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In quantum mechanics, it's well-known that observables are associated as the eigenvalue of a Hermitian operator.

My question is, is the converse also true? i.e. the eigenvalue of a Hermitian operator (which is real) must be an observable.

Jovan Alfian Djaja
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  • It depends on how you define things. Yes, in "standard" QM, an observable is represented by a hermitian operator, and its eigenvalues are said to be the possible measurement outcomes. Conversely, in the absence of superslection rules, you call every hermitian operator an observable, which in principle you can measure. Whether or not you actually can perform an experiment coresponding to an arbitrary operator is another story. – Tobias Fünke Jan 22 '24 at 17:10
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    See this post on the same topic: https://physics.stackexchange.com/q/479545/ – Léo Vacher Jan 22 '24 at 17:21
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    Possible duplicates: https://physics.stackexchange.com/q/373357/2451 , https://physics.stackexchange.com/q/27038/2451 and links therein. – Qmechanic Jan 22 '24 at 17:24

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