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.
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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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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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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