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## Why isn't momentum a function of position?

Quote by strangerep
 Are you talking about the Jabs paper
No.
So then where can I find out how to go from the superselection rule derivation presented in Ballentine to the spin-statistics theorem?
 So are you saying [...]
No.
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.
 Consider classical Hamiltonian dynamics. One can apply time-dependent canonical transformations (which include mapping one time parameter to another), obtaining a different description of the same dynamical physical situation.
But in the nonrelativistic case the Galilean transformation treats time as trivial, so what's the point of even fiddling with the time parameter?
 If you mean "why can't you have an interaction term in a momentum operator?", well you can: it's called the "point form" of dynamics.
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?

Recognitions:
 Quote by lugita15 I[...] you're also agreeing with me that for a dissipative system the time-dependent Hamiltonian operator generates the time translation group
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.

 But in the nonrelativistic case the Galilean transformation treats time as trivial, so what's the point of even fiddling with the time parameter?
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.

 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?
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.

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 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?

 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.

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Recognitions:
Homework Help
 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 .

 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).

Recognitions:
 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.

 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.

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 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?
 (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.

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 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?

 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.

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

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 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?