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A state in quantum mechanics I think is just a vector in a complex Hilbert space. As the physical properties are defined up to a phase $e^{i\theta}$ then this Hilbert space is invariant under the action of $S^1$ and the quotient which what is physically relevant is just an infinite complex projective space. Is this right?

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up vote 6 down vote accepted

A more modern view (see, e.g. Bengtsson and Zyczkowski, 2008): the quantum state space is the convex set of trace 1 positive semidefinite operators. That is $\rho \in L(\mathcal H)$ (space of linear operators acting on $\mathcal H$) which satisfy $$ \rho\geq 0, \text{Tr}(\rho) = 1. $$ The extreme points of this convex set are the projectors onto 1-dimensional subspaces. These pure states can be characterized as the states additionally satisfying $$ \text{Tr}(\rho^2) = 1. $$ These can be put in one-to-one correspondence with the elements of the complex projective Hilbert space $P(\mathcal H)$.

This point of view has the advantage of treating pure and mixed states on equal footing and allows one, for example, to study the geometric properties of quantum states.

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This answer is better than mine: +1. You might consider introducing the term "density operator" as well just as a matter of terminological completeness? –  joshphysics Mar 7 '13 at 5:54
    
This viewpoint of the projective Hilbert space as the true state space is often referred to as "Geometric Quantum Mechanics". See for example sciencedirect.com/science/article/pii/S0393044000000528, I think Ashtekar has some papers on it too. –  Tobias Diez Mar 8 '13 at 18:09
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Yea. Said another way, the space of physical states is just the space of equivalence classes of vectors (rays) in the Hilbert space with two states being equivalent provided they differ by multiplication by a nonzero complex number.

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