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I often read in books that an observable is represented by an Hermitean operator. But it is deceiving as operator isn't the observable.

As far as I've read the observable is denoted like $\langle \psi|\hat{x}|\psi\rangle$ which is equivalent of $\langle \psi |\hat{x} \psi\rangle$ or $\langle \hat{x}^\dagger\psi| \psi \rangle$. So I would say that an observable is represented by an inner product of a (1st) wavefunction with (2nd) an operator acted on a wavefunction. (If I look the second equation).

Is this even correct? What do you think about this?

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up vote 5 down vote accepted

Functionally speaking, an observable is a physical quantity with this property: you can design a physics experiment that measures the value of that quantity possessed by a particle. So, for example, you can measure the speed of an electron by checking the radius of motion as it goes through a magnetic field, you can measure a particle's spin with the Stern-Gerlach experiment, and so on. And, as it turns out, any quantity for which you can design these kinds of experiments corresponds to a Hermitian operator.

The inner product isn't the observable itself; rather, it's the expectation value of that observable. If I repeat the experiment I did many times over, the average value of the measurement will be $\langle\psi|\hat{x}|\psi\rangle$. This is why we say that, mathematically, the operator itself is the observable: it's the mathematical object that we use to make predictions regarding the actual measured numbers we get from experiments.

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