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cianfa72

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- TL;DR Summary
- About the eigenstates of particles with 1/2 spin (qbit)

A very basic doubt about a QM system (particle) with spin 1/2 (qbit).

From the Bloch sphere representation we know that a qbit's pure state is represented by a point on the surface of the sphere. Picking a base, for instance the pair of vector/states ##\ket{\uparrow}## and ##\ket{\downarrow}##, any qbit's pure state can be given as (normalized) complex linear combination of the above states $$\ket{\psi} = \alpha \ket{\uparrow} + \beta \ket{\downarrow}$$

Does that mean that for any qbit's pure state there always exist an Hermitian/self-adjoint operator (an observable) such that that state is eigenstate of ? Thanks.

From the Bloch sphere representation we know that a qbit's pure state is represented by a point on the surface of the sphere. Picking a base, for instance the pair of vector/states ##\ket{\uparrow}## and ##\ket{\downarrow}##, any qbit's pure state can be given as (normalized) complex linear combination of the above states $$\ket{\psi} = \alpha \ket{\uparrow} + \beta \ket{\downarrow}$$

Does that mean that for any qbit's pure state there always exist an Hermitian/self-adjoint operator (an observable) such that that state is eigenstate of ? Thanks.