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Let $\rho \in \mathfrak{L}(A)$ and $\sigma \in \mathfrak{L}(A)$. Now I want to know can we say that:

$\rho \geqslant 0 $ and $\sigma \geqslant 0 \quad \quad $ if and only if $\quad \quad Tr(\rho \sigma) \geqslant 0$

$\rho \geqslant 0 $ and $\sigma \geqslant 0 $ means that both of them are positive semi-definite operator and $Tr$ means trace. $A$ is a Hilbert space and $\mathfrak{L}(A)$ means all linear maps on $A$.

If the above theorem is true could you please help me how can I show it?

299792458
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1 Answers1

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Implication

$$\quad \quad Tr(\rho \sigma) \ge 0 \ \implies \ \rho \succcurlyeq 0 \ \text{and} \ \sigma \succcurlyeq 0$$

cannot hold; here is a counterexample:

$$\rho=I \ \text{and} \ \sigma=\begin{pmatrix}1&2\\2&0\end{pmatrix}.$$

We have $Tr(\rho \sigma)=1 > 0$, but $\sigma$ not positive definite.

For the other implication, see here.

Jean Marie
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  • You're right, But can we say anything about Trace of the product of two operators greater than or equal to zero and each of the operators? And was the back direction of my statement true? – 299792458 Oct 31 '20 at 11:17
  • See the reference I just added. – Jean Marie Oct 31 '20 at 11:22
  • Thank you so much. Can I ask one more question? If $Tr(AB) \geqslant 0$ and $B \geqslant 0$, then can we say that $A \geqslant 0$? If yes could you pleas give me a hint to show it? – 299792458 Oct 31 '20 at 12:05
  • In fact the counterexample I gave, taken with $A=\begin{pmatrix}1&2\2&0\end{pmatrix}$ and $B=I$ show we can not conclude that $A \succcurlyeq 0$. – Jean Marie Oct 31 '20 at 13:09