Entanglement destroying channels

In a previous post, we were concerned with channels of the form $\Phi\in\Channel(\X,\Y)$ such that  $\bigl(\Phi\otimes \I_{\Lin(\Z)}\bigr)(\rho) \in \Sep(\Y:\Z)$ for every complex Euclidean space $\Z$ and every density operator $\rho\in\Density(\X\otimes\Z)$. Channels of this form have the effect of destroying entanglement that exists between the register they act on and any other registers.

Theorem:
There exist two channels $\Phi_0,\Phi_1\in\Channel(\X,\Y)$, both having the property described above, such that
  \[
  \bigtriplenorm{\Phi_0 - \Phi_1}_1
  > \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1
  \]
  for every $\rho\in\Density(\X)$. (Channels like this have the strange property that they destroy entanglement, and yet evaluating them on an entangled state helps to distinguish them.)


Proof:

For $\lambda\in[0,1]$, consider the two channels $\Phi_0(X),\Phi_0(X)\in\Channel(X)$ defined by
\[\begin{align*}
\Phi_0(X)&=\frac{\lambda}{n+1}(\tr(X)\I_\X+X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X \\
\Phi_1(X)&=\frac{\lambda}{n-1}(\tr(X)\I_\X-X^T)+\frac{(1-\lambda)}{n}\tr(X)\I_\X
\end{align*}\]
Then for sufficiently small $\lambda\in[0,1]$ both of the Choi representations $J(\Phi_0(X))$ and $J(\Phi_1(X))$ are in a separable neighborhood of the maximally mixed state which implies that they are both separable by some theorem. Therefore, from the results in the previous post, we have that $\Phi_0(X)$ and $\Phi_0(X)$ are entanglement destroying as described in the problem statement.

Now considering that
\[
\Phi_0(\rho) - \Phi_1(\rho)=\frac{-2\lambda}{(n+1)(n-1)}\rho^T,
\]
it follows that
\[
\bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}\bignorm{\rho^T}_1=\frac{2\lambda}{(n+1)(n-1)},
\]
since $\rho\in\Density(\X)$.

Moreover, since
\[\begin{align*}
\bigtriplenorm{\Phi_0 - \Phi_1}_1&=max\{\bignorm{((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)}_1 \ : \ x\in S(\X\otimes\X)\} ,
\end{align*}\]
where $((\Phi_0(\rho) - \Phi_1(\rho))\otimes\I_{\Lin(\X)})(xx^\ast)$ gives the partial transpose of $xx^\ast$ (which is at most $n$ since $x\in S(\X\otimes\X)$) multiplied by the scalar quantity $\frac{2\lambda}{(n+1)(n-1)}$. Therefore $\bigtriplenorm{\Phi_0 - \Phi_1}_1=\frac{2\lambda n}{(n+1)(n-1)}$, which implies that
  \[
  \frac{2\lambda n}{(n+1)(n-1)}=\bigtriplenorm{\Phi_0 - \Phi_1}_1
  > \bignorm{\Phi_0(\rho) - \Phi_1(\rho)}_1=\frac{2\lambda}{(n+1)(n-1)}.
  \]

Separable channels decrease the entaglement of formation

The entanglement of formation of a density operator $\rho\in\Density(\X^{A}\otimes\X^{B})$ is defined as
  \[
  E_{f}(\rho) = \inf\Biggl\{\sum_{a\in\Sigma} p(a) E(u_a u_a^{\ast})
  \,:\, \rho = \sum_{a\in\Sigma} p(a) u_a u_a^{\ast} \Biggr\},
  \]
  where $E(u u^{\ast}) = S(\tr_{\X^{B}}(u u^{\ast}))$ denotes the entanglement entropy of the pure state $u u^{\ast}$ and the infimum is over all expressions of $\rho$ of the given form, where $\Sigma$ is any alphabet, $p\in\P(\Sigma)$ is a probability vector, and $\{u_a\,:\,a\in\Sigma\} \subset \X^{A}\otimes\X^{B}$ is a collection of unit vectors.

Theorem:

For every choice of complex Euclidean spaces $\X^{A}$, $\X^{B}$, $\Y^{A}$, and $\Y^{B}$, every density operator $\rho\in\Density(\X^{A}\otimes\X^{B})$, and every separable channel $\Phi\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$, it holds that
  \[
  E_{f}(\Phi(\rho)) \leq E_{f}(\rho).
  \]

Proof:

Assuming that $\Phi\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$ allows $\Phi$ to be expressed as
\[
\Phi(X)=\sum_{b\in\Gamma}(A_b\otimes B_b)X(A_b^\ast\otimes B_b^\ast),
\]
where $\Gamma$ is some alphabet and $\{A_b : b\in \Gamma\}\subset \Pos(\X^A)$ and $\{B_b : b\in \Gamma\}\subset \Pos(\X^B)$. For $\rho = \sum_{a\in\Sigma} p(a) u_a u_a^{\ast}$, the action of $\Phi$ on $\rho$ is specified by the action of $\Phi$ on each $u_au_a^\ast$ as
\[
\Phi(\rho)=\sum_{a\in\Sigma} p(a) \Phi(u_a u_a^{\ast})=\sum_{a\in\Sigma} p(a)\sum_{b\in\Gamma}(A_b\otimes B_b)u_au_a^*(A_b^\ast\otimes B_b^\ast).
\]
Therefore, represent $\Phi(u_au_a*)$ as
\[
\Phi(u_au_a*)=\sum_{b\in\Gamma}(A_b\otimes B_b)u_au_a^*(A_b^\ast\otimes B_b^\ast)=\sum_{b\in\Gamma}q_a(b) v_{ab} v_{ab}^{\ast},
\]
where $(A_b\otimes B_b)u_a=\sqrt{q_a(b)}v_{ab}$. Now let
\[
C_b=\frac{1}{\sqrt{q_a(b)}}(A_b\otimes B_b),
\]
so that $C_bu_au_a^\ast C_b^\ast=v_{ab}v_{ab}$.

Consider the channel $\Psi_{ab}\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$ defined by
\[
\Psi_{ab}(X)=C_bXC_b^\ast+(\tr(X)-\tr(C_bXC_b^\ast))\sigma,
\]
for some arbitrary $\sigma\in\Density(\Y^A\otimes\Y^B)$. Then $\Psi_{ab}$ is indeed a channel since it is completely positive because it is defined in terms of $C_b$ and $\Phi$ is assumed to be completely positive. Likewise, $\Psi_{ab}$ is separable since $\Phi$ is separable. Moreover, $\Psi_{ab}$ is trace preserving since
\[\begin{align*}
\tr(\Psi_{ab}(X))&=\tr(C_bXC_b^\ast)+(\tr(X)-\tr(C_bXC_b^\ast))\tr(\sigma) \\
&=\tr(C_bXC_b^\ast)+\tr(X)-\tr(C_bXC_b^\ast) \\
&=\tr(X).
\end{align*}\]
By construction $\Psi_{ab}(u_au_a^\ast)=v_{ab}v_{ab}^\ast$.
 Therefore, by a corollary (6.36) to Nielsen's theorem it follows that for every $a\in\Sigma$ with $\rho_a^A=\tr_{\X^B}(u_au_a^\ast)$ and $\sigma_a^A=\tr_{\X^B}(v_{ab}v_{ab}^\ast)$ and $r=min\{rank(\rho_a^A),rank(\sigma^A_a)\}$ it holds that
\[
\lambda_1(\rho_a^A)+\dots+\lambda_1(\rho_m^A)\leq \lambda_1(\sigma_a^A)+\dots+\lambda_1(\sigma_m^A)
\]
for every $m\in\{1,\dots, r\}$.

Thus, the von Neummann entropy satisfies $S(\sigma_a^A)\leq S(\rho_a^A)$, which implies that the entanglement entropy also satisfies $E(v_{ab}v_{ab}^\ast)\leq E(u_au_a)$ for all $a\in\Sigma$ and $b\in\Gamma$. Then by tracing out system $B$, and taking the weighted average that is described the original state $\rho$ and using the joint convexity of the von Neumann entropy it follows that 
\[\begin{align*}
\sum_{a\in\Sigma} p(a)\sum_{b\in\Gamma}q_a(b) E(v_{ab} v_{ab}^{\ast})
\leq \sum_{a\in\Sigma} p(a) E(u_a u_a^{\ast}).
\end{align*}\]

 Therefore, by definition $E_{f}(\Phi(\rho)) \leq E_{f}(\rho)$.

A nonlocal game

Consider the game where Alice and Bob are physically separated and their goal is to produce outputs that satisfy the winning conditions specified below. Alice and Bob receive $s,t \in \{0,1,2\}$ as input ($s$ to Alice and $t$ to Bob), at which point they are forbidden to communicate with each other. They each output a bit, $a$ for alice and $b$ for Bob. The winning conditions are

1. $a=b$ in the cases where $s=t$.
2. $a\neq b$ in the cases where $s\neq t$.

Consider the classical procedure in which Alice and Bob agree to output bits $a$ and $b$ upon receiving inputs $s$ and $t$, respectively, as described in the table displayed on the left below. Then observe the results of all possible game plays when Alice and Bob receive bits $(s,t)$ and output $(a,b)$ according to the table shown on the right:



 It can therefore be seen that in all cases where Alice and Bob receive inputs $(s,t)$ such that $s=t$ they always produce outputs $(a,b)$ such that $a=b$, and therefore win with probability $1$ in this case. However, in the case when Alice and Bob receive inputs $(s,t)$ such that $s\neq t$, then they also produce outputs $(a,b)$ such that $a\neq b$ in only $4$ out of the $6$ cases, and produce outputs such that $a=b$ in the remaining $2$ cases. Hence, they only when $4$ out of $6$ times and so succeed in the case when $s\neq t$ with probability $2/3$.

Suppose now that Alice and Bob are playing the game with the same winning conditions, but this time they each possess one qubit of the Bell pair
\[
 \ket{\beta}=\frac{1}{\sqrt{2}}(\ket{00}-\ket{11}),
\]
 where say Alice has the first qubit of the register and Bob the second. Moreover, Alice and Bob are allowed to perform local operations and measurements on their parts of the Bell pair.  They can then base their output bits in the game, $a$ and $b$, depending on the measurement outcomes of their qubits.

 Consider the single qubit (unitary) rotation operations by some integer multiple $n$ of $\pi/3$ given by
\[
R_n=\left(
\begin{array}{cc}
 \cos \left(\frac{\pi  n}{3}\right) & -\sin \left(\frac{\pi  n}{3}\right) \\
 \sin \left(\frac{\pi  n}{3}\right) & \cos \left(\frac{\pi  n}{3}\right) \\
\end{array}
\right).
\]
 Given input bits $s$ and $t$ to Alice and Bob, respectively,  Alice will perform the rotation $R_s$ on her qubit of the Bell state and Bob will perform the rotation $R_{-t}$ in the `negative' direction on his qubit of the Bell state. Then for $r,s\in\{0,1,2\}$, the tensor product of their operations will be given in general by
\[
 R_s\otimes R_{-t}=\left(
\begin{array}{cccc}
 \cos \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi  t}{3}\right) & \cos
   \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) & -\sin
   \left(\frac{\pi  s}{3}\right)\cos \left(\frac{\pi
   t}{3}\right) & -\sin \left(\frac{\pi  s}{3}\right) \sin
   \left(\frac{\pi  t}{3}\right) \\
 -\cos \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) & \cos
   \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi  t}{3}\right) & \sin
   \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) & -\sin
   \left(\frac{\pi  s}{3}\right)\cos \left(\frac{\pi
   t}{3}\right) \\
 \sin \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi  t}{3}\right) & \sin
   \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) & \cos
   \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi  t}{3}\right) & \cos
   \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) \\
 -\sin \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) & \sin
   \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi  t}{3}\right) & -\cos
   \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi  t}{3}\right) & \cos
   \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi  t}{3}\right) \\
\end{array}
\right), \]
and the resulting state will be $ (R_s\otimes R_{-t})\ket{\beta}$. At this point, Alice and Bob will then output bits $a$ and $b$ corresponding to their measurement outcomes upon measuring the state $ (R_s\otimes R_{-t})\ket{\beta}$.

 Now consider the resulting state in the various cases given inputs $s$ and $t$. In general, the state will be of the form
\[
 (R_s\otimes R_{-t})\ket{\beta}=\left(
\begin{array}{c}
 \frac{\sin \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}+\frac{\cos \left(\frac{\pi  s}{3}\right) \cos
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
 \frac{\sin \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}-\frac{\cos \left(\frac{\pi  s}{3}\right) \sin
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
 \frac{\sin \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}-\frac{\cos \left(\frac{\pi  s}{3}\right) \sin
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
 -\frac{\sin \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}-\frac{\cos \left(\frac{\pi  s}{3}\right) \cos
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
\end{array}
\right)
\]

If $s=t$, it can be seen that the resulting state is

\[
 (R_s\otimes R_{-t})\ket{\beta}=\left(
\begin{array}{c}
 \frac{\sin ^2\left(\frac{\pi  s}{3}\right)}{\sqrt{2}}+\frac{\cos
   ^2\left(\frac{\pi  s}{3}\right)}{\sqrt{2}} \\
 0 \\
 0 \\
 -\frac{\sin ^2\left(\frac{\pi  s}{3}\right)}{\sqrt{2}}-\frac{\cos
   ^2\left(\frac{\pi  s}{3}\right)}{\sqrt{2}} \\
\end{array}
\right)=\ket{\beta}.
\]
Therefore, upon measuring the state of their system, since $\ket{\beta}$ is an entangled state, the measurement outcome will either be $\ket{00}$ or $\ket{11}$ with equal probability. Regardless, they will both output either $a=b=0$ or $a=b=1$ and so their outputs always agree in the case when $s=t$.

In the case when $s\neq t$, the resulting state after Alice and Bob perform their rotations is of the form
\[
(R_s\otimes R_{-t})\ket{\beta}=\left(
\begin{array}{c}
 \frac{\sin \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}+\frac{\cos \left(\frac{\pi  s}{3}\right) \cos
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
 \frac{\sin \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}-\frac{\cos \left(\frac{\pi  s}{3}\right) \sin
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
 \frac{\sin \left(\frac{\pi  s}{3}\right) \cos \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}-\frac{\cos \left(\frac{\pi  s}{3}\right) \sin
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
 -\frac{\sin \left(\frac{\pi  s}{3}\right) \sin \left(\frac{\pi
   t}{3}\right)}{\sqrt{2}}-\frac{\cos \left(\frac{\pi  s}{3}\right) \cos
   \left(\frac{\pi  t}{3}\right)}{\sqrt{2}} \\
\end{array}
\right).
\]
Expressing the six cases when $s\neq t$ more explicitly:
\[\begin{align*}
(R_0\otimes R_{-1})\ket{\beta}&=\frac{1}{2\sqrt{2}}(\ket{00}-\sqrt{3}\ket{01}-\sqrt{3}\ket{10}-\ket{00})\\
(R_0\otimes R_{-2})\ket{\beta}&=\frac{1}{2\sqrt{2}}(-\ket{00}-\sqrt{3}\ket{01}-\sqrt{3}\ket{10}+\ket{00})\\
(R_1\otimes R_{-0})\ket{\beta}&=\frac{1}{2\sqrt{2}}(\ket{00}+\sqrt{3}\ket{01}+\sqrt{3}\ket{10}-\ket{00})\\
(R_1\otimes R_{-2})\ket{\beta}&=\frac{1}{2\sqrt{2}}(\ket{00}-\sqrt{3}\ket{01}-\sqrt{3}\ket{10}-\ket{00})\\
(R_2\otimes R_{-0})\ket{\beta}&=\frac{1}{2\sqrt{2}}(-\ket{00}+\sqrt{3}\ket{01}+\sqrt{3}\ket{10}+\ket{00})\\
(R_2\otimes R_{-1})\ket{\beta}&=\frac{1}{2\sqrt{2}}(\ket{00}+\sqrt{3}\ket{01}+\sqrt{3}\ket{10}-\ket{00}).\\
\end{align*}\]
For either of these cases, the probability that Alice and Bob will both measure either $\ket{00}$ or $\ket{11}$ is given by
\[
Pr(\ket{00} \ or \ \ket{11})=|\bra{00}(R_s\otimes R_{-t})\ket{\beta}|^2+|\bra{11}(R_s\otimes R_{-t})\ket{\beta}|^2=\frac{1}{8}+\frac{1}{8}=\frac{1}{4},
\]
and the probability that they will measure either $\ket{01}$ or $\ket{10}$ is given by
For either of these cases, the probability that Alice and Bob will both measure either $\ket{00}$ or $\ket{11}$ is given by
\[
Pr(\ket{01} \ or \ \ket{10})=|\bra{01}(R_s\otimes R_{-t})\ket{\beta}|^2+|\bra{10}(R_s\otimes R_{-t})\ket{\beta}|^2=\frac{3}{8}+\frac{3}{8}=\frac{3}{4}.
\]

Thus, in any of the cases when $s\neq t$, Alice and Bob will output different bits (either $a=0$ and $b=1$ or $a=1$ and $b=0$) with probability $3/4$, in which case they win the game. However, they lose the game in the case when $s\neq t$, when they both output the same bits (either $a=b=0$ or $a=b=1$) with probability $1/4$.

Entangled vs. Seperable states

If possible, each two-qubit state given below will be expressed as a product of two one-qubit states. Otherwise, it will be shown that such a factorization is impossible (in this case, the qubits are said to be entangled).


Case (i):
$\ket{\psi}=\frac{1}{2}\ket{00}+\frac{i}{2}\ket{01}-\frac{1}{2}\ket{10}-\frac{i}{2}\ket{11}$

Consider the one-qubit states $\ket{\phi_1}=\frac{1}{\sqrt2}\ket{0}-\frac{1}{\sqrt2}\ket{1}$ and $\ket{\phi_2}=\frac{1}{\sqrt2}\ket{0}+\frac{i}{\sqrt2}\ket{1}$. Then expanding the tensor product of the two states $\ket{\phi_1}\otimes\ket{\phi_2}=\ket{\phi_1}\ket{\phi_2}$ yields:
\[\begin{array}{rcl}\ket{\phi_1}\ket{\phi_2}&=&(\frac{1}{\sqrt2}\ket{0}-\frac{1}{\sqrt2}\ket{1})(\frac{1}{\sqrt2}\ket{0}+\frac{i}{\sqrt2}\ket{1}) \\
\\
&=&\frac{1}{\sqrt2}\frac{1}{\sqrt2}\ket{0}\ket{0}+\frac{1}{\sqrt2}\frac{i}{\sqrt2}\ket{0}\ket{1}-\frac{1}{\sqrt2}\frac{1}{\sqrt2}\ket{1}\ket{0}-\frac{1}{\sqrt2}\frac{i}{\sqrt2}\ket{1}\ket{1} \\
\\
&=&\frac{1}{2}\ket{00}+\frac{i}{2}\ket{01}-\frac{1}{2}\ket{10}-\frac{i}{2}\ket{11}.
\end{array}\]
Thus, $\ket{\psi}=\ket{\phi_1}\ket{\phi_2}$ can be expressed as the tensor product of two one-qubit states and is therefore a separable state.


Case (ii):
$\ket{\psi}=\frac{1}{2}\ket{00}+\frac{1}{2}\ket{01}+\frac{1}{2}\ket{10}-\frac{1}{2}\ket{11}$
It will be shown that $\ket{\psi}$ is an entangled state. For the sake of contradiction, suppose that $\ket{\psi}$ can be written as the tensor product of two one-qubit states expressed in their general form $\ket{\phi_1}=\alpha\ket{0}+\beta\ket{1}$ and $\ket{\phi_2}=\alpha'\ket{0}+\beta'\ket{1}$, where $\alpha,\beta,\alpha',\beta'\in\mathbb{C}$. Then,
\[\begin{array}{rcl}
\ket{\psi}&=&\ket{\phi_1}\ket{\phi_2} \\
\\
&=&(\alpha\ket{0}+\beta\ket{1})(\alpha'\ket{0}+\beta'\ket{1}) \\
\\
&=&\alpha\alpha'\ket{00}+\alpha\beta'\ket{01}+\beta\alpha'\ket{10}+\beta\beta'\ket{11}
\end{array}\]
Now compare these amplitudes to those actually provided in $\ket{\psi}$:
\[\alpha\alpha'=\frac{1}{2}, \alpha\beta'=\frac{1}{2}, \beta\beta'=\frac{1}{2}, \beta\beta'=\frac{-1}{2}.\]
Since these products are all nonzero numbers and $\alpha,\beta,\alpha',\beta'\in\mathbb{C}$, this implies that $\alpha,\beta,\alpha',\beta'\neq 0$. Consider the following relationships which must also hold:
\[\begin{array}{rl}
&\alpha\alpha'-\alpha\beta'=\frac{1}{2}-\frac{1}{2}=0\\
\implies&\alpha(\alpha'-\beta')=0 \\
\implies&\alpha'=\beta',
\end{array}\]
and likewise
\[\begin{array}{rl}
&\beta\alpha'+\beta\beta'=\frac{1}{2}+\frac{-1}{2}=0\\
\implies&\beta(\alpha'+\beta')=0 \\
\implies&\alpha'=-\beta',
\end{array}\]
but then $\beta'=-\beta'$ implies that $\beta'=0$ which contradicts the fact that $\beta'\neq0$. Therefore, the two-qubit state $\ket{\psi}$ cannot be expressed as the tensor product of two one-qubit states, and must be entangled.


Case (iii):
$\ket{\psi}=\frac{9}{25}\ket{00}+\frac{12}{25}\ket{01}+\frac{12}{25}\ket{10}-\frac{16}{25}\ket{11}$

Consider the two identical one-qubit states $\ket{\phi_1}=\ket{\phi_2}=\frac{3}{5}\ket{0}+\frac{4}{5}\ket{1}$. Then expanding the tensor product of the two states $\ket{\phi_1}\otimes\ket{\phi_2}=\ket{\phi_1}\ket{\phi_2}$ yields:
\[\begin{array}{rcl}\ket{\phi_1}\ket{\phi_2}&=&(\frac{3}{5}\ket{0}+\frac{4}{5}\ket{1})(\frac{3}{5}\ket{0}+\frac{4}{5}\ket{1}) \\
\\
&=&\frac{3}{5}\frac{3}{5}\ket{0}\ket{0}+\frac{3}{5}\frac{4}{5}\ket{0}\ket{1}+\frac{4}{5}\frac{3}{5}\ket{1}\ket{0}+\frac{4}{5}\frac{4}{5}\ket{1}\ket{1} \\
\\
&=&\frac{9}{25}\ket{00}+\frac{12}{25}\ket{01}+\frac{2}{25}\ket{10}+\frac{16}{25}\ket{11}.
\end{array}\]
Thus, $\ket{\psi}=\ket{\phi_1}\ket{\phi_2}$ can be expressed as the tensor product of two one-qubit states and is therefore a separable state.

Quantum Teleportation

Suppose Alice has a qubit in an arbitrary state \(\left|\psi\right>=\alpha\left|0\right>+\beta\left|1\right>\), and Bob wants to have an identical qubit \(\left|\psi\right>\) too.  If Alice knew enough information about the state \(\left|\psi\right>\) to either completely specify the numbers \(\alpha\) and \(\beta\), then these can be communicated to Bob. This would be worthwhile if Bob knew a unitary operator \(U\) such that, say,  \(U\left|0\right>=\left|\psi\right>\) that allows Bob to prepare the state. If the exact state \(\left|\psi\right>\) is unknown to Alice, then the situation is complicated because there is no amount of measurements Alice could make to the state \(\left|\psi\right>\) that provide sufficient information to uniquely determine the state \(\left|\psi\right>\) since any measurement made to \(\left|\psi\right>\) would only return some computational basis state. If Alice is willing to give up her qubit there is the option of physically sending the qubit to Bob if Bob is spatially displaced. This method entails a safe means of sending \(\left|\psi\right>\) through some quantum channel without any error, which may be rather difficult in practice. If the means of communicating quantum mechanically does not exist than the only alternative Alice and Bob have is to rely on classical means of communication. Fortunately, communicating classical bits is fairly easy to do in practice.

The quantum teleportation protocol allows two parties to send a qubit in an unknown state \(\left|\psi\right>\) without actually physically relocating the original qubit, and only having to communicating two classical bits of information. The protocol relies on the essential ability of the two parties involved to share an entangled Bell state between them.

Suppose Alice possesses an unknown state \(\left|\psi\right>=\alpha\left|0\right>+\beta\left|1\right>\) and shares an entangled pair of qubits in the Bell state \[\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>)\] with Bob. In the quantum teleportation protocol, Alice first performs a Bell measurement on her two qubits of the register. This will yield information of some basis state \(\left|\beta_{ab}\right>\), or equivalently some state \(\left|ab\right>\) in the computational basis. Using the information from what state Alice observes, Alice communicates  two classical bits \(ab\) to Bob who then performs the operation \(Z^aX^b\) on his qubit based off the values of the bits \(ab\)  to turn it into the desired state \(\left|\psi\right>\).

(A circuit for quantum teleportation.  The top two wires represent Alice's register, which begin with  some qubit in an unknown state \(\left|\psi\right>\), and one qubit of a Bell state \(\left|\beta_{00}\right>\) shared with Bob's register represented by the bottom wire. Alice first performs a Bell measurement obtaining a state  \(\left|ab\right>\), and then communicates two classical bits \(ab\) to Bob who performs the operation \(X^bZ^a\) that transforms his qubit to the state \(\left|\psi\right>\).)

In the circuit for the quantum teleportation protocol the top two wires represent Alice's register and the bottom wire represents Bob's register. Note that the operations and measurement that Alice performs constitutes a Bell measurement. At the start of the protocol the state representing the whole system is
\[\left|\psi_0\right>=\left|\psi\right>\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\alpha\left|0\right>+\beta\left|1\right>)(\left|00\right>+\left|11\right>),\]or when expanded
\[\left|\psi_0\right>=\frac{1}{\sqrt2}(\alpha\left|000\right>+\alpha\left|011\right>+\beta\left|100\right>+\beta\left|111\right>),\]
Since the first two qubits in the register represent Alice's system and the third qubit represents Bob's register the states are grouped with kets accordingly in order to make this more explicit in the notation as follows
\[\left|\psi_0\right>=\frac{1}{\sqrt2}(\alpha\left|00\right>\left|0\right>+\alpha\left|01\right>\left|1\right>+\beta\left|10\right>\left|0\right>+\beta\left|11\right>\left|1\right>).\]
Then the state after the controlled-NOT operation where the first qubit serves as the control and the second qubit as the target is
\[\left|\psi_1\right>=\frac{1}{\sqrt2}(\alpha\left|00\right>\left|0\right>+\alpha\left|01\right>\left|1\right>+\beta\left|11\right>\left|0\right>+\beta\left|10\right>\left|1\right>),\]
and after the Hadamard gate is applied to the first qubit the state becomes
\[\left|\psi_2\right>=\frac{1}{2}(\alpha(\left|00\right>+\left|10\right>)\left|0\right>+\alpha(\left|01\right>+\left|11\right>)\left|1\right>+\beta(\left|01\right>-\left|11\right>)\left|0\right>+\beta(\left|00\right>-\left|10\right>)\left|1\right>),\]
which can be regrouped as
\[\left|\psi_2\right>=\frac{1}{2}(\left|00\right>(\alpha\left|0\right>+\beta\left|1\right>)+\left|01\right>(\alpha\left|1\right>+\beta\left|0\right>)+\left|10\right>(\alpha\left|0\right>-\beta\left|1\right>)+\left|11\right>(\alpha\left|1\right>-\beta\left|0\right>)).\]
Examining Bob's register and noticing that
\[\begin{array}{l l}
\alpha\left|0\right>+\beta\left|1\right> = \left|\psi\right>, \\
\alpha\left|1\right>+\beta\left|0\right> = X\left|\psi\right>, \\
\alpha\left|0\right>-\beta\left|1\right> = Z\left|\psi\right>, \\
\alpha\left|1\right>-\beta\left|0\right> = XZ\left|\psi\right>, \\
\end{array}\]
allows the state before measurement to be expressed as
\[\left|\psi_2\right>=\frac{1}{2}(\left|00\right>\left|\psi\right>+\left|01\right>X\left|\psi\right>+\left|10\right>Z\left|\psi\right>+\left|11\right>XZ\left|\psi\right>),\]
or more concisely as
\[\left|\psi_2\right>=\frac{1}{2}\displaystyle\sum\limits_{i,j\in\{0,1\}}\left|ij\right>X^jZ^i\left|\psi\right>.\]
This is an entangled state! Therefore, when Alice measures the first two qubits where some basis state \(\left|ab\right>\) will be observed with equal probability,  Bob's qubit will be left in the state \(X^bZ^a\left|\psi\right>\). When Alice communicates the two bits \(ab\) that result as the outcome of the measurement, Bob can perform the  inverse operation \(Z^aX^b\) on the state \(X^bZ^a\left|\psi\right>\) turning the qubit in his register to the desired state \(Z^aX^bX^bZ^a\left|\psi\right>=\left|\psi\right>\) thereby completing the protocol.

The state of the joint system at the end of the protocol is \(\left|\psi_3\right>=\left|ab\right>\left|\psi\right>\), which leaves Bob's register in the state \(\left|\psi\right>\). Its important to note that as a consequence of this protocol Alice no longer possess the state \(\left|\psi\right>\). The original register that did contain \(\left|\psi\right>\) has turned into the state \(\left|a\right>\) after the measurement was performed. If this were not the case, and the joint system did happen to contain two copies of \(\left|\psi\right>\), then the proposed protocol would in fact contradict the no-cloning theorem. Moreover, it may seem that this teleportation protocol is in violation of the theory of special relativity, which prohibits information to be exchanged faster than the speed of light. In this regard, the protocol is actually void of contradiction because the state \(\left|\psi\right>\) is not effectively transmitted to Bob until after he receives the information of the two bits \(ab\) from Alice. Since these bits are communicated classically, they are indeed constrained from traveling faster than the speed of light avoiding any violation of the principles of special relativity.

The quantum teleportation protocol is another prime example of how entanglement can be used as a resource. In this sense, it can be said that a shared entangled pair and  the exchange of two bits of classical information is equatable to the exchange of one quantum bit of information.

Superdense Coding

Suppose we have two parties, Alice and Bob, and Alice would like to communicate some information to Bob. More specifically, Alice would like to communicate the values of two classical bits \(a\) and \(b\) to Bob. Quite trivially, Alice can do so by simply sending these two bits of information to Bob, but this requires Alice to send two bits to completely specify both. That is, communicating only a single bit is not sufficient information to determine the state of the second bit---at least classically.  Superdense coding is a quantum algorithm or protocol that allows Alice to send two bits of classical information to Bob by only sending a single quantum bit. This is accomplished by exploiting the entanglement that exists in the Bell state \(\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>)\).

In the superdense coding scenario, Alice and Bob share the entangled Bell state \[\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>),\] where Alice posses the first qubit in the register and Bob the second. Depending on which of the four possible combinations of two bits \(ab\in\{00,01,10,11\}\) Alice would like to communicate, Alice performs the operation \(Z^aX^b\) to her qubit of the Bell state and sends her qubit to Bob who then performs a Bell measurement on both of the qubits to get the state \(\left|ab\right>\) encoding the two classical bits Alice originally wanted to communicate. The circuit diagram describing this procedure is presented in the figure below, where the top wire of the circuit represents Alice's qubit and the bottom wire represents Bob's. Before the Bell measurement is performed, it is assumed that Alice has somehow provided Bob with her qubit so that Bob can jointly measure both qubits.


(A circuit for superdense coding. To communicate two classical bits \(ab\) to Bob, Alice applies the operation \(X^bZ^a\) to her qubit (represented by the top wire of the circuit) of the shared bell state \(\left|\beta_{00}\right>\), and then allows Bob to perform a joint Bell measurement on both qubits to yield the state \(\left|ab\right>\) encoding the two classical bits \(ab\).)

To observe why the superdense coding circuit accomplishes what its intended to note that applying \(Z^aX^b\) to \(\left|\beta_{00}\right>\), for the different choices of \(ab\in\{00,01,10,11\}\), transforms \(\left|\beta_{00}\right>\) to the other Bell states.
To see this, let \(Z^aX^b\left|\beta_{00}\right>=\left|\psi_1\right>\), then
\[\left|\psi_1\right>=\frac{1}{\sqrt2}(\overline b\left|00\right>+b\left|01\right>+(-1)^ab\left|10\right>+(-1)^a\overline b\left|11\right>)=\left|\beta_{ab}\right>,\]
where \(\overline b=(b+1)mod2\). Thus, by applying a Bell measurement to the state \(\left|\psi_1\right>=\left|\beta_{ab}\right>\), the state \(\left|ab\right>\) will be observed with certainty.

Superdense coding shows that it is possible to send two bits of classical information by sending only a single qubit. However, this can only be done provided that the qubit being sent is in an entangled state with another qubit. It is also implicit here that one party has the ability to successfully send a qubit to another receiving party. This is an example that shows how entanglement can be used a resource to accomplish something that would otherwise be impossible in the classical realm.

Quantum Entanglement

Although Bell states are systems of two qubits, neither of them can be expressed as the tensor product of any two states in \(\mathcal{H}^2\). That is, for each Bell state \(\left| \beta_{ij}\right>
\neq\left|\psi\right>\otimes \left|\phi\right>\) for any states \(\left|\psi\right>, \left|\phi\right>\in\mathcal{H}^2\). This has the implication of certain correlations that exist when measurements of the individual qubits are made on a Bell state. For instance, consider \[\left|\beta_{00}\right>=\frac{1}{\sqrt2}(\left|00\right>+\left|11\right>),\] and suppose the first qubit in the register is measured. This measurement will yield either \(\left|0\right> \) or \(\left|1\right>\) each with probability \(\frac{1}{2}\). A subsequent measurement on the second qubit in the register would then give a definite state \(\left|0\right>\) or \(\left|1\right>\) with probability \(1\) depending on the state of the first qubit being either \(\left|0\right> \) or \(\left|1\right>\), respectively. However, if it was the second qubit that was observed first, then either \(\left|0\right> \) or \(\left|1\right>\) would be observed with equal probability and the state of the first qubit would be correlated with the second in the same way.

The kind of correlations present in Bell states, where the measurement outcomes of subsystems of a larger composite system are inherently correlated is a property of entangled systems. A state \(\left|\psi\right>\in \mathcal{H}_A\otimes \mathcal{H}_B\) of some composite system is entangled if \(\left|\psi\right>\neq\left|\phi_A\right>\otimes\left|\phi_B\right>\) for any states \(\left|\phi_A\right>\in \mathcal{H}_A\) and \(\left|\phi_B\right>\in\mathcal{H}_B\). Otherwise, if it is possible to express a state \(\left|\psi\right>\in \mathcal{H}_A\otimes \mathcal{H}_B\) as a tensor product of two states \(\left|\psi\right>=\left|\phi_A\right>\otimes\left|\phi_B\right>\), then the state \(\left|\psi\right>\) is called a separable state. The existence of entangled states is a unique feature of quantum mechanics, and such states are exploited in various quantum algorithms and protocols.

Measurements in the Bell Basis

The state of a quantum system is generally represented in terms of the classical states that would result after the system is measured. These classical states have the property that they are  classically distinguishable from one another. The computational basis conventionally serves to index these classical states. The postulates of quantum mechanics allows non-classical states to be defined, and these states can be used to construct an arbitrary basis to describe the states in the system. Therefore, there ought to be a way to make measurements in an arbitrary basis. Abstractly speaking, this can always be performed. In practice however, these basis states must be related to the computational basis states in order to make better physical sense of the measurements.

 In the previous section an explicit circuit was constructed that output the Bell states given the computational basis states as input. This circuit can be thought of as performing the unitary operation \(U=(H\otimes I)(c-X)\). Then the inverse operation of \(U\) given by \(U^\dagger=(c-X)^\dagger(H\otimes I)^\dagger\) is the operation that takes as input some Bell state \(\left| \beta_{ij}\right>\) and outputs the corresponding classical basis state \(\left|ij\right>\). Therefore, in order to perform a measurement in the Bell basis it suffices to apply \(U^\dagger\) to some Bell state \(\left| \beta_{ij}\right>\), measure the system, and use information obtained from the observed state \(\left|ij\right>\) to infer \(\left| \beta_{ij}\right>\). This procedure can be generalized to make measurements in an arbitrary basis provided with a circuit that can implement the transformation of the classical basis to the desired one.

 In the figure below, a circuit for performing a measurement in the Bell basis is given along with the Bell measurement circuit to be interpreted as an equivalent circuit. Notice that this circuit before the measurements are performed is precisely the circuit that results from reversing the circuit for the construction of Bell sates. To construct the actual inverse of an arbitrary circuit not containing any measurements, simply reverse the order of the gates and replace each with its conjugate transpose. In the case of the circuit for measuring in the Bell basis, since \(H^\dagger=H\) and \(c-X^\dagger=c-X\), the circuit also happens to be the same circuit that results from merely reversing the order of the gates.

(A circuit for measuring in the Bell basis. This circuit takes some Bell state \(\left|\beta_{ij}\right>\) as input and applies a controlled\(-NOT\) gate \(c-X\) followed and a Hadamard gate \(H\) followed by a measurement to each each register to yield the corresponding computational basis state \(\left|ij\right>\). Shown below is the Bell measurement gate, which will be used as a shorthand symbol that represents the circuit shown above.)

Bell States

Vector spaces generally admit many choices for a suitable basis to represent its elements. For a Hilbert space describing a register of qubits \(\mathcal{H}= \bigotimes_{i=1}^n\mathcal{H}^2\)  the  canonical choice of basis is the computational basis given by the states expressed in the decimal representation as \(\left|0\right>, \left|1\right>, . . . ,\left|N-1\right>\), where \(N=2^n\), or equivalently, in the binary representation as states of the form \(\left|b_1b_2...b_n\right>\) for \(b_i\in\{0,1\}^n\).

A particularly important set of basis states for the Hilbert space \(\mathcal{H}^2\otimes\mathcal{H}^2\) of two qubits consists of the \(4\) Bell states. The Bell states can be defined in terms of the computational basis states \(\left|0\right>\) and \(\left|1\right>\) as
\[\left| \beta_{xy}\right>\equiv \frac{1}{\sqrt2}(\left|0y\right>+(-1)^x\left|1\overline y\right>),\]
where here \(\overline y=(y+1)mod2\) denotes the NOT operation which simply flips the state of the qubit to the opposite value. Therefore, this change of basis maps the computational basis as follows
\[ \begin{array}{r l}
\left|00\right> &\mapsto \frac{1}{\sqrt2}(\left|00\right>+\left|11\right>)=\left| \beta_{00}\right>, \\
 \left|01\right> &\mapsto \frac{1}{\sqrt2}(\left|01\right>+\left|10\right>)=\left| \beta_{01}\right>,  \\
 \left|10\right> &\mapsto \frac{1}{\sqrt2}(\left|00\right>-\left|11\right>)=\left| \beta_{10}\right>, \\
 \left|11\right> &\mapsto \frac{1}{\sqrt2}(\left|01\right>-\left|10\right>)=\left| \beta_{11}\right> 
 \end{array}\]
An explicit circuit implementing this transformation of basis states using a single Hadamard gate followed by a controlled\(-NOT\) gate, or \(c_X\), is given in the figure shown below

(A circuit for creating Bell states. A change of basis from the computational basis states \(\left|ij\right>\) to the Bell basis \(\left|\beta_{ij}\right>\) is accomplished using a single Hadamard gate \(H\) followed by a controlled\(-NOT\) gate \(c-X\).)