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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.

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2 Answers 2

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.

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For the sake of completeness, the theorem is called spectral theorem and it's quite technical for general operators but basically it's just a diagonalization of the operator in the "basis" of "eigenvectors" (scare quotes because operators on infinite-dimensional spaces need not have eigenvalues or eigenvectors). –  Marek Feb 23 '11 at 16:57
    
May be I was not clear in my question.I should have I asked some justification of self adjoint operators as observable.You may say we want real eigen values etc.But why do we assume eigen values to be set of observed values...Is this framework necessary--I mean can we deduce that it is necessary to deal with linear operators,eigen values from some assumptions...And my problem regarding momentum was why is the fourier representation related to momentum which was some time derivative in Classical Physics. –  Ket Feb 23 '11 at 17:49
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@Ket: In this case I have misunderstood your question. You can analyze an observable O through true-false questions of the form "Is the value of O in the set E?" for Borel sets E in $\mathbb R$. This leads to projection-valued measures on $\mathbb R$. Then the spectral theorem (again) gives the connections to self-adjoint operators. I find this natural, but that is subjective. To write this in detail would be more than a comment and as I said I understood your question differntly. –  MBN Feb 23 '11 at 18:01
    
Thanks for the reference.I will look at it. –  Ket Feb 23 '11 at 18:12

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.

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Just two nit-picks: 1) those "bases" are not quite bases and their vectors don't even belong to the Hilbert space. I think this is worth pointing out to someone who is just beginning with QM if only as a curiosity to keep in mind; 2) that these (and other) representations work so well is a consequence of Stone-von Neumann theorem which guaranties that these (and other) representations are unitarily equivalent. This theorem of course fails if some assumptions are removed and leads to interesting possibilities of inequivalent vacuua, etc. –  Marek Feb 23 '11 at 16:46
    
I understand that Ehrenfest theorem relates expected values..But my problem was to understand the necessity in the framework with operators and their eigen values representing physical observables,values..what requirements imposed the necessity..and why the classical results come out as the relations of expected values...And regarding momentum why is the fourier transform related to momentum..Everything is fine if I just follow axiomatic approach , all my problem was to understand the connections from classical to quantum physics.. –  Ket Feb 23 '11 at 18:06
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@Ket: are you comfortable with Hamiltonian mechanics? It's a formalism that resembles quantum mechanics very much. Only difference is that observables (which are just functions on the phase space) there commute. But in quantum physics we need non-commutativity (because of HUP) so we replace algebra of functions (which is commutative) with an algebra of operators which (which need not be). But except for this "detail", everything stays the same. –  Marek Feb 23 '11 at 18:16
    
@Marek:Yes,I have tried to study a little bit of symplectic geometry.So you say HUP dictates a noncommutative algebra of observables.Thanks for that.That seems to answer my problem.I will think about it.Thanks a lot once again. –  Ket Feb 23 '11 at 18:25
    
@Marek: I of course know of these nit-picks. We just seem to disagree on whether it is important to mention them to someone who has just begun to study QM :) @Ket: When you measure an observable in a physical state, you will obtain different results with certain probabilities. The expectation value is just the statistical average of these values. On some states you always get the same result, with probability one. The operator of the observable is naturally constructed so that these states are its eigenstates (and the corresponding values of the observable the eigenvalues). –  Tomáš Brauner Feb 23 '11 at 19:57

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