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Does the expectation value of an observable must be equal to an eigenvalue of the corresponding operator? I already know that 0 is not an eigenvalue, but is there any other examples?

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I would actually expect this to be rare, and only generically true when the state of the system corresponds to an eigenstate. This simply because, for a state $\psi = \sum a_{n}\lvert\phi_{n}\rangle$ with eigenvalues $V_{n}$, you would have $\langle V\rangle = \sum V_{n}\lvert a_{n}\rvert^{2}$, which is not constrained to be equal to one of the $V_{n}$. It's easy to check this for a two state system with the two values of $V_{n}$ different.

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Right, for a classical experiment to get the basic idea, consider the expected number of eyes when throwing a die, namely $\frac{7}{2}$. –  Qmechanic Sep 19 '12 at 19:13
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A specific quantum mechanical example to show the contrary is spin for spin 1/2 systems. If you are in an eigenstate of the Sz operator, the expectation value of Sx is 0, but it has eigenvalues h/2, -h/2, where those are h-bars. Not exactly sure how to do latex here..

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