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I should show that the density operator $\rho \in \text{Herm}(\mathbb C^d)$ is positive semi-definite if and only if $\text{Tr}[\rho A^\dagger A] \geq 0 \quad \forall A\in L(\mathbb C^d)$.

I don't know how to begin to proof this. I think I'm missing some properties of Traces. The only thing I notice is that the operator $A A^\dagger$ is positive semi-definite and hermitian. I hope some one can give me a hint on this.

Edit:

For the case that this expression leads to a positive semi-definite $\rho$ I have:

\begin{align} \text{Tr}(\rho A^\dagger A) &= \sum\limits_n \sum\limits_i p_i \langle n \left| \right. \psi_i \rangle \langle \psi_i \left| \right. A^\dagger A \left|\right. n \rangle \\ &= \sum\limits_n \sum\limits_i p_i \langle \psi_i \left| \right. A^\dagger A \left|\right. n \rangle \langle n \left| \right. \psi_i \rangle \\ &= \sum\limits_i p_i \langle \psi_i \left| \right. A^\dagger A \left| \right. \psi_i \rangle \geq 0 \end{align}

Now one knows that the expectation value of a positive semidefinite operator is positive, so it follows that $\sum_i p_i$ has to be positive and in general this is only fulfilled when all $p_i$ are positive. Is this solution right?

But how do you show the "only if" part?

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  • $\begingroup$ Use a generic one-dimensional projector for $A$... $\endgroup$ – Valter Moretti Nov 11 '20 at 12:19
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An operator $\rho$ is positive, i.e., $\rho \geq 0$ iff

  1. $\rho = |Q|^2 = Q^\ast Q$ for some $Q$ iff
  2. $\langle\psi,\rho\psi\rangle \geq 0$ for any vector $\psi$ iff
  3. $\rho$ is self-adjoint and has spectrum in $\left[0,\infty\right)$.

I assume the equivalence of these three conditions is understood.

For your question:

Assume $\rho$ obeys the following constraint: For any other operator $A$, $\mathrm{tr}(\rho|A|^2)\geq 0$. This implies condition 2. Indeed, let $\psi$ be a given vector and define $A := \psi\otimes\psi^\ast$. Then $\mathrm{tr}(\rho|A|^2) = \langle\psi,\rho\psi\rangle \geq 0$.

Conversely, assume condition 1. and let $A$ be any operator. We want to show that $\mathrm{tr}(\rho|A|^2)\geq 0$. We have $\mathrm{tr}(\rho|A|^2) =\mathrm{tr}(|Q|^2|A|^2) = \mathrm{tr}(Q^\ast Q A^\ast A) = \mathrm{tr}( Q A^\ast A Q^\ast) = \mathrm{tr}( |AQ^\ast|^2) $. But the trace of a positive operator is of course positive, so we are finished.

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