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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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  • It should be noted that the density operator $\rho$ (which you call the state operator) of a pure state is not an element of the physical Hilbert space as such, only the state |ψ⟩ is an element of the Hilbert space. – NDewolf Nov 07 '19 at 15:25
  • That's why I tried to be consistent in avoiding the term "state" (except in the first two sentences) and only talk about state operators (i.e. density operators) and state vectors – Übend Nov 07 '19 at 15:37
  • There is a way to represent also mixed states as unit vectors in a suitable Hilbert space, see my answer... – Valter Moretti Nov 09 '19 at 21:52

3 Answers3

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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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  • Oh, neat. B(H) here is Banach space? – PhysicsTeacher Nov 09 '19 at 22:45
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    $B(H)$ is the algebra of all bounded opeartors $A : H \to H$. Yes it is a Banach space, more precisely a Banach algebra which is also a $C^*$-algebra. – Valter Moretti Nov 09 '19 at 22:46
  • Okay I completely agree with your answer. However with mine I just tried to make clear that there is not sensible Hilbert space which contained only the pure and mixed states. On the other hand it would be interesting to find an interpretation of the vectors which do not belong the the norm-1 HS-operator subset. – NDewolf Nov 10 '19 at 11:05
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    Non-normalized Hilbert Schmidt operators can be simply normalized so that hey represent mixed states as well, exactly as non-normalized vectors represent states when normalized. From this viewpoint the situation is symmetric. – Valter Moretti Nov 10 '19 at 11:13
  • True, by going to the projectivization one would indeed obtain a similar framework as in ordinary quantum mechanics. – NDewolf Nov 10 '19 at 18:59
  • so, if I'm reading this correctly, you are essentially mapping each state $\rho$ in its purification, right? – glS Aug 16 '20 at 18:02
  • "purification" means several things in several different contexts. It depends on your notion of "purification"... – Valter Moretti Aug 16 '20 at 18:40
  • right, sorry. I mean that a vector state $\lvert\psi\rangle\in\mathcal H_1\otimes\mathcal H_2$ is a purification of a density matrix $\rho\in\mathrm{Lin}(\mathcal H_1)$ if $\operatorname{Tr}_2[\lvert\psi\rangle!\langle\psi\rvert]=\rho$. Given a state $\rho$, you can write the corresponding set of purifications as $\text{vec}(\sqrt\rho U)$ for any unitary $U$ (more generally, $U=V^\dagger$ with $V$ an isometry), where $\text{vec}$ denotes the vectorisation operation (I'm only considering finite-dimensional spaces here, I don't know how these things generalise to infinite-dimensional ones). – glS Aug 16 '20 at 18:45
  • I see. Indeed, everything you wrote extends to the infinite dimensional case. Referring to your definition of purification I can say that my formalism above does not coincide with a map associating a state with its purification. That is because the procedure I use is intrinsic whereas the purification needs to choose some arbitray mathematical objects as the second Hilbert space. – Valter Moretti Aug 16 '20 at 18:56
  • well, if I'm reading it correctly, you are essentially mapping $\rho\mapsto\sqrt\rho$, with an added (negligible) phase. Modulo a few modifications in how you have the handle the objects after the transformation, that is essentially the same as taking a specific purification of $\rho$, with the choice $U=I$ and thinking of the result as an operator rather than as a vector. What is the inner product in the set of purifications becomes the Hilbert-Schmidt inner product for $\sqrt\rho$ etc. And the same catch applies: the mapping is not unique: your $S$ is also defined up to a partial isometry. – glS Aug 16 '20 at 19:03
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    Yes, I agree: it is equivalent to use a specific (actually still arbitrary) purification. – Valter Moretti Aug 16 '20 at 19:11
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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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  • I disagree on the fact that non-pure states oprators are not vectors in a Hilbert space. If $T$ is a statistical operator, it is trace class an trace-class opeartors are Hilbert Schmidt. So $T$, not only $S$ (with my notation), is a vector of the Hilbert space $H_{HS}$. What it is true is that this Hilbert space does non coincide with the standard Hilbert space of the pure states of the quantum system. – Valter Moretti Nov 10 '19 at 09:27
  • Yes, but that's taking an advanced approach of thinking of abstract Hilbert spaces. I wanted to add a simpler-explanation in terms of the original ("physical") Hilbert space and emphasizing the difference between representing a state as a vector in it or operator on it. – PhysicsTeacher Nov 10 '19 at 11:04
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    Yes I did not want to criticize your answer, it was just to clarify a point. I know that this matter is technically subtle...and this is just the tip of an iceberg... For instance, the distinction between pure and properly mixed states in terms of vectors and statistical operators is a bit misleading as it strongly depends on the fact that the von Neumann algebra of observables coincides with $B(H)$. But it is false in presence of superselection rules or a gauge group... – Valter Moretti Nov 10 '19 at 11:15