# Is every observable a function of position and momentum?

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.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.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.