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\begin{document}

\lecture{8 --- September 21, 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 presented the notions of an ensemble of quantum states and a density operator. We also discussed unitary evolution of the density operator and measurement of a density operator.

In this lecture we continue our development of the noisy quantum theory, discussing a more general model of measurements, the POVM formalism, product states, separable states, and the partial trace operation.

The material is coming from Chapter~4 of
\verb+http://markwilde.com/qit-notes.pdf+ .

\section{Measurement in the Noisy Quantum Theory}

We have described measurement in the quantum theory using a set of projectors
that form a resolution of the identity. For example, the set $\left\{  \Pi
_{j}\right\}  _{j}$ of projectors that satisfy the condition $\sum_{j}\Pi
_{j}=I$ form a valid von Neumann quantum measurement. A projective measurement
is not the most general measurement that we can perform on a quantum
system\ (though it is certainly one valid type of quantum measurement).

There is a more general description of quantum measurements that follows from
allowing the system of interest to interact unitarily with a probe system that
we measure after the interaction occurs. So suppose that the system of
interest is in a state $|\psi\rangle_{S}$ and that the probe is in a state
$|0\rangle_{P}$, so that the overall state before anything happens is as
follows:%
\begin{equation}
|\psi\rangle_{S}\otimes|0\rangle_{P}.
\end{equation}
Let $\{|0\rangle_{P},|1\rangle_{P},\ldots,|d-1\rangle_{P}\}$ be an orthonormal
basis for the probe system (assuming that it has dimension $d$). Now suppose
that the system and the probe interact according to a unitary $U_{SP}$, and
then we perform a measurement of the probe system, described by measurement
operators $\{|j\rangle\langle j|_{P}\}$. The probability to obtain outcome $j$
is%
\begin{equation}
p_{J}(j)=\left(  \langle\psi|_{S}\otimes\langle0|_{P}U_{SP}^{\dag}\right)
\left(  I_{S}\otimes|j\rangle\langle j|_{P}\right)  \left(  U_{SP}|\psi
\rangle_{S}\otimes|0\rangle_{P}\right)  , \label{eq-nqt:more-gen-meas-prob}%
\end{equation}
and the post-measurement state upon obtaining outcome $j$ is%
\begin{equation}
\frac{1}{\sqrt{p_{J}(j)}}\left(  I_{S}\otimes|j\rangle\langle j|_{P}\right)
\left(  U_{SP}|\psi\rangle_{S}\otimes|0\rangle_{P}\right)  .
\label{eq-nqt:more-gen-meas-state}%
\end{equation}


We can rewrite the expressions above in a simpler way. Let us expand the
unitary operator $U_{SP}$ in the orthonormal basis of the probe system $P$ as
follows:%
\begin{equation}
U_{SP}=\sum_{j,k}M_{S}^{j,k}\otimes|j\rangle\langle k|_{P},
\label{eq-nqt:U_SP-rep}%
\end{equation}
where $\{M_{S}^{j,k}\}$ is a set of operators. Up to a permutation of the $S$
and $P$ systems and using the mathematics of the tensor product, this is the same as writing the
unitary $U_{SP}$\ as follows:%
\begin{equation}%
\begin{bmatrix}
M_{S}^{0,0} & M_{S}^{0,1} & \cdots & M_{S}^{0,d-1}\\
M_{S}^{1,0} & M_{S}^{1,1} & \cdots & M_{S}^{1,d-1}\\
\vdots & \vdots & \ddots & \vdots\\
M_{S}^{d-1,0} & M_{S}^{d-1,1} & \cdots & M_{S}^{d-1,d-1}%
\end{bmatrix}
. \label{eq-nqt:matrix-rep-U_SP}%
\end{equation}
This set $\{M_{S}^{j,k}\}$ needs to satisfy some constraints corresponding to
the unitarity of $U_{SP}$. In particular, consider the following operator:%
\begin{equation}
\sum_{j}M_{S}^{j,0}\otimes|j\rangle\langle0|_{P}, \label{eq-nqt:U_SP-part}%
\end{equation}
which corresponds to the first column of operator-valued entries in $U_{SP}$,
as illustrated in \eqref{eq-nqt:matrix-rep-U_SP}. In what follows, we employ
the shorthand $M_{S}^{j}\equiv M_{S}^{j,0}$. From the fact that $U_{SP}^{\dag
}U_{SP}=I_{SP}=I_{S}\otimes I_{P}$, we deduce that the following equality must
hold%
\begin{align}
I_{S}\otimes|0\rangle\langle0|_{P}  &  =\left(  \sum_{j^{\prime}}%
M_{S}^{j^{\prime}\dag}\otimes|0\rangle\langle j^{\prime}|_{P}\right)  \left(
\sum_{j}M_{S}^{j}\otimes|j\rangle\langle0|_{P}\right) \\
&  =\sum_{j^{\prime},j}M_{S}^{j^{\prime}\dag}M_{S}^{j}\otimes|0\rangle
\left\langle j^{\prime}|j\right\rangle \langle0|_{P}\\
&  =\sum_{j}M_{S}^{j\dag}M_{S}^{j}\otimes|0\rangle\langle0|_{P},
\end{align}
where the last line follows from the fact that we chose an orthonormal basis
in the representation of $U_{SP}$ in \eqref{eq-nqt:U_SP-rep}. So this implies
that the following condition holds%
\begin{equation}
\sum_{j}M_{S}^{j\dag}M_{S}^{j}=I_{S}.
\end{equation}


Plugging \eqref{eq-nqt:U_SP-rep} into \eqref{eq-nqt:more-gen-meas-prob} and
\eqref{eq-nqt:more-gen-meas-state}, a short calculation (similar to the above
one) reveals that they simplify as follows:%
\begin{align}
p_{J}(j)  &  =\left\langle \psi\right\vert M_{j}^{\dag}M_{j}\left\vert
\psi\right\rangle ,\\
\frac{1}{\sqrt{p_{J}(j)}}\left(  I_{S}\otimes|j\rangle\langle j|_{P}\right)
\left(  U_{SP}|\psi\rangle_{S}\otimes|0\rangle_{P}\right)   &  =\frac
{M_{j}\left\vert \psi\right\rangle _{S}\otimes|j\rangle_{P}}{\sqrt{p_{J}(j)}}.
\end{align}
Since the system and the probe are in a pure product state (and thus
independent of each other) after the measurement occurs, we can discard the
probe system and deduce that the post-measurement state is simply
$M_{j}\left\vert \psi\right\rangle _{S}/\sqrt{p_{J}(j)}$.

Motivated by the above development, we allow for a more general notion of
quantum measurement, saying that it consists of a set of measurement operators
$\left\{  M_{j}\right\}  _{j}$ that satisfy the following completeness
condition:%
\begin{equation}
\sum_{j}M_{j}^{\dag}M_{j}=I.
\end{equation}
Observe from the above development that this is the only constraint that the
operators $\{M_{j}\}$ need to satisfy. This constraint is a consequence of
unitarity, but can be viewed as a generalization of the completeness relation
for a set of projectors that constitute a projective quantum measurement.
Given a set of measurement operators of the above form, the probability for
obtaining outcome $j$ when measuring a state $\left\vert \psi\right\rangle
$\ is%
\begin{equation}
p_{J}(j)\equiv\left\langle \psi\right\vert M_{j}^{\dag}M_{j}\left\vert
\psi\right\rangle ,
\end{equation}
and the post-measurement state when we receive outcome $j$ is%
\begin{equation}
\frac{M_{j}\left\vert \psi\right\rangle }{\sqrt{p_{J}(j)}}.
\end{equation}


Suppose that we instead have an ensemble $\left\{  p_{X}(x),\left\vert
\psi_{x}\right\rangle \right\}  $ with density operator $\rho$. We can carry
out an analysis similar to that in the last lecture to
 conclude that the probability $p_{J}(j)$ for obtaining outcome $j$ is%
\begin{equation}
p_{J}(j)\equiv\operatorname{Tr}\{M_{j}^{\dag}M_{j}\rho\},
\end{equation}
and the post-measurement state when we measure result $j$ is%
\begin{equation}
\frac{M_{j}\rho M_{j}^{\dag}}{p_{J}(j)}.
\end{equation}
The expression $p_{J}(j)=\operatorname{Tr}\{M_{j}^{\dag}M_{j}\rho\}$ is a
restatement of the Born rule.


\subsection{POVM\ Formalism}

\label{sec-nqt:POVM}Sometimes, we simply may not care about the
post-measurement state of a quantum measurement, but instead we only care
about the probability for obtaining a particular outcome. This situation
arises in the transmission of classical data over a quantum channel. In this
situation, we are merely concerned with minimizing the error probabilities of
the classical transmission. The receiver does not care about the
post-measurement state because he no longer needs it in the quantum
information-processing protocol.

We can specify a measurement of this sort by some set $\left\{  \Lambda
_{j}\right\}  _{j}$ of operators that satisfy non-negativity and completeness:%
\begin{align}
\Lambda_{j}  &  \geq0,\\
\sum_{j}\Lambda_{j}  &  =I.
\end{align}
The set $\left\{  \Lambda_{j}\right\}  _{j}$ of operators is a positive
operator-valued measure%
\index{positive operator-valued measure}
(POVM). The probability for obtaining outcome $j$ is%
\begin{equation}
\left\langle \psi\right\vert \Lambda_{j}\left\vert \psi\right\rangle ,
\end{equation}
if the state is some pure state $\left\vert \psi\right\rangle $. The
probability for obtaining outcome $j$ is%
\begin{equation}
\operatorname{Tr}\left\{  \Lambda_{j}\rho\right\}  ,
\end{equation}
if the state is in a mixed state described by some density operator $\rho$.
This is another restatement of the Born rule.

\section{Composite Noisy Quantum Systems}

We are again interested in the behavior of two or more quantum systems when we
join them together. Some of the most exotic, truly \textquotedblleft
quantum\textquotedblright\ behavior occurs in joint quantum systems, and we
observe a marked departure from the classical world.

\subsection{Independent Ensembles}

Let us first suppose that we have two independent ensembles for quantum
systems $A$ and $B$. The first quantum system belongs to Alice and the second
quantum system belongs to Bob, and they may or may not be spatially separated.
Let $\{p_{X}( x) ,\vert\psi_{x}\rangle\}$ be the ensemble for the system $A$
and let $\{p_{Y}( y) ,|\phi_{y}\rangle\}$ be the ensemble for the system $B$.
Suppose for now that the state on system $A$ is $\left\vert \psi
_{x}\right\rangle $ for some $x$ and the state on system $B$ is $|\phi
_{y}\rangle$ for some $y$. Then, using the composite system postulate of the
noiseless quantum theory, the joint state for a given $x$ and $y$ is
$\vert\psi_{x}\rangle\otimes|\phi_{y}\rangle$. The density operator for the
joint quantum system is the expectation of the states $\vert\psi_{x}%
\rangle\otimes|\phi_{y}\rangle$ with respect to the random variables $X$ and
$Y$ that describe the individual ensembles:%
\begin{equation}
\mathbb{E}_{X,Y}\left\{  \left(  \left\vert \psi_{X}\right\rangle \otimes
|\phi_{Y}\rangle\right)  \left(  \left\langle \psi_{X}\right\vert
\otimes\langle\phi_{Y}|\right)  \right\}  .
\end{equation}
The above expression is equal to the following one:%
\begin{equation}
\mathbb{E}_{X,Y}\left\{  \left\vert \psi_{X}\right\rangle \left\langle
\psi_{X}\right\vert \otimes|\phi_{Y}\rangle\langle\phi_{Y}|\right\}  ,
\end{equation}
because $\left(  \vert\psi_{x}\rangle\otimes|\phi_{y}\rangle\right)  \left(
\langle\psi_{x}\vert\otimes\langle\phi_{y}|\right)  =\vert\psi_{x}%
\rangle\langle \psi_{x}\vert \otimes|\phi_{y}\rangle\langle\phi
_{y}|$. We then explicitly write out the expectation as a sum over
probabilities:%
\begin{equation}
\sum_{x,y}p_{X}( x) p_{Y}( y) \vert\psi_{x}\rangle\left\langle \psi
_{x}\right\vert \otimes|\phi_{y}\rangle\langle\phi_{y}|.
\end{equation}
We can distribute the probabilities and the sum because the tensor product
obeys a distributive property:%
\begin{equation}
\sum_{x}p_{X}( x) \vert\psi_{x}\rangle\langle \psi_{x}\vert
\otimes\sum_{y}p_{Y}( y) |\phi_{y}\rangle\langle\phi_{y}|.
\end{equation}
The density operator for this ensemble admits the following simple form:%
\begin{equation}
\rho\otimes\sigma, \label{eq-qt:product-state}%
\end{equation}
where $\rho = \sum_x p_X(x) \vert\psi_{x}\rangle\langle \psi
_{x}\vert$ is the density operator of the $X$ ensemble and $\sigma = \sum_y p_Y(y) |\phi_{y}\rangle\langle\phi_{y}| $ is the
density operator of the $Y$ ensemble. We can say that Alice's local density
operator is $\rho$ and Bob's local density operator is $\sigma$. The overall state is a tensor product of these two density operators.
\begin{definition}[Product State] \index{product state} A 
density operator which is equal to a tensor product of two or more density operators is called a \textit{product state.}
\end{definition}
We should expect
the density operator to factor as it does above because we assumed that the
ensembles are independent. There is nothing much that distinguishes this
situation from the classical world, except for the fact that the states in
each respective ensemble may be non-orthogonal to other states in the same
ensemble. But even here, there is some equivalent description of each ensemble
in terms of an orthonormal basis so that there is really not too much difference between
this description and a joint probability distribution that factors as two
independent distributions.

%\begin{exercise}
%Show that the purity $P(  \rho_{A})  $ is equal to the following
%expression:%
%\begin{equation}
%P(  \rho_{A})  =\operatorname{Tr}\left\{  \left(  \rho_{A}%
%\otimes\rho_{A^{\prime}}\right)  F_{AA^{\prime}}\right\}  .
%\end{equation}
%where system $A^{\prime}$ has a Hilbert space structure isomorphic to that of
%system $A$ and $F_{AA^{\prime}}$ is the swap operator that has the following
%action on kets in $A$ and $A^{\prime}$:%
%\begin{equation}
%\forall x,y\ \ \ \ \ F_{AA^{\prime}}\vert x\rangle_{A}\left\vert
%y\right\rangle _{A^{\prime}}=\vert y\rangle_{A}\left\vert x\right\rangle
%_{A^{\prime}}.
%\end{equation}
%(Hint:\ First show that $\operatorname{Tr}\left\{  f(  \rho_{A})
%\right\}  =\operatorname{Tr}\left\{  \left(  f(  \rho_{A})  \otimes
%I_{A^{\prime}}\right)  F_{AA^{\prime}}\right\}  $ for any function $f$ on the
%operators in system $A$.)
%\end{exercise}

\subsection{Separable States}

Let us now consider two systems $A$ and $B$ whose corresponding ensembles are
correlated in a classical way. We describe this correlated ensemble as the joint ensemble%
\begin{equation}
\left\{  p_{X}( x) ,\vert\psi_{x}\rangle\otimes\left\vert \phi_{x}%
\right\rangle \right\}  .
\end{equation}
It is straightforward to verify that the density operator of this correlated
ensemble has the following form:%
\begin{equation}
\mathbb{E}_{X}\left\{  \left(  \left\vert \psi_{X}\right\rangle \otimes
\left\vert \phi_{X}\right\rangle \right)  \left(  \left\langle \psi
_{X}\right\vert \otimes\left\langle \phi_{X}\right\vert \right)  \right\}
=\sum_{x}p_{X}( x) \vert\psi_{x}\rangle\langle \psi_{x}\vert
\otimes\vert \phi_{x}\rangle \langle \phi_{x}\vert .
\label{eq-nqt:sep-state-motiv-1}
\end{equation}
By ignoring Bob's system,
Alice's local density operator is of the form%
\begin{equation}
\mathbb{E}_{X}\left\{  \left\vert \psi_{X}\right\rangle \left\langle \psi
_{X}\right\vert \right\}  =\sum_{x}p_{X}( x) \vert\psi_{x}\rangle\langle
\psi_{x}\vert,
\end{equation}
and similarly, Bob's local density operator is%
\begin{equation}
\mathbb{E}_{X}\left\{  \left\vert \phi_{X}\right\rangle \left\langle \phi
_{X}\right\vert \right\}  =\sum_{x}p_{X}( x) \vert \phi_{x}\rangle
\langle \phi_{x}\vert .
\end{equation}

States of the form in \eqref{eq-nqt:sep-state-motiv-1} can be generated by a classical procedure. A third party generates a symbol $x$ according to the probability distribution $p_X(x)$ and sends the symbol $x$ to both Alice and Bob. Alice prepares the state $\vert\psi_{x}\rangle$ and Bob prepares the state $\vert\phi_{x}\rangle$. If they then discard the symbol $x$, the state of their systems is given by \eqref{eq-nqt:sep-state-motiv-1}.


We can generalize this classical preparation procedure one step further. Let
us suppose that we first generate a random variable $Z$ according to some
distribution $p_{Z}( z) $. We then generate two other ensembles, conditional
on the value of the random variable $Z$. Let $\{p_{X|Z}( x|z) ,\left\vert
\psi_{x,z}\right\rangle \}$ be the first ensemble and let $\{p_{Y|Z}( y|z)
,|\phi_{y,z}\rangle\}$ be the second ensemble, where the random variables $X$
and $Y$ are independent when conditioned on $Z$. Let us label the density
operators of the first and second ensembles when conditioned on a particular
realization $z$ by $\rho_{z}$ and$~\sigma_{z}$, respectively. It is then
straightforward to verify that the density operator of an ensemble created
from this classical preparation procedure has the following form:%
\begin{equation}
\mathbb{E}_{X,Y,Z}\left\{  \left(  \left\vert \psi_{X,Z}\right\rangle
\otimes|\phi_{Y,Z}\rangle\right)  \left(  \left\langle \psi_{X,Z}\right\vert
\otimes\langle\phi_{Y,Z}|\right)  \right\}  =\sum_{z}p_{Z}( z) \rho_{z}%
\otimes\sigma_{z}. \label{eq-qt:separable-state}%
\end{equation}


\begin{exercise}
By ignoring Bob's system, we can determine Alice's local density operator.
Show that%
\begin{equation}
\mathbb{E}_{X,Y,Z}\left\{  \left\vert \psi_{X,Z}\right\rangle \left\langle
\psi_{X,Z}\right\vert \right\}  =\sum_{z}p_{Z}( z) \rho_{z},
\label{eq-qt:local-state-separable}%
\end{equation}
so that the above expression is the density operator for Alice. It similarly
follows that the local density operator for Bob is%
\begin{equation}
\mathbb{E}_{X,Y,Z}\left\{  |\phi_{Y,Z}\rangle\langle\phi_{Y,Z}|\right\}
=\sum_{z}p_{Z}( z) \sigma_{z}.
\end{equation}

\end{exercise}

\begin{exercise}
\label{ex-nqt:separable-as-pure}Show that we can always write a
state of the form in \eqref{eq-qt:separable-state} as a convex combination of pure product states:%
\begin{equation}
\sum_{z}p_{Z}( z) \left\vert \phi_{z}\right\rangle \left\langle \phi
_{z}\right\vert \otimes\left\vert \psi_{z}\right\rangle \left\langle \psi
_{z}\right\vert ,
\end{equation}
by manipulating the general form in \eqref{eq-qt:separable-state}.
\end{exercise}

As a consequence of Exercise~\ref{ex-nqt:separable-as-pure}, we see that any state of the form in \eqref{eq-qt:separable-state} can be written as a convex
combination of pure product states. Such states are called  \textit{separable} states%
\index{separable states}, defined formally as follows:
\begin{definition}[Separable State]
\label{def-nqt:separable-state}
\index{separable states}
A bipartite density operator $\sigma_{AB}$ is a separable state if it can be written in the following form:
\begin{equation}
\sigma_{AB} = \sum_{x}p_{X}( x) \vert\psi_{x}\rangle\langle \psi_{x}\vert_A
\otimes\vert \phi_{x}\rangle \langle \phi_{x}\vert_B
\end{equation}
for some probability distribution $p_X(x)$ and sets
$\{ \vert\psi_{x}\rangle_A\}$ and $\{ \vert \phi_{x}\rangle_B\}$
of pure states.
\end{definition}
The term \textquotedblleft separable\textquotedblright\ implies that there
is no quantum entanglement in the above state, i.e., there is a completely
classical procedure that prepares the above state. In fact, this is the definition of entanglement for a general bipartite density operator:
\begin{definition}[Entangled State]
\label{def-nqt:entangled-state}
\index{entangled state}
A bipartite density operator $\rho_{AB}$ is entangled if it is not separable.
\end{definition}

\subsubsection{Separable States and the CHSH Game}

\index{CHSH game}

One motivation for Definitions~\ref{def-nqt:separable-state} and \ref{def-nqt:entangled-state} was already given above: for a separable state, there is a classical procedure that can be used to prepare it. Thus, for an entangled state, there is no such procedure. That is, a non-classical (quantum) interaction between the systems is necessary to prepare an entangled state.

Another related motivation is that separable states admit an explanation in terms of a classical strategy for the CHSH game. Recall from before that classical strategies $p_{AB|XY}(a,b|x,y)$ are of the following form:
\begin{equation}
p_{AB|XY}(a,b|x,y) = \int d\lambda \ p_{\Lambda}(\lambda)
\ p_{A|\Lambda X}(a|\lambda,x) \ p_{B|\Lambda Y}(b|\lambda,y) . \label{eq-nqt:classical-strat-CHSH-1}
\end{equation}
If we allow for a continuous index $\lambda$ for a separable state, then we can write such a state as follows:
\begin{equation}
\sigma_{AB} = \int d\lambda \ p_{\Lambda}( \lambda)\  \vert\psi_{\lambda}\rangle\langle \psi_{\lambda}\vert_A
\otimes\vert \phi_{\lambda}\rangle \langle \phi_{\lambda}\vert_B .
\end{equation}
Recall that in a general quantum strategy, there are measurements $\{\Pi^{(x)}_a \}$ and 
$\{\Pi^{(y)}_b \}$, giving output bits $a$ and $b$ based on the input bits $x$ and $y$ and leading to the following strategy:
\begin{align}
p_{AB|XY}(a,b|x,y) & = \operatorname{Tr} \{ 
(\Pi^{(x)}_a  \otimes \Pi^{(y)}_b) \sigma_{AB} \} \\
& = \operatorname{Tr} \left\{ 
(\Pi^{(x)}_a  \otimes \Pi^{(y)}_b) \left(
\int d\lambda \ p_{\Lambda}( \lambda)\  \vert\psi_{\lambda}\rangle\langle \psi_{\lambda}\vert_A
\otimes\vert \phi_{\lambda}\rangle \langle \phi_{\lambda}\vert_B \right) \right\} \\
& = \int d\lambda \ p_{\Lambda}( \lambda)\
\operatorname{Tr} \left\{ 
\Pi^{(x)}_a \vert\psi_{\lambda}\rangle\langle \psi_{\lambda}\vert_A \otimes \Pi^{(y)}_b 
\vert \phi_{\lambda}\rangle \langle \phi_{\lambda}\vert_B  \right\} \\
& = \int d\lambda \ p_{\Lambda}( \lambda)\
\langle \psi_{\lambda}\vert_A \Pi^{(x)}_a \vert\psi_{\lambda}\rangle_A  \ \langle \phi_{\lambda}\vert_B \Pi^{(y)}_b 
\vert \phi_{\lambda}\rangle_B .
\end{align}
By picking the probability distributions 
$p_{A|\Lambda X}(a|\lambda,x)$ and $ p_{B|\Lambda Y}(b|\lambda,y)$ in \eqref{eq-nqt:classical-strat-CHSH-1} as follows:
\begin{align}
p_{A|\Lambda X}(a|\lambda,x)
& = \langle \psi_{\lambda}\vert_A \Pi^{(x)}_a \vert\psi_{\lambda}\rangle_A, \\
p_{B|\Lambda Y}(b|\lambda,y) & =
\langle \phi_{\lambda}\vert_B \Pi^{(y)}_b 
\vert \phi_{\lambda}\rangle_B ,
\end{align}
we see that there is a classical strategy that can simulate
any quantum strategy which uses separable states in the CHSH game. Thus, the winning probability of quantum strategies involving separable states are subject 
to the classical bound of $3/4$ derived before. In this sense, such strategies are effectively classical.

\section{Local Density Operators and Partial Trace}

\subsection{A First Example}

\label{sec-nqt:example-partial-trace}

Consider the entangled Bell state $\left\vert \Phi^{+}\right\rangle _{AB}$
shared on systems $A$ and $B$. In the above analyses, we 
determined a local density operator description for both Alice and Bob. Now,
we are curious if it is possible to determine such a local density operator
description for Alice and Bob with respect to the state $\left\vert \Phi
^{+}\right\rangle _{AB}$ or more general ones.

As a first approach to this issue, recall that the density operator
description arises from its usefulness in determining the probabilities of the
outcomes of a particular measurement. We say that the density operator is
\textquotedblleft the state\textquotedblright\ of the system merely because it
is a mathematical representation that allows us to compute the probabilities
resulting from a physical measurement. So, if we would like to determine a
\textquotedblleft local density operator,\textquotedblright\ such a local
density operator should predict the result of a local measurement.

Let us consider a local POVM $\left\{  \Lambda^{j}\right\}  _{j}$ that Alice
can perform on her system. The global measurement operators for this local
measurement are $\{\Lambda_{A}^{j}\otimes I_{B}\}_{j}$ because nothing (the
identity) happens to Bob's system. The probability of obtaining outcome $j$ when performing this measurement on the state $\left\vert \Phi
^{+}\right\rangle _{AB}$ is%
\begin{align}
\left\langle \Phi^{+}\right\vert _{AB}\Lambda_{A}^{j}\otimes I_{B}\left\vert
\Phi^{+}\right\rangle _{AB} &  =\frac{1}{2}\sum_{k,l=0}^{1}\left\langle
kk\right\vert _{AB}\Lambda_{A}^{j}\otimes I_{B}\left\vert ll\right\rangle
_{AB}\label{eq-qt:partial-trace-bell-1}\\
&  =\frac{1}{2}\sum_{k,l=0}^{1}\left\langle k\right\vert _{A}\Lambda_{A}%
^{j}|l\rangle_{A}\left\langle k|l\right\rangle _{B}\\
&  =\frac{1}{2}\left(  \langle0|_{A}\Lambda_{A}^{j}|0\rangle_{A}+\langle
1|_{A}\Lambda_{A}^{j}\left\vert 1\right\rangle _{A}\right)  \\
&  =\frac{1}{2}\left(  \operatorname{Tr}\left\{  \Lambda_{A}^{j}%
|0\rangle\langle0|_{A}\right\}  +\operatorname{Tr}\left\{  \Lambda_{A}%
^{j}|1\rangle\langle1|_{A}\right\}  \right)  \\
&  =\operatorname{Tr}\left\{  \Lambda_{A}^{j}\frac{1}{2}\left(  |0\rangle
\langle0|_{A}+|1\rangle\langle1|_{A}\right)  \right\}  \\
&  =\operatorname{Tr}\left\{  \Lambda_{A}^{j}\pi_{A}\right\}
.\label{eq-qt:partial-trace-bell}%
\end{align}
The above steps follow by applying the rules of taking the inner product with
respect to tensor product operators. The last line follows by recalling the
definition of the maximally mixed state $\pi$, where $\pi$ here is a qubit maximally
mixed state.

The above calculation demonstrates that we can predict the result of any local
\textquotedblleft Alice\textquotedblright\ measurement using the density
operator $\pi$. Therefore, it is reasonable to say that Alice's local density
operator is $\pi$, and we even go as far to say that her \textit{local state}
is $\pi$. A symmetric calculation shows that Bob's local state is also $\pi$.


\subsection{Partial Trace}

In general, we would like to determine a local density operator that predicts
the outcomes of all local measurements. The general method for determining a
local density operator is to employ%
\index{partial trace}
the \textit{partial trace operation}, which we motivate and define here, as a
generalization of the example discussed at the beginning of
Section~\ref{sec-nqt:example-partial-trace}.

Suppose that Alice and Bob share a bipartite state $\rho_{AB}$ and that Alice
performs a local measurement on her system, described by a POVM\ $\{\Lambda
_{A}^{j}\}$. Then the overall POVM\ on the joint system is $\{\Lambda_{A}%
^{j}\otimes I_{B}\}$ because we are assuming that Bob is not doing anything to
his system. According to the Born rule, the probability for Alice to receive
outcome $j$ after performing the measurement is given by the following
expression:%
\begin{equation}
p_{J}(j)=\operatorname{Tr}\{(\Lambda_{A}^{j}\otimes I_{B})\rho_{AB}\}.
\label{eq-nqt:local-global-partial-trace}%
\end{equation}
In order to evaluate this trace, we can choose any orthonormal basis that we
wish (recall the definition of trace and subsequent statements). Taking
$\{|k\rangle_{A}\}$ as an orthonormal basis for Alice's Hilbert space and
$\{|l\rangle_{B}\}$ as an orthonormal basis for Bob's Hilbert space, the set
$\{|k\rangle_{A}\otimes|l\rangle_{B}\}$ constitutes an orthonormal basis for
the tensor product of their Hilbert spaces. So we can evaluate
\eqref{eq-nqt:local-global-partial-trace} as follows:%
\begin{align}
\operatorname{Tr}\{(\Lambda_{A}^{j}\otimes I_{B})\rho_{AB}\}  &  =\sum
_{k,l}\left(  \langle k|_{A}\otimes\langle l|_{B}\right)  \left[  (\Lambda
_{A}^{j}\otimes I_{B})\rho_{AB}\right]  \left(  |k\rangle_{A}\otimes
|l\rangle_{B}\right) \\
&  =\sum_{k,l}\langle k|_{A}\left(  I_{A}\otimes\langle l|_{B}\right)  \left[
(\Lambda_{A}^{j}\otimes I_{B})\rho_{AB}\right]  \left(  I_{A}\otimes
|l\rangle_{B}\right)  |k\rangle_{A}\\
&  =\sum_{k,l}\langle k|_{A}\Lambda_{A}^{j}\left(  I_{A}\otimes\langle
l|_{B}\right)  \rho_{AB}\left(  I_{A}\otimes|l\rangle_{B}\right)
|k\rangle_{A}\\
&  =\sum_{k}\langle k|_{A}\Lambda_{A}^{j}\left[  \sum_{l}\left(  I_{A}%
\otimes\langle l|_{B}\right)  \rho_{AB}\left(  I_{A}\otimes|l\rangle
_{B}\right)  \right]  |k\rangle_{A} .\label{eq-nqt:partial-trace-dev-last}%
\end{align}
The first equality follows from the definition of the trace  and using the orthonormal basis $\{|k\rangle
_{A}\otimes|l\rangle_{B}\}$. The second equality follows because%
\begin{equation}
|k\rangle_{A}\otimes|l\rangle_{B}=\left(  I_{A}\otimes|l\rangle_{B}\right)
|k\rangle_{A}.
\end{equation}
The third equality follows because%
\begin{equation}
\left(  I_{A}\otimes\langle l|_{B}\right)  (\Lambda_{A}^{j}\otimes
I_{B})=\Lambda_{A}^{j}\left(  I_{A}\otimes\langle l|_{B}\right)  .
\end{equation}
The fourth equality follows by bringing the sum over $l$ inside. Using the
definition of trace and the fact that
$\{|k\rangle_{A}\}$ is an orthonormal basis for Alice's Hilbert space, we can
rewrite \eqref{eq-nqt:partial-trace-dev-last} as%
\begin{equation}
\operatorname{Tr}\left\{  \Lambda_{A}^{j}\left[  \sum_{l}\left(  I_{A}%
\otimes\langle l|_{B}\right)  \rho_{AB}\left(  I_{A}\otimes|l\rangle
_{B}\right)  \right]  \right\}  . \label{eq-nqt:partial-trace-dev-last-1}%
\end{equation}
Our final step is to define the partial trace operation as follows:

\begin{definition}
[Partial Trace]Let $X_{AB}$ be a square operator acting on a tensor product
Hilbert space $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, and let $\{|l\rangle
_{B}\}$ be an orthonormal basis for $\mathcal{H}_{B}$. Then the partial trace
over the Hilbert space $\mathcal{H}_{B}$ is defined as follows:%
\begin{equation}
\operatorname{Tr}_{B}\{X_{AB}\}\equiv\sum_{l}\left(  I_{A}\otimes\langle
l|_{B}\right)  X_{AB}\left(  I_{A}\otimes|l\rangle_{B}\right)  .
\end{equation}
For simplicity, we often suppress the identity operators $I_{A}$ and write
this as follows:%
\begin{equation}
\operatorname{Tr}_{B}\{X_{AB}\}\equiv\sum_{l}\langle l|_{B}X_{AB}|l\rangle
_{B}.
\end{equation}

\end{definition}

For the same reason that the definition of the trace is invariant under the
choice of an orthonormal basis, the same is true for the partial trace
operation. We can also observe from the above definition that the partial
trace is a linear operation. Continuing with our development above, we can
define a local operator $\rho_{A}$, using the partial trace, as follows:%
\begin{equation}
\rho_{A}=\operatorname{Tr}_{B}\{\rho_{AB}\}.
\end{equation}
This then allows us to arrive at a rewriting of
\eqref{eq-nqt:partial-trace-dev-last-1} as%
\begin{equation}
\operatorname{Tr}\{\Lambda_{A}^{j}\rho_{A}\},
\end{equation}
which allows us to conclude that%
\begin{equation}
p_{J}(j)=\operatorname{Tr}\{(\Lambda_{A}^{j}\otimes I_{B})\rho_{AB}%
\}=\operatorname{Tr}\{\Lambda_{A}^{j}\rho_{A}\}.
\end{equation}
Thus, from the operator $\rho_{A}$, we can predict the outcomes of local
measurements that Alice performs on her system. Also important here is that
the global picture, in which we have a density operator $\rho_{AB}$ and a
measurement of the form $\{\Lambda_{A}^{j}\otimes I_{B}\}$, is consistent with
the local picture, in which the measurement is written as $\{\Lambda_{A}%
^{j}\}$ and the operator $\rho_{A}$ is used to calculate the probabilities
$p_{J}(j)$. The operator $\rho_{A}$ is itself a density operator, called the
\textit{local} or \textit{reduced density operator}, and the next exercise
asks you to verify that it is indeed a density operator.

\begin{exercise}
[Local Density Operator]Let $\rho_{AB}$ be a density operator acting on a
bipartite Hilbert space. Prove that $\rho_{A}=\operatorname{Tr}_{B}\{\rho
_{AB}\}$ is a density operator, meaning that it is positive semi-definite and
has trace equal to one.
\end{exercise}

In conclusion, given a density operator $\rho_{AB}$ describing the joint state
held by Alice and Bob, we can always calculate a local density operator
$\rho_{A}$, which describes the local state of Alice if Bob's system is
inaccessible to her.

There is an alternate way of describing partial trace, of which it is helpful
to be aware. For a simple state of the form%
\begin{equation}
|x\rangle\langle x|_{A}\otimes|y\rangle\langle y|_{B},
\end{equation}
with $|x\rangle_{A}$ and $|y\rangle_{B}$ each unit vectors, the partial trace
has the following action:%
\begin{equation}
\operatorname{Tr}_{B}\left\{  |x\rangle\langle x|_{A}\otimes|y\rangle\langle
y|_{B}\right\}  =|x\rangle\langle x|_{A}\ \operatorname{Tr}\left\{
|y\rangle\langle y|_{B}\right\}  =|x\rangle\langle x|_{A},
\end{equation}
where we \textquotedblleft trace out\textquotedblright\ the second system to
determine the local density operator for the first. If the partial trace acts
on a tensor product of rank-one operators (not necessarily corresponding to a
state)%
\begin{equation}
\left\vert x_{1}\right\rangle \left\langle x_{2}\right\vert _{A}
\otimes\left\vert y_{1}\right\rangle \left\langle y_{2}\right\vert _{B} ,
\end{equation}
its action is as follows:%
\begin{align}
\operatorname{Tr}_{B}\left\{  \left\vert x_{1}\right\rangle \left\langle
x_{2}\right\vert _{A}\otimes\left\vert y_{1}\right\rangle \left\langle
y_{2}\right\vert _{B}\right\}   &  =\left\vert x_{1}\right\rangle \left\langle
x_{2}\right\vert _{A}\ \operatorname{Tr}\left\{  \left\vert y_{1}\right\rangle
\left\langle y_{2}\right\vert _{B}\right\} \\
&  =\left\vert x_{1}\right\rangle \left\langle x_{2}\right\vert _{A}%
\ \left\langle y_{2}|y_{1}\right\rangle .
\end{align}
In fact, an alternate way of defining the partial is as above and to extend it
by linearity.

\begin{exercise}
Show that the two notions of the partial trace operation are consistent. That
is, show that%
\begin{align}
\operatorname{Tr}_{B}\left\{  \left\vert x_{1}\right\rangle \left\langle
x_{2}\right\vert _{A}\otimes\left\vert y_{1}\right\rangle \left\langle
y_{2}\right\vert _{B}\right\}   &  =\sum_{i}\left\langle i\right\vert
_{B}\left(  \left\vert x_{1}\right\rangle \left\langle x_{2}\right\vert
_{A}\otimes\left\vert y_{1}\right\rangle \left\langle y_{2}\right\vert
_{B}\right)  \vert i\rangle_{B}\\
&  =\left\vert x_{1}\right\rangle \left\langle x_{2}\right\vert _{A}%
\ \left\langle y_{2}|y_{1}\right\rangle ,
\end{align}
for some orthonormal basis $\{\vert i\rangle_{B}\}$ on Bob's system.
\end{exercise}

It can be helpful to see the alternate notion of partial trace worked out in
detail. The most general density operator on two systems $A$ and $B$ is some
operator $\rho_{AB}$ that is positive semi-definite with unit trace. We can
obtain the local density operator $\rho_{A}$ from $\rho_{AB}$ by tracing out
the $B$ system:%
\begin{equation}
\rho_{A}=\operatorname{Tr}_{B}\left\{  \rho_{AB}\right\}  .
\end{equation}
In more detail, let us expand an arbitrary density operator $\rho_{AB}$ with
an orthonormal basis $\{\vert i\rangle_{A}\otimes\left\vert j\right\rangle
_{B}\}_{i,j}$ for the bipartite (two-party) state:%
\begin{equation}
\rho_{AB}=\sum_{i,j,k,l}\lambda_{i,j,k,l}(\vert i\rangle_{A}\otimes\vert
j\rangle_{B})(\langle k\vert_{A}\otimes\langle l\vert_{B}).
\end{equation}
The coefficients $\lambda_{i,j,k,l}$ are the matrix elements of $\rho_{AB}$
with respect to the basis $\{\vert i\rangle_{A}\otimes\left\vert
j\right\rangle _{B}\}_{i,j}$, and they are subject to the constraint of
non-negativity and unit trace for $\rho_{AB}$ . We can rewrite the above
operator as%
\begin{equation}
\rho_{AB}=\sum_{i,j,k,l}\lambda_{i,j,k,l}\vert i\rangle\langle k\vert
_{A}\otimes\vert j\rangle\langle l\vert_{B}.
\end{equation}
We can now evaluate the partial trace:%
\begin{align}
\rho_{A}  &  =\operatorname{Tr}_{B}\left\{  \sum_{i,j,k,l}\lambda
_{i,j,k,l}\vert i\rangle\langle k\vert_{A}\otimes\vert j\rangle\langle
l\vert_{B}\right\} \\
&  =\sum_{i,j,k,l}\lambda_{i,j,k,l}\operatorname{Tr}_{B}\left\{  \left\vert
i\right\rangle \langle k\vert_{A}\otimes\vert j\rangle\langle l\vert
_{B}\right\} \\
&  =\sum_{i,j,k,l}\lambda_{i,j,k,l}\vert i\rangle\left\langle k\right\vert
_{A}\operatorname{Tr}\left\{  \vert j\rangle\langle l\vert_{B}\right\} \\
&  =\sum_{i,j,k,l}\lambda_{i,j,k,l}\vert i\rangle\left\langle k\right\vert
_{A}\left\langle j|l\right\rangle \\
&  =\sum_{i,j,k}\lambda_{i,j,k,j}\vert i\rangle\left\langle k\right\vert
_{A}\\
&  =\sum_{i,k}\left(  \sum_{j}\lambda_{i,j,k,j}\right)  \left\vert
i\right\rangle \langle k\vert_{A}.
\end{align}
The second equality exploits the linearity of the partial trace operation. The
last equality explicitly shows how the partial trace operation earns its
name---it is equivalent to performing a trace operation over the coefficients
corresponding to Bob's system.



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