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Why isn't momentum a function of position? 
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#37
Mar812, 12:11 PM

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#38
Mar812, 02:06 PM

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And thanks for the further references, but they are currently out of my scope. 


#39
Mar812, 03:07 PM

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#40
Mar812, 03:25 PM

P: 752




#41
Mar912, 10:16 PM

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P: 1,939

http://en.wikipedia.org/wiki/Hamiltonian_dynamics http://en.wikipedia.org/wiki/Canonical_transformation but they are no substitute for a textbook if you haven't already studied this stuff. 


#42
Mar1412, 12:00 PM

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#43
Mar1412, 12:30 PM

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#44
Mar1412, 01:46 PM

P: 22

(I was obviously referring to oneparameter groups)
If your system is a particle in a timedependent potential it is clearly not true that [itex]U(t_1)U(t_2) = U(t_1 + t_2) [/itex] in general, so you can't use Stone's theorem. 


#45
Mar1412, 02:09 PM

P: 1,583




#46
Mar1412, 03:21 PM

P: 22

The solution involves something like the Dyson operator, but I only know that is (rigorously) obtained by iterative method, as you say. The evolution from time t0 to time t1 [itex] U(t_0, t_1)[/itex] depends in general on t0 and t1 separately, not only on their difference [itex] U(t_1  t_0)[/itex]. All you can say is that [itex] U(0, t_1 + t_2) = U(t_1, t_1 + t_2) U(0, t_1)[/itex] and not that [itex] U(0, t_1 + t_2) = U(0, t_2) U(0, t_1)[/itex]. If [itex] U(t_2, t_1) =U(t_1  t_2) [/itex] holds for any t1,t2 you can apply Stone's theorem, so you have a timeindependent Hamiltonian. 


#47
Mar1412, 03:31 PM

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#48
Mar1412, 03:41 PM

P: 22

No, twoparameter group implies oneparameter group, the other parameter fixed.



#49
Mar1412, 05:29 PM

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#50
Mar1512, 11:42 AM

P: 22

For the exact solution I know "Quantum Mechanics I", GalindoPascual. In general they are just unitary operators, I don't know other structures. 


#51
Mar1512, 11:53 AM

P: 1,583

Here's what I found in Reed and Simon on the subject of timedependent Hamiltonians and the associated time evolution operators. They call the structure of the set of time evolution operators a unitary propagator.



#52
Mar1512, 12:28 PM

P: 255

Hey guys, I haven't read the whole thread, but I noticed that some people were saying that angle is a parameter (like time) in QM, and I don't see it like this at all, to me angles are observables in QM, I mean phi and theta are just components of the position operator in spherical coordinates, aren't they? Am I wrong here?



#53
Mar1512, 02:44 PM

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#54
Mar1512, 10:25 PM

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P: 1,939

(Sigh...) OK, I'll try once more to straighten all this out...
First, the independence of position and momentum (considered as dynamical quantities, not in the geometric sense where momentum generates spatial translations)... Consider the motion of a classical free particle. It follows a straight line, i.e., has zero acceleration. IOW, its equation of motion is ##\ddot x(t) = 0##. This is a 2nd order differential equation. To solve it we need two initial conditions: an initial position ##x_0 \equiv x(t_0)## and an initial velocity ##v_0 \equiv \dot x(t_0)##. These two conditions are independent  you can specify any combination of them and arrive at a valid solution of the equation of motion. Also, velocity is (in this case) conserved  it remains constant in time, whereas position keeps increasing with time. This is another way of seeing that velocity of a free particle is not a function of position. We express this mathematically as $$ \def\Pdrv#1#2{\frac{\partial #1}{\partial #2}} \Pdrv{\dot x}{x} ~=~ 0 $$ Now note that (in this case) momentum ##p = m \dot x## (mass times velocity), and we have the corollary that momentum is also independent of position. I.e., $$ \Pdrv{p}{x} ~=~ 0 $$ (and vice versa). Second, for more general cases involving forces acting on the particle, one can still express the dynamical problem in terms of generalized momentum p' and position x'  referred to as a "canonical pair" of dynamical variables for which $$ \Pdrv{p'}{x'} ~=~ 0 $$ (More generally, one expresses this via Poisson brackets, but I don't need that here.) For example, in the case of a charged particle in an external magnetic field, the canonical momentum is of the form $$ p' ~=~ p  eA $$ where e is the charge, and A is the vector potential for the applied magnetic field. Since ##A = A(x)##, the new canonical momentum p' is dependent on ordinary position x (through A), but we can (if we wish) find a new position variable x' which is independent of p'. (I.e., x' is "canonically conjugate" to p', and vice versa.) Thirdly, similar remarks apply to other pairs of canonically conjugate variables in more general dynamical situations  such as angular momentum and pose (angle). In the case of quantized intrinsic spin, things get trickier as one is now working in a Hilbert space. For a nonrelativistic particle of spin 1/2, we have a 2D Hilbert space and the spin angular momentum operators generate unitary transformations in this Hilbert space, parameterized by anglelike quantities. Bottom line: we choose the mathematical model which best models the physical situation and (out of those) the one which is most convenient for calculations. Often, that involves canonically conjugate pairs of variables which are generalizations of the (mutually independent) position and momentum that we found above in the simplest case of a free particle. [I really, really hope that helps...] 


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