# Why isn't momentum a function of position?

by lugita15
Tags: function, momentum, position
P: 1,574
 Quote by strangerep I didn't say that. I'm finding it quite hard to discuss this with you. Please try to phrase your future responses differently, in a way that doesn't put words in my mouth. I don't like trying to un-distort "so you're saying[...]" assertions.
Sorry about that strangerep, I'll try to phrase my questions more neutrally. First of all, is it true that for a dissipative system, the time-dependent Hamiltonian operator is the infinitesimal generator of the 1-parameter time translation group? By that I mean the following: is it true that U(t,dt)=1-iH(t)dt? (modulo a factor of h-bar)? If so, is it fair to say that the Lie algebra generators inherit the parametric dependence of the associated Lie group?
 In classical Hamiltonian dynamics the relevant group becomes a much larger symplectic group. How much classical Hamiltonian dynamics have you studied? Maybe I'm pitching my answers at the wrong level.
Yes, you may be. I've mainly studied advanced mechanics in the Lagrangian formalism. And I'm not really familiar with terms like "symplectic group". My knowledge of Hamiltonian mechanics mainly comes from allusions to it in quantum mechanics books. And just to have all my cards on the table, I know nonrelativistic QM at least at the level of Sakurai, and what I know of Lie groups and Lie algebra also largely originates from reading about QM.
 This is another "are you saying [...]" response that I find tedious to deconstruct. But it's bedtime for me now. Maybe I'll have more energy tomorrow.
OK, again let me make my question less loaded. Can the momentum operator have a parametric dependence on position, and can the spin angular momentum operator (in a given direction) have a parametric dependence on angle? If not, is the reason they can't a mathematical reason or a physical reason?
P: 613
 Quote by dextercioby In the same way spin is embedded in the Dirac equation, it is also embeded in a linearized version of the Schrödinger equation as shown by Levy-Leblond and explained neatly in one of Greiner's books (either the one on symmetries, or the one on wave equations, don't remember which).
This is really something new for me. Just to get you right: you say that the Schrödinger equation and the Klein-Gordon equation are on equal footing, as well as the Dirac equation and a first-order Schrödinger-like equation? This implies, that the Klein-Gordon equation describes the behaviour of electrons correctly, in cases where spin doesn't matter. So can I really calculate the hydrogen spectrum including relativistic effects (but obviously without spin-orbit coupling) from it?

And thanks for the further references, but they are currently out of my scope.
HW Helper
P: 11,717
 Quote by kith This is really something new for me. Just to get you right: you say that the Schrödinger equation and the Klein-Gordon equation are on equal footing, as well as the Dirac equation and a first-order Schrödinger-like equation?[...]
Yes.

 Quote by kith [...] This implies, that the Klein-Gordon equation describes the behaviour of electrons correctly, in cases where spin doesn't matter.[...]
Yes.

 Quote by kith So can I really calculate the hydrogen spectrum including relativistic effects (but obviously without spin-orbit coupling) from it?[...]
Of course you can. This is done in some books, where an exact formula for the bound states spectrum is obtained, as opposed to applying first order perturbation theory to the relativistic correction (due to energy varying with speed, the p^4 term) which is exposed in almost all books of quantum mechanics/atomic physics .
P: 613
 Quote by dextercioby This is done in some books, where an exact formula for the bound states spectrum is obtained[...]
Ok, thanks for clarifying (wikipedia says the same).
P: 1,472
 Quote by lugita15 I've mainly studied advanced mechanics in the Lagrangian formalism. And I'm not really familiar with terms like "symplectic group". My knowledge of Hamiltonian mechanics mainly comes from allusions to it in quantum mechanics books
I'm unlikely to have time to compose a detailed answer in the next few days, but I can say for sure that the fastest way to a better understanding of all this is to get hold of one of the textbooks I mentioned (Goldstein, or Jose & Saletan) to read up on Hamiltonian dynamics, canonical transformations, and how it all fits together. You could try these Wiki articles to get a (very brief) overview:

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.
P: 22
 Quote by lugita15 In quantum mechanics, the Hamilltonian operator is constructed as the infinitesimal generator of the time translation group, which is a 1-parameter group. Yet it can still depend on time. So you have a situation where the generator of a 1-parameter group can depend on the parameter. Yet the momentum operator, the generator of infinitesimal spatial translations, cannot depend on the parameter(s) of the spatial translation group, namely position. And similarly the angular momentum operator cannot depend on angle (or direction), the parameter(s) of the rotation group. Is there a fundamental reason for this, or is it simply that we happen to already know the properties of the dynamical variables from classical mechanics, so we don't bother with it?
I think it's simply because spatial translations and rotations form always a group, while time translations not. Stone's theorem says that the generator is fixed.
P: 1,574
 Quote by naffin I think it's simply because spatial translations and rotations form always a group, while time translations not. Stone's theorem says that the generator is fixed.
Under what circumstances do the unitary time evolution operators not form a group?
 P: 22 (I was obviously referring to one-parameter groups) If your system is a particle in a time-dependent potential it is clearly not true that $U(t_1)U(t_2) = U(t_1 + t_2)$ in general, so you can't use Stone's theorem.
P: 1,574
 Quote by naffin (I was obviously referring to one-parameter groups) If your system is a particle in a time-dependent potential it is clearly not true that $U(t_1)U(t_2) = U(t_1 + t_2)$ in general.
That's completely new to me. So you're saying that if H(t1)H(t2)≠H(t2)H(t1), then U(t1)U(t2)≠U(t2)U(t1)? I thought U(t1)U(t2)=U(t2)U(t1) regardless, but you just have to be more careful. In QM books, they often do an informal derivation where they say that the Hamiltonian is constant on each infinitesimal interval (t,t+dt), and so you can define U(t,dt) quite easily. Then just take an infinite product to find the time evolution operator for a finite time interval. Is there anything wrong with that kind of approach?
P: 22
 Quote by lugita15 In QM books, they often do an informal derivation where they say that the Hamiltonian is constant on each infinitesimal interval (t,t+dt), and so you can define U(t,dt) quite easily. Then just take an infinite product to find the time evolution operator for a finite time interval. Is there anything wrong with that kind of approach?
Hamiltonian is still defined by the Schrodinger equation, so it satisfies the equation $i \hbar \frac{dU}{dt} (t) = H(t) U(t)$.
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 $U(t_0, t_1)$ depends in general on t0 and t1 separately, not only on their difference $U(t_1 - t_0)$.
All you can say is that $U(0, t_1 + t_2) = U(t_1, t_1 + t_2) U(0, t_1)$ and not that $U(0, t_1 + t_2) = U(0, t_2) U(0, t_1)$.
If $U(t_2, t_1) =U(t_1 - t_2)$ holds for any t1,t2 you can apply Stone's theorem, so you have a time-independent Hamiltonian.
P: 1,574
 Quote by naffin Hamiltonian is still defined by the Schrodinger equation, so it satisfies the equation $i \hbar \frac{dU}{dt} (t) = H(t) U(t)$. 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 $U(t_0, t_1)$ depends in general on t0 and t1 separately, not only on their difference $U(t_1 - t_0)$.
I agree with this.
 All you can say is that $U(0, t_1 + t_2) = U(t_1, t_1 + t_2) U(0, t_1)$ and not that $U(0, t_1 + t_2) = U(0, t_2) U(0, t_1)$. If $U(t_2, t_1) =U(t_1 - t_2)$ holds for any t1,t2 you can apply Stone's theorem, so you have a time-independent Hamiltonian.
So then can you consider the time-evolution operators to form a two-parameter group or something?
 P: 22 No, two-parameter group implies one-parameter group, the other parameter fixed.
P: 1,574
 Quote by naffin No, two-parameter group implies one-parameter group, the other parameter fixed.
OK, so then what structure does the set of unitary time evolution operators possess?
P: 22
Ballentine, pag. 89.
For the exact solution I know "Quantum Mechanics I", Galindo-Pascual.

In general they are just unitary operators, I don't know other structures.
P: 1,574
Here's what I found in Reed and Simon on the subject of time-dependent Hamiltonians and the associated time evolution operators. They call the structure of the set of time evolution operators a unitary propagator.
Attached Files
 Reed.Simon Excerpt.pdf (342.3 KB, 8 views)
 P: 254 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?
P: 1,574
 Quote by Amok 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?
Certainly, the spherical coordinate position operators are observables in QM. But my question was rather about angle as a parameter of the intrinsic rotation group, i.e. the rotation operators which give rise to spin angular momentum (since spin angular momentum does not arise from any position operators). So the question was, why can't the spin angular momentum operator get the angular parameter of those rotation operators, just as the Hamiltonian operator gets the time parameter of the time evolution operators. Now I'm finding out that the relationship between a time-dependent Hamiltonian operators and the time evolution operators may not be as simple as it seems.
 Sci Advisor P: 1,472 (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 angle-like 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...]

 Related Discussions Advanced Physics Homework 4 Math & Science Software 2 Calculus & Beyond Homework 3 Introductory Physics Homework 0 Advanced Physics Homework 3