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This question follows up on this question on the ideal Von-Neumann Measurement scheme - also regarding the way Schlosshauer (978-3-540-35773-4) puts it on page 50 f.


In this scheme one has given a system in a state, described as a vector in a hilbert space, which has the basis $|\sigma_i\rangle$. A measurement apparatus then measures the system, making him end up in a state $|\mu_i\rangle$ - according to the state $|\sigma_i\rangle$ the system has. These apparatus states are orthogonal by definition, because one wants to be able to tell them apart.

The question is: Don't the states $|\sigma_i\rangle$ also need to be orthgonal? If not, why not? Because in the end, they respond to the eigenstates of a observable and those must be orthogonal, don't they?

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The $|\sigma_i\rangle$ are a basis for $\mathcal{H}$ so they need to be linearly indipendentement. They don't need to be eigenvectors of any observable, they are just a maximal set of linear indipendentement vectors in the space

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  • $\begingroup$ Why do they have to be a basis at all? $\endgroup$
    – manuel459
    Commented Jun 4, 2023 at 16:58
  • $\begingroup$ You said that it is a basis lol. Anyway, you have a state $|\alpha\rangle$ and on $\mathcal{H}$ you have always a basis. This means that your state $|\alpha\rangle$ can always be expanded (viewed) in that basis. If you have a second basis, your state can be thought as a linear combination of this basis too, the only important fact is that every state can be expanded into one or more basis, and that depending on what you are measuring, each state has different coefficients in different basis, which are your probability amplitudes $\endgroup$
    – LolloBoldo
    Commented Jun 5, 2023 at 10:20
  • $\begingroup$ Yes. But them not being orthonormal made me question even that ^^ ;) thank you, that makes sense. (if those basis states correspond to "what is measured", aren't they the eigenstates of the observable though? ) $\endgroup$
    – manuel459
    Commented Jun 5, 2023 at 10:25
  • $\begingroup$ Yes, if that basis also diagonalize an hermitian operator (the "what is measured"), then the basis is a set of observable eigenstates of that operator :) $\endgroup$
    – LolloBoldo
    Commented Jun 6, 2023 at 8:17

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