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Def: The partial transpose of a linear operator $\rho_{AB}$ over a Hilbert space $H_A \otimes H_B$ w.r.t A is defined for a linear operator $\rho_{AB}=\rho_A \otimes\rho_B$ as $\rho^{T_A}_{AB}=\rho_A^T \otimes\rho_B$ The definition can be extended to a general linear operator

I want to prove the Partial transpose test: If $\rho_{AB}$ is a separable (unentangled) then the partial transpose w.r.t to A is PSD (positive semidefinite)

My try:

My definition of PSD is that a hermitian operator is PSD if it has non negative eigenvalues

Assume that $\rho_{AB}=\rho_A \otimes\rho_B$. Then since $\rho_A^T$ and $\rho_A$ have the same eigenvalues, then they both have non-negative eigenvalues. I think I can say they are both PSD but they still have to be hermitian for that and it looks like the transpose is not hermitian: $(\rho_A^T)^\dagger=\overline{(\rho_A^T)^T}=\overline{\rho_A}$.

Furthermore how do I conclude that $\rho^{T_A}_{AB}=\rho_A^T \otimes\rho_B$ is hermitian and with nonnegative eigenvalues? I don't know what the eigenvalues of a tensor product are in terms of the eigenvalues of the initial spaces.

Finally I have to extend this to the general case. If $\rho_{AB}$ is a generic separable state then it a convex linear combination of product states: $\rho^{T_A}_{AB}=\sum_{i=1}^d c_i \rho_{A_i}^T \otimes\rho_{B_i}$ And then the eigenvalues are the sum of the eigenvalues of the addends, so they are non negative .To prove it is PSD again I need to prove that it is hermitian but I guess that if the addends $\rho_A^T \otimes\rho_B$ are prove to be hermitian, then a convex linear combination of hermitian matrices is hermitian

How do I complete or fix the proof?

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1 Answer 1

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The transpose of a Hermitian matrix is again Hermitian, and your attempt got quite close. All you have to do is, essentially, the same calculation again: $$ (\rho^T)^\dagger=\overline{(\rho^T)^T}=\overline{\rho}=\overline{\rho^\dagger}=\overline{\overline{\rho^T}}=\rho^T $$ (In the third step we used that $\rho$ is Hermitian)

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