## Friday, 24 January 2014

### Some more facts concerning the von Neumann entropy

Let $\reg{X}$, $\reg{Y}$, and $\reg{Z}$ be registers, assume that the classical state set of $\reg{X}$ is $\Sigma$, and let $n = \abs{\Sigma}$.

Theorem:
For every state $\rho\in\Density(\X\otimes\Y\otimes\Z)$ of $(\reg{X},\reg{Y},\reg{Z})$ it holds that
$S(\reg{X},\reg{Y} : \reg{Z}) \leq S(\reg{Y}:\reg{X},\reg{Z}) + 2\log(n).$

Proof:

From the result proved in a previous post, it holds that for every choice of registers $\reg{X}$ and $\reg{Z}$, and for any state of $\Density(\X\otimes\Z)$, $S(\reg{Z})\leq S(\reg{X})+S(\reg{X}, \reg{Z})$, or equivalently that
$0\leq S(\reg{X})+S(\reg{X}, \reg{Z})-S(\reg{Z}).$
Also, by sub-additivity $S(\reg{X},\reg{Y})\leq S(\reg{X})+S(\reg{Y})$, or equivalently
$0\leq S(\reg{X})+S(\reg{Y})-S(\reg{X},\reg{Y}).$
Then by adding these two inequalities, it must also hold that
$0\leq S(\reg{X})+S(\reg{X}, \reg{Z})-S(\reg{Z})+S(\reg{X})+S(\reg{Y})-S(\reg{X},\reg{Y}),$
and since in general $S(\reg{X})\leq \log(n)$ or $2S(\reg{X})\leq 2\log(n)$,
$0\leq S(\reg{X}, \reg{Z})-S(\reg{Z})+S(\reg{Y})-S(\reg{X},\reg{Y})+2\log(n).$
Therefore,
$S(\reg{Z})+S(\reg{X},\reg{Y})\leq S(\reg{X}, \reg{Z})+S(\reg{Y})+2\log(n),$
Adding $-S(\reg{X},\reg{Y}, \reg{Z})$ to both sides of this inequality yields
$S(\reg{Z})+S(\reg{X},\reg{Y})-S(\reg{X},\reg{Y}, \reg{Z})\leq S(\reg{X}, \reg{Z})+S(\reg{Y})-S(\reg{X},\reg{Y}, \reg{Z})+2\log(n),$
or equivalently
$S(\reg{X},\reg{Y} : \reg{Z})\leq S(\reg{Y}:\reg{X},\reg{Z}) + 2\log(n).$

Here is an example, for $\Sigma = \{0,1\}$, of a state $\rho$ for which this inequality becomes an equality.

Consider the three qubit pure state
$\left|\psi\right>_{\reg{X},\reg{Y},\reg{Z}}=\frac{1}{\sqrt{2}}(\left|0\right>_{\reg{X}}\left|0\right>_{\reg{Y}}\left|0\right>_{\reg{Z}}+\left|1\right>_{\reg{X}}\left|0\right>_{\reg{Y}}\left|1\right>_{\reg{Z}}).$
Then the states of  the following particular subystems are also pure :
\begin{align*} \left|\psi\right>_{\reg{Y}}&=\left|0\right>_{\reg{Y}}\\ \left|\psi\right>_{\reg{X},\reg{Z}}&=\frac{1}{\sqrt{2}}(\left|0\right>_{\reg{X}}\left|0\right>_{\reg{Z}}+\left|1\right>_{\reg{X}}\left|1\right>_{\reg{Z}}). \end{align*}

However, the following subsystems are in the maximally mixed state:
\begin{align*} \rho_{\reg{X}}=\frac{1}{2}(\left|0\right>\left<0\right|+\left|1\right>\left<1\right|) \\ \rho_{\reg{Z}}=\frac{1}{2}(\left|0\right>\left<0\right|+\left|1\right>\left<1\right|). \end{align*}
Moreover, the state of the subsystem $\reg{X},\reg{Y}$ is in the tensor product state
\begin{align*} \rho_{\reg{X},\reg{Y}}&=\rho_{\reg{X}}\otimes \rho_{\reg{Y}}\\ &=\frac{1}{2}\bigl(\left|0\right>\left<0\right|+\left|1\right>\left<1\right|\bigr)\otimes \left|0\right>\left<0\right| \end{align*}

The entropy of a pure state is zero and the entropy of a maximally entangled state in this case is $\log(n)=\log(2)$. Then the entropies of the states listed above are
\begin{align*} S(\reg{Y})=S(\reg{X},\reg{Z})&=0 \\ S(\reg{X})=S(\reg{Z})&=\log(2) \\ S(\reg{X},\reg{Y})=S(\reg{X})+S(\reg{Y})&=\log(2). \end{align*}

Therefore,
\begin{align*} S(\reg{X},\reg{Y} : \reg{Z}) - S(\reg{Y}:\reg{X},\reg{Z})&=S(\reg{X},\reg{Y})-S(\reg{Y})-S(\reg{X},\reg{Z})+S(\reg{Z})+\left(S(\reg{X},\reg{Y},\reg{Z})-S(\reg{X},\reg{Y},\reg{Z})\right) \\ &=S(\reg{X},\reg{Y})-S(\reg{Y})-S(\reg{X},\reg{Z})+S(\reg{Z}) \\ &=S(\reg{X}+S(\reg{Y})-S(\reg{Y})-S(\reg{X},\reg{Z})+S(\reg{Z}) \\ &=S(\reg{X}-S(\reg{X},\reg{Z})+S(\reg{Z}) \\ &=\log(2)-0+\log(2) \\ &=2\log(2), \end{align*}
which implies that
$S(\reg{X},\reg{Y} : \reg{Z}) = S(\reg{Y}:\reg{X},\reg{Z})+2\log(2).$

Theorem:

Let $p\in\P(\Sigma)$ be a probability vector, let $\{\sigma_a\,:\,a\in\Sigma\} \subset \Density(\Y\otimes\Z)$ be a collection of density operators, and let
$\rho = \sum_{a\in \Sigma} p(a) E_{a,a}\otimes \sigma_a.$
In other words, $\rho$ is a state of $(\reg{X},\reg{Y},\reg{Z})$ in which we view $\reg{X}$ as a classical register. With respect to the state $\rho$, it holds that
$S(\reg{X},\reg{Y} : \reg{Z}) \leq S(\reg{Y}:\reg{X},\reg{Z}) + \log(n).$

Proof:

First, observe that
\begin{align*} S(\reg{X}|\reg{Y})-S(\reg{X}|\reg{Z})&=S(\reg{X},\reg{Y})-S(\reg{Y})-S(\reg{X},\reg{Z})+S(\reg{Z})\\ &=S(\reg{X},\reg{Y})-S(\reg{Y})-S(\reg{X},\reg{Z})+S(\reg{Z})+\left(S(\reg{X},\reg{Y},\reg{Z})-S(\reg{X},\reg{Y},\reg{Z})\right) \\ &=S(\reg{X},\reg{Y} : \reg{Z}) - S(\reg{Y}:\reg{X},\reg{Z}). \end{align*}

Now consider the individual bounds on the quantities $S(\reg{X}|\reg{Y})$ and $S(\reg{X}|\reg{Z})$ in order to infer a bound on the difference $S(\reg{X}|\reg{Y})-S(\reg{X}|\reg{Z})$. In this case, since the state of register $\reg{X}$ is classical the conditional entropies  are at most $S(\reg{X}|\reg{Y})\leq \log(n)$ and likewise $S(\reg{X}|\reg{Z})\leq \log(n)$. On the contrary, it could be the case that $S(\reg{X}|\reg{Y})\leq 0$ or $S(\reg{X}|\reg{Z})\leq 0$ in the presence of stronger entanglement correlations in which case $S(\reg{Y})\leq S(\reg{X},\reg{Y})$ or $S(\reg{Z})\leq S(\reg{X},\reg{Z})$. Therefore, the largest the difference of the two could be is when $S(\reg{X}|\reg{Y})=\log(n)$ and $S(\reg{X}|\reg{Z})=0$. Hence, $S(\reg{X}|\reg{Y})-S(\reg{X}|\reg{Z})\leq \log(n)$, or equivalently $S(\reg{X},\reg{Y} : \reg{Z})-S(\reg{Y}:\reg{X},\reg{Z}) \leq \log(n)$ implying that  $S(\reg{X},\reg{Y} : \reg{Z}) \leq S(\reg{Y}:\reg{X},\reg{Z}) + \log(n)$.