# When does Hermitian operator with unit trace become a density operator?

The definition of density operators is that (i) positive semidefinite; and (ii) unit trace. Given a Hermitian matrix $$\rho$$ (say, the size is $$2\times 2$$) with unit trace, I know that such matrix may not be positive semidefinite. However, I find some lecture notes on the internet states that such $$\rho$$ is a density operator iff the bloch vector $$\textbf{r}$$ of $$\rho$$ has $$\ell_2$$ norm $$\leq 1$$. How to prove that?

My attempt: $$\text{Tr}(\rho^2)=\frac{1}{2^2}\left((I+\textbf{r}\cdot E)(I+\textbf{r}\cdot E)\right)=\frac{1}{2}(1+\|\textbf{r}\|^2)$$. But I can't go further. (I'm quite unfamiliar with the trace properties in linear algebra)

• A Hermitian operator is PSD iff its eigenvalues are nonnegative. May 23 at 15:31

Let $$\mathbf{r} = (r_x,r_y,r_z)$$. Then the eigenvalues of your state are $$\frac12\left(1 \pm \sqrt{r_x^2 + r_y^2 + r_z^2}\right).$$ For $$\rho$$ to be PSD we need both eigenvalues to be nonnegative and this is satisfied iff $$\|\textbf{r}\| = \sqrt{r_x^2 + r_y^2 + r_z^2} \leq 1$$.
• Thanks for the help!! To see the eigenvalues of $\rho$, the simplest way is to compute the characteristic polynomial of that matrix $\rho$ right? (Or is there a more clever way that I am ignored?) May 23 at 16:34