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Tuesday, 15 October 2013

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.

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