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\noindent {\bf Assignment 2: Concurrence, a Measure of Entanglement}\\
\noindent DUE: Thursday, 2/23/2023, work in pairs\\

A little background: Let's consider an arbitrary 2-qubit state
\begin{eqnarray*}
\ket{\Psi} = \alpha \ket{00} + \beta \ket{01} + \gamma \ket{10} + \delta \ket{11}.
\end{eqnarray*}
 On page 18, Mermin remarks in passing that the
state $\ket{\Psi}$ is separable iff $\alpha\delta - \beta \gamma = 0$ (actually, 
at that point he's using the notation $\alpha_{00}$ for $\alpha$,
$\alpha_{01}$ for $\beta$, etc., but for our purposes, I expect you will welcome
the typographical convenience of dispensing with subscripts). This is not as obvious as the casual remark may indicate! It's also quite an interesting fact. The quantity $2|\alpha\delta - \beta \gamma|$ is called the {\it concurrence}. We will derive some of its important properties, and then apply it to some cases.
This leads to the important fact that some states are more entangled than others; it's not quite a dichotomy! 

The first two problems are a little challenging, so I'll provide you with what I found
to be some crucial insights (full disclosure: I struggled to find these myself!). Of course, if you find a better way to prove these
facts, I'd like to hear it.\\

\begin{enumerate}
\item Show that the concurrence ranges between 0 and 1. That is, prove that
\begin{eqnarray*}
  0 \le 2|\alpha\delta - \beta \gamma| \le 1,
\end{eqnarray*}
i.e., $0 \le |\alpha \delta - \beta\gamma| \le 1/2$.

HINTS: Consider the related state $\ket{\Phi} = \delta^*\ket{00} - \gamma^*\ket{01} - \beta^*\ket{10} + \alpha^*\ket{11}$. Ask yourself: (a) What upper and lower bounds can you put on
$|\bracket{\Phi}{\Psi}|^2$ (and indeed for any two single- or multiple- qubit states
$\ket{\Phi}$ and $\ket{\Psi}$, quite independent of this problem!)? (b) What do I get if I compute $\bracket{\Phi}{\Psi}$ in terms of $\alpha, \beta, \gamma, \delta$?
(c) From the answers I get to (a) and (b), what can I conclude about the
upper and lower bounds on $|\alpha\delta - \beta\gamma|$?\\

\hrule
In the next two problems, we will prove that
$\ket{\Psi}$ is separable {\it if and only if}
$\alpha \delta - \beta \gamma = 0$. That is, it
 is possible to write $\ket{\Psi} = (a\ket{0} + b\ket{1})\otimes(c\ket{0} + d\ket{1})$, where $|a|^2 + |b|^2 = |c|^2 + |d|^2 = 1$, iff $\alpha \delta - \beta \gamma = 0$.
 For the ``iff" we of course need two logical directions,
 the ``if" and the ``only if."

\item The ``only if" direction: Prove that if $\ket{\Psi}$ is separable, then $\alpha \delta - \beta \gamma = 0$.
 This is relatively easy, with the below hints.\\
\noindent{HINTS}: If $\ket{\Psi}$ is separable, then we
can write it as a tensor product, e.g.,
$(a\ket{0} + b\ket{1})\otimes(c\ket{0} + d\ket{1})$.
Expand this, work out that $\alpha, \beta, \gamma, \delta$
are in terms of $a, b, c, d$, and show that the concurrence
is then $0$.

\item Now prove the ``if" direction: that is, if 
$\alpha \delta - \beta \gamma = 0$, then $\ket{\Psi}$ is separable. 
 This is not so easy, so more
extensive hints here.\\
\noindent{HINTS}: First prove that if any one of $\alpha, \beta, \gamma, \delta$ is $0$, then (assuming, as we do, that
$\alpha \delta - \beta \gamma = 0$)
 the state is separable. There are 4 cases here; it is enough to prove one, and if your argument is sufficiently air-tight, I will accept the assertion that ``the other cases are similar." Now turn to the remaining possibility that $\alpha, \beta, \gamma, \delta$
are all non-zero. I hope it's obvious to you that,
in such a case, we can write
 $\ket{\Psi} = \frac{\alpha}{\gamma}\gamma\ket{00} 
                  + \frac{\beta}{\delta}\delta\ket{01}
                  + \gamma\ket{10} + \delta\ket{11}$. This is a more
 significant observation than you may imagine!
 Starting with this simple observation, {\it use the hypothesis} (i.e., $\alpha\delta = \beta\gamma$)
 to change this into a form that can be written as a (tensor) product of two states.
Now that might sound like you're done, but the problem is that if you do this naively, you
will find that the two ``states" aren't normalized to 1, so they're not really qubits. 
At that point, you must show how to manipulate quantities so that you get a tensor product
of two genuine normalized qubit states. (Caution: proving the two qubits in the product
are normalized will likely require another use of the hypothesis.) At that point, you're done.

\item Let's {\it do} something with the concurrence! Since $0 \le 2|\alpha\delta -\beta\gamma| \le 1$, where the ``$0$" value corresponds to ``separable" (as we saw in the
last result), it makes sense to regard concurrence as a {\it measure of entanglement}.
We say a state with concurrence 1 is {\it maximally entangled}. Let's say if it's
somewhere strictly between 0 and 1 it's {\it partially entangled}; and of 
course if it is $0$, the state is not at all entangled, i.e., separable. For each of 
the following 2-qubit states, identify $\alpha, \beta, \gamma$, and $\delta$, compute the concurrence, and state whether the
state is separable, partially entangled, or maximally entangled. {\it For the separable states, express the state as a tensor product of 2 single-qubit states.} Of those 
states here that are entangled, which one(s) is (are) the {\it least} 
entangled?
\begin{enumerate}
\item $\frac{1}{2}\left(\ket{00}+\ket{01}+\ket{10}+\ket{11}\right)$
\item $\frac{1}{2}\left(\ket{00}+\ket{01}+\ket{10}-\ket{11}\right)$
\item $\frac{1}{2}\left(\ket{00}+\ket{01}+\ket{10}-i\ket{11}\right)$
\item $\frac{1}{\sqrt{3}}\left(\ket{00}+\ket{01}+\ket{11}\right)$
\item $\frac{1}{\sqrt{2}}\left(\ket{00}-\ket{10}\right)$
\item $\frac{1}{\sqrt{2}}\left(\ket{01}-\ket{10}\right)$
\end{enumerate}




\end{enumerate}


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