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				\hbox to 5.78in { {\bf PHYS 7895: Gaussian Quantum Information } \hfill #2 }
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\begin{document}
	
	\lecture{11}{}{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 reviewed a method to find the symplectic eigenvalues of a positive definite matrix. We then derived a relation between the Hamiltonian matrix and the covariance matrix corresponding to a faithful Gaussian state. Finally,	 we reviewed the conditions for the purity of a Gaussian state and found an expression for the  von Neumann entropy of a Gaussian state 
	
	In this lecture, we find the quantum relative entropy and the R\'enyi entropies for  faithful Gaussian states. We point readers to \cite{AS17} for background on topics covered in this lecture. 


\section{Relative entropy of faithful Gaussian states}

The quantum relative entropy $D(\rho\Vert \tau)$ of a density operator $\rho$ and a positive definite operator $\tau$ is defined as follows:
\begin{align}
D(\rho \Vert \tau) = \tr\{\rho (\ln \rho - \ln \tau   )    \}~. 
\end{align}
This is the formula for the finite-dimensional case, and it turns out to be legitimate for faithful Gaussian states.

In the last lecture we showed that 
\begin{align}
\tr\{ \rho \ln\rho  \} = - \frac{1}{2} \ln \text{Det}[ (\sigma_{\rho} + i \Omega)/2  ] - \frac{1}{4} \tr\{ H_{\rho} \sigma_{\rho}  \}
\end{align}

We now calculate $- \tr\{  \rho \ln\tau  \}.$ Consider that 
\begin{align}
\rho = \hat{D}_{- \bar{r}_{\rho}  } \rho_0 \hat{D}_{\bar{r}_{\rho}}, 
\end{align}
where $\rho_0$ has zero mean and the covariance matrix is  $\sigma_{\rho}$. Then using cyclicity of trace and functional calculus of $\ln(\cdot)$, we find that 
\begin{align}
- \tr\{ \rho \ln \tau   \} & = - \tr\{ \rho_0 \ln \hat{D}_{ \bar{r}_{\rho}  }   \tau \hat{D}_{ -\bar{r}_{\rho }  } \}. 
\end{align}

Let
\begin{align}
\tau = \frac{\exp[{  - (1/2)  (\hat{r} - \bar{r}_{\tau}  )^T H_{\tau} ( \hat{r} - \bar{r}_{\tau}  )         }]}{\sqrt{\text{Det}[ (\sigma_{\tau}  +i \Omega)/2  ]}}. 
\end{align}
Then
\begin{align}
\hat{D}_{ \bar{r}_{\rho} } \tau \hat{D}_{ - \bar{r}_{\rho}  } = \frac{\exp[ -(1/2) (\hat{r} - \delta)^T H_{\tau} (\hat{r} - \delta )    ] }{\sqrt{\text{Det} [ (\sigma_{\tau}+ i \Omega)/2   ] }}
\end{align}
with $\delta = \bar{r}_{\tau} - \bar{r}_{\rho}$. Therefore, 
\begin{align}
& - \tr\{\rho_0 \ln \hat{D}_{ \bar{r}_{\rho} } \tau \hat{D}_{- \bar{r}_{\rho}}    \} \notag\\
&\qquad = - \tr\{ \rho_0 \ln \frac{1}{\sqrt{\text{Det}[( \sigma_{\tau} + i \Omega    )/2]   }}  \} + \tr\{\rho_0 \bigg( \frac{1}{2} (\hat{r} - \delta)^T H_{\tau} ( \hat{r} - \delta  ) \bigg)   \}\\
&\qquad  = \frac{1}{2} \ln\text{Det}[( \sigma_{\tau} + i \Omega  )/2] + \frac{1}{2} \tr\{ \rho_0 (\hat{r} - \delta )^T H_{\tau} (\hat{r} - \delta)   \} ~. \label{eq:2nd-term-rel-ent} 
\end{align}
We now focus on the second term. 
\begin{align}
\frac{1}{2} \sum_{j, k} \tr\{\rho_0 &( \hat{r}_j - \delta_j  )(  \hat{r}_k -  \delta_k  )      \}H^{\tau}_{j, k} \nonumber\\
&= \frac{1}{2} \sum_{j, k} ( \tr\{ \rho_0 \hat{r}_j \hat{r}_k  \} - \tr\{\rho_0 \hat{r}_k  \}\delta_j - \tr\{\rho_0 \hat{r}_j  \}\delta_k + \tr\{ \rho_0 \}\delta_j \delta_k       ) H^{\tau}_{j, k}\\
& = \frac{1}{2} \sum_{j, k} (\tr\{  \rho_0 \hat{r}_j \hat{r}_k \} + \delta_j \delta_k   ) H^{\tau}_{j, k}\\
& = \frac{1}{2}  \sum_{j, k} \tr\{\rho_0 \hat{r}_j \hat{r}_k    \} H^{\tau}_{j, k}  + \frac{1}{2} \delta^T H_{\tau} \delta\\
& = \frac{1}{4}  \tr\{\sigma_{\rho} H_{\tau}  \}  + \frac{1}{2} \delta^T H_{\tau} \delta~. \label{eq:term-rel-ent}
\end{align}

From \eqref{eq:2nd-term-rel-ent} and \eqref{eq:term-rel-ent}, we get
\begin{align}
-\tr\{ \rho \ln \tau  \} = \frac{1}{2}\ln\text{Det}[ (\sigma_{\tau} + i \Omega)/2   ] + \frac{1}{4} \tr\{  \sigma_{\rho} H_{\tau} \} + \frac{1}{2} \delta^T H_{\tau} \delta~,
\end{align}
where $\delta = \bar{r}_{\tau} - \bar{r}_{\rho}$. 

Therefore, the quantum relative entropy of two Gaussian states $\rho$ and $\tau$ is given by
\begin{align}
D(\rho \Vert \tau) & = \frac{1}{2}\bigg[ \ln\bigg( \frac{\text{Det} [ (\sigma_{\tau}  + i \Omega)/2   ] }{\text{Det} [ (\sigma_{\rho}  + i \Omega)/2   ]  }   \bigg)    + \frac{1}{2} \tr\{  \sigma_{\rho} ( H_{\tau} - H_{\rho}  ) \}  + \delta^T H_{\tau} \delta       \bigg]\\
& = \frac{1}{2}\bigg[ \ln\bigg( \frac{\text{Det} [ \sigma_{\tau}  + i \Omega   ] }{\text{Det} [ \sigma_{\rho}  + i \Omega   ]  }   \bigg)    + \frac{1}{2} \tr\{  \sigma_{\rho} ( H_{\tau} - H_{\rho}  ) \}  + \delta^T H_{\tau} \delta       \bigg]~. 
\end{align}
The aforementioned expression is finite whenever $\tau$ is faithful. 

\section{Computing R\'enyi entropies and powers of Gaussian states}

In this section, we first find R\'enyi entropies of Gaussian states in terms of symplectic eigenvalues. We then find the power of Gaussian states in terms of the mean vector and the covariance matrix. 

\subsection{R\'eny entropies of Gaussian states}

The quantum R\'enyi entropy of a quantum state $\rho$ is defined as 
\begin{align}\label{eq:Renyi-ent}
S_{\alpha}(\rho) = \frac{1}{1 - \alpha} \ln\tr\{  \rho^{\alpha} \},
\end{align}
for $ \alpha \in (0, 1) \cup (1, \infty)$. Moreover, 
\begin{align}
\lim_{\alpha \to 1} S_{\alpha}(\rho)  =S(\rho). 
\end{align}

Our goal is to find $\tr\{  \rho^{\alpha} \}$. Using the fact that 
\begin{align}
\rho = \hat{D}_{- \bar{r}} \hat{S} \left(   \bigotimes_{j =1}^n \theta(\bar{n}_j)  \right) \hat{S}^{\dagger} \hat{D}_{\bar{r}}~,
\end{align}
we find that
\begin{align}
\tr\{\rho^{\alpha}   \} = \prod_{j =1}^n \tr\{ \theta(\bar{n}_j)^{\alpha}  \}~. 
\end{align}

Consider the following chain of equalities: 
\begin{align}
\tr\{ \theta(\bar{n})^{\alpha}   \} &= \frac{1}{(\bar{n}+1)^{\alpha}} \sum_{n = 0}^{\infty} \bigg( \frac{\bar{n}}{\bar{n}+1}  \bigg)^{\alpha n}\\
& = \frac{1}{(\bar{n}+1)^{\alpha}} \frac{1}{1 - (\bar{n}/(\bar{n}+1))^{\alpha}}\\
& = \frac{1}{ ( \bar{n} + 1  )^{\alpha} - \bar{n}^{\alpha}  }~,
\end{align}
which implies that 
\begin{align}
\tr\{ \rho^{\alpha} \} = \prod_{j =1}^n  \frac{1}{  ( \bar{n}_j + 1  )^{\alpha} - \bar{n}_j^{\alpha}      }. 
\end{align}

In terms of symplectic eigenvalues 
$\nu_j = 2 \bar{n}_j +1$, $\tr\{ \rho^{\alpha}  \}$ is given by
\begin{align}
\tr\{ \rho^{\alpha}  \} = \prod_{j =1}^n \frac{2^{\alpha}}{  (\nu_j +1)^{\alpha} - (\nu_j-1)^{\alpha}   }. \label{eq:tr-rhoalpha}
\end{align}

Therefore, from \eqref{eq:Renyi-ent} and \eqref{eq:tr-rhoalpha}, it follows that
\begin{align}
S_{\alpha} (\rho)  = \frac{1}{1- \alpha} \sum_{j =1}^n \ln\bigg(\frac{2^{\alpha}}{  (\nu_j +1)^{\alpha} - (\nu_j-1)^{\alpha}   }\bigg)~. 
\end{align}

The R\'enyi entropy can also be expressed as
 \begin{align}
 S_{\alpha}(\rho) &= \frac{\alpha}{1- \alpha} \ln\tr\{  \rho^{\alpha}  \}^{  1/\alpha }\\
 & = \frac{\alpha}{1- \alpha} \ln \left\Vert \rho \right\Vert_{\alpha}~. 
 \end{align}
 Therefore, 
 \begin{align}
 S_{\infty}(\rho ) = - \ln \Vert \rho \Vert_{\infty} \equiv  S_{\operatorname{min}}(\rho)
 \end{align}
 
We now find  $ S_{\infty}(\rho ) $ using the fact that $ \left\Vert \theta(\bar{n})\right\Vert_{\infty} = 1/(\bar{n}+1)$. Consider the following chain of equalities:
 \begin{align}
 S_{\infty}(\rho ) &=  - \ln \left\Vert \bigotimes_{j = 1}^n \theta(\bar{n}_j) \right\Vert_{\infty}\\
 & = - \ln \prod_{j=1}^n \left\Vert \theta(\bar{n}_j) \right\Vert_{\infty}\\
 & = \sum_{j =1}^n -\ln( 1/(\bar{n}_j +1) )\\
 & = \sum_{j=1}^{n} \ln(\bar{n}_j +1  )\\
 & = \sum_{j =1}^n \ln[(\nu_j+1)/2   ]~. 
 \end{align}
 
 
 In general,  the following relation holds for the R\'enyi entropy:
 \begin{align}
 S_{\alpha}(\rho) \geq S_{\beta} (\rho), 
 \end{align}
for $ \alpha \leq \beta $. 

We now find the difference between $S(\rho)$ and $S_{\infty}(\rho)$. Consider the following chain of inequalities:
\begin{align}
S(\rho) - S_{\infty} (\rho) &=  \sum_{j =1}^n g(\bar{n}_j) - \ln(\bar{n}_j +1)\\
& = \sum_{j =1}^n ( \bar{n}_j + 1  )\ln( \bar{n}_j +1  ) - \bar{n}_j \ln \bar{n}_j - \ln(\bar{n}_j +1  )\\
& = \sum_{j =1}^n \ln [ ( (\bar{n}_j + 1)/\bar{n}_j  )^{\bar{n}_j}     ]\\
&\leq \sum_{j =1}^n \ln(e)\\
& = n~. 
\end{align}
Therefore, the difference between $S(\rho)$ and $S_{\infty}(\rho)$ never exceeds the number of modes. 


\subsection{Power of Gaussian state}

 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}

Therefore, 
 \begin{align}
 \rho^{\alpha} \propto \hat{D}_{- \bar{r}} \exp( -(1/2) \hat{r}^T \alpha H \hat{r}   ) \hat{D}_{\bar{r}}. 
 \end{align}
 
 Let $H_{(\alpha)} = \alpha H$. Then there exists a corresponding $\sigma_{(\alpha)}$ such that 
 \begin{align}
 \frac{\rho^{\alpha}}{\tr\{ \rho^{\alpha}   \}}  &= \frac{\hat{D}_{-\bar{r}} \exp(-(1/2)\hat{r}^T H_{(\alpha  )} \hat{r}) \hat{D}_{\bar{r}}   }{\sqrt{ \text{Det}[ (\sigma_{(\alpha)}+i \Omega)/2   ]   }} \\
 & = \frac{\exp(-(1/2)(\hat{r} - \bar{r})^T H_{ (\alpha )} (\hat{r}- \bar{r}))}{\sqrt{ \text{Det}[ (\sigma_{(\alpha)}+i \Omega)/2   ]   }}~.
 \end{align}
 
To determine $\sigma_{(\alpha)}$ in terms of $\sigma$, we use the following formulas derived in the previous lecture: 
 \begin{align}
 \sigma &= \coth(i \Omega H /2) i \Omega, \\
 H & = 2 \arccoth( i \Omega \sigma) i \Omega~.
 \end{align}
Consider that
 \begin{align}
 \sigma_{(\alpha)} &= \coth( i \Omega H_{(\alpha)}/2  ) i \Omega\\
 & = \coth(  i \Omega \alpha H/2 ) i \Omega\\
 & = \coth( i \Omega \alpha/2  [  2  \arccoth( \sigma i \Omega  )  i \Omega]   )i \Omega\\
 & = \coth(  \alpha \arccoth(\sigma i \Omega)  )i \Omega. 
 \end{align}
 
For $\vert x \vert >1$, we have that 
 \begin{align}\label{eq:arccoth-approx}
 \coth(\alpha \arccoth(x)) = \frac{(1+1/x)^{\alpha} + (1-1/x)^{\alpha}  }{ (1+1/x)^{\alpha} - (1-1/x)^{\alpha}   }~. 
 \end{align}
 
 Since eigenvalues of $\sigma i \Omega$ are either greater than 1 or less than $-1$, by using \eqref{eq:arccoth-approx} we find that  
 \begin{align}
\sigma_{(\alpha)}  = \frac{ (I + (\sigma i \Omega)^{-1})^{\alpha}  +(I - (\sigma i \Omega)^{-1})^{\alpha  }}{(I + (\sigma i \Omega)^{-1})^{\alpha}- (I - (\sigma i \Omega)^{-1})^{\alpha }}i \Omega~,
\end{align}
which implies that 
\begin{align}
\frac{\rho^{\alpha}}{\tr\{  \rho^{\alpha} \}} = \frac{\exp( -(1/2) (\hat{r} - \bar{r})^T H_{(\alpha)} (\hat{r} - \bar{r} )) }{\sqrt{ \text{Det}[ (\sigma_{(\alpha)}+i \Omega)/2   ]  }}~. 
\end{align}
 Moreover, 
\begin{align}
\tr\{ \rho^{\alpha}  \} &= \tr\left\{\bigg(\frac{\exp( -(1/2) (\hat{r} - \bar{r})^T H (\hat{r} - \bar{r} )) }{\sqrt{ \text{Det}[  (\sigma+i \Omega)/2   ]   }}\bigg)^{\alpha}\right\} \\
& = \frac{1}{(  \text{Det}[(\sigma+ i \Omega)/2)    ])^{\alpha/2}}\tr\{ \exp( -(1/2) ( \hat{r} - \bar{r})^T  \alpha H(\hat{r} - \bar{r}   )  )    \}\\
& =  \frac{1}{(  \text{Det}[(\sigma+ i \Omega)/2]   )^{\alpha/2}}\sqrt{\text{Det}[( \sigma_{(\alpha)} + i \Omega )/2]     }~. 
\end{align} 
 
We now focus on two special cases: $\alpha =2$ and $\alpha =1/2$. For $\alpha =2$, we have
 \begin{align}
 \frac{ (1+1/x)^{\alpha} + (1-1/x)^{\alpha} }{(1+1/x)^{\alpha} - (1-1/x)^{\alpha}} = \frac{1}{2} (x+ x^{-1})
 \end{align}
 Therefore, 
 \begin{align}
 \sigma_{(2)} &= \frac{1}{2}( \sigma i \Omega + (\sigma i \Omega)^{-1}   )i \Omega\\
 & = \frac{1}{2} ( \sigma + i \Omega \sigma^{-1} i \Omega   )\\
 & = \frac{1}{2} (\sigma + \Omega \sigma^{-1} \Omega^T   )~, 
 \end{align}
which implies that 
\begin{align}
\frac{\rho^2}{ \tr\{ \rho^2  \}   }  = \frac{\exp(  -(1/2)(\hat{r} - \bar{r})^T2H (\hat{r} - \bar{r})        )}{\sqrt{ \text{Det}[  (  (1/2)(\sigma + \Omega \sigma^{-1} \Omega^T )  + i \Omega     )/2    ]    }   }~.
\end{align}
 Moreover, the purity of the Gaussian state $\rho$ is given by
 \begin{align}
 \tr\{ \rho^2  \} = \frac{1}{\text{Det}[(\sigma+ i \Omega)/2    ]  }   \sqrt{   \text{Det}[(   (1/2)(\sigma + \Omega \sigma^{-1} \Omega^T )  + i \Omega     )/2]    }  ~, 
 \end{align}
 which further reduces (after many steps) to 
  \begin{align}
 \tr\{ \rho^2   \} = \frac{1}{\sqrt{\text{Det}(\sigma)}}~. 
 \end{align}
We note that the same expression for the purity of a Gaussian state was derived in the previous lecture by using a different approach. 
 
 
Let $\alpha = 1/2$. Consider that 
\begin{align}
\frac{\rho^{1/2}}{\tr\{  \rho^{1/2} \}} = \frac{\exp( -(1/2) (\hat{r} - \bar{r})^T H_{(1/2)} (\hat{r} - \bar{r} )) }{\sqrt{ \text{Det}[(  \sigma_{(1/2)}+i \Omega)/2   ]   }}, 
\end{align}
where $H_{(1/2)} = 1/2 H$. Moreover, for $\alpha = 1/2$, we have
 \begin{align}
\frac{ (1+1/x)^{\alpha} + (1-1/x)^{\alpha} }{(1+1/x)^{\alpha} - (1-1/x)^{\alpha}} = (1+ \sqrt{1 - 1/x^2})x~, 
\end{align} 	
which implies that 
\begin{align}
\sigma_{(1/2)} & =  (I + \sqrt{I -   (\sigma i \Omega)^{-2} }   ) (\sigma i \Omega)i \Omega    \\
& = (  \sqrt{ I + (\sigma \Omega)^{-2} }+ I    )\sigma~. 
\end{align}
 
Therefore, 
\begin{align}
\tr\{ \rho^{1/2}  \} = \frac{1}{( \text{Det}[(\sigma + i \Omega)/2   ]          )^{1/4}  }   \sqrt{ \text{Det}[    (  (\sqrt{I  + (\sigma \Omega)^{-2} }+I)\sigma   + i \Omega    )/2    ]	       }~.
\end{align} 



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