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I understand that states are associated with a state operator with certain mathematical properties. There is a subset of states, called pure states, that are represented by a state operator of the form $\rho = \lvert \psi \rangle \langle \psi \rvert$ where $\lvert \psi \rangle$ is a pure state vector. So the set of all state operators is a superset of the set of pure state operators.

These pure state vectors are elements of the Hilbert Space. And every element of the Hilbert space is a pure state vector.

Does the conclusion hold – in this use of terms – that there is no thing like a non-pure state vector that can be understood as an element of some superset of the Hilbert space?

If not, what is this superset?

glS
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Übend
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There is actually a Hilbert space where mixed states (pure and non-pure) are represented by unit vectors up to phases. This Hilbert space is not the same Hilbert space where pure states are represented by all unit vectors (up to phases). However it includes the set of pure states (the vectors of the initial Hilbert space up to multiplication with scalars) as a proper subset.

Every statistical operator $T$, i.e., a positive, trace-class, unit-trace operator representing a mixed state, over the Hilbert space $H$ can be written as $$T= S^\dagger S\tag{1}$$ where $S= e^{ia} \sqrt{T}$ and $a$ an arbitrary real number. (Actually $S$ is determined from $T$ up to a partial isometry with initial space given by the range of $\sqrt{T}$ as follows from the polar decomposition theorem.)

As $T$ is trace class, $S$ is Hilbert-Schmidt. These operators form another Hilbert space $H_{HS}\subset B(H)$ (the latter is the full algebra of all bounded operators $A: H \to H$) in their own right, whose scalar product is $$\langle S, S'\rangle := tr(S^\dagger S')$$ Every element of $T\in H_{HS}$ with unit HS norm defines a statistical operator as $T=S^\dagger S$.

If $A\in B(H)$ is a bounded observable (selfadjoint operator in $H$) and the mixed state $T$ is represented by the vector $S$ of $H_{HS}$, namely a Hilbert-Schmidt operator, the expectation value $$tr(TA)$$ can be re-written using the Hilbert-Schmidt scalar product. In fact, (1), the fact that $SA$ is Hilbert-Schmidt is $S$ is Hilbert-Schmidt, and $A$ is bounded, and the cyclic property of the trace imply $$tr(TA) = \langle S, A S\rangle\:.$$

Every pure state is represented by a one-dimensional orthogonal projector $T$. These operators satisfy $TT=T$ and are Hilbert-Schmidt, so we can set $S:=T$.

Valter Moretti
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So AFAIK there does not exist a straightforward generalization of the Hilbert space to mixed (nonpure) states. The density matrices only form a convex subset of the space of linear operators and hence do not even form a vector space themselves. Hence it is not possible to get a Hilbert space out of this.

The following post might be interesting to you: Density matrix formalism

NDewolf
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Allow me to add a lower-level explanation:

One can represent the pure states by vectors in the Hilbert space.

Alternatively, one can represent states by density operators. These include pure states, as a subset. These are operators on the Hilbert space, however, not vectors in it.

So we don't have a "non-pure state vector in Hilbert space", instead we have a "non-pure state operator in the [density-operator] set", which is not a vector space and certainly not the original Hilbert space we started with.

PhysicsTeacher
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