Is the set of density matrices on a $d$-dimensional Hilbert space compact?

Question

The set of the density matrices is a set with elements $\rho$ satisfying the following two conditions: $0\preceq \rho $ and $\mathrm{tr}\left( \rho \right) =1$. I wonder whether the set is compact?

Answer 1

Yes, it is compact (see however this for a discussion of the infinite-dimensional case).

Consider a finite-dimensional complex Hilbert space $H$ and the space of all (bounded) linear operators on $H$, denoted by $\mathcal B(H)$. With the inner product defined by $\langle A,B\rangle:=\mathrm{Tr}A^\dagger B$ for $A,B\in \mathcal B(H)$, this is a (finite-dimensional, complex) Hilbert space.

The (sub-)set of density matrices is $$\mathcal S(H):=\left\{\rho\in \mathcal B(H)\,|\, \mathrm{Tr}\rho=1, \, \rho \geq 0\right\} \tag 1\quad .$$

Since $\mathcal S(H)$ is a subset of a finite-dimensional Hilbert space, it suffices to show that it is closed and bounded to prove compactness.

To show that $\mathcal S(H)$ is closed, consider a convergent sequence of density operators $(\rho_n)_{n\in \mathbb N}$ with limit $\rho \in \mathcal B(H)$. We have to show that $\rho$ is a density operator. To this end, recall that inner products are continuous and thus $$0\leq \mathrm{Tr}\rho_nA \to \mathrm{Tr}\rho A\quad ,\tag 2$$

for all positive semi-definite $A\in \mathcal B(H)$, which implies that $\rho \geq 0$. Choosing $A=\mathbb I_H$, the identity operator on $H$, also shows that $\mathrm{Tr}\rho=1$, so $\rho\in\mathcal S(H)$.

Boundedness of $\mathcal S(H)$ in turn follows from the fact that for all $\rho \in S(H)$ it holds that

$$\mathrm{Tr}\rho^2 \leq 1 \tag 3\quad .$$

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