Momentum-Representations in Quantum Mechanics Why do we get information about position and momentum when we go to different representations. Why is momentum, which was related to time derivative of position in classical physics, now in QM just a different representation brought about by some unitary transformation. Is Ehrenfest's theorem the only link?
I just started studying QM. So please suggest some references explaining the structural aspects and different connections.I don't want to start with noncommutative geometry. I would like something of an introductory nature and motivating.
 A: You can get information for all observables in any representation. The reason to go to different ones is that it is easier to work with them depending on what you are doing. They are all equivalent by the Stone-von Neumann theorem, so it is a matter of convenience. 
There is a theorem (mathematical) that roughly says that for any operator, that is of interest in QM, there is a representation of the operator as a multiplication operator, in it it acts as a multiplication by a function . In the coordinate space the position operators are multiplication by the coordinates. In momentum space it is the momentum that is represented as a multiplication. It is true for any of the QM observables. Unfortunately (or fortunately) since they do not commute there isn't a single representation for all of them. Hence people use more than one.  
Edit: In response to the comment. This is probably written in many books, but here is a reference. Look at Folland's "Quantum Field Theory. A Tourist Guide for Mathematicians". The first section of chapter 3 gives a nice motivation for the use of self-adjoint operators for modeling QM observables.
A: 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.
