In Quantum mechanics, observables are represented by hermitian operator. But does every hermitian operator represent a observable? If not , how do we know that whether a hermitian operator represent observable or not? What is the precise definition of the term "observable"?
[Physics] Does every hermitian operator represent a measurable quantity
observablesoperatorsquantum mechanics
Best Answer
Given a quantum system with associated Hilbert space $\mathcal H$, the set of all self-adjoint bounded operators is $\newcommand{\bh}{\mathcal B(\mathcal H)_\text{sa}}\bh$. In general, only a small subset of $\bh$ will represent physically observable operators. For infinite-dimensional systems, $\bh$ is huge and there's no hope ever finding experiments for all its members; even in finite-dimensional systems it is very challenging to find experimental schemes sensitive to even a vector-space basis for $\bh$.
The physical approach to this is to begin with a finite set of operators which you know you can measure. For a single free particle, for example, you'd take position and momentum; for a finite set of spins you'd take all their Pauli matrices. You then form the set $\mathcal A$ of all operators that can be formed from them via products and linear combinations, which has the structure of a $\mathcal C^\ast$ algebra, and that is your set of physical observables. The $\mathcal C^\ast$algebra itself is the really fundamental description of the system; the Hilbert space is simply one possible representation.
In this formalism, states are functionals on $\mathcal A$: they are functions $$\rho:\mathcal A\rightarrow \mathbb C $$ that take an observable and give its measured value (or probable measured value, etc.) in that state. (In a Hilbert space representation, each such functional is associated with a density matrix $\hat\rho$, a trace-class positive operator such that $\rho(A)=\text{Tr}(\hat\rho\hat A)$ for $\hat A$ the Hilbert space operator associated with an arbitrary $A\in\mathcal A$.