Dear Ket, momentum in QM is not "just a different representation brought about by some unitary transformation". As you probably already know, a physical state in quantum mechanics cannot have simultaneously a well-defined (sharp) value of momentum and position; this the Heisenberg uncertainty principle. However, you can still measure the expectation values of both momentum and position in the same state. It is at the level of expectation values that momentum and position satisfy exactly the same relation as in classical physics; this is Ehrenfest's theorem.
When you talk about representations and unitary transformations, you probably mean the choice of basis in the Hilbert space of physical states. But this is merely a mathematical tool: to be able to work with vectors from the Hilbert space, it is suitable to choose a basis and work with the coordinates in this basis instead of the abstract vectors. Should you choose the basis of eigenstates of the position operator, the "coordinates" will be what is called the wave function. But you can choose any other basis as well. You can work in momentum representation, corresponding to the basis of eigenstates of the momentum operator, which is indeed related to the coordinate representation by a unitary transformation (called the Fourier transform in mathematics). This is because both bases are orthogonal, being formed by eigenstates of self-adjoint (Hermitian) operators. However, you can as well use any basis, not related to any operator of an observable. What is physical are the expectation values of observables (which are independent of the choice of basis) and relations between them, which are via Ehrenfest's theorem equivalent to classical equations of motion.