# Significance of eigenvalues of an observable Of a wavefunction [closed]

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?

## closed as off-topic by ZeroTheHero, stafusa, John Rennie, glS, By SymmetryOct 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.