In quantum mechanics, we generally take about "expectation values of dynamical variables". However, by the postulates of quantum mechanics, every dynamical variable in quantum theory is represented by its corresponding operator.
Is it therefore, incorrect to talk about "expectation value of an operator" (rather than "expectation value a dynamical variable")? Is is just semantics or is something more going on here?
In other words is it incorrect to write:
$$<\widehat{A} > =\int \psi \widehat{A}\psi^{*}\mathbf{ d^{3}r}$$
instead of
$$<A > =\int \psi \widehat{A}\psi^{*}\mathbf{ d^{3}r}?$$
Best Answer
Remember, operators are nothing but maps. Expectation value of an operator is pretty much defined (I guess) in general operator theory. It just turns out that in QM (Hermitian) operators correspond to dynamical variables. In general you can also calculate expectation values of operators like $L_+$ and $a^{\dagger}$ etc., which don't have any dynamical variables associated !!
On a more general note, if you want to get a representation for the operator (given some basis, which could be the eigenset of an operator), the expectation values would be the diagonal elements of the representation.
It is related to Linear algebra, This might throw some light on Operator theory. Linear operators (bounded of course) are maps defined in LVS that take one vector to another vector in the same LVS.
$$ \hat O : V \rightarrow V$$
$$ \hat O \left|\psi\right> = \left|\phi\right> $$ Then you have the inner product defined on the LVS, that takes two vectors to one complex number : $$ \left<\psi|\phi\right> : V^* \times V \rightarrow \mathbb{C} $$
Using these two, one can define the expectation value of an operator to be,
$$ <\hat O> = \left<\psi\right|\; \big(\hat O\left|\psi\right>\big) = \left<\psi|\phi\right> \in \mathbb{C} $$