# Show that the density operator is positive semidefinite

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?

• Use a generic one-dimensional projector for $A$... – Valter Moretti Nov 11 '20 at 12:19

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.

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.