# 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

## 2 Answers

The statement is precise because "observable" is itself a technical term and does not just mean "thing that is measurable".

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.

Of course there are many things that are "measurable" (in the sense of "can be measured in experiment, not in the sense of mathematical measures) that are not observables in this strict, formal sense; you already list some in your question. This does not mean that the statement about observables is "imprecise", it means that to understand a technical statement in physics (just like in any other field) you must know the precise definition of the terms that occur in it. That many terms have both (at least) one formal and one informal meaning is a fact of life we have to accept.

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.