Originally Posted by gravenewworld
For example one postulate states that for every observable in classical mechanics there is a linear QM operator. The operator is found from the classical mechanical expression for the observable written in terms of cartesian coordinates and conjugate momenta by replacing each coordinate q by itself and the conjugate component by -i h/2pi d/dq (the d's should be partial deriv. symbols, but I can't write them.)
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Yes, that's how one usually first gets into contact with QM. And although one should go through that process, and first learn to use the tools to calculate simple things and get a feel for it, afterwards, it is a good idea to come back to the basics and see where they come from. One can simplify a lot the postulates of QM, and derive what you write above as their consequences. The way to do so is the following:
step1:
choose what is a set of maximally compatible observables. If you have a classical equivalent, these are simply the variables defining the configuration space, but you might consider non-classical systems too.
So you DEFINE yourself that variables: A, B, C and D are the things you'll be able to measure together. You put this in by hand. For a point particle, you say that its position X Y and Z are the maximal set of compatible observables (if the particle is spinless).
step2:
define what are the possible combinations of values that these maximally compatible observables can take on. For our point particle, it is simple: all 3-tuples (x,y,z) are possible outcomes, if we consider a particle in free euclidean space. Now associate with each of these possible combinations, one single dimension of a hilbert space. So each ket |x,y,z> is a basis vector of the Hilbert space of states of the point particle.
In our other example of A, B, C, and D, to each allowed combination (a,b,c,d) corresponds a basis ket |a,b,c,d>. Mind you, it is again BY HAND that the 'allowed values' are put into the theory.
step3:
the observable operators corresponding to our variables A,B,C and D are defined to be diagonal in the basis introduced in step2, and have as eigenvalues the corresponding allowed value of the set of values that corresponds to it. So, the eigenvalue of B for the ket |a1,b2,c1,d4> is b2.
Because we have defined these linear operators over a basis, they are defined in the entire hilbert space. Also, because they are diagonal and have real values in an orthonormal basis, they correspond to hermitean operators.
For our point particle, X |x,y,z> = x |x,y,z>
step4:
Relate, in one way or another, OTHER observables to these "defining" observables. For example, when there is a classical equivalent, the conjugate momenta correspond to the infinitesimal translation operators of the original defining observables.
So we have, for our point particle, Px, which is the conjugate momentum to X, and it corresponds to the infinitesimal translation operator which maps the state |x,y,z> onto |x+dx, y,z>.
It is a bit long but rather straightforward that this then implies the commutation relations [X, Px] = i hbar.
If we don't have a classical equivalent, introducing other observables is less obvious, but can be accomplished by considering symmetries of the setup. For example, for a spin-1/2 system, if we took Sz as the single defining observable with the two allowed values -1/2 and 1/2 (so having a 2-dim hilbert space), the spin observable Sx and Sy can be found by considering rotations of 90 degrees around the x,y and z axes.
In the case of a classical system, the commutation relation [X,Px] = ihbar AUTOMATICALLY implies that the representation of the Px operator is hbar/i d/dx.
cheers,
Patrick.
