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A pure state is separable (unentangled) if it can be written as a tensor product of states of each qubit.

A mixed state is separable, if it can be written as a probability distribution over separable pure states.

Consider the mixed state which is $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ with the probability of $\frac{1}{2}$, and $\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$ with the probability of $\frac{1}{2}$.

How to show that this mixed state is separable?

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  • $\begingroup$ For the two qubit mixed state that you have given, this may be of interest to you en.wikipedia.org/wiki/Peres%E2%80%93Horodecki_criterion. Essentially, if the approach described there is taken, and negative eigenvalues are obtained, then it is not separable. $\endgroup$ Commented Jan 4, 2021 at 14:52

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Begin by writing down the density matrix

$$ \begin{align} \rho_{AB} &= \frac{1}{4}\left[(|00\rangle+|11\rangle)(\langle00|+\langle11|)\right] + \frac{1}{4}\left[(|01\rangle+|10\rangle)(\langle01|+\langle0|)\right] \\ &=\frac{1}{4}\left(\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\right) \\ &= \frac{1}{4}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}. \end{align} $$

At this stage we already see that $\rho_{AB}$ satisfies the Peres–Horodecki criterion: if we transpose all four $2\times 2$ blocks then we obtain

$$ (I \otimes T)\rho_{AB} = \frac{1}{4}\begin{pmatrix} \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}^T & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^T \\ \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}^T & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}^T \\ \end{pmatrix} = \frac{1}{4}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} = \rho_{AB} $$

which is a state. In case of two qubit (and qubit-qutrit) pairs the criterion is both necessary and sufficient, so $\rho_{AB}$ is separable.

Note that for larger systems, the criterion is necessary, but not sufficient.


In this case, it is also not too difficult to see how $\rho_{AB}$ may be expressed as a convex combination of product states

$$ \begin{align} \rho_{AB} &= \frac{1}{4}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \\ &= \frac{1}{8}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} + \frac{1}{8}\begin{pmatrix} 1 & -1 & -1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & 1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{pmatrix} \\ &= \frac{1}{2}\left(|+\rangle\langle+| \otimes |+\rangle\langle+| \,+\, |-\rangle\langle-| \otimes |-\rangle\langle-|\right). \end{align} $$

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