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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)

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    $\begingroup$ A Hermitian operator is PSD iff its eigenvalues are nonnegative. $\endgroup$
    – Rammus
    May 23 at 15:31
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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 $\|\textbf{r}\| = \sqrt{r_x^2 + r_y^2 + r_z^2} \leq 1$.

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  • $\begingroup$ 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?) $\endgroup$
    – Shara
    May 23 at 16:34
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    $\begingroup$ Yes they're the zeros of the characteristic polynomial. $\endgroup$
    – Rammus
    May 23 at 16:45

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