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\begin{document}

\lecture{9 --- September 23, 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 last lecture we discussed more general models of measurements, the POVM formalism, product states, separable states, and the partial trace operation.

The evolution of a quantum state is never perfect. In
this lecture, we discuss the most general approach to understanding quantum
evolutions:\ the \textit{axiomatic approach}. In this powerful approach, we
start with three physically reasonable axioms that should hold for any quantum
evolution and then deduce a set of mathematical constraints that any quantum evolution
should satisfy (this is known as the \textit{Choi-Kraus theorem}). For the rest of the course, we will refer to quantum evolutions satisfying these constraints as
\textit{quantum channels}.


\section{Axiomatic Approach to Noisy Quantum Evolutions}

\label{sec-nqt:axiomatic-approach}We now discuss a powerful approach to
understanding quantum physical evolutions called the \textit{axiomatic
approach}. Here we make three physically reasonable assumptions that any
quantum evolution should satisfy and then prove that these axioms imply
mathematical constraints on the form of any quantum physical evolution.

All of the constraints we impose are motivated by the reasonable requirement
for the output of the evolution to be a quantum state (density operator) if
the input to the evolution is a quantum state (density operator). An important
assumption to clarify at the outset is that we are viewing a quantum physical
evolution as a \textquotedblleft black box,\textquotedblright\ meaning that
Alice can prepare any state that she wishes before the evolution begins,
including pure states or mixed states. Critically, we even allow her to input
one share of an entangled state. This is a standard assumption in quantum
information theory, but one could certainly question whether this assumption
is reasonable. If we do accept this criterion as physically reasonable, then
the Choi-Kraus representation theorem for quantum evolutions follows as a consequence.

\begin{notation}
[Density Operators and Linear Operators]Let $\mathcal{D}(\mathcal{H})$ denote
the space of density operators acting on a Hilbert space $\mathcal{H}$, let
$\mathcal{B}(\mathcal{H})$ denote the space of square linear operators acting
on $\mathcal{H}$, and let $\mathcal{B}(\mathcal{H}_{A},\mathcal{H}_{B})$
denote the space of linear operators taking a Hilbert space $\mathcal{H}_{A}$
to a Hilbert space $\mathcal{H}_{B}$.
\end{notation}

Throughout this development, we let $\mathcal{N}$ denote a map which takes
density operators in $\mathcal{D}(\mathcal{H}_{A})$ to those in $\mathcal{D}%
(\mathcal{H}_{B})$. In general, the respective input and output Hilbert spaces
$\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ need not be the same. Implicitly, we
have already stated a first physically reasonable requirement that we impose
on $\mathcal{N}$, namely, that\ $\mathcal{N}(\rho_{A})\in\mathcal{D}%
(\mathcal{H}_{B})$ if $\rho_{A}\in\mathcal{D}(\mathcal{H}_{A})$. Extending
this requirement, we demand that $\mathcal{N}$ should be \textit{convex
linear} when acting on $\mathcal{D}(\mathcal{H}_{A})$:%
\begin{equation}
\mathcal{N}(\lambda\rho_{A}+(1-\lambda)\sigma_{A})=\lambda\mathcal{N}(\rho
_{A})+(1-\lambda)\mathcal{N}(\sigma_{A}),
\label{eq-nqt:convex-linearity-channels}%
\end{equation}
where $\rho_{A},\sigma_{A}\in\mathcal{D}(\mathcal{H}_{A})$ and $\lambda
\in\left[  0,1\right]  $.

The physical interpretation of this convex-linearity requirement is in terms
of repeated experiments. Suppose a large number of experiments are conducted
in which identical quantum systems are prepared in the state $\rho_{A}$ for a
fraction $\lambda$ of the experiments and in the state $\sigma_{A}$ for the
other fraction $1-\lambda$ of the experiments. Suppose further that it is not
revealed which states are prepared for which experiments. Before you are
allowed to perform measurements on each system, the evolution $\mathcal{N}$ is
applied to each of the systems. The density operator characterizing the state
of each system for these experiments is then $\mathcal{N}(\lambda\rho
_{A}+(1-\lambda)\sigma_{A})$. You are then allowed to perform a measurement on
each system, which after a large number of experiments allow you to infer that
the density operator is $\mathcal{N}(\lambda\rho_{A}+(1-\lambda)\sigma_{A})$.
Now, in principle, it could have been revealed which fraction of the
experiments had the state $\rho_{A}$ prepared and which fraction had
$\sigma_{A}$ prepared. In this case, the density operator describing the
$\rho_{A}$ experiments would be $\mathcal{N}(\rho_{A})$ and that describing
the $\sigma_{A}$ experiments would be $\mathcal{N}(\sigma_{A})$. So, it is
reasonable to expect that the statistics observed in your measurement outcomes
in the first scenario would be consistent with those observed in the second
scenario, and this is the physical statement that the requirement \eqref{eq-nqt:convex-linearity-channels}\ makes.

Now, it is mathematically convenient to extend the domain and range of the
quantum channel to apply not only to density operators but to all linear
operators. To this end, it is possible to find a unique linear extension
$\widetilde{\mathcal{N}}$ of any quantum evolution $\mathcal{N}$ defined as
above (originally defined exclusively by its action on density operators and
satisfying convex linearity). See Section~\ref{app:unique-linear-ext}\ for a
full development of this idea. Thus, it is reasonable to associate this unique
linear extension $\widetilde{\mathcal{N}}$\ to the quantum physical evolution
$\mathcal{N}$ mathematically, and in what follows (and for the rest of the
book), we simply identify a physical evolution $\mathcal{N}$\ with its unique
linear extension $\widetilde{\mathcal{N}}$, and this is what we call a
\textit{quantum channel}. For these reasons, we now impose that any quantum
channel $\mathcal{N}$ is linear:

\begin{criterion}
[Linearity]\label{crit-nqt:linearity}A quantum channel $\mathcal{N}$ is a
linear map:%
\begin{equation}
\mathcal{N}(\alpha X_{A}+\beta Y_{A})=\alpha\mathcal{N}(X_{A})+\beta
\mathcal{N}(Y_{A}),
\end{equation}
where $X_{A},Y_{A}\in\mathcal{B}(\mathcal{H}_{A})$ and $\alpha,\beta
\in\mathbb{C}$.
\end{criterion}

We have already demanded that quantum physical evolutions should take density
operators to density operators. Combining with linearity (in particular, scale
invariance) implies that quantum channels should preserve the class of
positive semi-definite operators. That is, they should be positive maps, as
defined below:

\begin{definition}
[Positive Map]A linear map $\mathcal{M}:\mathcal{B}(\mathcal{H}_{A}%
)\rightarrow\mathcal{B}(\mathcal{H}_{B})$ is positive if $\mathcal{M}(X_{A})$
is positive semi-definite for all positive semi-definite $X_{A}\in
\mathcal{B}(\mathcal{H}_{A})$.
\end{definition}

If we were dealing with classical systems, then positivity would be sufficient
to describe the class of physical evolutions. However, above we argued that we
are working in the \textquotedblleft black box\textquotedblright\ picture of
quantum physical evolutions, and here, in principle, we allow for Alice to
prepare the input system $A$ to be one share of an arbitrary two-party state
$\rho_{RA}\in\mathcal{D}(\mathcal{H}_{R}\otimes\mathcal{H}_{A})$, where $R$ is
a reference system of arbitrary size. So this means that the evolution
consisting of the identity acting on the reference system $R$ and the map
$\mathcal{N}$ acting on system $A$ should take $\rho_{RA}$ to a density
operator on systems $R$ and $B$. Let $\operatorname{id}_{R}\otimes
\mathcal{N}_{A\rightarrow B}$ denote this evolution, where $\operatorname{id}%
_{R}$ denotes the identity superoperator acting on the system $R$.

How do we describe the evolution $\operatorname{id}_{R}\otimes\mathcal{N}%
_{A\rightarrow B}$ mathematically? Let $X_{RA}$ be an arbitrary operator
acting on $\mathcal{H}_{R}\otimes\mathcal{H}_{A}$, and let $\{|i\rangle_{R}\}$
be an orthonormal basis for $\mathcal{H}_{R}$. Then we can expand $X_{RA}$
with respect to this basis as follows:%
\begin{equation}
X_{RA}=\sum_{i,j}|i\rangle\langle j|_{R}\otimes X_{A}^{i,j},
\end{equation}
and the action of $\operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow B}$
on $X_{RA}$ (for linear $\mathcal{N}$) is defined as follows:%
\begin{align}
\left(  \operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow B}\right)
\left(  X_{RA}\right)   &  =\left(  \operatorname{id}_{R}\otimes
\mathcal{N}_{A\rightarrow B}\right)  \left(  \sum_{i,j}|i\rangle\langle
j|_{R}\otimes X_{A}^{i,j}\right) \\
&  =\sum_{i,j}\left(  \operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow
B}\right)  \left(  |i\rangle\langle j|_{R}\otimes X_{A}^{i,j}\right) \\
&  =\sum_{i,j}\operatorname{id}_{R}\left(  |i\rangle\langle j|_{R}\right)
\otimes\mathcal{N}_{A\rightarrow B}\left(  X_{A}^{i,j}\right) \\
&  =\sum_{i,j}|i\rangle\langle j|_{R}\otimes\mathcal{N}_{A\rightarrow
B}\left(  X_{A}^{i,j}\right)  . \label{eq-nqt:action-id-N}%
\end{align}
That is, the identity superoperator $\operatorname{id}_{R}$ has no effect on
the $R$ system. The above development leads to the notion of a linear map
being \textit{completely positive} and our next criterion for any quantum
physical evolution:

\begin{definition}
[Completely Positive Map]\label{def-nqt:completely-positive}A linear map
$\mathcal{M}:\mathcal{B}(\mathcal{H}_{A})\rightarrow\mathcal{B}(\mathcal{H}%
_{B})$ is completely positive if $\operatorname{id}_{R}\otimes\mathcal{M}$ is
a positive map for a reference system $R$ of arbitrary size.
\end{definition}

\begin{criterion}
[Complete Positivity]\label{crit-nqt:CP}A quantum channel is a completely
positive map.
\end{criterion}

There is one last requirement that we impose for quantum physical evolutions,
known as \textit{trace preservation}. This requirement again stems from the
reasonable constraint that $\mathcal{N}$ should map density operators to
density operators. That is, it should be the case that $\operatorname{Tr}%
\{\rho_{A}\}=\operatorname{Tr}\{\mathcal{N}(\rho_{A})\}=1$ for all input
density operators $\rho_{A}$. However, now that have argued for linearity of
every quantum physical evolution, trace preservation on density operators
combined with linearity implies that quantum channels are trace preserving on
the set of all operators. This is due to the fact that there are sets of
density operators that form a basis for $\mathcal{B}(\mathcal{H}_{A})$.
Indeed, one such basis of density operators is as follows:%
\begin{equation}
\rho_{A}^{x,y}=\left\{
\begin{array}
[c]{cc}%
|x\rangle\langle x|_{A} & \text{if }x=y\\
\frac{1}{2}\left(  |x\rangle_{A}+|y\rangle_{A}\right)  \left(  \langle
x|_{A}+\langle y|_{A}\right)  & \text{if }x<y\\
\frac{1}{2}\left(  |x\rangle_{A}+i|y\rangle_{A}\right)  \left(  \langle
x|_{A}-i\langle y|_{A}\right)  & \text{if }x>y
\end{array}
\right.  . \label{eq-nqt:density-op-basis}%
\end{equation}
Consider that for all $x,y$ such that $x<y$, the following holds%
\begin{align}
|x\rangle\langle y|_{A}  &  =\left(  \rho_{A}^{x,y}-\frac{1}{2}\rho_{A}%
^{x,x}-\frac{1}{2}\rho_{A}^{y,y}\right)  -i\left(  \rho_{A}^{y,x}-\frac{1}%
{2}\rho_{A}^{x,x}-\frac{1}{2}\rho_{A}^{y,y}\right)  ,\\
|y\rangle\langle x|_{A}  &  =\left(  \rho_{A}^{x,y}-\frac{1}{2}\rho_{A}%
^{x,x}-\frac{1}{2}\rho_{A}^{y,y}\right)  +i\left(  \rho_{A}^{y,x}-\frac{1}%
{2}\rho_{A}^{x,x}-\frac{1}{2}\rho_{A}^{y,y}\right)  ,
\end{align}
so that we can represent any operator $X_{A}$ as a linear combination of
density operators from the set $\left\{  \rho_{A}^{x,y}\right\}  $. This leads
to our final criterion for quantum channels:

\begin{criterion}
[Trace Preservation]\label{crit-nqt:TP}A quantum channel is trace preserving,
in the sense that $\operatorname{Tr}\{X_{A}\}=\operatorname{Tr}\{\mathcal{N}%
(X_{A})\}$ for all $X_{A}\in\mathcal{B}(\mathcal{H}_{A})$.
\end{criterion}

\begin{definition}
[Quantum Channel]A quantum channel is a linear, completely positive, trace
preserving map, corresponding to a quantum physical evolution.%
\index{completely positive trace-preserving map}%

\end{definition}

Criteria~\ref{crit-nqt:linearity}, \ref{crit-nqt:CP}, and \ref{crit-nqt:TP}
detailed above lead naturally to the Choi-Kraus representation theorem, which
states that a map satisfies all three criteria if and only if it takes a
particular form according to a Choi-Kraus decomposition:

\begin{theorem}
[Choi-Kraus]\label{thm-nqt:choi-kraus-theorem}A map $\mathcal{N}%
:\mathcal{B}(\mathcal{H}_{A})\rightarrow\mathcal{B}(\mathcal{H}_{B})$ (denoted
also by $\mathcal{N}_{A\rightarrow B}$) is linear, completely positive, and
trace-preserving if and only if it has a Choi-Kraus decomposition as follows:%
\begin{equation}
\mathcal{N}_{A\rightarrow B}(X_{A})=\sum_{l=0}^{d-1}V_{l}X_{A}V_{l}^{\dag},
\label{eq-nqt:kraus-decomp}%
\end{equation}
where $X_{A}\in\mathcal{B}(\mathcal{H}_{A})$, $V_{l}\in\mathcal{B}%
(\mathcal{H}_{A},\mathcal{H}_{B})$ for all $l\in\left\{  0,\ldots,d-1\right\}
$,%
\begin{equation}
\sum_{l=0}^{d-1}V_{l}^{\dag}V_{l}=I_{A}, \label{eq-nqt:trace-preserve-kraus}%
\end{equation}
and $d$ need not be any larger than $\dim(\mathcal{H}_{A})\dim(\mathcal{H}%
_{B})$.
\end{theorem}

Before we delve into a proof, it is helpful to give a sketch. There is an
easier part and a more challenging part of the proof. For the more challenging
part, a helpful tool is an operator called the Choi operator:

\begin{definition}
[Choi Operator]\label{def-nqt:choi-op}Let $\mathcal{H}_{R}$ and $\mathcal{H}%
_{A}$ be isomorphic Hilbert spaces, and let $\{|i\rangle_{R}\}$ and
$\{|i\rangle_{A}\}$ be orthonormal bases for $\mathcal{H}_{R}$ and
$\mathcal{H}_{A}$, respectively. Let $\mathcal{H}_{B}$ be some other Hilbert
space, and let $\mathcal{N}:\mathcal{B}(\mathcal{H}_{A})\rightarrow
\mathcal{B}(\mathcal{H}_{B})$ be a linear map (abbreviated as $\mathcal{N}%
_{A\rightarrow B}$). The Choi operator corresponding to $\mathcal{N}%
_{A\rightarrow B}$ and the bases $\{|i\rangle_{R}\}$ and $\{|i\rangle_{A}\}$
is defined as the following operator:%
\begin{equation}
\left(  \operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow B}\right)
\left(  \left\vert \Gamma\right\rangle \left\langle \Gamma\right\vert
_{RA}\right)  =\sum_{i,j=0}^{d_{A}-1}\vert i\rangle\left\langle j\right\vert
_{R}\otimes\mathcal{N}_{A\rightarrow B}(|i\rangle\left\langle j\right\vert
_{A}), \label{eq:choi-matrix}%
\end{equation}
where $d_{A}\equiv\dim(\mathcal{H}_{A})$ and $\left\vert \Gamma\right\rangle
_{RA}$ is an unnormalized maximally entangled vector, as defined in
(\ref{eq-qt:unnorm-max-ent}):%
\begin{equation}
\left\vert \Gamma\right\rangle _{RA}\equiv\sum_{i=0}^{d_{A}-1}\left\vert
i\right\rangle _{R}\otimes|i\rangle_{A}.
\end{equation}

\end{definition}

If $\mathcal{N}_{A\rightarrow B}$ is a completely positive map, then the Choi
operator is positive semi-definite. This follows as a direct consequence of
Definition~\ref{def-nqt:completely-positive} and the fact that $\left\vert
\Gamma\right\rangle \left\langle \Gamma\right\vert _{RA}$ is positive
semi-definite. The converse is true as well, and Exercise~\ref{ex-nqt:choi-CP}%
\ asks you to verify this. The converse is in some sense a much more powerful
statement. Definition~\ref{def-nqt:completely-positive} suggests that we would
have to check a seemingly infinite number of cases in order to verify whether
a given linear map is completely positive, but the converse statement
establishes that we need to check only one condition: whether the Choi
operator is positive semi-definite.

Why else is the Choi operator a useful tool? One other important reason is
that it encodes how a quantum channel acts on any possible input operator
$X_{A}$, and thus specifies the channel completely. Consider that we can
expand the Choi operator as a matrix of matrices (of total size $d_{A}%
d_{B}\times d_{A}d_{B}$) in the following way, by exploiting properties of the
tensor product:%
\begin{equation}%
\begin{bmatrix}
\mathcal{N}_{A\rightarrow B}(|0\rangle\langle 0\vert _{A}) &
\mathcal{N}_{A\rightarrow B}(|0\rangle\langle 1\vert _{A}) & \cdots
& \mathcal{N}_{A\rightarrow B}(|0\rangle\left\langle d_{A}-1\right\vert
_{A})\\
\mathcal{N}_{A\rightarrow B}(|1\rangle\langle 0\vert _{A}) &
\mathcal{N}_{A\rightarrow B}(|1\rangle\langle 1\vert _{A}) & \cdots
& \mathcal{N}_{A\rightarrow B}(|1\rangle\left\langle d_{A}-1\right\vert
_{A})\\
\vdots & \vdots & \ddots & \vdots\\
\mathcal{N}_{A\rightarrow B}(|d_{A}-1\rangle\langle 0\vert _{A}) &
\mathcal{N}_{A\rightarrow B}(|d_{A}-1\rangle\langle 1\vert _{A}) &
\cdots & \mathcal{N}_{A\rightarrow B}(|d_{A}-1\rangle\left\langle
d_{A}-1\right\vert _{A})
\end{bmatrix}
.
\end{equation}
So if we would like to figure out how the channel $\mathcal{N}_{A\rightarrow
B}$ acts on an input operator $X_{A}$, we can first expand $X_{A}$ with
respect to the orthonormal basis $\{|i\rangle_{A}\}$ as $X_{A}=\sum
_{i,j}x^{i,j}|i\rangle\langle j|_{A}$ and then apply the channel, using
linearity:%
\begin{equation}
\mathcal{N}_{A\rightarrow B}(X_{A})=\mathcal{N}_{A\rightarrow B}\left(
\sum_{i,j}x^{i,j}|i\rangle\langle j|_{A}\right)  =\sum_{i,j}x^{i,j}%
\mathcal{N}_{A\rightarrow B}(|i\rangle\langle j|_{A}).
\label{eq-nqt:linear-action-of-channel}%
\end{equation}
So the procedure is to expand $X_{A}$ as above, multiple the $\left(
i,j\right)  $ coefficient $x^{i,j}$ with the $(  i,j)  $ entry in
the Choi operator, and then sum these operators over all indices $i$ and $j$.

\bigskip

\begin{proof}
[Proof of Theorem~\ref{thm-nqt:choi-kraus-theorem}]We first prove the easier
\textquotedblleft if-part\textquotedblright\ of the theorem. So let us suppose
that $\mathcal{N}_{A\rightarrow B}$ has the form in
\eqref{eq-nqt:kraus-decomp} and that the condition in
\eqref{eq-nqt:trace-preserve-kraus} holds as well. Then $\mathcal{N}%
_{A\rightarrow B}$ is clearly a linear map. It is completely positive because
$(\operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow B})(X_{RA})\geq0$ if
$X_{RA}\geq0$ when $\mathcal{N}_{A\rightarrow B}$ has the form in
\eqref{eq-nqt:kraus-decomp}, and this holds for a reference system $R$ of
arbitrary size. That is, consider from \eqref{eq-nqt:action-id-N} that
$\{I_{R}\otimes V_{l}\}$ is a set of Kraus operators for the extended channel
$\operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow B}$\ and thus%
\begin{align}
(\operatorname{id}_{R}\otimes\mathcal{N}_{A\rightarrow B})(X_{RA})  &
=\sum_{l=0}^{d-1}(I_{R}\otimes V_{l})X_{RA}(I_{R}\otimes V_{l}^{\dag})\\
&  =\sum_{l=0}^{d-1}(I_{R}\otimes V_{l})X_{RA}(I_{R}\otimes V_{l})^{\dag}.
\end{align}
We know that $(I_{R}\otimes V_{l})X_{RA}(I_{R}\otimes V_{l})^{\dag}\geq0$ for
all $l$ when $X_{RA}\geq0$, and the same is true for the sum. Trace
preservation follows because%
\begin{align}
\operatorname{Tr}\left\{  \mathcal{N}_{A\rightarrow B}(X_{A})\right\}   &
=\operatorname{Tr}\left\{  \sum_{l=0}^{d-1}V_{l}X_{A}V_{l}^{\dag}\right\} \\
&  =\operatorname{Tr}\left\{  \sum_{l=0}^{d-1}V_{l}^{\dag}V_{l}X_{A}\right\}
\\
&  =\operatorname{Tr}\left\{  X_{A}\right\}  ,
\end{align}
where the second line is from linearity and cyclicity of trace and the last
line follows from the condition in \eqref{eq-nqt:trace-preserve-kraus}.

We now prove the more difficult \textquotedblleft only-if\textquotedblright%
\ part. Let $d_{A}\equiv\dim(\mathcal{H}_{A})$ and $d_{B}\equiv\dim
(\mathcal{H}_{B})$. Consider that we can diagonalize the Choi operator as
given in Definition~\ref{def-nqt:choi-op}, because it is positive
semi-definite:%
\begin{equation}
\mathcal{N}_{A\rightarrow B}\left(  \left\vert \Gamma\right\rangle
\left\langle \Gamma\right\vert _{RA}\right)  =\sum_{l=0}^{d-1}\left\vert
\phi_{l}\right\rangle \left\langle \phi_{l}\right\vert _{RB},
\label{eq:decomposition-choi}%
\end{equation}
where $d\leq d_{A}d_{B}$ is the Choi rank of the map $\mathcal{N}%
_{A\rightarrow B}$. (This decomposition does not necessarily have to be such
that the vectors $\left\{  \left\vert \phi_{l}\right\rangle _{RB}\right\}  $
are orthonormal, but keep in mind that there is always a choice such that
$d\leq d_{A}d_{B}$.) Consider by inspecting (\ref{eq:choi-matrix}) that%
\begin{equation}
\left(  \langle i\vert_{R}\otimes I_{B}\right)  \left(  \mathcal{N}%
_{A\rightarrow B}\left(  \left\vert \Gamma\right\rangle \left\langle
\Gamma\right\vert _{RA}\right)  \right)  \left(  \left\vert j\right\rangle
_{R}\otimes I_{B}\right)  =\mathcal{N}_{A\rightarrow B}\left(  |i\rangle
\langle j\vert\right)  .
\end{equation}


Now, consider that for any bipartite vector $\left\vert \phi\right\rangle
_{RB}$, we can expand it in terms of an orthonormal basis $\left\{  \left\vert
j\right\rangle _{B}\right\}  $ and the basis $\left\{  \left\vert
i\right\rangle _{R}\right\}  $ given above:%
\begin{equation}
\left\vert \phi\right\rangle _{RB}=\sum_{i=0}^{d_{A}-1}\sum_{j=0}^{d_{B}%
-1}\alpha_{ij}|i\rangle_{R}\otimes|j\rangle_{B}.
\end{equation}
Let $V_{A\rightarrow B}$ denote the following linear operator:%
\begin{equation}
V_{A\rightarrow B}\equiv\sum_{i=0}^{d_{A}-1}\sum_{j=0}^{d_{B}-1}\alpha
_{i,j}|j\rangle_{B}\langle i\vert_{A}, \label{eq-qt:V-lin-op}%
\end{equation}
where $\left\{  |i\rangle_{A}\right\}  $ is the orthonormal basis given above.
Then we see that%
\begin{align}
\left(  I_{R}\otimes V_{A\rightarrow B}\right)  \left\vert \Gamma\right\rangle
_{RA}  &  =\sum_{i=0}^{d_{A}-1}\sum_{j=0}^{d_{B}-1}\alpha_{i,j}\left\vert
j\right\rangle _{B}\langle i\vert_{A}\sum_{k=0}^{d_{A}-1}\left\vert
k\right\rangle _{R}\otimes\vert k\rangle_{A}%
\label{eq-nqt:bipartite-vec-trick-1}\\
&  =\sum_{i=0}^{d_{A}-1}\sum_{j=0}^{d_{B}-1}\sum_{k=0}^{d_{A}-1}\alpha
_{i,j}\vert k\rangle_{R}\otimes|j\rangle_{B}\left\langle i|k\right\rangle
_{A}\\
&  =\sum_{i=0}^{d_{A}-1}\sum_{j=0}^{d_{B}-1}\alpha_{ij}\left\vert
i\right\rangle _{R}\otimes|j\rangle_{B}\\
&  =\left\vert \phi\right\rangle _{RB}.
\label{eq-nqt:bipartite-vec-trick-last}%
\end{align}
So this means that for all bipartite vectors $\left\vert \phi\right\rangle
_{RB}$, we can find a linear operator $V_{A\rightarrow B}$ such that $\left(
I_{R}\otimes V_{A\rightarrow B}\right)  \left\vert \Gamma\right\rangle
_{RA}=\left\vert \phi\right\rangle _{RB}$. Consider also that%
\begin{align}
\langle i\vert_{R}\left\vert \phi\right\rangle _{RB}  &  =\langle i\vert
_{R}\left(  I_{R}\otimes V_{A\rightarrow B}\right)  \left\vert \Gamma
\right\rangle _{RA}\\
&  =V_{A\rightarrow B}|i\rangle_{A}.
\end{align}
Applying this to our case of interest, for each $l$, we can write%
\begin{equation}
\left\vert \phi_{l}\right\rangle _{RB}=I_{R}\otimes\left(  V_{l}\right)
_{A\rightarrow B}\left\vert \Gamma\right\rangle _{RA},
\end{equation}
where $\left(  V_{l}\right)  _{A\rightarrow B}$ is some linear operator of the
form in (\ref{eq-qt:V-lin-op}). After making this observation, we realize that
it is possible to write%
\begin{align}
\mathcal{N}_{A\rightarrow B}\left(  |i\rangle\langle j\vert\right)   &
=\left(  \langle i\vert_{R}\otimes I_{B}\right)  \left(  \mathcal{N}%
_{A\rightarrow B}\left(  \left\vert \Gamma\right\rangle \left\langle
\Gamma\right\vert _{RA}\right)  \right)  \left(  |j\rangle_{R}\otimes
I_{B}\right) \\
&  =\left(  \langle i\vert_{R}\otimes I_{B}\right)  \sum_{l=0}^{d-1}\left\vert
\phi_{l}\right\rangle \left\langle \phi_{l}\right\vert _{RB}\left(
|j\rangle_{R}\otimes I_{B}\right) \\
&  =\sum_{l=0}^{d-1}\left[  \left(  \langle i\vert_{R}\otimes I_{B}\right)
\left\vert \phi_{l}\right\rangle _{RB}\right]  \left[  \left\langle \phi
_{l}\right\vert _{RB}\left(  |j\rangle_{R}\otimes I_{B}\right)  \right] \\
&  =\sum_{l=0}^{d-1}V_{l}|i\rangle\langle j\vert_{A}V_{l}^{\dag}.
\end{align}
By linearity of the map $\mathcal{N}_{A\rightarrow B}$, exploiting the above
result, and the development in \eqref{eq-nqt:linear-action-of-channel}, it
follows that the action of $\mathcal{N}_{A\rightarrow B}$ on any input
operator $X_{A}$ can be written as follows:%
\begin{equation}
\mathcal{N}_{A\rightarrow B}(X_{A})=\sum_{l=0}^{d-1}V_{l}X_{A}V_{l}^{\dag}.
\end{equation}


To prove the condition in \eqref{eq-nqt:trace-preserve-kraus}, let us begin by
exploiting the fact that the map $\mathcal{N}_{A\rightarrow B}$ is trace
preserving, so that%
\begin{equation}
\operatorname{Tr}\left\{  \mathcal{N}_{A\rightarrow B}\left(  \vert
i\rangle\langle j\vert_{A}\right)  \right\}  =\operatorname{Tr}\left\{
|i\rangle\langle j\vert_{A}\right\}  =\delta_{ij}.
\label{eq-qt:kraus-consistency}%
\end{equation}
for all operators $\left\{  |i\rangle\langle j\vert_{A}\right\}  _{i,j}$. But
consider also that%
\begin{align}
\operatorname{Tr}\left\{  \mathcal{N}_{A\rightarrow B}\left(  \vert
i\rangle\langle j\vert_{A}\right)  \right\}   &  =\operatorname{Tr}\left\{
\sum_{l}V_{l}\left(  |i\rangle\langle j\vert_{A}\right)  V_{l}^{\dag}\right\}
\\
&  =\operatorname{Tr}\left\{  \sum_{l}V_{l}^{\dag}V_{l}\left(  \vert
i\rangle\langle j\vert_{A}\right)  \right\} \\
&  =\langle j\vert_{A}\sum_{l}V_{l}^{\dag}V_{l}\left\vert i\right\rangle _{A}.
\end{align}
Thus, in order to have consistency with (\ref{eq-qt:kraus-consistency}), we
require that $\langle j\vert_{A}\sum_{l}V_{l}^{\dag}V_{l}\vert i\rangle
_{A}=\delta_{i,j}$, or equivalently, for \eqref{eq-nqt:trace-preserve-kraus}
to hold.
\end{proof}

\begin{remark}
If the decomposition in \eqref{eq:decomposition-choi} is a spectral
decomposition, then it follows that the Kraus operators $\left\{
V_{l}\right\}  $ are orthogonal with respect to the Hilbert--Schmidt inner
product:%
\begin{equation}
\operatorname{Tr}\left\{  V_{l}^{\dag}V_{k}\right\}  =\operatorname{Tr}%
\left\{  V_{l}^{\dag}V_{l}\right\}  \delta_{l,k}.
\end{equation}
This follows from the fact that%
\begin{align}
\delta_{l,k}\left\langle \phi_{l}|\phi_{l}\right\rangle  &  =\left\langle
\phi_{l}|\phi_{k}\right\rangle \\
&  =\left\langle \Gamma\right\vert _{RB}\left[  I_{R}\otimes\left(
V_{l}^{\dag}V_{k}\right)  _{B}\right]  \left\vert \Gamma\right\rangle _{RB}\\
&  =\operatorname{Tr}\left\{  V_{l}^{\dag}V_{k}\right\}  ,
\end{align}
where in the third line we have applied the result of
Exercise~\ref{ex-qt:alt-trace-max-ent}.
\end{remark}

\begin{exercise}
\label{ex-nqt:choi-CP}Prove that a linear map $\mathcal{N}$ is completely
positive if its corresponding Choi operator, as defined in
Definition~\ref{def-nqt:choi-op}, is a positive semi-definite operator. (Hint:
Use the fact that any positive semi-definite operator can be diagonalized, the
fact that $\operatorname{id}_{R}\otimes\mathcal{N}$ is linear, and use
something similar to \eqref{eq-nqt:bipartite-vec-trick-1}--\eqref{eq-nqt:bipartite-vec-trick-last}).
\end{exercise}


\section{Unique Linear Extension of a Quantum Physical Evolution}

\label{app:unique-linear-ext}Recall in
Section~\ref{sec-nqt:axiomatic-approach}\ that we argued on physical grounds
how any quantum physical evolution $\mathcal{N}$\ should be convex linear when
acting on the space $\mathcal{D}(\mathcal{H}_{A})$\ of density operators:%
\begin{equation}
\mathcal{N}(\lambda\rho_{A}+(1-\lambda)\sigma_{A})=\lambda\mathcal{N}(\rho
_{A})+(1-\lambda)\mathcal{N}(\sigma_{A}),
\end{equation}
where $\rho_{A},\sigma_{A}\in\mathcal{D}(\mathcal{H}_{A})$ and $\lambda
\in\left[  0,1\right]  $. Here we show how to construct a unique linear
extension $\widetilde{\mathcal{N}}$\ of $\mathcal{N}$, whose action is well
defined on the space of all operators $X_{A}\in\mathcal{B}(\mathcal{H}_{A})$.
The development follows the approach given in Proposition~2.30 of ``The Mathematical Language of Quantum Theory:
From Uncertainty to Entanglement,'' by Teiko Heinosaari and Mario Ziman.

We first define $\widetilde{\mathcal{N}}(0)\equiv0$, where the inputs and
outputs are understood to be the zero operator. We next extend the action of
$\mathcal{N}$ to all positive semi-definite operators $P_{A}\neq0$ as follows:%
\begin{equation}
\widetilde{\mathcal{N}}(P_{A})\equiv\operatorname{Tr}\{P_{A}\}\mathcal{N}%
(\left[  \operatorname{Tr}\{P_{A}\}\right]  ^{-1}P_{A}),
\end{equation}
where it is clear that this is well defined from $\mathcal{N}$ because
$\left[  \operatorname{Tr}\{P_{A}\}\right]  ^{-1}P_{A}$ is a density operator.
Now consider for a constant $s>0$ that we have scale invariance:%
\begin{align}
\widetilde{\mathcal{N}}(sP_{A})  &  =\operatorname{Tr}\{sP_{A}\}\mathcal{N}%
(\left[  \operatorname{Tr}\{sP_{A}\}\right]  ^{-1}sP_{A})\\
&  =s\operatorname{Tr}\{P_{A}\}\mathcal{N}(\left[  \operatorname{Tr}%
\{P_{A}\}\right]  ^{-1}P_{A})\\
&  =s\widetilde{\mathcal{N}}(P_{A}).
\end{align}
Furthermore, for two non-zero positive semi-definite operators $P_{A}$ and
$Q_{A}$, we have the following additivity relation:%
\begin{equation}
\widetilde{\mathcal{N}}(P_{A}+Q_{A})=\widetilde{\mathcal{N}}(P_{A}%
)+\widetilde{\mathcal{N}}(Q_{A}), \label{eq-app-linear:additive-PSD}%
\end{equation}
which follows because%
\begin{align}
&  \widetilde{\mathcal{N}}(P_{A}+Q_{A})\nonumber\\
&  =\operatorname{Tr}\{P_{A}+Q_{A}\}\mathcal{N}(\left[  \operatorname{Tr}%
\{P_{A}+Q_{A}\}\right]  ^{-1}(P_{A}+Q_{A}))\\
&  =\operatorname{Tr}\{P_{A}+Q_{A}\}\mathcal{N}\left(  \frac{1}%
{\operatorname{Tr}\{P_{A}+Q_{A}\}}P_{A}+\frac{1}{\operatorname{Tr}%
\{P_{A}+Q_{A}\}}Q_{A}\right) \\
&  =\operatorname{Tr}\{P_{A}+Q_{A}\}\mathcal{N}\left(  \frac{\operatorname{Tr}%
\{P_{A}\}}{\operatorname{Tr}\{P_{A}+Q_{A}\}}\frac{P_{A}}{\operatorname{Tr}%
\{P_{A}\}}+\frac{\operatorname{Tr}\{Q_{A}\}}{\operatorname{Tr}\{P_{A}+Q_{A}%
\}}\frac{Q_{A}}{\operatorname{Tr}\{Q_{A}\}}\right) \\
&  =\operatorname{Tr}\{P_{A}\}\mathcal{N}\left(  \frac{P_{A}}%
{\operatorname{Tr}\{P_{A}\}}\right)  +\operatorname{Tr}\{Q_{A}\}\mathcal{N}%
\left(  \frac{Q_{A}}{\operatorname{Tr}\{Q_{A}\}}\right) \\
&  =\widetilde{\mathcal{N}}(P_{A})+\widetilde{\mathcal{N}}(Q_{A}),
\end{align}
where in the fourth equality, we exploited convex linearity of the quantum
physical evolution $\mathcal{N}$.

For the next step, recall that any Hermitian operator $T_{A}$\ can be written
as a linear combination of a positive part and a negative part:\ $T_{A}%
=T_{A}^{+}-T_{A}^{-}$, where both $T_{A}^{+}$ and $T_{A}^{-}$ are positive
semi-definite operators. So we define the action of $\widetilde{\mathcal{N}}$
on any Hermitian operator $T_{A}$ as follows:%
\begin{equation}
\widetilde{\mathcal{N}}(T_{A})\equiv\widetilde{\mathcal{N}}(T_{A}%
^{+})-\widetilde{\mathcal{N}}(T_{A}^{-}).
\label{eq-app-linear:define-hermitian}%
\end{equation}
To see that the following additivity relation holds for all Hermitian $S_{A}$
and $T_{A}$%
\begin{equation}
\widetilde{\mathcal{N}}(S_{A}+T_{A})=\widetilde{\mathcal{N}}(S_{A}%
)+\widetilde{\mathcal{N}}(T_{A}), \label{eq-app-linear:additive-hermitian}%
\end{equation}
consider that%
\begin{equation}
S_{A}+T_{A}=(S_{A}+T_{A})^{+}-(S_{A}+T_{A})^{-},
\end{equation}
while also%
\begin{equation}
S_{A}+T_{A}=S_{A}^{+}+T_{A}^{+}-S_{A}^{-}-T_{A}^{-}.
\end{equation}
Equating both sides, we find that%
\begin{equation}
(S_{A}+T_{A})^{+}+S_{A}^{-}+T_{A}^{-}=(S_{A}+T_{A})^{-}+S_{A}^{+}+T_{A}^{+}.
\end{equation}
Now we exploit this equality, \eqref{eq-app-linear:additive-PSD}, and the
definition in \eqref{eq-app-linear:define-hermitian} to establish \eqref{eq-app-linear:additive-hermitian}.

The final step is to extend the action of $\widetilde{\mathcal{N}}$ to all
operators $X_{A}\in\mathcal{B}(\mathcal{H}_{A})$. Here, we recall that any
linear operator can be written in terms of a real and imaginary part as
follows:%
\begin{equation}
X_{A}^{R}\equiv\frac{1}{2}\left(  X_{A}+X_{A}^{\dag}\right)
,\ \ \ \ \ \ \ \ X_{A}^{I}\equiv\frac{1}{2i}\left(  X_{A}-X_{A}^{\dag}\right)
,
\end{equation}
where by inspection, $X_{A}^{R}$ and $X_{A}^{I}$ are Hermitian operators. So
we define%
\begin{equation}
\widetilde{\mathcal{N}}(X_{A})\equiv\widetilde{\mathcal{N}}(X_{A}%
^{R})+i\widetilde{\mathcal{N}}(X_{A}^{I}).
\end{equation}
This completes the development of a well defined linear extension
$\widetilde{\mathcal{N}}$\ of the quantum physical evolution $\mathcal{N}$.

To show that it is unique, recall that any operator $X_{A}$ can be expanded as
a linear combination of density operators from the basis $\{\rho_{A}^{x,y}\}$,
defined in \eqref{eq-nqt:density-op-basis}, as follows:%
\begin{equation}
X_{A}=\sum_{x,y}\alpha_{x,y}\rho_{A}^{x,y},
\end{equation}
where $\alpha_{x,y}\in\mathbb{C}$ for all $x$ and $y$. It is straightforward
to show from the above development that%
\begin{equation}
\widetilde{\mathcal{N}}(X_{A})=\sum_{x,y}\alpha_{x,y}\mathcal{N}(\rho
_{A}^{x,y}).
\end{equation}
Now suppose that $\mathcal{N}^{\prime}$ is some other linear map for which
$\mathcal{N}^{\prime}(\rho_{A})=\mathcal{N}(\rho_{A})$ for all $\rho_{A}%
\in\mathcal{D}(\mathcal{H}_{A})$. Then the following equality holds for all
$X_{A}\in\mathcal{B}(\mathcal{H}_{A})$:%
\begin{equation}
\mathcal{N}^{\prime}(X_{A})=\sum_{x,y}\alpha_{x,y}\mathcal{N}^{\prime}%
(\rho_{A}^{x,y})=\sum_{x,y}\alpha_{x,y}\mathcal{N}(\rho_{A}^{x,y}%
)=\widetilde{\mathcal{N}}(X_{A}).
\end{equation}
As a result, $\mathcal{N}^{\prime}=\widetilde{\mathcal{N}}$, given that they
have the same action on every operator $X_{A}\in\mathcal{B}(\mathcal{H}_{A})$.

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\end{document}
