The whole question is in the title. This seems to be the case based on what I've read, but I don't know if I've ever seen this stated explicitly.
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1You might need to define what a "physical state" is. After all, some rays might correspond to zero physical states, because the state they correspond to is unphysical (possible example: momentum eigenstates, which have infinite extent). – probably_someone Jan 30 '20 at 03:12
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@probably_someone Momentum eigenstates aren't actually elements of the Hilbert space, because they aren't normalizable. – tparker Feb 06 '20 at 01:35
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@tparker From Shankar: The first postulate states that a particle is described by a ket $\left|\Psi\right>$ in a Hilbert space which, you will recall, contains proper vectors normalizable to unity as well as improper vectors, normalizable only to the Dirac delta functions. – Feb 10 '20 at 18:22
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@PiKindOfGuy Mathematically, the Hilbert space does not and can not contain the vectors that Shankar calls improper. If one identifies the Hilbert space with $L^2(\mathbb{R}^d)$ (as it is usual in QM with finitely many degrees of freedom), then it is well known that $L^2(\mathbb{R}^d)\subset \mathscr{S}'(\mathbb{R}^d)$, the space of tempered distributions (and $\delta\in \mathscr{S}'\setminus L^2$). However, the topology on tempered distributions differs completely from the Hilbert topology on square-integrable functions, and "improper vectors" are essentially useless in concrete situations. – yuggib Feb 11 '20 at 09:17
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Let me be clearer: it is not possible to consistently define an inner product on tempered distributions, or on any subset of those containing $\delta$ distributions, and thus it is not possible to have an Hilbert space containing the "improper vectors". Therefore, if the quote of Shankar is truly literal, it is also mistaken. – yuggib Feb 11 '20 at 09:23
3 Answers
Short answer: No.
Long answer: The concept of state "pre-exists" the concept of Hilbert space, in the following sense. The starting point for the mathematical description of a physical system is the collection of observables, and such collection forms a C*-algebra.
States are continuous functionals of the C*-algebra, associating to each observable its expected value in that particular system's configuration.
Now, every C*-algebra can be represented as an algebra of (bounded) operators acting on a Hilbert space. Such representations can be constructed starting from each state $\omega$ by a procedure called GNS conctruction. In the GNS construction built on $\omega$, the latter is identified with a ray $[\psi_\omega]$ in the corresponding Hilbert space $\mathscr{H}_\omega$. Every other ray $[\psi]$ also defines a (unique) state for the C*-algebra, as well as density matrices of the form $$\sum_{i\in\mathbb{N}} \rho_i \lvert \psi_i\rangle\langle \psi_i\rvert$$ with $\sum_{i\in\mathbb{N}} \rho_i=1$.
However, the set of all Hilbert rays on $\mathscr{H}_\omega$ plus the corresponding density matrices does not exaust in general all possible physical states associated to a given C*-algebra of observables. In fact, there may be states $\tilde{\omega}$ that are disjoint from $\omega$, i.e. that cannot be written as rays or density matrices in $\mathscr{H}_{\omega}$ (and then vice-versa, $\omega$ cannot be written as a ray or a density matrix in $\mathscr{H}_{\tilde{\omega}}$).
Examples of disjoint states that are extremely relevant in QFT are:
Two ground states (or vacuum states) corresponding to free scalar or spin one-half theories with different mass (disjoint with respect to the algebra of Canonical (Anti)Commutation Relations for time-zero fields);
The ground state of a free theory, and the ground state of a relativistic invariant interacting theory with the same bare mass (e.g. a free scalar field $\varphi$ and the $(\varphi^4)_{1+2}$ theory), that are disjoint by Haag's theorem with respect to the CCR algebra of relativistic fields.
Therefore, to sum up, there is not a one-to-one correspondence between rays (and more in general density matrices) on a given Hilbert representation of a quantum theory and the set of all states of the given theory.
Edit: As suggested in the comments, it may also be interesting to consider the particular case of systems with only $2d$ finitely many classical degrees of freedom (positions and momenta of the particles in the system).
In this case, it is possible to prove that there is a 1-1 correspondence between pure states and rays in the Hilbert space $L^2(\mathbb{R}^d)$, the latter seen as the Schrödinger representation of the algebra of canonical commutation relations (i.e. where the multiplication by the variable $x$ is the position operator, and $-i\hslash\nabla$ is the momentum operator).
In fact, for systems with finitely many degrees of freedom there is only one, up to unitary equivalence, irreducible representation of the algebra of canonical commutation relations (this is the so-called Stone-von-Neumann theorem). Now since every pure state yields an irreducible representation by GNS construction (and vice-versa any state that yields an irreducible GNS representation is pure), it follows that every pure state is identifiable with only one ray in the aforementioned Hilbert space $L^2(\mathbb{R}^d)$ (uniqueness follows from standard considerations, e.g. observing that the trace norm is non-degenerate).

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I think the question is mostly in the context of closed systems with finite number of degrees of freedom. So you don't have issues like mixed states or subtleties from AQFT. – Heidar Feb 04 '20 at 13:55
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While I care about the most general case, it would be informative to consider closed systems with a finite number of degrees of freedom which @Heidar brought up. – Feb 05 '20 at 15:33
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1In my opinion, all that this result tells us is that the Hilbert space $\mathcal{H}_\omega$ that the GNS construction builds from a given state $\omega$ isn't the full physical Hilbert space $\mathcal{H}$, but is just a proper subspace. – tparker Feb 06 '20 at 01:42
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@PiKindOfGuy For closed systems, you don't have mixed states and rays are 1-to-1 to pure states, essentially by the axioms of quantum physics. But as tparker points out, you have to be careful about what is meant by the (full) Hilbert space. – Heidar Feb 06 '20 at 02:39
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1@tparker I would not call the direct sum over all disjoint representations of the algebra the "physical" Hilbert space. That is because in that way, the representation is reducible, and each irreducible subrepresentation is essentially a superselection sector. And the system shall be in just one of these sectors. My point of view is that given a relativistic field theory, the existence of multiple different vacua (either because one is trying to consider interacting and free theories at the same time, or because the theory at hand has degenerate vacua (mexican hat potential)) – yuggib Feb 06 '20 at 07:41
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implies that it is not possible to describe the theory with a single Hilbert space in a meaningful way, but the description (i.e. the vacuum) shall be superselected. Nonetheless, all disjoint vacua are possible physical states for the system. – yuggib Feb 06 '20 at 07:44
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@Heidar : If one considers only a system with finitely many degrees of freedom, there is a unique irreducible representation of the algebra of canonical commutation relations. Nonetheless, there are still other possible states that cannot be written as rays or density matrices in such representation (that is the usual Schrodinger one). These states, that are called non-normal, however are not very physical in my opinion, and I am not aware of any physical situation, with finitely many degrees of freedom, in which they play a role. – yuggib Feb 06 '20 at 07:52
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Nonetheless, mathematically these are still possible (and so the "one-on-one" correspondence cannot be proved even in that case, unless one defines the states of the system to be just the normal pure states). – yuggib Feb 06 '20 at 07:53
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1Do you have a good reference about these non-normal states and their construction? Not sure I'm familiar with them. – Heidar Feb 06 '20 at 12:39
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@Heidar I did not phrase my comment properly, I'm sorry. A non-normal state (w.r.t. a representation) is a state that cannot be written as a (pure or mixed) density matrix in that given representation. The specific case I had in mind in systems with finitely many degrees of freedom are the so-called non-regular states. Without entering into details, these states lack a weak continuity property when the expectation with Weyl operators is taken. Therefore, they cannot be normal with respect to the usual Schrödinger representation. – yuggib Feb 06 '20 at 14:01
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However, it also follows that their induced representation is reducible, and so probably not very physical (the physical parts would be the irreducible sectors). Regular states are introduced in a paper by I. Segal of the end of the 50s, I will give you the precise reference as soon as I have it at hand. – yuggib Feb 06 '20 at 14:02
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@Heidar Actually, after thinking about it, in the end there is indeed a one-to-one correspondence between pure states for systems with finitely many degrees of freedom and rays in the Hilbert space of the Schrödinger representation. I have edited the answer accordingly. In light of these considerations that you find in the edit, the concept of regular states is essentially useless in systems with finitely many degrees of freedom, and only relevant in QFT. – yuggib Feb 06 '20 at 14:32
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I see, that makes more sense to me. I'm only aware of Stone-von-Neumann theorem failing for QFT's, which then introduces many of the subtleties you highlight in your answer. – Heidar Feb 06 '20 at 22:22
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@yuggib I don't quite understand why you start your definition of a theory from $C^*$ algebras instead of Hilbert spaces as is usually presented in a QFT class. Maybe this could be tautological because then by definition the Hilbert space contains all the physical states you have. – Gonenc Feb 06 '20 at 23:19
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1@Gonenc In a standard QFT class, the Hilbert space used is usually the Fock space. However that space is the GNS space of the non-interacting ground state. By Haag's theorem, any interacting theory would be in an inequivalent representation of the canonical (anti)commutation relations, that you cannot construct easily from the free one (in fact, it is known how to do it only in very few, and unphysical, cases). The advantage of the algebraic point of view is that it allows to consider all possible vacua, and it means all possible scalar theories with a given (bare) mass all at once. – yuggib Feb 07 '20 at 18:12
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1Of course this does not mean that finding explicitly the ground states for the interacting theories is any easier, but in my opinion at least it helps in understanding the structure of states, induced representations, et cetera in relativistic quantum theories (and thus it clarifies "where the problems are", so to speak). – yuggib Feb 07 '20 at 18:13
No, mixed states cannot be represented by a ray in Hilbert space. But yes, every pure state corresponds to exactly one ray in Hilbert space and vice versa. This is more or less true by the definition of the term "pure state"; pure states are the special case of states that naturally correspond to rays in Hilbert space.
However, you need to be careful about exactly how you define "Hilbert space", as the term is often abused in the context of gauge theory. In gauge theory, you'll often hear people sloppily talking about "negative-norm states in the Hilbert space", which is a complete contradiction in terms as the norm is positive by definition. When these people use the term "Hilbert space", they're actually referring to a superspace of the actual Hilbert space which does not actually have an inner product, but instead just a nondegenerate conjugate-symmetric sesquilinear form whose restriction to the physical Hilbert space agrees with its inner product. This superspace is often more mathematically convenient to work with than the Hilbert space. Rays in this superspace do not necessarily correspond to physical states (e.g. ghosts do not). More info at https://physics.stackexchange.com/a/333020/92058.

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I don't actually understand what a mixed state is. How do I know if, say, a spin-up state of silver atom is a pure state or a mixed state? According to what I've learned every state is a pure state by your definition. Are mixed states even bona fide states? – Feb 10 '20 at 09:51
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@PiKindOfGuy Yes, mixed states are bona fide states. And system $A$ that is entangled with an external system $B$ is in a mixed state (even if the composite system $AB$ is in a pure state). Under the epistemic interpretations of quantum mechanics, you can also have "classical" mixed states that simply reflect observer ignorance. (Philosophers of physics sometimes distinguish these two types of mixed states as "improper" and "proper" mixed states.) – tparker Feb 10 '20 at 14:53
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@PiKindOfGuy As for how to test if a state is pure or mixed, it's very difficult in practice. If you have access to the external system $B$ as well as the primary system $A$, then you can sometimes test whether they are entangled via e.g. Bell tests, but that usually requires you to have a candidate state of the composite system to be testing. And of course, if you do something (presumed to be) classically impossible like quantum computing, then that's very strong a posteriori evidence that your individual qubits were entangled and therefore in mixed states. – tparker Feb 10 '20 at 14:57
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@PiKindOfGuy That question is getting into kind of murky philosophical waters. According to the ontic interpretations of quantum mechanics, basically yes, every isolated system is presumed to be in a pure state. According to the epistemic interpretations of quantum mechanics, the purity of a state (or lack thereof) partially reflects the observer's knowledge, so the answer is not as clear-cut. – tparker Feb 11 '20 at 00:45
Let us consider first the case where a physical system is mapped to two different rays in Hilbert Space. This means that there exist some basis set $n$ in which our physical system can be expressed as two distinct rays $|n\rangle$ and $|n’\rangle$. Now consider the expectation value of an arbitrary hermitian operator $A$ in the two states. There exists at least one such operator for which the expectation values of the two rays are unequal. Because if there are no such operators then the two rays are equal, which we have assumed they are not. Now if the expectation of some hermitian operator are distant, the physical states are distinct. Clearly a contradiction. Thus each physical state is mapped uniquely to a ray in Hilbert Space.
Now consider the reverse. Let the same ray correspond to two distinct physical states. Distinct physical states means there exists some measurable whose expectation in the two states are unequal. But the expectation of any arbitrary hermitian operator is equal since the ray is the same. But this contradicts the fact that there are two distinct physical states.
Thus the mapping of a ray in Hilbert Space to a physical system is unique (as far as hermitian operators are concerned).

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You haven't considered the case in which a ray doesn't correspond to a physical state or the case in which a physical state doesn't correspond to a ray. – Feb 03 '20 at 11:35