Hermitian operators: observables and dynamical variables

In the postulates of quantum mechanics, it is sometimes said that dynamical variables are associated with hermitian operators. And sometimes, it is said that observables are associated with hermitian operators.

But observables means measurable. And there are many other measurable quantities which are not directly associated with operators such as scattering cross-section, transition amplitude etc.

So isn't the statement "observables are associated with hermitian operators" is a bit imprecise?

• In general, no, it is not imprecise. "Observables are associated with hermitian operators" is a tautology because that is the definition of observable in the context of QM. – AccidentalFourierTransform Jan 4 '17 at 18:05
• – Qmechanic Jan 4 '17 at 18:05
• If you have basic knowledge in maths and physics, I suggest you reading first few paragraphs of these Lectures by Faddeev and Yakoboskii. – mavzolej Jan 4 '17 at 18:08

A classical observable is a (sufficiently smooth) function on phase space, a quantum observable is a self-adjoint operator on a Hilbert space or abstractly a self-adjoint element of the $C^\ast$-algebra of observables. This is a definition.
The connection to dynamical variables is that the "dynamical variables" $x,p$ are the coordinates of phase space, which of course are very elementary classical observables because they are the simplest possible functions on phase space, which then get sent to elementary quantum observables by your preferred process of quantization. Quantum mechanics does have observables, like spin, not arising from dynamical variables this way, but many operators, in particular position and momentum, do arise as the quantization of dynamical variables.
The more precise statement is "In principle, every observable (i.e. dynamical variable) can be represented by a Hermitian operator." Now, by definition, a dynamical variable is a function of $\mathbf{r}$ and $\mathbf{p}$. Thus, all you need is to have the position and momentum operators to be well defined, and you can get by construction any operator of the form $$\hat{\Omega}(\hat{x},\hat{p}),$$
where $\hat{\Omega}$ is Hermitian.