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philo324

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philo324

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Hurkyl

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Conversely, every pure state can be represented by a normalized vector.

Also, every vector (or nonzero vector, depending on the details of how you define things) is indeed an eigenvector for some observable.

There are other states, though. These are not formed by superimposing, but by mixing -- if you represent states as density matrices, a linear combination of states (with positive real coefficients that add to 1) usually gives a mixed state.

Maybe I learned things oddly, but I've never heard of a "pure state for an observable".

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philo324

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philo324

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Hurkyl

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Vectors in the Hilbert space (and in the domain of the Hamiltonian) evolve to vectors in the Hilbert space, right? So....Clear. I'm thinking about a state that evolves in time, for example.

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A. Neumaier

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is EVERY allowable superposition in quantum mechanics, every state the system could end up in, a pure state for some observable, even if this observable is very complex to realize?

Yes. If \psi is a pure state then it is an eigenstate of the projector P = \psi\psi^* with eigenvalue 1, and P is an observable according to the standard definition (self-adjoint). Moreover, if the Hilbert space is low-dimensional, P can be realized quite well.

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