# If $\rho_{AB}$ is a separable then the partial transpose w.r.t to A is PSD

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

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

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.

• 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. Mar 4 at 20:52
• If $U$ and $V$ diagonalise $\rho_A$ and $\rho_B$, then $U\otimes V$ diagonalises $\rho_A\otimes\rho_B$. Mar 5 at 8:46