity of a Gaussian state is that all symplectic eigenvalues are equal to one. 



\section{Entropy of a Gaussian state}

In this section, we find an expression for the von Neumann entropy of a Gaussian state. 


The von Neumann entropy of a quantum state $\rho$ is defined as
\begin{align}
S(\rho ) \equiv - \tr\{  \rho\ln\rho \}~. 
\end{align}

We begin by expressing a thermal state with the mean photon number $\bar{n}$ in the following form:
\begin{align}
\theta(\bar{n}) & = \frac{1}{\bar{n}+1} \sum_{n =0}^{\infty} \bigg( \frac{\bar{n}}{\bar{n}+1}  \bigg)^n \ket{n} \bra{n}\\
& = \frac{1}{\bar{n}+ 1}\bigg(\frac{\bar{n}}{\bar{n}+1}  \bigg)^{ \hat{n}}~. 
\end{align}

Consider the following chain of equalities:
\begin{align}
 -\tr\{   \theta(\bar{n}) \ln\theta(\bar{n})   \} 
&  = - \tr\left\{ \theta(\bar{n}) \ln\frac{1}{\bar{n}+1} \bigg(\frac{\bar{n}}{\bar{n}+1}\bigg)^{\hat{n}}   \right\}\\
& = -\tr\left\{ \theta(\bar{n})\ln\bigg( \frac{1}{\bar{n}+1} \bigg)  \right \} - \tr\left\{  \theta(\bar{n}) \hat{n} \ln\bigg(  \frac{\bar{n}}{\bar{n}+1}  \bigg)  \right\}\\
& = \ln(\bar{n}+1) - \ln\bigg( \frac{\bar{n}}{\bar{n}+1}  \bigg) \tr\{ \theta(\bar{n}) \hat{n}  \}\\
& = \ln(\bar{n}+1) - \ln\bigg(  \frac{\bar{n}}{\bar{n}+1} \bigg) \bar{n}\\
& = (\bar{n}+1) \ln(\bar{n}+1) - \bar{n} \ln\bar{n}\\
& \equiv g(\bar{n})~. 
\end{align}

From unitary invariance and additivity of the von Neumann entropy, we get 
\begin{align}
S(\rho) = S\!\left( \bigotimes_{j=1}^n \theta(\bar{n}_j)  \right),
\end{align}
where $\rho$ is an $n$-mode Gaussian state as defined in \eqref{eq:Will-decomp-nmodeGstate}. Therefore, 
\begin{align}
S(\rho) &=  \sum_{j =1}^n S(\theta(\bar{n}_j))\\
& = \sum_{j=1}^n g(\bar{n}_j) 
\end{align}

We now derive an alternative formula for the von Neumann entropy of faithful Gaussian states. Let 
\begin{align}
\rho &= \frac{1}{\sqrt{\text{Det}[(\sigma + i \Omega)/2   )]}} \exp\bigg({ -\frac{1}{2} (  \hat{r} - \bar{r}  )^T H ( \hat{r}  - \bar{r} )    }\bigg)\\
& = \hat{D}_{ -\bar{r}  }\Bigg[   \frac{\exp\big({ - \frac{1}{2}   \hat{r}^T H \hat{r}    }\big)}{\sqrt{\text{Det}[(\sigma + i \Omega)/2   )]} } \Bigg]\hat{D}_{\bar{r}}~
\end{align}
and let
\begin{align}
\rho_0 = \frac{\exp\big({- \frac{1}{2}   \hat{r}^T H \hat{r}    }\big)}{\sqrt{\text{Det}[(\sigma + i \Omega)/2   )]} }~. 
\end{align}
Then from unitary invariance of the von Neumann entropy, we get
\begin{align}
S(\rho) & = S(\rho_0)\\
& = -\tr\{ \rho_0  \ln \rho_0   \}\\
& = -\tr\{ \rho_0 \ln\frac{\exp({-\frac{1}{2}\hat{r}^T H \hat{r}})}{\sqrt{\text{Det}[ (\sigma+i \Omega)/2   ]} }   \}\\
& = -\tr\{   \rho_0 \ln\frac{1}{\sqrt{\text{Det}[ (\sigma+i \Omega)/2   ]} }   \} - \tr\{ \rho_0 \ln  \exp({-\frac{1}{2}\hat{r}^T H \hat{r}}) \}\\
& = \frac{1}{2} \ln\text{Det}[ (\sigma+i \Omega)/2   ] + \frac{1}{2} \tr\{  \rho_0 \hat{r}^T H \hat{r}  \}~.
\end{align}
We now focus on the second term of the aforementioned equation. 
\begin{align}
\tr\{\rho_0 \hat{r}^T H \hat{r}   \} & = \tr\{   \rho_0 \sum_{j, k} \hat{r}_j H_{j, k} \hat{r}_k   \}\\
& = \sum_{j, k} H_{j, k} \tr\{  \rho_0 \hat{r}_j \hat{r}_k  \}\\
& = \frac{1}{2} \sum_{j, k} H_{j, k} \tr\{  \rho_0 (  \{ \hat{r}_j, \hat{r}_k  \} + [\hat{r}_j , \hat{r}_k  ]        )    \}\\
& = \frac{1}{2}  \sum_{j, k} H_{j, k}  (  \sigma_{j, k} + i \Omega_{j, k } )\\
& = \frac{1}{2} \sum_{j, k} H_{j, k} \sigma_{j, k} - \frac{i}{2} \sum_{j, k} H_{j, k} \Omega_{k, j }\\
& = \frac{1}{2} \tr\{H \sigma   \} - \frac{i}{2} \tr\{  H \Omega  \}\\ % H symmetric and omega anti, and trace invarinat under transpose
& = \frac{1}{2} \tr\{  H \sigma  \},
\end{align}
where we  used the fact that $\tr\{H \Omega  \}  = 0$, which holds because $H$ is symmetric and $\Omega$ is antisymmetric. 

Therefore, 
\begin{align}
S(\rho) = \frac{1}{2} \ln\text{Det}[ (\sigma+i \Omega)/2   ]+ \frac{1}{4} \tr\{ H\sigma  \}~. 
\end{align}
Moreover, from \eqref{eq:H-sigma} it follows that
\begin{align}
S(\rho) =  \frac{1}{2} \ln\text{Det}[ (\sigma+i \Omega)/2   ]+ \frac{1}{2} \tr\{ \arccoth( i \Omega \sigma) i \Omega \sigma  \}~.
\end{align}
This latter expression is valid for pure Gaussian states, with the expression $\tr\{ \arccoth( i \Omega \sigma) i \Omega \sigma  \}$ understood in a limiting sense.



 
 
\bibliography{Ref}





\bibliographystyle{alpha}

\end{document}
