## 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.