My poorly written lecture notes say that any Hermitian operator does have a complete set of orthogonal eigenstates with real corresponding eigenvalues and is therefore an observable.
In the article Observables, it is said that in order for a Hermitian operator to be observable its eigenvectors must form a complete set.
According to the postulates of quantum mechanics, each observable $p$ quantity is associated with an operator $\hat{p}$ that acts on the wavefunction $\psi$.
The relationship is given by the eigenvalue equation: $$ \hat{p}\psi = p\psi. $$
$\hat{p}$ is an operator, which means nothing on its own. $p$ is the eigenvalues, the observable which is a number.
For instance, if $p$ is the momentum:
$\hat{p} = \frac{\hbar}{i}\nabla $, i.e. a functional operator so quite useless on its own;
Acting of a plane wave $\psi = e^{ikx}$, $\hat{p}\psi = \hbar k\,\psi $. I.e. the observable momentum is $p=\hbar k$.
An operator need not be hermitian. For instance, the harmonic oscillator creation operator $\hat a^\dagger$ is not hermitian, and neither is the angular momentum lowering operator $\hat L_-$. Yet both are perfectly legitimate (linear) operators, i.e. they act linearly on a state and produce a different state.
Setting aside subtle points about domains of operators and self-adjointness, observables must be hermitian (in the sense that their matrix representations are hermitian matrices) because eigenvalues of hermitian matrices are real, which is good since in a lab we measure real (rather than complex) quantities. Moreover, hermitian matrices have a complete set of eigenvectors that spans the entire space.
Note that it is important to realize that this doesn’t imply that non-hermitian operators cannot have eigenvalues or eigenvectors, just that there’s no guarantee the eigenvalues are real and the eigenvectors for a complete set. The usual example of this is the harmonic oscillator coherent state $\vert \alpha\rangle$ (where $\alpha$ is any complex number) which is an eigenvector of the annihilation operator $\hat a$, with complex eigenvalue $\alpha$. The eigenvalue need NOT be real since $\alpha$ can be complex, and the coherent states form an overcomplete set of vectors for the Hilbert space of the harmonic oscillator.
First, a mathematical subtlety. In finite dimensional Hilbert spaces spaces, every self-adjoint operator can be represented by a Hermitian matrix, i.e. one that is equal to its conjugate transpose, and every Hermitian matrix corresponds to a self-adjoint operator. However, in infinite dimensional spaces, as it is often the case for the Hilbert spaces describing quantum systems, a symmetric operator $A$—which in finite dimensions is equivalent to a Hermitian matrix—is only self-adjoint if its domain is the same as the one of its adjoint $A^\dagger$, namely $\mathrm{dom}\ A = \mathrm{dom}\ A^\dagger$. A self-adjoint operator, though, is always symmetric. This being said, every observable corresponds to a self-adjoint operator.
Nevertheless, it is generally false to suppose the converse: not every self-adjoint operator is an observable, and a typical example of such is the density operator $\hat{\rho}$. More on the subject can be read here: not every self-adjoint operator is an observable.
