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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?


This question has been crossposted on MathSE

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  • $\begingroup$ please mark cross-posts as such $\endgroup$
    – glS
    Mar 4 at 11:43

1 Answer 1

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The main result that you need to complete all the steps that you mention is that if $\rho$ is a density matrix, then $\rho^T$ is also a density matrix.

So, what are the key properties of a density matrix?

  • trace 1
  • Hermitian
  • positive semi-definite

Since trace only applies to the diagonal elements, which are unchanged under transpose, the trace is clearly not changed. Thus, since $\text{tr}(\rho)=1$, $\text{Tr}(\rho^T)=1$.

Now the Hermitian conjugate: $\rho={\rho^\star}^T$. From this, it follows that $\rho^T=\rho^\star=(\rho^T)^\dagger$. So it's Hermitian.

Positive semi-definite. Since $\rho$ is Hermitian, there exists a unitary $U$ such that $\rho=UDU^\dagger$ with $D$ diagonal and positive semi-definite. Thus, $$ \rho^T=U^\star DU^T. $$ $U^\star$ is still a unitary ($U^\star U^T=(UU^\dagger)^\star=I$), and hence $\rho^T$ has the same eigenvalues as $\rho$.

So, now you know that $(\rho_A\otimes \rho_B)^{T_A}=\rho_A^T\otimes\rho_B$ is a separable density matrix. This shows that any $\rho^{T_A}$ for which $\rho$ is separable also has the separable structure.

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  • $\begingroup$ I also need that the tensor product of density matrices is a density matrix, how do I see that? I don't know how to relate eingevalues of matrices with eigenvalues of the tensor product. $\endgroup$ Mar 4 at 20:52
  • $\begingroup$ If $U$ and $V$ diagonalise $\rho_A$ and $\rho_B$, then $U\otimes V$ diagonalises $\rho_A\otimes\rho_B$. $\endgroup$
    – DaftWullie
    Mar 5 at 8:46

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