# Is every observable a function of position and momentum?

1. In classical physics, the definition of an observable is that it is a (reasonably smooth) function of position and momentum. Therefore, quantum systems obtained purely by quantization of a classical system also have that all their observables are functions of position and momentum.

2. In quantum physics, an observable is just an operator designate to belong to the "algebra of observables". There is no a priori reason to even have position and momentum operators, and indeed, systems with finite-dimensional state spaces (such as qubits or other system without a positional degree of freedom) don't have a position or momentum operator to begin with because the CCR $[x,p] = \mathrm{i}\hbar$ cannot hold on a finite-dimensional space.

3. In relativistic quantum field theory (which is a type of quantum theory, in the end), there are no position operators, the closest are the so-called Newton-Wigner operators, so the question dissolves because it doesn't make sense anymore. Additionally, the momentum operator is a function of the field operators, but not vice versa - you cannot restore the field from the total momentum operator alone, like you cannot restore a function from the value of a definite integral.