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\begin{document}

\lecture{20 --- November 2, 2015}{Fall 2015}{Prof.\ Mark M.\ Wilde}{Mark M.~Wilde}

This document is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

\section{Overview}



In the previous lecture, we discussed a quantum entropy inequality called the monotonicity of quantum relative entropy. This has several important consequences, such as the strong subadditivity of quantum entropy, concavity of conditional quantum entropy, joint convexity of quantum relative entropy, etc. We also discussed how the conditional quantum entropy is continuous and gave an explicit bound for continuity with respect to the topology induced by the trace norm.


In this lecture, we 
prove the monotonicity of quantum relative entropy using a very recent proof technique. This method gives a physically meaningful improvement of the entropy inequality which has an interpretation in terms of ``recoverability.''

\section{Introduction}

The quantum entropy inequalities discussed 
previously lie at the core of quantum Shannon theory and in fact underly
some important principles of physics such as the uncertainty principle. In fact, we will use these entropy
inequalities to prove the converse parts of every coding theorem in this course. Their prominence in both quantum Shannon
theory and other areas of physics motivates us to study them in more detail.
The book delves into more depth regarding many
of the classical entropy inequalities, and in the process,  establishes
necessary and sufficient conditions for the saturation of the inequalities, while also understanding the near
saturation of the entropy inequalities. The aim of this lecture is
to carry out a similar program for all of the quantum entropy inequalities
presented in the previous chapter. The outcome will be a proof for the
monotonicity of quantum relative entropy,
with the added benefit of an understanding of the saturation and near
saturation of this quantum entropy inequality.

The main result in this chapter can be summarized informally as follows: if
the decrease in quantum relative entropy between two quantum states after a
quantum channel is relatively small, then it is possible to perform a recovery
channel, such that we can perfectly recover one state while approximately
recovering the other. This can be interpreted as quantifying how well one can
reverse the action of a quantum channel. Throughout, we take $\rho$, $\sigma$,
and $\mathcal{N}$ as given in the following definition:

\begin{definition}
\label{def:rho-sig-N}Let $\rho\in\mathcal{D}(\mathcal{H})$ and let $\sigma
\in\mathcal{L}(\mathcal{H})$ be positive semi-definite, such that
$\operatorname{supp}(\rho)\subseteq\operatorname{supp}(\sigma)$. Let
$\mathcal{N}:\mathcal{L}(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H}%
^{\prime})$ be a quantum channel.
\end{definition}

The formal statement of the theorem is as follows:

\begin{theorem}
\label{thm-qeir:main-theorem}Given $\rho$, $\sigma$, and $\mathcal{N}$ as in
Definition~\ref{def:rho-sig-N}, there exists a recovery channel $\mathcal{R}%
_{\sigma,\mathcal{N}}:\mathcal{L}(\mathcal{H}^{\prime})\rightarrow
\mathcal{L}(\mathcal{H})$, depending only on $\sigma$ and $\mathcal{N}$, such
that%
\begin{equation}
D(\rho\Vert\sigma)-D(\mathcal{N}(\rho)\Vert\mathcal{N}(\sigma))\geq-\log
F(\rho,(\mathcal{R}_{\sigma,\mathcal{N}}\circ\mathcal{N})(\rho)).
\label{eq-qeir:main-inequality}%
\end{equation}

\end{theorem}

Given that the quantum fidelity $F$ takes values between zero and one, we can
immediately conclude that%
\begin{equation}
-\log F(\rho,(\mathcal{R}_{\sigma,\mathcal{N}}\circ\mathcal{N})(\rho))\geq0,
\end{equation}
so that the above theorem implies the monotonicity of quantum relative entropy as a consequence. Furthermore, the
recovery channel satisfying \eqref{eq-qeir:main-inequality}\ has the property
that it perfectly recovers $\sigma$ from $\mathcal{N}(\sigma)$:%
\begin{equation}
(\mathcal{R}_{\sigma,\mathcal{N}}\circ\mathcal{N})(\sigma)=\sigma,
\end{equation}
which is something that we show later.
%To see this, pick $\rho=\sigma^{\prime}\equiv\sigma/\operatorname{Tr}%
%\{\sigma\}$, implying that the upper bound in \eqref{eq-qeir:main-inequality}
%is equal to zero and thus that%
%\begin{equation}
%F(\sigma^{\prime},(\mathcal{R}_{\sigma,\mathcal{N}}\circ\mathcal{N}%
%)(\sigma^{\prime}))=1.
%\end{equation}
%But the fidelity is equal to one if and only if its two density operator
%arguments are equal, so that $\sigma^{\prime}=(\mathcal{R}_{\sigma
%,\mathcal{N}}\circ\mathcal{N})(\sigma^{\prime})$. Multiplying both sides of
%this last equation by $\operatorname{Tr}\{\sigma\}$ gives that the recovery
%channel $\mathcal{R}_{\sigma,\mathcal{N}}$ satisfying
%\eqref{eq-qeir:main-inequality}\ perfectly recovers $\sigma$ from
%$\mathcal{N}(\sigma)$.

The proof given here for Theorem~\ref{thm-qeir:main-theorem} relies on the
method of complex interpolation and the notion of a R\'{e}nyi generalization
of a relative entropy difference.\ We review this background first before
going through the proof. One of the consequences of
Theorem~\ref{thm-qeir:main-theorem} is to provide physically meaningful
improvements to many quantum entropy inequalities discussed in the previous
lecture, such as strong subadditivity, joint convexity of quantum relative
entropy, and concavity of conditional quantum entropy. We explore one of these
consequences in Section~\ref{sec-qeir:corollaries} and point to the book for the rest.

\section{Schatten Norms and Complex Interpolation}

The proof of Theorem~\ref{thm-qeir:main-theorem}\ given here requires a bit of
mathematical background before we can delve into it. So we first begin by
defining the Schatten norms and several of their properties. We then review
some essential results from complex analysis, that lead to a complex
interpolation theorem known as the Stein--Hirschman interpolation theorem.

\subsection{Schatten Norms and Duality}

An important technical tool in the proof given here is the Schatten $p$-norm
of an operator $A$, defined as%
\begin{equation}
\left\Vert A\right\Vert _{p}\equiv\left[  \operatorname{Tr}\left\{  \left\vert
A\right\vert ^{p}\right\}  \right]  ^{1/p},
\end{equation}
where $A\in\mathcal{L}(\mathcal{H})$,$\ |A|\equiv\sqrt{A^{\dag}A}$, and
$p\geq1$. We have already studied two special cases of this norm, which are
the trace norm when $p=1$  and the
Hilbert--Schmidt norm when $p=2$.
One can show, along the same lines as we did for the trace norm, that $\left\Vert
A\right\Vert _{p}$ is equal to the $p$-norm of the singular values of $A$.
That is, if $\sigma_{i}(A)$ is the vector of singular values of $A$, then%
\begin{equation}
\left\Vert A\right\Vert _{p}=\left[  \sum_{i}\sigma_{i}(A)^{p}\right]  ^{1/p}.
\label{eq-qeir:p-norm-sing-vals}%
\end{equation}
The convention is for $\left\Vert A\right\Vert _{\infty}$ to be defined as the
largest singular value of $A$ because $\left\Vert A\right\Vert _{p}$ converges
to this in the limit as $p\rightarrow\infty$. In the proof of
Theorem~\ref{thm-qeir:main-theorem}, we repeatedly use the fact that
$\left\Vert A\right\Vert _{p}$ is unitarily invariant. That is, $\left\Vert
A\right\Vert _{p}$ is invariant with respect to linear isometries, in the
sense that $\left\Vert A\right\Vert _{p}=\left\Vert UAV^{\dag}\right\Vert
_{p}$, where $U,V\in\mathcal{L}(\mathcal{H},\mathcal{H}^{\prime})$ are linear
isometries satisfying $U^{\dag}U=I_{\mathcal{H}}$ and $V^{\dag}%
V=I_{\mathcal{H}}$. Isometric invariance follows from
\eqref{eq-qeir:p-norm-sing-vals} and because these isometries do not change
the singular values of $A$. From these norms, one can define information
measures relating quantum states and channels, with the main one used here
known as a R\'{e}nyi generalization of a relative entropy difference.

Extending the Cauchy--Schwarz inequality is an important inequality known as
the H\"{o}lder inequality:%
\begin{equation}
\left\vert \left\langle A,B\right\rangle \right\vert =\left\vert
\operatorname{Tr}\{A^{\dag}B\}\right\vert \leq\left\Vert A\right\Vert
_{p}\left\Vert B\right\Vert _{q}, \label{eq-qeir:holder}%
\end{equation}
holding for $p,q\in\left[  1,\infty\right]  $ such that $\frac{1}{p}+\frac
{1}{q}=1$ and $A,B\in\mathcal{L}(\mathcal{H})$. When $p,q\in\left[
1,\infty\right]  $ and $\frac{1}{p}+\frac{1}{q}=1$, $p$ and $q$ are said to be
H\"{o}lder conjugates of each other. One can see that Cauchy--Schwarz is a
special case by picking $p=q=2$. Observe that equality is achieved in
\eqref{eq-qeir:holder}\ if $A$ and $B$ are such that $A^{\dag}=a\left\vert
B\right\vert ^{q/p}U^{\dag}$ for some constant $a\geq0$ and where $U$ is a
unitary such that $B=U\left\vert B\right\vert $ is a left polar decomposition
of $B$. The H\"{o}lder inequality along with the sufficient equality condition
is enough for us to conclude the following variational expression for the
$p$-norm in terms of its H\"{o}lder dual $q$-norm:%
\begin{equation}
\left\Vert A\right\Vert _{p}=\max_{\left\Vert B\right\Vert _{q}\leq
1}\operatorname{Tr}\{A^{\dag}B\}.
\end{equation}
This expression can be very useful in calculations.

\begin{exercise}
Prove that $\left\Vert AB\right\Vert _{1}\leq\left\Vert A\right\Vert
_{p}\left\Vert B\right\Vert _{q}$ for $p,q\in\left[  1,\infty\right]  $ such
that $\frac{1}{p}+\frac{1}{q}=1$ and $A,B\in\mathcal{L}(\mathcal{H})$.
\end{exercise}

Throughout we adopt the usual convention and define $f(A)$ for a
function $f$ and a positive semi-definite operator $A$ as follows:
$f(A)\equiv\sum_{i:\lambda_{i}\neq0}f(\lambda_{i})|i\rangle\langle i|$, where
$A=\sum_{i}\lambda_{i}|i\rangle\langle i|$ is a spectral decomposition of $A$.
We denote the support of $A$ by $\operatorname{supp}(A)$, and we let $\Pi_{A}$
denote the projection onto the support of $A$.

\subsection{Complex Analysis}

We now review a few concepts from complex analysis. We will not prove these
results in detail, but the purpose instead is to recall them, and the
interested reader can follow references to books on complex analysis for
details of proofs. The culmination of the development is the Stein--Hirschman
complex interpolation theorem (Theorem~\ref{thm-qeir:op-hirschman}).

The derivative of a complex-valued function $f:\mathbb{C}\rightarrow
\mathbb{C}$ at a point $z_{0}\in\mathbb{C}$\ is defined in the usual way as%
\begin{equation}
\left.  \frac{df(z)}{dz}\right\vert _{z=z_{0}}=\lim_{z\rightarrow z_{0}}%
\frac{f(z)-f(z_{0})}{z-z_{0}}.
\end{equation}
In order for this limit to exist, it must be the same for all possible
directions that one could take in the complex plane to approach $z_{0}$, and
this requirement demarcates a substantial difference between differentiability
of real functions and complex ones. Complex differentiability shares several
properties with real differentiability:\ it is linear and obeys the product
rule, the quotient rule, and the chain rule. If $f$ is complex differentiable
at every point $z_{0}$ in an open set $U$, then we say that $f$ is
\textit{holomorphic} on $U$.

There is a connection between real differentiability and complex
differentiability, given by the Cauchy--Riemann equations. Let
$f(x+iy)=u(x,y)+iv(x,y)$ where $x,y\in\mathbb{R}$ and $u,v:\mathbb{R}%
\rightarrow\mathbb{R}$. If $f$ is holomorphic, then $u$ and $v$ have first
partial derivatives with respect to $x$ and $y$ and satisfy the
Cauchy--Riemann equations:%
\begin{equation}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\ \ \ \ \ \ \frac
{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.
\end{equation}
The converse is not always true. However, if the first partial derivatives of
$u$ and $v$ are continuous and satisfy the Cauchy--Riemann equations, then $f$
is holomorphic. An important holomorphic function for our purposes is given in
the following exercise:

\begin{exercise}
\label{ex-qeir:x^z-holo}Verify that $f(z)=a^{z}=e^{\left[  \ln a\right]  z}$,
where $a>0$ and $z\in\mathbb{C}$, is a holomorphic function everywhere in the
complex plane. (Hint:\ Use that $e^{z}=e^{x}\left[  \cos(y)+i\sin(y)\right]  $
for $z=x+iy$ and $x,y\in\mathbb{R}$.)
\end{exercise}

Holomorphic functions have good closure properties. That is, the sums,
products, and compositions of holomorphic functions are holomorphic as well,
given that complex differentiation is linear and satisfies the product,
quotient, and chain rules. Note that the quotient of two holomorphic functions
is holomorphic wherever the denominator is not equal to zero.

The \textit{maximum modulus principle} is an important principle that
holomorphic functions obey. Formally, it is the following statement:\ let
$f:\mathbb{C}\rightarrow\mathbb{C}$ be a function holomorphic on some
connected, bounded open subset $U$ of $\mathbb{C}$. If $z_{0}\in U$ is such
that $\left\vert f(z_{0})\right\vert \geq\left\vert f(z)\right\vert $ for all
$z$ in a neighborhood of $z_{0}$, then the function $f$ is constant on $U$. A
consequence of this is that if $f$ is not constant on a bounded, connected,
open subset $U$ of $\mathbb{C}$, then it achieves its maximum on the boundary
of $U$.

The maximum modulus principle has an extension to an unbounded strip in
$\mathbb{C}$, which we call the \textit{maximum modulus principle on a strip}.
Let $S$ denote the standard strip in $\mathbb{C}$, $\overline{S}$ its closure,
and $\partial\overline{S}$ its boundary:%
\begin{align}
S  &  \equiv\left\{  z\in\mathbb{C}:0<\operatorname{Re}\{z\}<1\right\}  ,\\
\overline{S}  &  \equiv\left\{  z\in\mathbb{C}:0\leq\operatorname{Re}%
\{z\}\leq1\right\}  ,\\
\partial\overline{S}  &  \equiv\left\{  z\in\mathbb{C}:\operatorname{Re}%
\{z\}=0\vee\operatorname{Re}\{z\}=1\right\}  .
\end{align}
Let $f:\overline{S}\rightarrow\mathbb{C}$ be bounded on $\overline{S}$,
holomorphic on $S$, and continuous on $\partial\overline{S}$. Then the
supremum of $\left\vert f\right\vert $ is attained on $\partial\overline{S}$.
That is, $\sup_{z\in\overline{S}}\left\vert f(z)\right\vert =\sup
_{z\in\partial\overline{S}}\left\vert f(z)\right\vert $.

The maximum modulus principle on a strip implies a result known as the
Hadamard three-lines theorem:

\begin{theorem}
[Hadamard Three-Lines]Let $f:\overline{S}\rightarrow\mathbb{C}$ be a function
that is bounded on $\overline{S}$, holomorphic on $S$, and continuous on the
boundary $\partial\overline{S}$. Let $\theta\in\left(  0,1\right)  $ and
$M(\theta)\equiv\sup_{t\in\mathbb{R}}\left\vert f(\theta+it)\right\vert $.
Then $\ln M(\theta)$ is a convex function on $\left[  0,1\right]  $, implying
that%
\begin{equation}
\ln M(\theta)\leq\left(  1-\theta\right)  \ln M(0)+\theta\ln M(1).
\end{equation}

\end{theorem}

There is a strengthening of the Hadamard three-lines theorem due to Hirschman,
which in fact implies the Hadamard three-lines theorem:

\begin{theorem}
[Hirschman]\label{thm-qeir:hirschman}Let $f(z):\overline{S}\rightarrow
\mathbb{C}$ be a function that is bounded on $\overline{S}$, holomorphic on
$S$, and continuous on the boundary $\partial\overline{S}$. Then for
$\theta\in(0,1)$, the following bound holds%
\begin{equation}
\ln\left\vert f(\theta)\right\vert \leq\int_{-\infty}^{\infty}dt\ \left(
\alpha_{\theta}(t)\ln\left[  \left\vert f(it)\right\vert ^{1-\theta}\right]
+\beta_{\theta}(t)\ln\left[  \left\vert f(1+it)\right\vert ^{\theta}\right]
\right)  ,
\end{equation}
where%
\begin{align}
\alpha_{\theta}(t)  &  \equiv\frac{\sin(\pi\theta)}{2(1-\theta)\left[
\cosh(\pi t)-\cos(\pi\theta)\right]  },\label{eq-qeir:alpha-t}\\
\beta_{\theta}(t)  &  \equiv\frac{\sin(\pi\theta)}{2\theta\left[  \cosh(\pi
t)+\cos(\pi\theta)\right]  }. \label{eq-qeir:beta-t}%
\end{align}

\end{theorem}

For a fixed $\theta\in(0,1)$, we have that $\alpha_{\theta}(t),\beta_{\theta
}(t)\geq0$ for all $t\in\mathbb{R}$ and%
\begin{equation}
\int_{-\infty}^{\infty}dt\ \alpha_{\theta}(t)=\int_{-\infty}^{\infty}%
dt\ \beta_{\theta}(t)=1\ ,
\end{equation}
so that $\alpha_{\theta}(t)$ and
$\beta_{\theta}(t)$ can be interpreted as probability density functions.
Furthermore, we have that%
\begin{equation}
\lim_{\theta\searrow0}\beta_{\theta}(t)=\frac{\pi}{2}\left[  \cosh(\pi
t)+1\right]  ^{-1}\equiv\beta_{0}(t)\ , \label{eq:dist-limit}%
\end{equation}
where $\beta_{0}$ is also a probability density function on $\mathbb{R}$. With
these observations, we can see that Hirschman's theorem implies the Hadamard
three-lines theorem, given that an expectation can never exceed a supremum.

\subsection{Complex Interpolation of Schatten Norms}

We can extend much of the development above to operator-valued functions,
which is needed to prove Theorem~\ref{thm-qeir:main-theorem}. Let
$G:\mathbb{C}\rightarrow\mathcal{L}(\mathcal{H})$ be an operator-valued
function. We say that $G(z)$ is holomorphic if every function mapping $z$ to a
matrix entry is holomorphic. For our purposes in what follows, we are
interested in operator-valued functions of the form $A^{z}$, where $A$ is a
positive semi-definite operator. In this case, we apply the usual convention and take $A^{z}=\sum
_{i:\lambda_{i}\neq0}\lambda_{i}^{z}|i\rangle\langle i|$, where $A=\sum
_{i}\lambda_{i}|i\rangle\langle i|$ is an eigendecomposition of $A$ with
$\lambda_{i}\geq0$ for all $i$. Given the result of
Exercise~\ref{ex-qeir:x^z-holo}\ combined with the closure properties of
holomorphic functions mentioned above, we can conclude that $A^{z}$ is
holomorphic if $A$ is positive semi-definite.

We can now establish a version of the Hirschman theorem which applies to
operator-valued functions and allows for bounding their Schatten norms. This
is one of the main technical tools that we need to establish
Theorem~\ref{thm-qeir:main-theorem}.

\begin{theorem}
[Stein--Hirschman]\label{thm-qeir:op-hirschman} Let $G:\overline{S}\rightarrow
L(\mathcal{H})$ be an operator-valued function that is bounded on
$\overline{S}$, holomorphic on $S$, and continuous on the boundary
$\partial\overline{S}$. Let $\theta\in(0,1)$ and define $p_{\theta}$ by%
\begin{equation}
\frac{1}{p_{\theta}}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}\ ,
\end{equation}
where $p_{0},p_{1}\in\left[  1,\infty\right]  $. Then the following bound
holds%
\begin{equation}
\ln\left\Vert G(\theta)\right\Vert _{p_{\theta}}\leq\int_{-\infty}^{\infty
}dt\ \left(  \alpha_{\theta}(t)\ln\left[  \left\Vert G(it)\right\Vert _{p_{0}%
}^{1-\theta}\right]  +\beta_{\theta}(t)\ln\left[  \left\Vert
G(1+it)\right\Vert _{p_{1}}^{\theta}\right]  \right)  \ ,
\label{eq:oper-hirschman}%
\end{equation}
where $\alpha_{\theta}(t)$ and $\beta_{\theta}(t)$ are defined in \eqref{eq-qeir:alpha-t}--\eqref{eq-qeir:beta-t}.
\end{theorem}

\begin{proof}
For fixed $\theta\in(0,1)$, let $q_{\theta}$ be the H\"{o}lder conjugate of
$p_{\theta}$, defined by%
\begin{equation}
\frac{1}{p_{\theta}}+\frac{1}{q_{\theta}}=1\ .
\end{equation}
Similarly, let $q_{0}$ and $q_{1}$ be H\"{o}lder conjugates of $p_{0}$ and
$p_{1}$, respectively. From the sufficient equality condition for the
H\"{o}lder inequality, we can find an operator $X$ such that $\left\Vert
X\right\Vert _{q_{\theta}}=1$ and $\operatorname{Tr}\{XG(\theta)\}=\left\Vert
G(\theta)\right\Vert _{p_{\theta}}$. We can write the singular value
decomposition for $X$ in the form $X=UD^{1/q_{\theta}}V$ (implying
$\operatorname{Tr}\{D\}=1$). For $z\in S$, define%
\begin{equation}
X(z)\equiv UD^{\frac{1-z}{q_{0}}+\frac{z}{q_{1}}}V\ .
\end{equation}
As a consequence, $X(z)$ is bounded on $\overline{S}$, holomorphic on $S$, and
continuous on the boundary $\partial\overline{S}$. Also, observe that
$X(\theta)=X$. Then the following function satisfies the requirements needed
to apply Theorem~\ref{thm-qeir:hirschman}:%
\begin{equation}
g(z)\equiv\operatorname{Tr}\{X(z)G(z)\}\ .
\end{equation}
Indeed, we have that%
\begin{align}
\ln\left\Vert G(\theta)\right\Vert _{p_{\theta}}  &  =\ln\left\vert
g(\theta)\right\vert \\
&  \leq\int_{-\infty}^{\infty}dt\ \left(  \alpha_{\theta}(t)\ln\left[
\left\vert g(it)\right\vert ^{1-\theta}\right]  +\beta_{\theta}(t)\ln\left[
\left\vert g(1+it)\right\vert ^{\theta}\right]  \right)  .
\label{eq:app-hirschman}%
\end{align}
Now, from applying H\"{o}lder's inequality and the facts that $\left\Vert
X(it)\right\Vert _{q_{0}}=1=\left\Vert X(1+it)\right\Vert _{q_{1}}$, we find
that%
\begin{equation}
\left\vert g(it)\right\vert =\left\vert \operatorname{Tr}%
\{X(it)G(it)\}\right\vert \leq\left\Vert X(it)\right\Vert _{q_{0}}\left\Vert
G(it)\right\Vert _{p_{0}}=\left\Vert G(it)\right\Vert _{p_{0}},
\end{equation}
and
\begin{align}
\left\vert g(1+it)\right\vert  &  =\left\vert \operatorname{Tr}%
\{X(1+it)G(1+it)\}\right\vert \\
&  \leq\left\Vert X(1+it)\right\Vert _{q_{1}}\left\Vert G(1+it)\right\Vert
_{p_{1}}\\
&  =\left\Vert G(1+it)\right\Vert _{p_{1}}.
\end{align}
Bounding \eqref{eq:app-hirschman} from above using these inequalities then
gives \eqref{eq:oper-hirschman}.
\end{proof}

The theorem above is known as a complex interpolation theorem because it
allows us to obtain estimates on the \textquotedblleft
intermediate\textquotedblright\ norm in terms of other norms which might be
available. Furthermore, we are interpolating through the holomorphic family of
operators given by $G(z)$.

\section{Petz Recovery Map}

The channel appearing in the lower bound of
Theorem~\ref{thm-qeir:main-theorem}$\ $has an explicit form and is constructed
from a map known as the Petz recovery map, which we define as follows:

\begin{definition}
Let $\sigma\in\mathcal{L}(\mathcal{H})$ be positive semi-definite, and let
$\mathcal{N}:\mathcal{L}(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H}%
^{\prime})$ be a quantum channel. The Petz recovery map $\mathcal{P}%
_{\sigma,\mathcal{N}}:\mathcal{L}(\mathcal{H}^{\prime})\rightarrow
\mathcal{L}(\mathcal{H})$ is a completely positive, trace-non-increasing
linear map defined as follows for $Q\in\mathcal{L}(\mathcal{H}^{\prime})$:%
\begin{equation}
\mathcal{P}_{\sigma,\mathcal{N}}\left(  Q\right)  \equiv\sigma^{1/2}%
\mathcal{N}^{\dag}\left(  \left[  \mathcal{N}(\sigma)\right]  ^{-1/2}Q\left[
\mathcal{N}(\sigma)\right]  ^{-1/2}\right)  \sigma^{1/2}.
\label{eq:Petz-channel-Rel-ent}%
\end{equation}

\end{definition}

The Petz recovery map $\mathcal{P}_{\sigma,\mathcal{N}}$ is linear, and it is
completely positive because it is equal to a serial concatenation of three
completely positive maps: $Q\rightarrow\left[  \mathcal{N}(\sigma)\right]
^{-1/2}Q\left[  \mathcal{N}(\sigma)\right]  ^{-1/2}$, $Q\rightarrow
\mathcal{N}^{\dag}(Q)$, and $M\rightarrow\sigma^{1/2}M\sigma^{1/2}$ for
$M\in\mathcal{L}(\mathcal{H})$. It is trace-non-increasing because the
following holds for positive semi-definite $Q$:%
\begin{align}
\operatorname{Tr}\{\mathcal{P}_{\sigma,\mathcal{N}}\left(  Q\right)  \}  &
=\operatorname{Tr}\left\{  \sigma\mathcal{N}^{\dag}\left(  \left[
\mathcal{N}(\sigma)\right]  ^{-1/2}Q\left[  \mathcal{N}(\sigma)\right]
^{-1/2}\right)  \right\} \\
&  =\operatorname{Tr}\left\{  \mathcal{N}(\sigma)\left[  \mathcal{N}%
(\sigma)\right]  ^{-1/2}Q\left[  \mathcal{N}(\sigma)\right]  ^{-1/2}\right\}
\\
&  =\operatorname{Tr}\{\Pi_{\mathcal{N}(\sigma)}Q\}\\
&  \leq\operatorname{Tr}\{Q\}.
\end{align}


An important special case of the Petz recovery map occurs when $\sigma$ and
$\mathcal{N}$ are effectively classical. That is, suppose that $\mathcal{N}$
is a classical-to-classical channel with Kraus operators $\{\sqrt
{N(y|x)}|y\rangle\langle x|\}$,
where $N(y|x)$ is a conditional probability distribution. Suppose further that
$\sigma=\sum_{x}q(x)|x\rangle\langle x|$, with $q(x)\geq0$ for all $x$. In
this case, one can check that the Petz recovery map is a
classical-to-classical channel with Kraus operators $\{\sqrt{R(x|y)}%
|x\rangle\langle y|\}$, where $R(x|y)$ is a conditional probability
distribution given by the Bayes theorem, satisfying%
\begin{equation}
R(x|y)(Nq)(y)=N(y|x)q(x),
\end{equation}
for all $x$ and $y$, where $(Nq)(y)=\sum_{x}N(y|x)q(x)$. We leave the details
of this calculation as an exercise for the reader and point out that this
recovery channel appears in the refinement of the monotonicity of classical
relative entropy in the book.

We can also define a partial isometric map $\mathcal{U}_{\sigma,t}$ in the
following way:%
\begin{equation}
\mathcal{U}_{\sigma,t}(M)\equiv\sigma^{it}M\sigma^{-it},
\label{eq-qeir:unitaries}%
\end{equation}
and where $\sigma^{it}\sigma^{-it}=\Pi_{\sigma}$. In the case that $\sigma$ is
positive definite, $\mathcal{U}_{\sigma,t}$ is a unitary channel. We can then
define a rotated or \textquotedblleft swiveled\textquotedblright\ Petz map,
which plays an important role in the construction of a recovery channel
satisfying the lower bound in Theorem~\ref{thm-qeir:main-theorem}.

\begin{definition}
[Rotated Petz Map]\label{def-qeir:rotated-petz}Let $\sigma\in\mathcal{L}%
(\mathcal{H})$ be positive semi-definite, and let $\mathcal{N}:\mathcal{L}%
(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H}^{\prime})$ be a quantum
channel. A rotated Petz map is defined as follows for $Q\in\mathcal{L}%
(\mathcal{H}^{\prime})$:%
\begin{equation}
\mathcal{R}_{\sigma,\mathcal{N}}^{t}\left(  Q\right)  \equiv(\mathcal{U}%
_{\sigma,-t}\circ\mathcal{P}_{\sigma,\mathcal{N}}\circ\mathcal{U}%
_{\mathcal{N}(\sigma),t})(Q).
\end{equation}

\end{definition}

\begin{exercise}
[Perfect Recovery]\label{ex-qeir:perfect-recovery}Verify that a rotated Petz
map perfectly recovers $\sigma$ from $\mathcal{N}(\sigma)$:%
\begin{equation}
\mathcal{R}_{\sigma,\mathcal{N}}^{t}(\mathcal{N}(\sigma))=\sigma.
\end{equation}
(To show this equality when $\sigma$ is not invertible, first show that
$\mathcal{N}^{\dag}(\Pi_{\mathcal{N}(\sigma)})\geq\Pi_{\sigma}$.)
\end{exercise}



\section{R\'{e}nyi Information Measure}

Given $\rho$, $\sigma$, and $\mathcal{N}$ as in Definition~\ref{def:rho-sig-N}%
, we define a R\'{e}nyi information measure known as a R\'{e}nyi
generalization of a relative entropy difference:%
\begin{equation}
\widetilde{\Delta}_{\alpha}(\rho,\sigma,\mathcal{N})\equiv\frac{2\alpha
}{\alpha-1}\ln\left\Vert \left(  \left[  \mathcal{N}(\rho)\right]
^{\frac{1-\alpha}{2\alpha}}\left[  \mathcal{N}(\sigma)\right]  ^{\frac
{\alpha-1}{2\alpha}}\otimes I_{E}\right)  U\sigma^{\frac{1-\alpha}{2\alpha}%
}\rho^{1/2}\right\Vert _{2\alpha}, \label{eq:renyi-diff}%
\end{equation}
where $\alpha\in(0,1)\cup(1,\infty)$ and $U:\mathcal{H}\rightarrow
\mathcal{H}^{\prime}\otimes\mathcal{H}_{E}$ is an isometric extension of the
channel $\mathcal{N}$. That is, $U$ is a linear isometry satisfying
$\operatorname{Tr}_{E}\{U(\cdot)U^{\dag}\}=\mathcal{N}(\cdot)$ and $U^{\dag
}U=I_{\mathcal{H}}$. Recall that all isometric extensions of a channel are
related by an isometry acting on the environment system~$E$, so that the
definition in \eqref{eq:renyi-diff} is invariant under any such choice. Recall
 that the adjoint
$\mathcal{N}^{\dag}$\ of a channel is given in terms of an isometric extension
$U$ as $\mathcal{N}^{\dag}(\cdot)=U^{\dag}\left(  (\cdot)\otimes I_{E}\right)
U$.

The following lemma is one of the main reasons that we say that $\widetilde
{\Delta}_{\alpha}(\rho,\sigma,\mathcal{N})$ is a R\'{e}nyi generalization of a
relative entropy difference. A proof is available in the book.

\begin{lemma}
The following limit holds for $\rho$, $\sigma$, and $\mathcal{N}$ as given in
Definition~\ref{def:rho-sig-N}:%
\begin{equation}
\frac{1}{\ln2}\lim_{\alpha\rightarrow1}\widetilde{\Delta}_{\alpha}(\rho
,\sigma,\mathcal{N})=D(\rho\Vert\sigma)-D(\mathcal{N}(\rho)\Vert
\mathcal{N}(\sigma)). \label{eq:rel-ent-diff-a-1}%
\end{equation}

\end{lemma}



For $\alpha=1/2$, observe that%
\begin{align}
\widetilde{\Delta}_{1/2}(\rho,\sigma,\mathcal{N})  &  =-\ln\left\Vert \left(
\left[  \mathcal{N}(\rho)\right]  ^{1/2}\left[  \mathcal{N}(\sigma)\right]
^{-1/2}\otimes I_{E}\right)  U_{S\rightarrow BE}\sigma^{1/2}\rho
^{1/2}\right\Vert _{1}^{2}\\
&  =-\ln F\left(  \rho,\mathcal{P}_{\sigma,\mathcal{N}}(\mathcal{N}%
(\rho))\right)  .
\end{align}
where $F(\rho,\sigma)\equiv\left\Vert \sqrt{\rho}\sqrt{\sigma}\right\Vert
_{1}^{2}$ is the quantum fidelity. Thus, if $\widetilde{\Delta}_{\alpha}%
(\rho,\sigma,\mathcal{N})$ were monotone non-decreasing with respect to
$\alpha$, we could combine these observations to conclude that%
\begin{align}
D(\rho\Vert\sigma)-D(\mathcal{N}(\rho)\Vert\mathcal{N}(\sigma))  &  =\frac
{1}{\ln2}\widetilde{\Delta}_{1}(\rho,\sigma,\mathcal{N})\\
&  \overset{?}{\geq}\frac{1}{\ln2}\widetilde{\Delta}_{1/2}(\rho,\sigma
,\mathcal{N})\\
&  =-\log F\left(  \rho,\mathcal{P}_{\sigma,\mathcal{N}}(\mathcal{N}%
(\rho))\right)  .
\end{align}
If this were true, then we could conclude that
Theorem~\ref{thm-qeir:main-theorem}\ would be true with the recovery channel
taken to be the Petz recovery map. However, it is not known whether this is
true, and we will instead invoke the Stein--Hirschman theorem to conclude that
a convex combination of rotated Petz maps satisfies the bound stated in
Theorem~\ref{thm-qeir:main-theorem}.

\section{Proof of the Recoverability Theorem}

\label{sec:main-result}This section presents the proof of
Theorem~\ref{thm-qeir:main-theorem}. In fact, we prove the following stronger
statement, which implies Theorem~\ref{thm-qeir:main-theorem} for the following
recovery channel%
\begin{equation}
\mathcal{R}_{\sigma,\mathcal{N}}=\int_{-\infty}^{\infty}dt\ \beta
_{0}(t)\ \mathcal{R}_{\sigma,\mathcal{N}}^{t/2},
\end{equation}
due to the concavity of both the logarithm and the fidelity.

\begin{theorem}
\label{thm-qeir:rel-ent-better-main}Let $\rho$, $\sigma$, and $\mathcal{N}$ be
as given in Definition~\ref{def:rho-sig-N}. Then the following inequality
holds%
\begin{equation}
D(\rho\Vert\sigma)-D(\mathcal{N(}\rho)\Vert\mathcal{N(}\sigma))\geq
-\int_{-\infty}^{\infty}dt\ \beta_{0}(t)\ \log\left[  F\left(  \rho
,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N})(\rho)\right)
\right]  , \label{eq:rel-ent-ineq}%
\end{equation}
where $\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}$ is a rotated Petz recovery map
from Definition~\ref{def-qeir:rotated-petz}.
\end{theorem}

\begin{proof}
We can prove this result by employing Theorem~\ref{thm-qeir:op-hirschman}. We
first establish the inequality in (\ref{eq:rel-ent-ineq}). Let $U:\mathcal{H}%
\rightarrow\mathcal{H}^{\prime}\otimes\mathcal{H}_{E}$ be an isometric
extension of the channel $\mathcal{N}$. Pick%
\begin{equation}
G(z)\equiv\left(  \left[  \mathcal{N}(\rho)\right]  ^{z/2}\left[
\mathcal{N}(\sigma)\right]  ^{-z/2}\otimes I_{E}\right)  U\sigma^{z/2}%
\rho^{1/2},
\end{equation}
for $z\in\overline{S}$, $p_{0}=2$, $p_{1}=1$, and $\theta\in\left(
0,1\right)  $, which fixes $p_{\theta}=\frac{2}{1+\theta}$. The operator
valued-function $G\left(  z\right)  $ satisfies the conditions needed to apply
Theorem~\ref{thm-qeir:op-hirschman}. For the choices above, we find that%
\begin{align}
\left\Vert G(\theta)\right\Vert _{2/\left(  1+\theta\right)  }  &  =\left\Vert
\left(  \left[  \mathcal{N}(\rho)\right]  ^{\theta/2}\left[  \mathcal{N}%
(\sigma)\right]  ^{-\theta/2}\otimes I_{E}\right)  U\sigma^{\theta/2}%
\rho^{1/2}\right\Vert _{2/\left(  1+\theta\right)  },\\
\left\Vert G\left(  it\right)  \right\Vert _{2}  &  =\left\Vert \left(
\left[  \mathcal{N}(\rho)\right]  ^{it/2}\left[  \mathcal{N}(\sigma)\right]
^{-it/2}\otimes I_{E}\right)  U\sigma^{it}\rho^{1/2}\right\Vert _{2}%
\nonumber\\
&  \leq\left\Vert \rho^{1/2}\right\Vert _{2}\nonumber\\
&  =1,\label{eq:M0}\\
\left\Vert G\left(  1+it\right)  \right\Vert _{1}  &  =\left\Vert \left(
\left[  \mathcal{N}(\rho)\right]  ^{1+it/2}\left[  \mathcal{N}(\sigma)\right]
^{-\left(  1+it\right)  /2}\otimes I_{E}\right)  U\sigma^{\left(  1+it\right)
/2}\rho^{1/2}\right\Vert _{1}\nonumber\\
&  =\left\Vert \left(  \left[  \mathcal{N}(\rho)\right]  ^{\frac{it}{2}%
}\left[  \mathcal{N}(\rho)\right]  ^{\frac{1}{2}}\left[  \mathcal{N}%
(\sigma)\right]  ^{-\frac{it}{2}}\left[  \mathcal{N}(\sigma)\right]
^{-\frac{1}{2}}\otimes I_{E}\right)  U\sigma^{\frac{1}{2}}\sigma^{\frac{it}%
{2}}\rho^{\frac{1}{2}}\right\Vert _{1}\nonumber\\
&  =\left\Vert \left(  \left[  \mathcal{N}(\rho)\right]  ^{1/2}\left[
\mathcal{N}(\sigma)\right]  ^{-it/2}\left[  \mathcal{N}(\sigma)\right]
^{-1/2}\otimes I_{E}\right)  U\sigma^{1/2}\sigma^{it/2}\rho^{1/2}\right\Vert
_{1}\nonumber\\
&  =\sqrt{F}\left(  \rho,\left(  \mathcal{U}_{\sigma,-t/2}\circ\mathcal{P}%
_{\sigma,\mathcal{N}}\circ\mathcal{U}_{\mathcal{N}(\sigma),t/2}\right)
(\mathcal{N}(\rho))\right) \nonumber\\
&  =\sqrt{F}(\rho,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N}%
)(\rho)). \label{eq:M1}%
\end{align}
Then we can apply Theorem~\ref{thm-qeir:op-hirschman} to conclude that%
\begin{multline}
\ln\left\Vert \left(  \left[  \mathcal{N(}\rho)\right]  ^{\theta/2}\left[
\mathcal{N(}\sigma)\right]  ^{-\theta/2}\otimes I_{E}\right)  U\sigma
^{\theta/2}\rho^{1/2}\right\Vert _{2/\left(  1+\theta\right)  }%
\label{eq:leads-to-improved-bnd}\\
\leq\int_{-\infty}^{\infty}dt\ \beta_{\theta}(t)\ln\left[  F\left(
\rho,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N})(\rho)\right)
^{\theta/2}\right]  .
\end{multline}
This implies that%
\begin{multline}
-\frac{2}{\theta}\ln\left\Vert \left(  \left[  \mathcal{N(}\rho)\right]
^{\theta/2}\left[  \mathcal{N(}\sigma)\right]  ^{-\theta/2}\otimes
I_{E}\right)  U\sigma^{\theta/2}\rho^{1/2}\right\Vert _{2/\left(
1+\theta\right)  }\label{eq:critical-inequality}\\
\geq-\int_{-\infty}^{\infty}dt\ \beta_{\theta}(t)\ \ln\left[  F\left(
\rho,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N})(\rho)\right)
\right]  .
\end{multline}
Letting $\theta=\left(  1-\alpha\right)  /\alpha$, we see that this is the
same as%
\begin{equation}
\widetilde{\Delta}_{\alpha}(\rho,\sigma,\mathcal{N})\geq-\int_{-\infty
}^{\infty}dt\ \beta_{\left(  1-\alpha\right)  /\alpha}(t)\ \ln\left[  F\left(
\rho,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N})(\rho)\right)
\right]  . \label{eq:alpha-bound}%
\end{equation}
Since the inequality in \eqref{eq:critical-inequality} holds for all
$\theta\in\left(  0,1\right)  $ and thus \eqref{eq:alpha-bound} holds for all
$\alpha\in\left(  1/2,1\right)  $, we can take the limit as $\alpha\nearrow1$
and apply \eqref{eq:rel-ent-diff-a-1} and the dominated convergence theorem to
conclude that \eqref{eq:rel-ent-ineq} holds.
\end{proof}

As a corollary of the above theorem, we obtain equality conditions for the
monotonicity of quantum relative entropy:

\begin{corollary}
[Equality Conditions]Let $\rho$, $\sigma$, and $\mathcal{N}$ be as given in
Definition~\ref{def:rho-sig-N}. Then%
\begin{equation}
D(\rho\Vert\sigma)=D(\mathcal{N}(\rho)\Vert\mathcal{N}(\sigma))
\end{equation}
if and only if all rotated Petz recovery maps perfectly recover $\rho$ from
$\mathcal{N}(\rho)$:%
\begin{equation}
\forall t\in\mathbb{R}:(\mathcal{R}_{\sigma,\mathcal{N}}^{t}\circ
\mathcal{N})(\rho)=\rho. \label{eq-qeir:perfect-recovery-rho}%
\end{equation}

\end{corollary}

\begin{proof}
Recall from Exercise~\ref{ex-qeir:perfect-recovery}\ that, independent of the
conditions in the statement of the corollary, we always have that
$(\mathcal{R}_{\sigma,\mathcal{N}}^{t}\circ\mathcal{N})(\sigma)=\sigma$ for
all $t\in\mathbb{R}$.

We start by proving the \textquotedblleft only if\textquotedblright\ part.
Suppose that $\forall t\in\mathbb{R}:(\mathcal{R}_{\sigma,\mathcal{N}}%
^{t}\circ\mathcal{N})(\rho)=\rho$. Then for a particular $t\in\mathbb{R}$, the
monotonicity of quantum relative entropy implies that%
\begin{align}
D(\rho\Vert\sigma)  &  \geq D(\mathcal{N(}\rho)\Vert\mathcal{N(}\sigma))\\
D(\mathcal{N(}\rho)\Vert\mathcal{N(}\sigma))  &  \geq D((\mathcal{R}%
_{\sigma,\mathcal{N}}^{t}\circ\mathcal{N})(\rho)\Vert(\mathcal{R}%
_{\sigma,\mathcal{N}}^{t}\circ\mathcal{N})(\sigma))\\
&  =D(\rho\Vert\sigma),
\end{align}
which in turn imply that $D(\rho\Vert\sigma)=D(\mathcal{N}(\rho)\Vert
\mathcal{N}(\sigma))$.

We now prove the \textquotedblleft if part\textquotedblright\ of the theorem.
Suppose that $D(\rho\Vert\sigma)=D(\mathcal{N}(\rho)\Vert\mathcal{N}(\sigma
))$. By Theorem~\ref{thm-qeir:rel-ent-better-main}, we can conclude that%
\begin{equation}
\int_{-\infty}^{\infty}dt\ \beta_{0}(t)\ \left[  -\log\left[  F\left(
\rho,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N})(\rho)\right)
\right]  \right]  =0.
\end{equation}
Since $\beta_{0}(t)$ is a positive definite function for all $t\in\mathbb{R}$
and $-\log F\geq0$, we can conclude that%
\begin{equation}
-\log\left[  F\left(  \rho,(\mathcal{R}_{\sigma,\mathcal{N}}^{t/2}%
\circ\mathcal{N})(\rho)\right)  \right]  =0
\end{equation}
for all $t\in\mathbb{R}$, which is the same as $F(\rho,(\mathcal{R}%
_{\sigma,\mathcal{N}}^{t/2}\circ\mathcal{N})(\rho))=1$ for all $t\in
\mathbb{R}$. We can then conclude that \eqref{eq-qeir:perfect-recovery-rho}
holds because the fidelity between two states is equal to one if and only if
the states are the same.
\end{proof}

\section{Refinements of Quantum Entropy Inequalities}

\label{sec-qeir:corollaries}Theorem~\ref{thm-qeir:main-theorem} leads to a
strengthening of many quantum entropy inequalities, including strong
subadditivity of quantum entropy, concavity of conditional entropy, joint
convexity of relative entropy, non-negativity of quantum discord, and the
Holevo bound. We go through the refinement of strong subadditivity and point to the book for the others.

\subsection{Strong Subadditivity}

Recall the conditional quantum mutual information of a tripartite state
$\rho_{ABC}$:%
\begin{equation}
I(A;B|C)_{\rho}\equiv H(AC)_{\rho}+H(BC)_{\rho}-H(C)_{\rho}-H(ABC)_{\rho}.
\end{equation}
Strong subadditivity is the statement that $I(A;B|C)_{\rho}\geq0$ for all
tripartite states $\rho_{ABC}$.

Corollary~\ref{thm-qeir:CMI} below gives an improvement of strong
subadditivity. It is a direct consequence of
Theorem~\ref{thm-qeir:main-theorem}\ after choosing%
\begin{equation}
\rho=\rho_{ABC},\ \ \ \ \sigma=\rho_{AC}\otimes I_{B},\ \ \ \ \mathcal{N}%
=\operatorname{Tr}_{A},
\end{equation}
so that%
\begin{equation}
\mathcal{N}(\rho)=\rho_{BC},\ \ \ \ \mathcal{N}(\sigma)=\rho_{C}\otimes
I_{B},\ \ \ \ \mathcal{N}^{\dag}(\cdot)=(\cdot)\otimes I_{A},
\end{equation}
and%
\begin{align}
D(\rho\Vert\sigma)-D(\mathcal{N}(\rho)\Vert\mathcal{N}(\sigma))  &
=D(\rho_{ABC}\Vert\rho_{AC}\otimes I_{B})-D(\rho_{BC}\Vert\rho_{C}\otimes
I_{B})\nonumber\\
&  =I(A;B|C)_{\rho},\\
\mathcal{P}_{\sigma,\mathcal{N}}(\cdot)  &  =\sigma^{1/2}\mathcal{N}^{\dag
}\left(  \left[  \mathcal{N}(\sigma)\right]  ^{-1/2}(\cdot)\left[
\mathcal{N}(\sigma)\right]  ^{-1/2}\right)  \sigma^{1/2}\nonumber\\
&  =\rho_{AC}^{1/2}\left[  \rho_{C}^{-1/2}(\cdot)\rho_{C}^{-1/2}\otimes
I_{A}\right]  \rho_{AC}^{1/2}. \label{eq:Petz-channel-CMI}%
\end{align}


\begin{corollary}
\label{thm-qeir:CMI}Let $\rho_{ABC}\in\mathcal{D}(\mathcal{H}_{A}%
\otimes\mathcal{H}_{B}\otimes\mathcal{H}_{C})$. Then the following inequality
holds%
\begin{equation}
I(A;B|C)_{\rho}\geq-\log\left[  F(\rho_{ABC},\mathcal{R}_{C\rightarrow
AC}(\rho_{BC}))\right]  , \label{eq:main-result}%
\end{equation}
where the recovery channel $\mathcal{R}_{C\rightarrow AC}=\int_{-\infty
}^{\infty}dt\ \beta_{0}(t)\ \mathcal{R}_{\rho_{AC},\operatorname{Tr}_{A}%
}^{t/2}$ perfectly recovers $\rho_{AC}$ from $\rho_{C}$.
\end{corollary}



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