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What really is meant by eigenvalue of an observable? Does it mean that everytime we measure a value of an observable the result obtained is equal to the eigenvalue of the observable?

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closed as off-topic by ZeroTheHero, stafusa, John Rennie, glS, By Symmetry Oct 1 '18 at 16:01

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Yes. If you measure a specific value $V$, and $O$ is the operator that corresponds to the observable, then $O \psi = V \psi$ implies (by definition) that $\psi$ is an eigenvector that corresponds to the eigenvalue $V$.

On the other hand, if you don't measure this particular observable, then $\psi$ can be any (normalized) linear combination of eigenvectors of $O$ since they form a basis for the vector space. In the general case that's called a superposition, although the exact same $\psi$ could very well be an eigenvector of some other operator.

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