A mixed state is mathematically represented by a *bounded*, *positive* *trace-class* operator with *unit trace* $\rho : \cal H \to \cal H$.
Here $\cal H$ denotes the complex Hilbert space of the system (it may be nonseparable). The set of mixed states $S(\cal H)$ is a **convex body** in the complex linear space of trace class operators $B_1(\cal H)$ which is a two-side $*$-ideal of the $C^*$-algebra of bounded operators $B(\cal H)$.

*Convex* means that if $\rho_1,\rho_2 \in S(\cal H)$ then a **convex combination** of them, i.e. $p\rho_1 + q\rho_2$ if $p,q\in [0,1]$ with $p+q=1$, satisfies $p\rho_1 + q\rho_2 \in S(\cal H)$.

*two-side $*$-ideal* means that linear combinations of elements of $B_1(\cal H)$
belong to that space (the set is a subspace), the adjoint of an element of $B_1(\cal H)$ stays in that space as well and $AB, BA \in B_1(\cal H)$ if $A\in B_1(\cal H)$ and $B \in B(\cal H)$.

I stress that, instead, the subset of states $S(\cal H)\subset B_1(\cal H)$ is **not** a vector space since only convex combinations are allowed therein.

The **extremal** elements of $S(\cal H)$, namely the elements which cannot be decomposed as a nontrivial convex combinations of other elements, are all of the pure states. They are of the form $|\psi \rangle \langle \psi|$ for some unit vector of $\cal H$. (Notice that, since phases are physically irrelevant the operators $|\psi \rangle \langle \psi|$ biunivocally determine the pure states, i.e. $|\psi\rangle$ up to a phase.)

The space $B_1(\cal H)$ and thus the set $S(\cal H)$ admits at least three relevant normed topologies induced by corresponding norms. One is the standard operator norm $||T||= \sup_{||x||=1}||Tx||$ and the remaining ones are:
$$||T||_1 = || \sqrt{T^*T} ||\qquad \mbox{the trace norm}$$
$$||T||_2 = \sqrt{||T^*T||} \qquad \mbox{the Hilbert-Schmidt norm}\:.$$
It is possible to prove that:
$$||T|| \leq ||T||_2 \leq ||T||_1 \quad \mbox{if $T\in B_1(\cal H)$.}$$
Moreover, it turns out that $B_1(\cal H)$ is a Banach space with respect to $||\cdot||_1$ (it is not closed with respect the other two topologies, in particular, the closure with respect to $||\cdot||$ coincides to the ideal of compact operators $B_\infty(\cal H)$).

As
$S(\cal H)$ is closed with respect to $||\cdot ||_1$, it is a **complete metric space** with respect to the distance $d_1(\rho,\rho'):= ||\rho-\rho'||_1$.
When $dim(\cal H)$ is finite the three topologies coincide (though the norms do not), as a general result on finite dimensional Banach spaces.

Concerning your last question, there are many viewpoints. My opinion is that a density matrix is **physical** exactly as pure states are. It is disputable whether or not a mixed state encompasses a sort of physical ignorance, since there is no way to distinguish between "classical probability" and "quantum probability" in a quantum mixture as soon as the mixture is created. See my question Classical and quantum probabilities in density matrices and, in particular Luboš Motl's answer.
See also my answer to Why is the application of probability in QM fundamentally different from application of probability in other areas?

**ADDENDUM**. In finite dimension, barring the trivial case $dim({\cal H})=2$ where the structure of the space of the states is pictured by the **Poincaré-Bloch ball** as a **manifold with boundary**, $S(\cal H)$ has a structure which generalizes that of a manifold with boundary. A **stratified space**. Roughly speaking, it is not a manifold but is the union of (Riemannian) manifolds with different dimension (depending on the range of the operators) and the intersections are not smooth. When the dimension of $\cal H$ is infinite, one should deal with the notion of infinite dimensional manifold and things become much more complicated.

This post imported from StackExchange Physics at 2014-06-25 21:05 (UCT), posted by SE-user V. Moretti