Why isn't momentum a function of position?

lugita15

So then where can I find out how to go from the superselection rule derivation presented in Ballentine to the spin-statistics theorem? I'm a little confused. You said that the generators of a Lie algebra can never inherit the parametric dependence of the associated Lie group, but you're also agreeing with me that for a dissipative system the time-dependent Hamiltonian operator generates the time translation group, which is a group whose parameter is time. But in the nonrelativistic case the Galilean transformation treats time as trivial, so what's the point of even fiddling with the time parameter? So are you saying that you can have the momentum operator have a parametric dependence on position, and the spin angular momentum operator have a parametric dependence on angle?

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.

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.

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.

lugita15

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? 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. 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?

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? 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.

dextercioby

Yes.

Yes.

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 .

kith

Ok, thanks for clarifying (wikipedia says the same).

strangerep

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.

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.

lugita15

Under what circumstances do the unitary time evolution operators not form a group?

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 [itex]U(t_1)U(t_2) = U(t_1 + t_2) [/itex] in general, so you can't use Stone's theorem.

lugita15

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?

naffin

Hamiltonian is still defined by the Schrodinger equation, so it satisfies the equation [itex] i \hbar \frac{dU}{dt} (t) = H(t) U(t)[/itex].
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 time-independent Hamiltonian.

lugita15

Do you have any details about this, or do you know where I can find more information? I agree with this.So then can you consider the time-evolution operators to form a two-parameter group or something?

naffin

No, two-parameter group implies one-parameter group, the other parameter fixed.

lugita15

OK, so then what structure does the set of unitary time evolution operators possess?

naffin

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.

lugita15

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)