\leq i,j \leq d_A \\ 1 \leq \mu, \nu \leq d_B}} a_{ij}^{\mu\nu} (|j\rangle\langle i|)_A \otimes (|\mu\rangle\langle \nu|)_B.$$

A similar definition exists for the partial transposition of $\rho_{AB}$ with respect to Bob’s subsystem. Notice that $\rho_{AB}^{T_{B}}\ =(\rho_{AB}^{T_{A}})^{T}$. Although the partial transposition depends upon the choice of the basis in which $\rho_{AB}$ is written, its eigenvalues are basis independent. We say that a state has positive partial transposition (PPT) , whenever $\rho_{AB}^{T_A} \geq 0$, i.e. the eigenvalues of $\rho_{AB}^{T_{A}}$ are non-negative. Otherwise, the state is said to be non-positive under partial transposition (NPT).

**Theorem 2** [@PeresPPT]\
*If a state $\rho_{AB}$ is separable, then $\rho_{AB}^{T_{A}}\ \ge\ 0$ and $\rho_{AB}^{T_{B}}\ =\left(\rho_{AB}^{T_{A}}\right)^{T}\ \ge\ 0$.*

Proof:\
Since $\rho_{AB}$ is separable, it can be written as $$\begin{aligned}
\rho_{AB} &
 = \sum_{i=1}^{K}\ p_i |e_i \rangle \langle e_i |\otimes |f_i\rangle \langle f_i | \ge 0.\end{aligned}$$ Now performing the partial transposition w.r.t. A, we have $$\begin{aligned}
\rho_{AB}^{T_{A}} &=& \sum_{i=1}^{K}\ p_i\left(|e_i\rangle \langle e_i |
\right)^{T_{A}}\otimes |f_i \rangle \langle f_i |\nonumber\\
&=& \sum_{i=1}^{K}\ p_i |e_i^*\rangle \langle e_i^*| \otimes | f_i\rangle \langle f_i | \ge 0.\end{aligned}$$ Note that in the second line, we have used the fact that $A^{\dagger}=\left(A^*\right)^{T}$. $\square$

The *partial transposition criterion*, for detecting entanglement is simple: Given a bipartite state 