DarMM said:
A mixed state may be a vector state in the GNS Hilbert space, but not a pure state.
Secondly a sum of kets is not a pure state precisely in the case where both belong to different superselection sectors. Unless you are have a certain precise definition of Hilbert space in mind.
A. Neumaier said:
A vector state in the GNS Hilbert space is a pure state in the Hilbert space sense, but not in the ##C^*## algebra sense.
It is a pure state in the Hilbert space defined by the direct sum of the superselection sectors. What I don't understand is in which sense it can be a mixed state.
DarMM said:
Pure and Mixed refer to a state's entropy though. Thus a state is pure if it has vanishing entropy.
In many (most) cases of course we represent pure states with vector states. Thus the two are often, but not always, the same.
It's a vector state, not a pure state as its entropy does not vanish.
vanhees71 said:
It thought in a model with superselection rules the above superposition is simply not describing a physical preparable state. If it were preparable mathematically it'd by definition represent a pure state (or rather in the usual sense of ##\hat{\rho}=|\psi \rangle \langle \psi|## representing this pure state to be pedantic).
Morbert said:
We have the ket ##|\psi\rangle = \frac{1}{\sqrt{2}}|a\rangle+\frac{1}{\sqrt{2}}|b\rangle## mentioned previously. Thus, we have a quantum state ##\omega_\psi## such that $$\omega_\psi(A) = \frac{1}{2}\omega_a(A)+\frac{1}{2}\omega_b(A)+ \frac{1}{2}\langle a,Ab\rangle+\frac{1}{2}\langle b,Aa\rangle$$ for all observables ##A##. If ##|a\rangle## and ##|b\rangle## are subject to superselection rules such that ##\langle a,Ab\rangle = \langle b,Aa\rangle = 0## for all observables ##A##, then the above reduces to the mixed state $$\omega_\psi(A) = \frac{1}{2}\omega_a(A)+\frac{1}{2}\omega_b(A)$$ But ##\omega_\psi## is still a vector state, since ##\psi\in\mathcal{H},||\psi||=1##.
DarMM said:
Well the typical terminology in the most general applications of quantum theory, such as in algebraic field theory and quantum information, is that ##|\psi\rangle## is a vector state which represents ##\omega_{\psi}##. However it is to ##\omega_{\psi}## that the mixed/pure distinction refers, since regardless of the representation ##\omega## has finite entropy and all the other characteristics of a mixed state. Thus it is a mixed state represented by a vector state in this particular representation.
The misunderstandings in this thread related to the above are due to the fact that the notion of a state and the distinction between pure and mixed states are both ambiguous because both depend on the algebra on which they are considered. That's why we have been talking past each other.
A good open source general discussion is in Section 6.3 of the paper
The essence is here:
In standard quantum physics (i.e., that not using the ##C^*## algebra terminology), a state is a density operator (i.e., a Hermitian positive semidefinite trace 1 operator) ##\rho## on a Hilbert space ##H##, and it is pure when it has rank one and mixed otherwise. In the pure case (only) it can be represented by a state vector ##\psi## as ##\rho=\psi\psi^*##, uniquely up to a phase. Hence state vectors always represent pure states.
On the other hand, algebraic quantum physics uses the ##C^*## algebra terminology. There a state is a positive linear functional ##\omega## on a ##C^*## algebra ##A## (called the observable algebra) satisfying ##\omega(1)=1##, and the state is pure iff it is not decomposable into a convex combination of two different states. This is a
generalization of the notion from standard quantum physics, which is the special case where the observable algebra is the algebra ##B(H)## of bounded linear operators on a Hilbert space ##H## and the density operator ##\rho## is identified with the positive linear functional ##\omega## defined by $$\omega(A):=Tr~ \rho A.~~~~~~~~(1)$$ This defines a 1-1 correspondence, making the identification possible. With this identification the notion of pure and mixed states agree.
One says that there are superselection rules when the observable algebra has a center consisting not only of the multiples of the identity. When working in a fixed Hilbert space this implies that ##A## is a proper *-subalgebra of ##B(H)##.
In this case, (1) no longer defines a 1-1 correspondence between density operators and positive linear functionals, and
the notions both of state and of the pure/mixed distinction differ algebraically.
This is easiest to see if one takes the Hilbert space as ##C^n## and the observable algebra to be the algebra of diagonal matrices. Then ##\omega(A)=\omega(Diag(A))##, where ##Diag(A)## is the matrix obtained from ##A## by setting the off-diagonal elements to zero. The pure states are now only those whose density matrix have a single 1 and otherwise zeros on the diagonal. In particular, most pure states become mixed.
The nonuniqueness also implies that the Schrödinger equation becomes meaningless; only the von Neumann equation for the dynamics of the density operator remains meaningful, and only for Hamiltonians belonging to the observable algebra. In particular, this dynamics does not lead outside of a superselection sector; if the initial state is pure (and hence belongs to a single superselection sector, as it is often assumed), the state remains pure and lies for all times in the same superselection sector.
DarMM said:
Well a very dumb example, but it could be fleshed out with a more realistic one, but imagine a "trap" of some kind that can only hold a single particle. Either spin-1/2 or spin-1. It's in contact with a reservoir of spin-1/2 and spin-1 particles with which it can exchange particles. The trap starts off with a single spin-1/2 particle. Its state after a finite time would be described by such a state.
This can work only in a setting where the initial state of the system of trap plus particle is already mixed. The reduced dynamics of the particle alone is no longer Hamiltonian. Thus it is no surprise that the initially pure state becomes mixed.
