Why isn't momentum a function of position?

[strangerep]

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.

 

[lugita15]

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?

 

[kith]

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.

 

[dextercioby]

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.

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

Yes.

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 .

 

[kith]

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).

 

[strangerep]

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.

 

[naffin]

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.

 

[lugita15]

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?

 

[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, so you can't use Stone's theorem.

 

[lugita15]

(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?

 

[naffin]

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.

 

[lugita15]

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.

Do you have any details about this, or do you know where I can find more information?

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?

 

[naffin]

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

 

[lugita15]

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?

 

[naffin]

Do you have any details about this, or do you know where I can find more information?

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.

 

[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?

 

[lugita15]

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.

 

[strangerep]

(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...]

 

[lugita15]

strangerep, do you have any thoughts on the relationship between time-dependent Hamiltonian operators and the associated family of unitary time evolution operators (especially in the case where the Hamiltonian operator at time t1 does not commute with the Hamiltonian operator at time t2)? And why, whatever this relationship is, a similar relationship cannot exist between spin angular momentum operators and the intrinsic rotation group/

 

[strangerep]

strangerep, do you have any thoughts on the relationship between time-dependent Hamiltonian operators and the associated family of unitary time evolution operators (especially in the case where the Hamiltonian operator at time t1 does not commute with the Hamiltonian operator at time t2)?

A classical Hamiltonian (function on phase space) is mapped to a quantum Hamiltonian (operator on Hilbert space) as part of the procedure of quantizing that particular dynamical case. I don't know what more you expect. That's pretty much it (except possibly for some ambiguities with higher order classical functions not mapping uniquely to quantum operators -- but that's often handled ok by symmetrization).

BTW, have you tried to get hold of either of the classical mechanics textbooks I suggested?

And why, whatever this relationship is, a similar relationship cannot exist between spin angular momentum operators and the intrinsic rotation group

? Spin is intrinsic angular momentum. I don't know what you mean by "intrinsic rotation group". SO(3) is SO(3).

 

[lugita15]

A classical Hamiltonian (function on phase space) is mapped to a quantum Hamiltonian (operator on Hilbert space) as part of the procedure of quantizing that particular dynamical case. I don't know what more you expect.

That's not really what I want to know. I want to know the relationship between the time-dependent Hamiltonian operator and the time evolution operators. Clearly the idea of the time evolution operator U being the exponential exp(iHt) of the Hamiltonian operator only works for time-independent Hamiltonian. And the idea of U being the exponential of the integral of the Hamiltonian operator only works if the Hamiltonian operators at different times commute. So what do we do in the general case? In QM textbooks what is often done is to say that the Hamiltonian operator is approximately time independent on each interval (t,t+dt), so the U(dt) is approximately equal to exp(iHdt). Then to find the time evolution operator over a finite time, we take the infinite product of the operators for infinitesimal times. Can this be made more rigorous?

BTW, have you tried to get hold of either of the classical mechanics textbooks I suggested?

I already have Goldstein; I read the earlier chapters years ago. I just need to find time to read the two chapters on Hamiltonian mechanics and canonical transformations.

? Spin is intrinsic angular momentum. I don't know what you mean by "intrinsic rotation group". SO(3) is SO(3).

Well, there are two groups of Hilbert space operators which are isomorphic to a rotation group, the one generated by the spin angular momentum operator and the one generated by the orbital angular momentum operator.

 

[Ilmrak]

That's not really what I want to know. I want to know the relationship between the time-dependent Hamiltonian operator and the time evolution operators. Clearly the idea of the time evolution operator U being the exponential exp(iHt) of the Hamiltonian operator only works for time-independent Hamiltonian. And the idea of U being the exponential of the integral of the Hamiltonian operator only works if the Hamiltonian operators at different times commute. So what do we do in the general case? In QM textbooks what is often done is to say that the Hamiltonian operator is approximately time independent on each interval (t,t+dt), so the U(dt) is approximately equal to exp(iHdt). Then to find the time evolution operator over a finite time, we take the infinite product of the operators for infinitesimal times. Can this be made more rigorous?

The procedure you found is almost correct, it could be made rigorous introducing a time-ordering:

$U(t_1,t_2)= T\{exp{ {\large (} -\frac{i}{\hslash} \int_{t_1}^{t_2} \textbf{d}t \, H(t) {\large)} }\} \equiv \sum_{n=0...\infty} (-\frac{i}{\hslash})^n \frac{1}{n!} \int_{t_1}^{t_2} \textbf{d}\tau_1 \cdots \,\int_{t_1}^{t_2} \textbf{d}\tau_n \, T\{{H(\tau_1)\cdots H(\tau_n)} \}$

where $T\{{H(\tau_1)\cdots H(\tau_n)} \}$ is the product obtained disposing hamiltonians evaluated in lower times to the right. Obviously if hamiltonians commutes at different times the time ordering has no effect.

I already have Goldstein; I read the earlier chapters years ago. I just need to find time to read the two chapters on Hamiltonian mechanics and canonical transformations.
Well, there are two groups of Hilbert space operators which are isomorphic to a rotation group, the one generated by the spin angular momentum operator and the one generated by the orbital angular momentum operator.

Actually the spin group structure is isomorphic to $SU(2)$, which is the universal covering group of $O(3)$ but I don't think this matters here.

If I interpreted it right I think we could reprase your question this way: could exist a representation of an abstract Lie group on a Hilbert space such that the representation of the generators of this group depends on the group parameter?

If this was the question I think the answer would be no, we can't. But I'm a bit confused thinking about the time dependent hamiltonian

Ilm

 

[strangerep]

That's not really what I want to know. I want to know the relationship between the time-dependent Hamiltonian operator and the time evolution operators. [...]

This would be easier to discuss with a specific example of a time-dependent Hamiltonian.

Well, there are two groups of Hilbert space operators which are isomorphic to a rotation group, the one generated by the spin angular momentum operator and the one generated by the orbital angular momentum operator.

This is not quite right. The operators on those Hilbert spaces are particular representations of the so(3) generators $J_i$, but they correspond to different values of the Casimir operator $J^2$.

If one seeks a Hilbert space representation of the "pure" so(3) algebra, one finds that the Casimir values are 0, 1/2, 1, ... -- cf. Ballentine sect 7.1.

But if one imposes the additional restriction that $J_i = \epsilon_{ijk} q_j p_k$, (i.e., orbital angular momentum), there's a consequential restriction on the eigenvalues of $J^2$ to integer values only. Cf. Ballentine 7.3.

This is because we're really dealing with two distinct algebras in each case.

 

[lugita15]

This would be easier to discuss with a specific example of a time-dependent Hamiltonian.

I think a good example would be one where the Hamiltonian operators at different times don't commute, but I'm afraid I can't think of one off the top of my head. Do you know of a good one?

This is not quite right. The operators on those Hilbert spaces are particular representations of the so(3) generators $J_i$, but they correspond to different values of the Casimir operator $J^2$.

If one seeks a Hilbert space representation of the "pure" so(3) algebra, one finds that the Casimir values are 0, 1/2, 1, ... -- cf. Ballentine sect 7.1.

But if one imposes the additional restriction that $J_i = \epsilon_{ijk} q_j p_k$, (i.e., orbital angular momentum), there's a consequential restriction on the eigenvalues of $J^2$ to integer values only. Cf. Ballentine 7.3.

This is because we're really dealing with two distinct algebras in each case.

You're right, I neglected the difference between SO(3) and SU(2), which as you said have a slight difference in their representation theory.

Anyway, now that I've clarified what I mean by the intrinsic rotation group, i.e. the family of rotation operators generated by the spin angular momentum operator, let me repeat my question: why can't the spin angular momentum operator bear the same kind of nontrivial relationship to the intrinsic rotation group that a time-dependent Hamiltonion operator bears to the unitary time evolution group? The reason I raised this a bit earlier in this thread is that like the time evolution group and unlike the spatial translation group, the intrinsic rotation group has a parameter which does not correspond to an observable. So it gets around tom.stoer's objection that giving momentum a position dependence imperils the canonical commutation relation.

 

[lugita15]

The procedure you found is almost correct, it could be made rigorous introducing a time-ordering:

$U(t_1,t_2)= T\{exp{ {\large (} -\frac{i}{\hslash} \int_{t_1}^{t_2} \textbf{d}t \, H(t) {\large)} }\} \equiv \sum_{n=0...\infty} (-\frac{i}{\hslash})^n \frac{1}{n!} \int_{t_1}^{t_2} \textbf{d}\tau_1 \cdots \,\int_{t_1}^{t_2} \textbf{d}\tau_n \, T\{{H(\tau_1)\cdots H(\tau_n)} \}$

where $T\{{H(\tau_1)\cdots H(\tau_n)} \}$ is the product obtained disposing hamiltonians evaluated in lower times to the right. Obviously if hamiltonians commutes at different times the time ordering has no effect.

Where can I find more information about the time-ordering? Is there any way to write this expression as an infinite product rather than an infinite sum?

If I interpreted it right I think we could reprase your question this way: could exist a representation of an abstract Lie group on a Hilbert space such that the representation of the generators of this group depends on the group parameter?

If this was the question I think the answer would be no, we can't. But I'm a bit confused thinking about the time dependent hamiltonian

I asked almost this exact question earlier this thread. I gather from naffin that the answer is that the family of time evolution operators does not in general form a one-parameter Lie group, so it doesn't have to be generated by a parameter-independent Lie algebra. So then the question becomes, why can't similar things occur for other operators?

 

[strangerep]

I think a good example would be one where the Hamiltonian operators at different times don't commute, but I'm afraid I can't think of one off the top of my head. Do you know of a good one?

Not really. I normally work with conservative systems. But in any case, it's likely to be simpler to try and understand the situation in the classical case first. Jose & Saletan give an example of a dissipative system with friction, but not in the Hamiltonian formalism. In such case, one must work with a semigroup (i.e., not necessarily having inverses), since such dissipative dynamics are not reversible.

For a time-dependent Hamiltonian in a conservative system, one can presumably find a canonical transformation to a new set of dynamical variables, and a new time variable $\tau$, such that the new Hamiltonian is independent of $\tau$.

why can't the spin angular momentum operator bear the same kind of nontrivial relationship to the intrinsic rotation group that a time-dependent Hamiltonion operator bears to the unitary time evolution group? The reason I raised this a bit earlier in this thread is that like the time evolution group and unlike the spatial translation group, the intrinsic rotation group has a parameter which does not correspond to an observable.

Even in the classical case, the dynamics of a rotating rigid body are far trickier than the ordinary cases. The configuration manifold is in fact the SO(3) matrices themselves, and the obvious choices for momenta -- the angular momentum generators -- cannot be used as canonical momenta in the usual Hamiltonian sense because they do not mutually commute in the Poisson bracket sense. (One can still write a Hamiltonian in terms of them, and an inertia vector, but this Hamiltonian is not expressed in terms of canonical momenta.) So one must work with other formalisms of dynamics if one wants to get anywhere.

BTW, I had another look through Goldstein and J+S. I now don't think Goldstein covers enough ground for this discussion. J+S cover more.

 

[Ilmrak]

Where can I find more information about the time-ordering? Is there any way to write this expression as an infinite product rather than an infinite sum?

I don't actually know where to find more on time ordering in QM, I would say any QM book as Sakurai, but I'm not sure. For QFT though, any book covers this (Peskin for example).

Anyway I'll give some sketch on how to find the expression I wrote.
Discretize the time between $t_1$ and $t_2$ dividing it in intervals $\Delta t = (t_2 - t_1)/N$. Now use the propagators in $\Delta t$, $U(t_1 + k \Delta t , t_1 + (k+1)\Delta t) ≈ \exp{ \frac{-i}{ \hslash } H(t_1 + k \Delta t) \Delta t } ≈ 1 - \frac{i}{\hslash} H (t_1 + k \Delta t) \Delta t$, to evolve the system in each time interval (each hamiltonian is commuting, at first order in $\Delta t$, with itself in the time interval it's propagating). The full propagator is the product of $N$ of such propagators and it's naturally time-ordered. Now take the continuous limit and you're done.

I asked almost this exact question earlier this thread. I gather from naffin that the answer is that the family of time evolution operators does not in general form a one-parameter Lie group, so it doesn't have to be generated by a parameter-independent Lie algebra. So then the question becomes, why can't similar things occur for other operators?

Please tell me if this could be plausible.

There are transformation defined to represent a symmetry and transformations that instead makes sense for their own.

Time propagation is a transformation of the second type: it always makes sense. The system could be symmetric under time translation, so that time propagation forms a group, or it can be non conservative, so that time propagation is not a group. In the latter case hamiltonian could depend on the time.

Rotations in QM are instead defined to form a group. The definition of angular momentum do not depends on the system symmetries, it's always the same. If a system is rotationally invariant, then hamiltonian commutes with rotations. In QM angular momentum can't, by definition, depend on the angles.

This difference is because in QM there is only one real space dimension: the time. Coordinates are operators, they are more like fields then like space dimensions.

In QFT there are instead 3+1 spacetime dimensions. Rotation in the 3-dimensional space now makes sense fro their own, even when we write a non-rotational invariant hamiltonian.
So we can, in this case, find that angular momentum are "represented" (not in the group theory sense) on the fields with operators that depends on the angles and then do not forms a group.
We probably never see this case because we assume our theory is Lorentz invariant.

Ilm

 

[lugita15]

Anyway I'll give some sketch on how to find the expression I wrote.
Discretize the time between $t_1$ and $t_2$ dividing it in intervals $\Delta t = (t_2 - t_1)/N$. Now use the propagators in $\Delta t$, $U(t_1 + k \Delta t , t_1 + (k+1)\Delta t) ≈ \exp{ \frac{-i}{ \hslash } H(t_1 + k \Delta t) \Delta t } ≈ 1 - \frac{i}{\hslash} H (t_1 + k \Delta t) \Delta t$, to evolve the system in each time interval (each hamiltonian is commuting, at first order in $\Delta t$, with itself in the time interval it's propagating). The full propagator is the product of $N$ of such propagators and it's naturally time-ordered. Now take the continuous limit and you're done.

How is the continuous limit of a product rigorously defined?

 

[strangerep]

Oh duh. I don't know why I didn't think of this before...

Some features of time-dependent Hamiltonians can be seen by studying the damped harmonic oscillator with external time-dependent applied force. Cf. J+S, sect 4.2.4. Goldstein also has some discussion of it, but J+S seems more extensive.) Instead of the usual solutions like $e^{i\omega t}$ one gets something like
$$
e^{(-\beta + i\omega) t}
$$
(plus a forced term which I haven't shown). The above can be interpreted in terms of a complex energy (where the imaginary part determines growth or decay of the oscillations).

An quantum example that's (sort of) related is the modeling of the formation and decay of unstable particles (resonances). Here, one uses a Hamiltonian which does not have strictly-real eigenvalues, so $e^{iHt}$ is not a unitary operator in the usual sense.

Instead of the usual state vector space, one uses a space of so-called Gamow vectors. Typically one needs a proper rigged Hilbert space, or a pair of Hardy spaces to represent all this satisfactorily. http://arxiv.org/abs/quant-ph/0201091

EDIT: Just found this recent paper:

Chandrasekar, Senthilvelan, Lakshmanan,
"On the Lagrangian and Hamiltonian description of the damped
linear harmonic oscillator",
Available as: http://arxiv.org/abs/nlin/0611048

Abstract:
Using the modified Prelle-Singer approach, we point out that explicit time independent first integrals can be identified for the damped linear harmonic oscillator in different parameter regimes. Using these constants of motion, an appropriate Lagrangian and Hamiltonian formalism is developed and the resultant canonical equations are shown to lead to the standard dynamical description. Suitable canonical transformations to standard Hamiltonian forms are also obtained. It is also shown that a possible quantum mechanical description can be developed either in the coordinate or momentum representations using the Hamiltonian forms.

However, the Lagrangian and Hamiltonian they obtain are quite complicated -- see p7.

 

[Ilmrak]

How is the continuous limit of a product rigorously defined?

You essentially write that time-ordered product as a time ordered exponential of a sum, then take the limit for $N \rightarrow \infty$ to obtain an integral from the sum.

I'm sorry but I can't find anything better then http://en.wikipedia.org/wiki/Path-ordering. If I should find something more complete I'll eventually post that reference.

Ilm

 

[Ilmrak]

Oh duh. I don't know why I didn't think of this before...

Some features of time-dependent Hamiltonians can be seen by studying the damped harmonic oscillator with external time-dependent applied force. Cf. J+S, sect 4.2.4. Goldstein also has some discussion of it, but J+S seems more extensive.) Instead of the usual solutions like $e^{i\omega t}$ one gets something like
$$
e^{(-\beta + i\omega) t}
$$
(plus a forced term which I haven't shown). The above can be interpreted in terms of a complex energy (where the imaginary part determines growth or decay of the oscillations).

An quantum example that's (sort of) related is the modeling of the formation and decay of unstable particles (resonances). Here, one uses a Hamiltonian which does not have strictly-real eigenvalues, so $e^{iHt}$ is not a unitary operator in the usual sense.

Instead of the usual state vector space, one uses a space of so-called Gamow vectors. Typically one needs a proper rigged Hilbert space, or a pair of Hardy spaces to represent all this satisfactorily. http://arxiv.org/abs/quant-ph/0201091

EDIT: Just found this recent paper:

Chandrasekar, Senthilvelan, Lakshmanan,
"On the Lagrangian and Hamiltonian description of the damped
linear harmonic oscillator",
Available as: http://arxiv.org/abs/nlin/0611048

Abstract:
Using the modified Prelle-Singer approach, we point out that explicit time independent first integrals can be identified for the damped linear harmonic oscillator in different parameter regimes. Using these constants of motion, an appropriate Lagrangian and Hamiltonian formalism is developed and the resultant canonical equations are shown to lead to the standard dynamical description. Suitable canonical transformations to standard Hamiltonian forms are also obtained. It is also shown that a possible quantum mechanical description can be developed either in the coordinate or momentum representations using the Hamiltonian forms.

However, the Lagrangian and Hamiltonian they obtain are quite complicated -- see p7.

A non-unitary representation of the time evolution would describe a dissipative system, dissipation depending on the anti-hermitian component of energy.
The group structure would tough be preserved, so I suspect a system described by an hamiltonian not commuting with itself at different times would be something different and more complicated?

Also I never fully understood resonances. Time evolution for decaying particles, if I'm not wrong, is still unitary on the full Fock space, it's only its projection on the initial particle subspace that isn't unitary (is from anti hermitian component of the energy projection on the decaying particle subspace that resonance width come from?).

I'll try to read your references, hoping to understand something

Ilm

 

[lugita15]

You essentially write that time-ordered product as a time ordered exponential of a sum, then take the limit for $N \rightarrow \infty$ to obtain an integral from the sum.

I'm sorry but I can't find anything better then http://en.wikipedia.org/wiki/Path-ordering. If I should find something more complete I'll eventually post that reference.

Ilmrak, have you found any references on how to properly define a continuous product? What is essentially required is some way to define a Type II product integral, except for general Lie algebras rather than just real numbers, where the exponential refers to the exponential map connecting the Lie algebra and the Lie group.

 

[lugita15]

Does anyone know anything further about the group structure known as the unitary propagator defined in the Reed and Simon excerpt in my post #51? Specifically, what can we say about the Lie algebra associated with a unitary propagator? Also, what can be said about the representation theory of a unitary propagator?

 

[Ilmrak]

Sorry, I didn't see your link to Reed and Simons...

The funny thing is that that article is a nice reference about the time-ordered expression for the propagator i wrote in post #58
Note in fact that the definition (X.129) is equivalent to mine in that post.
They then demonstrate that the operator $U(t_1 , t_2)$ defined that way is exactly the propagator.

I think that's no need to warry too much on the continuous limit, simply take post #58 as a sloppy way do "derive" the right definition of the time evolution operator ^^

Ilm