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\title{Notes for Gaussian information 1}
\author{eneet2311 }

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      \hbox to 5.78in { {\bf PHYS 7895: Gaussian Quantum Information } \hfill #2 }
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\begin{document}

\lecture{6}{}{Lecturer: Mark M.\ Wilde}{Eneet Kaur}

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

\section{Overview}
In the last lecture, we developed the background required to study single-mode bosonic systems. We studied creation, annihilation, position, and momentum operators and their properties. We then extended the above for multiple-mode bosonic systems, and introduced the canonical symplectic form. 

In this lecture, we will introduce the mean vector in Section~\ref{sec:mean-vector} and the covariance matrix of a bosonic state in Section~\ref{sec:covariance-vector}. We will then derive constraints that are fulfilled by a covariance matrix of a bosonic state in Section~\ref{sec:constraints-cov-matrix}. 

\section{Mean vector and covariance matrix}

Consider the vector $\hat{r}$ of canonical quadrature operators for an $m$-mode bosonic system: 
\begin{equation}
    \hat{r}\equiv \left(\hat{x}_1,\hat{p}_1\ldots,\hat{x}_m,\hat{p}_m\right)^T,
\end{equation}
where $\hat{x}$ refers to the position-quadrature operator and $\hat{p}$ refers to the momentum-quadrature operator.

\subsection{Mean vector} \label{sec:mean-vector}

For a state $\rho$ of multiple modes, the mean vector $\overline{r}$ is given by 
\begin{equation}
    \overline{r}= \left(\overline{x}_1,\overline{p}_1,\ldots,\overline{x}_m,\overline{p}_m\right),
\end{equation}
where the components of the mean vector are defined as follows:
\begin{align}
    \overline{x}_1&=\operatorname{Tr}\left[\hat{x}_1\rho\right]= \operatorname{Tr}\left[\left(\hat{x}_1\otimes\hat{I}\otimes \ldots\hat{I}\right)\rho\right],\\
     \overline{p}_1&=\operatorname{Tr}\left[\hat{p}_1\rho\right]= \operatorname{Tr}\left[\left(\hat{I}\otimes\hat{p}_1\otimes \hat{I}\otimes\ldots\hat{I}\right)\rho\right],\\
     \overline{x}_j &=\operatorname{Tr}[\hat{x}_j\rho]= \langle \hat{x}_j\rangle_{\rho},\\ 
     \overline{p}_j &=\operatorname{Tr}[\hat{p}_j\rho]=\langle \hat{p}_j\rangle_{\rho},
\end{align}
where $\hat{I}$ is the identity operator and $j\in \left\{1,2,\ldots m\right\}$.
Then, as a shorthand we can write the mean vector as 
\begin{equation}
    \overline{r}=\operatorname{Tr}\left[\hat{r}\rho\right]=\left(\operatorname{Tr}[\hat{x}_1\rho],\ \operatorname{Tr}[\hat{p}_1\rho],\ldots, \operatorname{Tr}[\hat{x}_n\rho],\ \operatorname{Tr}[\hat{p}_n\rho]\right)^T
\end{equation}
Just like classical probability distributions need not have a finite mean, a quantum state need not have a finite mean. 

\subsection{Covariance matrix}\label{sec:covariance-vector}

Let us denote the covariance matrix of a quantum state by $\sigma$, and let the entries be given by $\sigma_{jk}$. Let $\hat{r}_j$ be the $j$th element of $\hat{r}$, where $j\in \left\{1,\ldots 2m\right\}$, and $m$ is the number of modes of the quantum state considered. Let us define $\hat{r}^c_j= \hat{r}_j - \langle \hat{r}_j\rangle_{\rho}$. Then, the covariance matrix elements are defined as
\begin{align}
    \sigma_{jk}&= \operatorname{Tr}\left[\left(\hat{r}_j^c\hat{r}_k^c+\hat{r}_k^c\hat{r}_j^c\right)\rho\right]\\
    &=\operatorname{Tr}\left[\left\{\hat{r}_j^c,\hat{r}_k^c\right\}\rho\right]\\
    &= \langle\left\{\hat{r}_j^c,\hat{r}_k^c\right\}\rangle_{\rho},
\end{align}
where $\sigma_{jk} \in \mathbb{R}$ and $k\in \left\{1,\ldots, 2m\right\}$.

Now, consider the total photon number operator 
\begin{equation}
    \hat{N} = \sum_{j=1}^{m} \hat{n}_j,
\end{equation}
where $\hat{n}_j=\hat{a}_j^\dagger \hat{a}_j$. Let us define finite-energy state as the states that fulfill the following constraint: 
\begin{equation}
    \operatorname{Tr}\left[\hat{N}\rho\right]< \infty.
\end{equation}
\begin{proposition}
A state has finite energy iff the elements of $\overline{r}$ and $\sigma$ are finite, that is $\overline{r}_j < \infty$ and $\sigma_{jk}< \infty$.
\end{proposition}
\begin{proof}
Let us first prove that if the state $\rho$ has finite energy then the elements of its mean vector $\overline{r}$ and covariance matrix $\sigma$ are finite. 
The definition of a finite-energy state implies 
\begin{equation}
    \operatorname{Tr}\left[\hat{n}_j \rho\right]< \infty.
\end{equation}
Then observe that, 
\begin{equation}
    \operatorname{Tr}\left[\hat{n}_j\rho\right]=
    \frac{1}{2} \operatorname{Tr}\left[\left(\hat{x}_j^2+\hat{p}_j^2-1\right)\rho\right]< \infty.
\end{equation}
This implies, $\operatorname{Tr}\left[\hat{x}_j^2 \rho\right],\operatorname{Tr}\left[\hat{p}_j^2\rho\right]<\infty$.
Then, we conclude the following: 
\begin{align}
    \vert \overline{x}_j \vert &= \vert \operatorname{Tr}\left[\hat{x}_j \rho\right]\vert\\
    &= \vert \operatorname{Tr} \left[\hat{x} \sqrt{\rho}\sqrt{\rho}\right]\vert\\
    &\leq \sqrt{\operatorname{Tr}\left[\hat{x}_j\sqrt{\rho}\sqrt{\rho}\hat{x}_j\right]\cdot \operatorname{Tr}\left[\sqrt{\rho}\sqrt{\rho}\right]}\\
    &= \sqrt{\operatorname{Tr}\left[\hat{x}_j^2\rho\right]}< \infty.
\end{align}
The first inequality follows from the Cauchy--Schwarz inequality. Similarly, $\vert \overline{p}_j \vert = \vert \operatorname{Tr}\left[\hat{p}_j\rho\right]\vert < \infty $. Therefore, we can conclude that finite-energy states have finite mean vector. 
\par Now, let us prove that the elements of a covariance matrix of finite-energy states are finite. First let us consider the diagonal terms: 
\begin{align}
    \sigma_{jj} &= 2\operatorname{Tr}\left[\left(\hat{r}_j^c\right)^2 \rho\right] \label{eqn:diagonal_finite}\\
    &= 2 \operatorname{Tr}\left[\left(\hat{r}_j-\langle \hat{r}_j\rangle\right)^2 \rho\right]\\
    &= 2 \operatorname{Tr}\left[\hat{r}_j^2\rho + \langle \hat{r}_j \rangle^2 \rho - 2 \hat{r}_j \langle \hat{r}_j \rangle \rho\right]\\
    &= 2 \operatorname{Tr}\left[\hat{r}_j^2\rho-\langle \hat{r}_j\rangle^2 \rho\right]\\
    &= 2\left[\langle \hat{r}_j^2\rangle_\rho-\langle \hat{r}_j\rangle_\rho^2\right] \label{eqn:final1}\\
    & < \infty.
\end{align}
Now, the first term of \eqref{eqn:final1} is finite as seen previously, and the second term is finite since the mean vector of the finite-energy state is finite. Therefore, we conclude that the diagonal elements of a covariance vector of a finite-energy state are finite. Now, we consider the off-diagonal elements $\sigma_{jk}$, where $j\neq k$.
\begin{align}
\vert \sigma_{jk}\vert &= \vert \langle \hat{r}_j^c\hat{r}_k^c+\hat{r}_k^c\hat{r}_j^c\rangle_{\rho}\vert\\
&\leq \vert \langle \hat{r}_j^c\hat{r}_k^c\rangle_{\rho}\vert+ \vert \langle \hat{r}_k^c\hat{r}_j^c\rangle_{\rho}\vert
\end{align}
Now, consider
\begin{align}
\vert \langle \hat{r}_j^c\hat{r}_k^c\rangle_{\rho}\vert &= \vert\operatorname{Tr}\left[\sqrt{\rho}\, \hat{r}_j^c\, \hat{r}_k^c\, \sqrt{\rho}\right]\vert \\
& \leq \sqrt{\operatorname{Tr}\left[\left(\hat{r}_j^c\right)^2\rho\right]\ \operatorname{Tr}\left[\left(\hat{r}_k^c\right)^2\rho\right]}\\ 
&< \infty
\end{align} 
The first inequality follows from the Cauchy--Schwarz inequality, and the second inequality follows from \eqref{eqn:diagonal_finite}.
Now, let us prove the converse. That is, if the state is a finite-energy state, then the covariance matrix is finite.

To prove the opposite implication, consider that
\begin{align}
\operatorname{Tr}\left(\hat{N}\rho\right)
&= \sum_{j=1}^{m}\operatorname{Tr}\left[\hat{n}_j \rho\right]\\
&= \sum_{j=1}^{m}\left[\operatorname{Tr}\left[\hat{x}_j^2 \rho\right]+\operatorname{Tr}\left[\hat{p}_j^2 \rho\right]-1\right]\\
&< \infty
\end{align}
The last inequality follows from the assumed finiteness of the elements of the mena vector and covariance matrix.
\end{proof}
\par Instead of writing all the $2m \times 2m$ elements of the covariance matrix, we condense it to write the covariance matrix as follows:
\begin{equation}
\sigma = \operatorname{Tr}\left[\left\{\left(\hat{r}-\overline{r}\right),\left(\hat{r}-\overline{r}\right)^{\dagger}\right\}\rho\right],
\end{equation}
where, 
\begin{equation}\left\{\left(\hat{r}-\overline{r}\right),\left(\hat{r}-\overline{r}\right)^{\dagger}\right\}= 
\begin{bmatrix}
\left\{\hat{r}_1-\overline{r}_1,\hat{r}_1-\overline{r}_1\right\}&\left\{\hat{r}_1-\overline{r}_1,\hat{r}_2-\overline{r}_2\right\}& \ldots\\ 
\left\{\hat{r}_2-\overline{r}_2,\hat{r}_1-\overline{r}_1\right\}&\left\{\hat{r}_2-\overline{r}_2,\hat{r}_2-\overline{r}_2\right\} & \ldots\\
\vdots & \vdots & \ddots
\end{bmatrix}.
\end{equation}
Then, 
\begin{equation}\sigma = \begin{bmatrix}
\operatorname{Tr}\left[\left\{\hat{r}_1-\overline{r}_1,\hat{r}_1-\overline{r}_1\right\}\rho\right]&\operatorname{Tr}\left[\left\{\hat{r}_1-\overline{r}_1,\hat{r}_2-\overline{r}_2\right\}\rho\right]& \ldots\\ 
\operatorname{Tr}\left[\left\{\hat{r}_1-\overline{r}_1,\hat{r}_2-\overline{r}_2\right\}\rho\right]&\operatorname{Tr}\left[\left\{\hat{r}_2-\overline{r}_2,\hat{r}_2-\overline{r}_2\right\}\rho\right]& \ldots\\
\vdots & \vdots & \ddots
\end{bmatrix}.
\end{equation}

\section{Constraints on covariance matrix}\label{sec:constraints-cov-matrix}

In this section, we establish certain properties of the covariance matrix. We first prove that the covariance matrix (CM) of a vector of random variables is Hermitian and positive semi-definite (PSD). Next, we prove that the covariance matrix of a quantum state fulfills a stronger constraint, that is $\sigma + i \Omega \geq 0$, and that the covariance matrix is positive definite.

\subsection{CM of vector of random variables is PSD}

Consider a covariance matrix $\Sigma$ for a vector of random variables. We now prove that the covariance matrix is positive semi-definite. 

\begin{proposition}
The covariance matrix of a vector of random variables is Hermitian and PSD, that is, $\Sigma = \Sigma^{\dagger}$ and $\Sigma \geq 0$.  
\end{proposition}

\begin{proof}
That the covariance matrix is Hermitian follows from the definition. We now give a proof that the covariance matrix is PSD. 
Let $X$ be a vector of random variables. Then, $X = \left[X_1, X_2, \ldots, X_m\right]^T$, where $X_i$ is a random variable and has realizations in $\mathbb{C}$. Then, 
\begin{equation}
\Sigma = \mathbb{E}\left[\left(X-\mathbb{E}(X)\right)\ \left(X-\mathbb{E}(X)\right)^{\dagger}\right].
\end{equation}
Now, let $\underline{w}$ be a constant vector in $\mathbb{C}^m$. Consider then 
\begin{align}
\underline{w}^\dagger\Sigma \underline{w} &= \underline{w}^{\dagger} \mathbb{E}\left[\left(X-\mathbb{E}(X)\right)\ \left(X-\mathbb{E}(X)\right)^{\dagger}\right]\underline{w}\\
&=   \mathbb{E}\left[\underline{w}^{\dagger}\left(X-\mathbb{E}(X)\right)\ \left(X-\mathbb{E}(X)\right)^{\dagger}\underline{w}\right]\\
&= \mathbb{E}\left[\vert w^\dagger \left(X-\mathbb{E}(X)\right)\vert^2\right] \geq 0.
\end{align}
Since this holds for all $w \in \mathbb{C}^m$, it follows that $\Sigma \geq 0$. 
\end{proof}
 
\subsection{Uncertainity principle of covariance matrix}

Now, we derive an important constraint on the covariance matrix of a quantum state. This is the \textbf{uncertainity principle} for the covariance matrix. 
\begin{theorem}
The covariance matrix $\sigma$ of a quantum state fulfills the following constraint:
\begin{equation}
\sigma + i \Omega \geq 0. 
\end{equation}
\end{theorem}
\begin{proof}
Consider the following $\left(2m \times 2m\right)$ complex matrix given by 
\begin{equation}
\tau = 2 \operatorname{Tr} \left[\left(\hat{r}-\overline{r}\right)\ \left(\hat{r}-\overline{r}\right)^{\dagger}\rho\right]. 
\end{equation}
We first prove that $\tau$ is PSD, and then deduce the statement of the theorem. Let $\underline{w} \in \mathbb{C}^{2m}$. Then, 
\begin{align}
\underline{w}^{\dagger} \tau \underline{w}
&= 2  \operatorname{Tr} \underline{w}^{\dagger}\left[\left(\hat{r}-\overline{r}\right)\ \left(\hat{r}-\overline{r}\right)^{\dagger}\rho\right]\underline{w}\\
& =  2  \operatorname{Tr} \left[\underline{w}^{\dagger}\left(\hat{r}-\overline{r}\right)\ \left(\hat{r}-\overline{r}\right)^{\dagger}\underline{w}\,\rho\right]\\
& = 2\operatorname{Tr}\left[\hat{O}\hat{O}^{\dagger}\rho\right]\\
& \geq 0,
\end{align}
where $\hat{O} = \underline{w}^{\dagger}\left(\hat{r}-\overline{r}\right)$. Since $\hat{O}^\dagger \hat{O}$ is PSD and so is $\rho$, we arrive at the last inequality. Now, the above argument holds for all $\underline{w}\in \mathbb{C}^{2m}$, and so we conclude that $\tau$ is PSD. 
\par Now, consider that
\begin{align}
2 \hat{r}_j \hat{r}_k = \left\{\hat{r}_j,\hat{r}_k\right\}+ \left[\hat{r}_j, \hat{r}_k\right].
\end{align}
This implies, 
\begin{align}
2 \left(\hat{r}-\overline{r}\right)\left(\hat{r}-\overline{r}\right)^{\dagger}&= \left\{\left(\hat{r}-\overline{r}\right),\left(\hat{r}-\overline{r}\right)^{\dagger}\right\}+ \left[\left(\hat{r}-\overline{r}\right),\left(\hat{r}-\overline{r}\right)^{\dagger}\right]\\
& = \left\{\left(\hat{r}-\overline{r}\right),\left(\hat{r}-\overline{r}\right)^{\dagger}\right\}+ \left[\hat{r},\hat{r}^{\dagger}\right]
\end{align}
Then, we obtain the following: 
\begin{align}
\tau &= 2 \operatorname{Tr}\left[\left( \left(\hat{r}-\overline{r}\right)\left(\hat{r}-\overline{r}\right)^{\dagger}\right)\rho\right]\\
&= \operatorname{Tr}\left[ \left\{\left(\hat{r}-\overline{r}\right),\left(\hat{r}-\overline{r}\right)^{\dagger}\right\}\rho\right]+\operatorname{Tr}\left[ \left[\hat{r},\hat{r}^{\dagger}\right]\rho\right]\\
&= \sigma+ i \Omega \geq 0. \label{eqn:constraint} 
\end{align}
The last inequality follows from $\tau \geq 0$. 
\end{proof}
Now, we prove that $\sigma$ is PSD. Note that the eigenvalues of a matrix do not change under a transpose. So, if they are positive, then they remain positive after the transpose of the matrix. Then, 
\begin{equation}\label{eq:constraint}
    \sigma+i\Omega \geq 0,
\end{equation} 
implies
\begin{equation} 
\left(\sigma + i \Omega\right)^T \geq 0.
\end{equation}
which in turn implies, 
\begin{equation}\label{eq:constraint1}
\sigma - i \Omega \geq 0.
\end{equation}
Then combining \eqref{eq:constraint1} with \eqref{eq:constraint}, we obtain that $\sigma \geq 0$. That is, $\sigma$ is a PSD. 

\subsection{CM of quantum states is positive definite}

We now prove that a quantum covariance matrix is in fact positive definite. This makes them more special and easier to work mathematically than classical covariance matrices. 
\begin{proposition}
A quantum covariance matrix is positive definite.
\end{proposition}
\begin{proof}
We prove this statement by contradiction. Let us assume that the quantum covariance matrix is not positive definite. That is, $\exists$ a real, non-zero vector $\psi \in \mathbb{R}^{2m}$, such that $\sigma \ket{\psi}=0$. Then, for $\varepsilon \in \mathbb{R}$, set $\psi (\varepsilon) = \left(I +\varepsilon i \Omega\right)\psi$. By invoking the following assumption $\sigma \psi =0$, and the following facts: $\psi^T \Omega \psi =0 \, \forall \psi \in \mathbb{R}^{2m}$ and $(i\Omega)^2 = I$, we find that
\begin{align}
    &\psi(\varepsilon)^{\dagger}\left(\sigma+ i \Omega\right)\psi(\varepsilon) \notag \\
    & = \psi^T\left(I+\varepsilon i \Omega\right)\left(\sigma + i \Omega\right)\left(I+\varepsilon i \Omega\right)\psi\\
    & = \psi^T\left(I+\varepsilon i \Omega\right)\left(i \Omega + \varepsilon \sigma i \Omega + \varepsilon I\right)\psi \\
    & = \psi^T\left(i\Omega + \varepsilon \sigma i \Omega + \varepsilon I + \varepsilon i \Omega \left(i\Omega+\varepsilon\sigma i \Omega +\varepsilon I\right)\right)\psi\\
    & = \psi^T \left(i\Omega + \varepsilon\sigma i \Omega + 2 \varepsilon I +\varepsilon^2 \Omega^T \sigma\Omega +\varepsilon^2 i\Omega\right)\psi\\
    & = \psi^T\left(2 \varepsilon I + \varepsilon^2 \Omega^T \sigma \Omega\right)\psi\\
    & = 2\varepsilon\psi^T\psi + \varepsilon^2 \left(\Omega \psi\right)^{T}\sigma \left(\Omega \psi\right)
\end{align}
Now, suppose that $\left(\Omega \psi\right)^{T}\sigma \left(\Omega \psi\right) =0$. Then picking $\varepsilon<0$, implies that $2\varepsilon \psi^T\psi < 0$, which contradicts the fact that $\sigma + i \Omega \geq 0$ for any quantum covariance matrix $\sigma$. 

Now, suppose that $\left(\Omega \psi\right)^{T}\sigma \left(\Omega \psi\right) > 0$. Then pick 
$\varepsilon <0$ and such that 
\begin{equation}
    \vert \varepsilon \vert \leq \frac{2 \psi^T\psi}{\left(\Omega \psi\right)^{T}\sigma \left(\Omega \psi\right)}.
\end{equation}
This implies, 
\begin{equation}
    2\varepsilon\psi^T\psi + \varepsilon^2 \left(\Omega \psi\right)^{T}\sigma \left(\Omega \psi\right)<0,
\end{equation}
and there exists $\psi(\varepsilon)$ such that 
\begin{equation}
    \psi(\varepsilon)^{\dagger}\left(\sigma + i \Omega\right)\psi(\varepsilon) <0,
\end{equation}
again contradicting the assumption that $\sigma + i \Omega \geq 0$. 
Hence, $\sigma$ must be positive definite. 
\end{proof}

\subsection{Uncertainity principle for a single-mode bosonic state}

The covariance matrix of a single-mode bosonic state is given as
\begin{equation} \sigma = 
\begin{bmatrix}
2\langle \left( \hat{x}^c\right)^2\rangle_\rho & \langle \left\{\hat{x}^c,\hat{p}^c\right\}\rangle_\rho \\
\langle \left\{\hat{x}^c,\hat{p}^c\right\}\rangle_\rho & 2\langle \left( \hat{p}^c\right)^2\rangle_\rho
\end{bmatrix} = 
\begin{bmatrix}
\sigma_{11} & \sigma_{12}\\
\sigma_{21} & \sigma_{22}
\end{bmatrix}.
\end{equation}

The $2\times 2$ matrix $\sigma$ is the covariance matrix of a single-mode bosonic system if and only if the following constraint holds
\begin{equation}
 \sigma + i\Omega \geq 0 \iff \det(\sigma) \geq 1 \ \mathrm{and} \ \sigma >0.   
\end{equation}
The forward direction of the above statement is easy to prove. We have already shown 
\begin{equation}
    \sigma + i \Omega \geq 0 \implies \sigma >0.
\end{equation}
Now we prove that 
\begin{equation}
    \sigma+i \Omega\geq 0 \implies \det(\sigma) \geq 1.
\end{equation}
The constraint $\sigma +i \Omega \geq 0$ implies that
\begin{equation}
    \det(\sigma+i \Omega) = \sigma_{11}\sigma_{22}- (\sigma_{12}^2+1) \geq  0.
\end{equation}
We thus see that $\det(\sigma) \geq 1$. 
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