\documentclass[landscape]{seminar}
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\usepackage{amsfonts}
\usepackage{epsf}
\newcommand{\Tr}{{\mathrm{Tr}}}
\newcommand{\ket}[1]{{\left| #1 \right \rangle}}
\newcommand{\bra}[1]{{\left\langle #1 \right |}}
\newcommand{\braket}[2]{{\left\langle #1 | #2 \right \rangle}}
\newcommand{\proj}[1]{{\left| #1 \right\rangle\!\left\langle #1 \right| }}
\begin{document}
\thispagestyle{empty}
\begin{slide}

{\large Properties of an EPR pair}

If you measure the two halves of an EPR pair using the same basis, you always get
orthogonal measurement outcomes.
\begin{eqnarray*}
\ket{0}\ket{1} - \ket{1}\ket{0} &=& \vspace{1em}(\alpha \ket{0} - \beta \ket{1})
(\beta^* \ket{0} + \alpha^* \ket{1}) \\
&& - (\beta^* \ket{0} + \alpha^* \ket{1})(\alpha \ket{0} - \beta \ket{1})
\end{eqnarray*}

Physical explanation for spin $\frac{1}{2}$ particles:\\
$\frac{1}{\sqrt{2}}
(\ket{\uparrow}  \ket{\downarrow} 
- \ket{\downarrow} \ket{\uparrow} )
$
has spin 0, no matter what basis it's expressed in.

Does one of the qubits ``know'' what basis it should be in
after you measure the other one?
\end{slide}

\begin{slide}
{\large Entanglement:}

Einstein, Podolsky, Rosen (1935)\\
Quantum mechanics is a very strange theory, in that it doesn't allow
for local descriptions of local reality, and thus it must be incomplete.

Schr\"odinger (1935)\\
Motivated by EPR, he introduced the term {\em entanglement (verschr\"ankung)}.

Bell (1964)\\ 
Bell's inequalities --- completing quantum mechanics won't make the strangeness
go away; probability distributions arising in quantum mechanics can't be 
explained classically without some kind of nonlocality.

Greenberger, Horne, Shimony, Zeilinger (1990)\\
``Bell's Theorem without Inequalities.''

\end{slide}

\begin{slide}
GHZ state (Greenberger, Horne, Zeilinger)

\[
\frac{1}{\sqrt{2}} (\left| 000 \right \rangle + \left| 111 \right \rangle )
\]

In the GHSZ experiment, we measure each qubit either in the basis
$\frac{1}{\sqrt{2}} (\left| 0 \right \rangle \pm \left| 1 \right \rangle )$ or the basis
$\frac{1}{\sqrt{2}} (\left| 0 \right \rangle \pm i \left| 1 \right \rangle ).$

What do we get?
\end{slide}

\begin{slide}
Let 
\begin{eqnarray*}
\ket{+} &=& \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}),\\
\ket{-} &=& \frac{1}{\sqrt{2}}(\ket{0}-\ket{1}).
\end{eqnarray*}

The state
\begin{eqnarray*}
\ket{\phi_{\mathrm{GHZ}}} &=& \frac{1}{\sqrt{2}}(\ket{000}+\ket{111})\\
&=& 
\frac{1}{2}(\ket{\,+++}+\ket{+--} \\
&&+\,\, \ket{-+-} + \ket{--+}\,)
\end{eqnarray*}

Thus, if you measure $\ket{\phi_{\mathrm{GHZ}}}$ in the $\{\ket{+},\ket{-} \}$ basis, you 
always get an even number of $\ket{-}$'s.

(Since 
$\ket{0} = \frac{1}{\sqrt{2}}(\ket{+}+\ket{-})$
and
$\ket{1} = \frac{1}{\sqrt{2}}(\ket{+}-\ket{-})$, coefficients of terms
with an odd number of $\ket{-}$s cancel in the expansion).
\end{slide}

\begin{slide}
\begin{eqnarray*}
\ket{+} = \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}), &&
\ket{-} = \frac{1}{\sqrt{2}}(\ket{0}-\ket{1}),\\
\ket{C^+} = \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}),&&
\ket{C^-} = \frac{1}{\sqrt{2}}(\ket{0}-i\ket{1}).
\end{eqnarray*}

If you measure $\ket{\phi_{\mathrm{GHZ}}}$ using the $\ket{C^\pm}$ basis for two of the qubits,
and using the $\ket{\pm}$ basis for the other qubit, you always get an {\em odd} 
total number of $\ket{-}$ $\cup$ $\ket{C^-}$s.

\begin{eqnarray*}
\ket{\phi_{\mathrm{GHZ}}} &=& \frac{1}{\sqrt{2}}(\ket{000}+\ket{111})\\
&=& 
\frac{1}{2}(\,\ket{\,+\, C^+C^-}+\ket{\,+\,C^-C^+} \\
&&+ \ket{\,-\,C^+C^+} + \ket{\,-\,C^-C^-}\,)
\end{eqnarray*}
(Almost same calculation as previous slide; but an odd total because $i\cdot i = -1$.)

\end{slide}

\begin{slide}

Now, we consider the results of measuring $\ket{\phi_{\mathrm{GHZ}}}$ using the
four bases given in the table below, keeping track of the parity of the number
of $\{\ket{-},\ket{C^{-}}\}$ in the outcome.  Here, D means the basis $\{\ket{\pm}\}$
and C the basis $\{\ket{C^\pm}\}$.

{\small
\[
\begin{array}{cccc}
{\mathrm{qubit\ 1}} &{\mathrm{qubit\ 2}} &{\mathrm{qubit\ 3}} &{\mathrm{parity\ of\ }}\{-,C^-\}\\
D&D&D&0\\
D&C&C&1\\
C&D&C&1\\
C&C&D&1
\end{array}
\]
}
No deterministic set of outcomes that specify the outcomes of each measurement
for every qubit can give this, since the total
parity of the outcomes in each of the first
three columns would have to be 0.

Thus, any local realistic theory must disagree with quantum mechanics 
for at least one quarter of its predictions.
\end{slide}

\begin{slide}
Applications of entanglement include 
\begin{itemize}
\item quantum teleportation 
\item  superdense coding
\item quantum cryptography
\item quantum computing
\end{itemize}

\end{slide}

\begin{slide}
When are two quantum systems {\em entangled}?

Given a joint density matrix $\rho_{AB}$, we could ask

\begin{itemize}
\item
Can we create $\rho_{AB}$ in two quantum laboratories using only a 
telephone line to communicate between them?
\item
Can we make measurements whose outcomes do not satisfy Bell's inequality or
a generalization (e.g., CHSH) of it?
\item
Can we make EPR pairs from many copies of $\rho_{AB}$ (i.e., $\rho_{AB}^{\otimes n}$)?
\item
Can we use a classical channel from Alice to Bob and a supply of shared states 
$\rho_{AB}$ to teleport quantum information?
\end{itemize}
\end{slide}

\begin{slide}
$\rho_{AB}$ is called {\em separable} if it can be created in two quantum laboratories
which only use a telephone line to communicate.

A separable state can be made by Alice (in one laboratory) flipping a many-sided coin, 
sending the results to Bob, and then both Alice and Bob making a specified pure 
state depending on the outcome.  

That is, a separable state can be expressed as
\[
\rho_{AB} = \sum_{i=1}^k p_i \ket{v_i}\bra{v_i} \otimes \ket{w_i}\bra{w_i}.
\]

Any state which is not separable is called entangled.
\end{slide}

\begin{slide}
Consider a pure state $\Psi_{AB}$ shared between Alice and Bob.

{\bf Theorem} (Schmidt decomposition)\\
Let $\lambda_i, \ket{v_i}$ be the eigenvectors and eigenvalues of
$\Tr_B \ket{\Psi_{AB} }\bra{\Psi_{AB} }$. Then
\[
\Psi_{AB} = \sum_i \sqrt{\lambda_i} \ket{v_i} \ket{w_i}
\]
where 
\[
\braket{v_i}{v_j} = \delta_{ij}
\]
\[
\braket{w_i}{w_j} = \delta_{ij}
\]

Thus, the $\ket{w_i}$ are the eigenvectors of the matrix $\Tr_A \ket{\Psi_{AB} }\bra{\Psi_{AB} }$,
and 
\[
H(\Tr_A \ket{\Psi_{AB} }\bra{\Psi_{AB} }) = H(\Tr_B \ket{\Psi_{AB} }\bra{\Psi_{AB} }).
\]
\end{slide}

\begin{slide}
If $\rho_{AB}$ is a pure state, then we have a good definition of entanglement.

\[
E_P(\rho_{AB}) = H(\Tr_A \rho_{AB}) =H(\Tr_B \rho_{AB}) 
\]

Theorem {\small (Bennett, Bernstein, Popescu, Schumacher)}
\begin{itemize}
\item
$n E_P(\rho_{AB}) -o(n)$ EPR pairs can be obtained from $n$ copies of $\rho_{AB}$,
using local quantum operations, with high probability.
\item
$n$ copies of a state very close to $\rho_{AB}$ can be obtained from 
$n E_P(\rho_{AB}) + o(n)$ EPR pairs, using local quantum operations and 
classical communication.
\end{itemize}
\end{slide}

\begin{slide}
To prove this theorem, we will use Nielsen's non-asymptotic characterization
of when one pure entangled states can be created from another pure
entangled state.

{\bf Theorem} (Nielsen)\\
If we have a pure entangled state $\rho_{AB}$ and a pure entangled state
$\sigma_{AB}$, both shared between Alice and Bob, then using local
operations on Alice's and Bob's states, 
we can create $\rho_{AB}$ from $\sigma_{AB}$ if and only the eigenvalues
$\lambda_\sigma$ of $\Tr_B\sigma$ are majorized by the eigenvalues 
$\lambda_\rho$ of $\Tr_B \rho$.

Definition:
$\lambda_\rho$ majorizes $\lambda_\sigma$  if and only if
\[\sum_{i=1}^k \lambda_{\rho,i} \geq 
\sum_{i=1}^k \lambda_{\sigma,i} \]
when the eigenvalues are given in decreasing order.  
 
\end{slide}

\begin{slide}
Proof (one direction) for Alice and Bob both holding qubits.

Let us take without loss of generality
\[
\psi_{\rho} = {\alpha}\ket{00} + {\beta} \ket{11}
\]
where $\alpha \geq \beta$, $\alpha^2 + \beta^2 = 1$.  The eigenvalues of
$\Tr_B\rho$ are $\alpha^2$ and $\beta^2$.  

Alice can measure her state with the quantum instrument derived from matrices
$\frac{1}{\sqrt{2}} A_+$ and $\frac{1}{\sqrt{2}} A_-$, where
\[
A_+ = \left(\begin{array}{cc}x& y\\y & x\end{array}\right)\qquad
A_- = \left(\begin{array}{cc}x& -y\\-y & x\end{array}\right)
\]
where $x^2 + y^2 = 1$.  This is a valid instrument since

\[\frac{1}{2}\left(A_+^\dag A_+  + A_-^\dag A_-\right) = x^2+y^2 = 1.\]
\end{slide}

\begin{slide}
\[
\mathrm{Recall\ \ \ } \psi_{\rho} = {\alpha}\ket{00} + {\beta} \ket{11}
\]
\[
A_+ = \left(\begin{array}{cc}x& y\\y & x\end{array}\right)\qquad
A_- = \left(\begin{array}{cc}x& -y\\-y & x\end{array}\right)\qquad
\]
After Alice's measurement (suppose she obtains $A_+$), the state is
\[
\ket{\Psi_+} =
(A_+ \otimes I ) {\small{\left(\begin{array}{c}\alpha\\0\\0\\
\beta\end{array}\right) }}
= x\alpha\ket{00} + y\alpha\ket{10} + y\beta\ket{01} + x\beta \ket{11}.
\]
\vspace{-4ex}
\[\Tr_B\proj{\Psi_+} = 
\left(\begin{array}{cc} \alpha^2x^2 + \beta^2 y^2 & xy \\ xy &
\alpha^2y^2+\beta^2x^2\end{array}\right)
\]
This has eigenvalues $\frac{1}{2} \pm
\frac{1}{2}\sqrt{1-4\alpha^2\beta^2(x^2-y^2)}$.
These range between ($\alpha^2$, $\beta^2$) ($x=y=1/\sqrt{2}$) and 
(1,0) ($x=1,y=0$).
\end{slide}
\begin{slide}
Thus, if Alice and Bob want to go from $\psi_\rho$ to $\psi_\sigma$, where
the eigenvaules of $\Tr_B \sigma $ majorize $\Tr_B \rho$, she can do
it when they have qubits.  

Now, suppose they have higher dimensional states.  They can alter two
of the eigenvalues at a time by using the quantum instrument with
matrices 
\[
\left(\begin{array}{cc}A_+ &0\\0 & I_{d-2}\end{array}\right)
\mbox{\ \ \ and \ \ \ }
\left(\begin{array}{cc}A_- &0\\0 & I_{d-2}\end{array}\right)
\]
By repeatedly using such quantum instruments, we can go from any
$\rho$ to $\sigma$ if the eigenvalues of $\sigma$ majorize those of $\rho$.
\end{slide}

\begin{slide}
Suppose we have many copies of a pure state $\ket{\psi}^{\otimes m}$, and
we want to make $\ket{\phi}^{\otimes n}$, where
\[
\ket{\phi} = \frac{1}{\sqrt{2}}(\ket{00}+\ket{11}).
\]
Expanding, we get
\begin{eqnarray*}
\ket{\phi} &=& \frac{1}{2^{n/2}}(\ket{00}+\ket{11})^{\otimes n}\\
&=&\frac{1}{2^{n/2}}\sum_{i=0}^{2^n-1}\ket{i_A}\otimes \ket{i_B}
\end{eqnarray*}

This is an equal superposition of $2^n$ vectors $\ket{i_A}\otimes \ket{i_B}$
where the $\ket{i_A}$ are orthogonal vectors in Alice's subsystem, and the
$\ket{i_B}$ are orthogonal vectors in Bob's subsystem.

This is called a {\em maximally entangled state} in $2^n$ dimensions.

\end{slide}


\begin{slide}
{\large Asyptotic pure state entanglement}\\
Let
\[
\ket{\psi} = \sum_{i=1}^d \sqrt{\lambda_i} \ket{v_i} \ket{w_i}
\]
Then
\[
\ket{\psi}^{\otimes n}  = \hspace{-.3in}
\sum_{n=k_1+...+k_d} \hspace{-3ex} \sqrt{\lambda_1^{k_1}\lambda_2^{k_2}\ldots \lambda_d^{k_d}}
\hspace{-1ex}
\sum_{I \in {n \choose k_1 \ldots k_d}} \ket{v_I} \otimes \ket{w_I}
\]

Nearly all the weight is in eigenvectors with eigenvalues close to 
$2^{-n H(\lambda_1, \lambda_2, \ldots, \lambda_d})$.  If we take the
state very close to this which drops all other eigenvalues, this 
is majorized by the state consisting of 
$n H(\lambda_1, \lambda_2, \ldots, \lambda_d) - o(n)$ EPR pairs,
and majorizes  the state consisting of 
$n H(\lambda_1, \lambda_2, \ldots, \lambda_d) + o(n)$ EPR pairs,

Thus LOCC operations can asymptotically convert any pure state with E
entanglement to any other pure state with E entanglement.
\end{slide}

\begin{slide}
How much communication do you need to create a state $\ket{\phi}^{\otimes n}$ 
starting from around $n E(\phi)$ EPR pairs?  It turns out you 
need $O(\sqrt{n})$ bits of classical communication  (Upper bound: Lo and 
Popsecu; Lower bound: Harrow and Lo). 
\end{slide}

\begin{slide}
How to create EPR pairs from $n$ copies of $\ket{\phi}$ using
local operations but without any communication.\\
We have state
\[
\ket{\psi}^{\otimes n}  = \hspace{-.3in}
\sum_{n=k_1+...+k_d} \hspace{-3ex} \sqrt{\lambda_1^{k_1}\lambda_2^{k_2}\ldots \lambda_d^{k_d}}
\hspace{-1ex}
\sum_{I \in {n \choose k_1 \ldots k_d}} \ket{v_I} \otimes \ket{w_I}
\]

Measure the number of times each eigenvector appears, 
i.e., the values $k_1 \ldots k_d$.\\
This projection measurement
gives us  a maximally entangled state in $n \choose k_1 \ldots k_d$ dimensions.
Can obtain $\log_2 {n \choose k_1 \ldots k_d}$ EPR pairs from this.\\
With high probability, 
\[
\log_2 {n \choose k_1 \ldots k_d} \approx H(\lambda_1, \ldots \lambda_d) n.
\]

\end{slide}

\begin{slide}
What about measuring entanglement of mixed states?

Entanglement cost: $E_C(\rho)$
\[
\lim_{n \rightarrow \infty} \frac{1}{n} \left( \parbox{3.5in}{Number EPR pairs needed to make
approximation of $\rho^{\otimes n}$.}\right) 
\]

Distillable entanglement: $E_D(\rho)$
\[
\lim_{n \rightarrow \infty} \frac{1}{n} \left( \parbox{3.5in}{Number approximate EPR pairs obtainable from $\rho^{\otimes n}$.}\right)
\]

The protocols above may use only local operations and classical communication.

Clearly, $E_C \geq E_D$.
For mixed states, we can have $E_C > E_D$.

\end{slide}

\begin{slide}
Conjectures:

\begin{itemize}
\item[(1)]
Entanglement cost, $E_C$, is additive:
\[
E_F(\rho_1 \otimes \rho_2)=
E_F(\rho_1) + E_F(\rho_2).
\]
\item[(2)]
$E_D$ is not additive\\
(A conjectured counterexample exists)
\item[(3)]
Entanglement of formation, 
$E_F$, can be thought of as one-shot entanglement cost $E_{F}$,
where
\[
E_{F} = \min_{\rho = \sum_i p_i \rho_i} \sum_i p_i E_P(\rho_i)
\]
and the $\rho_i$ are pure states (rank 1).
\end{itemize}
\end{slide}

\begin{slide}
Another Conjecture

It turns out that conjectures (1) and (3) on the previous slide would 
follow from a proof of additivity of entanglement of formation, $E_{F}$.
\[
E_{F}(\rho_1 \otimes \rho_2)=
E_{F}(\rho_1) + E_{F}(\rho_2).
\]
where
\[
E_{F} = \min_{\rho = \sum_i p_i \rho_i} \sum_i p_i E_P(\rho_i)
\]
and the $\rho_i$ are pure states (rank 1).

This follows from the theorem
\[
E_C(\rho) = \lim_{n \rightarrow \infty} \frac{1}{n} E_F(\rho^{\otimes n})
\]
\end{slide}

\begin{slide}
There is a characterization of entangled states in $2\times 2$ and $2 \times 3$ dimensions.

Partial transpose: take transpose of Bob's half.

What happens for $\frac{1}{\sqrt{2}} (\ket{00}+\ket{11})$?
\[
\rho = \left(
\begin{array}{cc|cc}
\frac{1}{2}&0&0&\frac{1}{2}\\
0&0&0&0\\
\hline
0&0&0&0\\
\frac{1}{2}&0&0&\frac{1}{2}
\end{array}
\right)
\]

Take the transpose of Bob's part:
\[
\rho^{PT} = \left(
\begin{array}{cc|cc}
\frac{1}{2}&0&0&0\\
0&0&\frac{1}{2}&0\\*[.05in]
\hline
0&\frac{1}{2}&0&0\\
0&0&0&\frac{1}{2}
\end{array}
\right)
\]

Eigenvalues: $\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$, $-\frac{1}{2}$.
\end{slide}
\begin{slide}
If $\rho$ is not entangled, the partial transpose has positive eigenvalues.

Proof:
\[
\rho = \sum_i p_i \ket{v_i}\bra{v_i} \otimes \ket{w_i}\bra{w_i}
\]
Then
\[
\rho^{PT} = \sum_i p_i \ket{v_i}\bra{v_i} \otimes \ket{w_i^*}\bra{w_i^*}
\]

\end{slide}

\begin{slide}
Theorem (Peres, Horodecki$^3$):

If $\rho^{PT} \geq 0$, then $E_D = 0$: no distallable entanglement.   \\
Proof idea: LOCC operations cannot create negative eigenvalues

If $\rho \in {\mathbb{C}}^2 \otimes {\mathbb{C}}^3$, 
$\rho^{PT} \geq 0$ if and only if
\[
\rho = \sum p_i \ket{v_i} \bra{v_i} \otimes \ket{w_i} \bra{w_i},
\]
i.e., $\rho$ is not entangled.

not true if $\rho$ in any larger entangled system.
\end{slide}

\begin{slide}
\epsfxsize=.9in
\hspace*{-.3in}
\begin{parbox}{3.2in}{
There are {\em bound entangled} states, so that
$\rho^{PT} \geq 0$, implying $E_D =0$, but so $\rho$ cannot be created
without using entanglement.} \end{parbox} 
\hspace*{\fill} \raisebox{-.2in}{\epsfbox{befig.eps}}

\vspace{-2ex}

{\small 
\begin{eqnarray*}
\ket{v_1} &=& {\textstyle{\frac{1}{3}}}\left( \ket{0} - \ket{1} + \ket{2}\right) \otimes
\left( \ket{0} - \ket{1} + \ket{2}\right) \\
\ket{v_2} &=& {\textstyle{\frac{1}{\sqrt{2}}}}\left( \ket{0} + \ket{1} \right) \otimes
\left(  \ket{0}\right) \\
\ket{v_3} &=& {\textstyle{\frac{1}{\sqrt{2}}}}\left( \ket{2} \right) \otimes
\left(  \ket{0} + \ket{1}\right) \\
\ket{v_4} &=& {\textstyle{\frac{1}{\sqrt{2}}}}\left( \ket{0} \right) \otimes
\left(  \ket{1} + \ket{2}\right) \\
\ket{v_5} &=& {\textstyle{\frac{1}{\sqrt{2}}}}\left( \ket{1} + \ket{2} \right) \otimes
\left(  \ket{2}\right) 
\end{eqnarray*}
}
$\ket{v_i}$ and $\ket{v_{i+1}}$ orthogonal in first coordinate.\\
\hspace*{1ex}$\ket{v_i}$ and $\ket{v_{i+2}}$ orthogonal in second.\\
No product state is orthogonal to all five vectors (it would have to be orthogonal
to three vectors in one of the coordinates).

\end{slide}

\begin{slide}
{\large Bound Entanglement}

Consider the state 
{\small
\[
\rho = \frac{1}{4}\left( 
I 
- \ket{v_1}\bra{v_1}
- \ket{v_2}\bra{v_2}
- \ket{v_3}\bra{v_3}
- \ket{v_4}\bra{v_4}
- \ket{v_5}\bra{v_5}
\right)
\]
}

Since $I^{PT} = I$, and $(\ket{v_i}\bra{v_i})^{PT} = (\ket{v_i}\bra{v_i})$, we have
that $\rho^{PT} = \rho$ is positive, so $E_D(\rho)=0$.
Thus no EPR pairs are distillable from it.

${\mathrm{support}}(\rho)$ has no 
tensor product vectors in it, so $\rho$ is not separable.
Thus, $\rho$ cannot be created without entanglement.

In fact, one can show that $E_F(\rho)>0$, so that a linear 
number of EPR pairs are
required to create $\rho$.
\end{slide}

\begin{slide} 
Theorem (Yang, Horodecki, Horodecki, Synak-Radtke):\\
All non-separable states
have $E_C > 0$. \\
The proof takes several steps.  
The key is a definition called the classical correlation
of two quantum states.
\begin{eqnarray*}
C_{\rightarrow}(\rho_{AB}) &=& \max_{A_i^\dag A_i}
H(\rho_B) - \sum_i p_i H(\rho_B^i)\\
C_{\leftarrow}(\rho_{AB}) &=& \max_{A_i^\dag A_i}
H(\rho_B) - \sum_i p_i H(\rho_B^i)\\
C_{\leftrightarrow}(\rho_{AB}) &=& \max \{ 
C_{\rightarrow}(\rho_{AB}), C_{\leftarrow}(\rho_{AB}) \}
\end{eqnarray*}
Intuitively, $C_\leftarrow$ is how much classical information 
Alice can learn about Bob's state by performing a measurement
on her state.\\
$C_\leftarrow(\rho_{AB}) =  0$ if and only if $\rho_{AB} = \rho_A \otimes
\rho_B$.
\end{slide}

\begin{slide}
We now define
\begin{eqnarray*}
G_{\rightarrow}(\rho_{AB}) &=& \inf \sum_i p_i C_\rightarrow (\rho_{AB}^i)\\
G_{\leftarrow}(\rho_{AB}) &=& \inf \sum_i p_i C_\leftarrow (\rho_{AB}^i)
\qquad \mbox{where} \qquad \sum_i p_i \rho_{AB}^i = \rho_{AB}\\
G_{\leftrightarrow}(\rho_{AB}) &=& \inf \sum_i p_i 
C_\leftrightarrow (\rho_{AB}^i)\\
\end{eqnarray*}
 
Theorem:
$G = 0$ if and only if $\rho_{AB}$ is separable.\\
Proof: Follows from $C = 0$ if and only if $\rho_{AB}$ is a tensor
product, continuity of $G$, and a compactness argument.  
\end{slide}

\begin{slide}
Duality:\\
Let $\ket{\phi}_{ABC}$ be a pure state.
\[
H(\rho_A) = E_F(\rho_{AC}) + C_\leftarrow(\rho_{AB})
\]

This holds because $C_\leftarrow$ is a minimization over all measurements
of $\rho_{AB}$ on Bob's half.  These measurements have an infimum where
Bob's measurements is a POVM with just rank one matrices.  This induces
a decomposition of $\rho_{AC}$ into pure states, which is the same
decomposition of entanglement of formation.
\end{slide}

\begin{slide}
Lemma: For any fourpartite pure state $\ket{\phi}_{AA'BB'}$, we
have 
\[
E_F(\phi_{AA':BB'}) \geq
E_F(\rho_{A:B}) + C_\leftrightarrow(\rho_{A':B'})
\]
Proof: Use duality
\begin{eqnarray*}
E_F(\phi_{AA':BB'}) = H(\rho_{AB}) &=& E_F(\rho_{AA':B}) +
C_\leftarrow(\rho_{AA':B'})\\
&\geq &
E_F(\rho_{A:B}) + C_\leftarrow(\rho_{A':B'})
\end{eqnarray*}
and a similar argument gives 
\[
E_F(\phi_{AA':BB'}) = H(\rho_{A'B'})
\geq 
E_F(\rho_{A:B}) + C_\rightarrow(\rho_{A':B'})
\]
\end{slide}

\begin{slide}
Theorem: For a four-partite state $\rho_{AA'BB'}$,
\[
E_F(\rho_{AA':BB'}) \geq E_F(\rho_{A:B}) + G_\leftrightarrow(\rho_{A':B'}).
\]

Proof:
\begin{eqnarray*}
E_F(\rho_{AA':BB'}) &=& \sum_i p_i H(\rho^i_{AA'}) \\
&\geq& \sum_i p_i E_f(\rho^i_{A:B}) + 
\sum_i p_i C_{\leftrightarrow} (\rho_{A'B'}^i) \\
&\geq& E_F(\rho_{A:B}) + G_{\leftrightarrow}(\rho_{A':B'}) 
\end{eqnarray*}
\end{slide}
\begin{slide}
Theorem: For a four-partite state $\rho_{AA'BB'}$,
\[
E_F(\rho_{AA':BB'}) \geq E_F(\rho_{A:B}) + G_\leftrightarrow(\rho_{A':B'}).
\]
Corollary:
\begin{eqnarray*}
n E_C(\rho) &\geq& E_F(\rho^{\otimes n})\\
&\geq& E_F(\rho^{\otimes n-1}) +  G_{\leftrightarrow}(\rho)\\
&\geq& E_F(\rho^{\otimes n-2}) +  2G_{\leftrightarrow}(\rho)\\
&\geq& E_F(\rho) +  (n-1) G_{\leftrightarrow}(\rho).
\end{eqnarray*}
So this gives
\[
E_C(\rho) \geq G_{\leftrightarrow}(\rho) \geq 0. 
\]

\end{slide}


\begin{slide}
{\bf Other Measures Useful for Entanglement}
\begin{itemize}
\item
Relative entropy of entanglement
\[
\inf_{\sigma\in D} \Tr \rho (\log \rho - \log \sigma)
\]
where $D$ is the set of separable matrices.
\item
Logarithmic negativity
\[
\log || \rho^{PT} ||_1 = \log (1 + 2 N(\rho))
\]
where $N(\rho)$ is the sum of the negative eigenvalues.
\end{itemize}

Relative entropy of entanglement is between $E_D(\rho)$ and $E_F(\rho)$,
and is not additive.

Logarithmic negativity is an upper bound on $E_D(\rho)$, and is additive.
\end{slide}

\begin{slide}
{\bf Even more measures}
\begin{itemize}
\item Bounds from positive partial transpose distillable entanglement\\
(Rains), (Audenaert, de Moor, Vollbrecht, Werner)
\item
Squashed entanglement (Christandl and Winter)
\[
E_{sq}(\rho_{AB}) = \inf \left\{\textstyle\frac{1}{2} I(A;B|E) \mbox{\
where\ } \rho_{ABE} \mbox{\ is\ an\ extension\ of\ } \rho_{AB}\right\}
\]
\[
I(A;B|E) = H(\rho_{AE}) + H(\rho_{BE}) -H(\rho_{ABE}) - H(\rho_E)
\]
\end{itemize}
The Rains bound is computabable via a semi-definite program, and always
larger than distillable entanglement.\\
It is related to bounds by Audenaert et al.\\
Squashed entanglement is additive, and $E_C \geq E_{sq} \geq E_D$.
\end{slide}

\end{document}