Why isn't momentum a function of position?

[lugita15]

Why would you linearize the Klein-Gordon equation (as Dirac did in 1928), if the square rooted equation was enough to cure the <negative norm issue> ?

Probably because Dirac wasn't aware of this solution to the problem, so the only available resolution he saw was linearization. But the problem with the Klein-Gordon equation doesn't arise for the Schrodinger equation, so you wouldn't feel a need to linearize it, unless you knew in advance that it would yield spin.

 

[martinbn]

Klein-Gordon and Schrodinger's equations are linear. Surely you don't mean linearization but finding degree one equation.

 

[dextercioby]

Yes, linearization = linear dependence of first derivatives.

 

[strangerep]

Are you talking about the Jabs paper

No.

So are you saying [...]

No.

[...] but you only deal with different time parameters in relativity, don't you? (And in relativity, isn't time not a parameter anyway, but rather part of the position 4-vector operator?) In nonrelativistic QM all the time parameters are the same.

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.

[...]what Ballentine has to say [...] seems rather hand-wavy to me. Why is the Hamiltonian operator singled out to redefined as "the generator of dynamic evolution in time, rather than merely a geometric displacement along the time axis"? Why can't you have, akin to a dissipative time-dependent Hamiltonian operator, a position-dependent momentum operator or a angle-dependent angular momentum operator?

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. In principle, one can find canonical transformations between one form of dynamics and another (if they correspond to the same physical situation).

One can sometimes think more clearly about this stuff by reviewing some classical dynamics theory, and concentrating on canonical coordinates and momenta, and canonical transformations which preserve the dynamics (Hamilton's equations). E.g., if we start with a canonical pair (q,p) we can find transformations to a new pair (q'(q,p), p'(q,p)) such that canonical commutations relations (or Poisson brackets) still hold for (q',p'). But p' is a function of the old coordinate and momentum -- this is not important, because we also have a new canonical coordinate q'.

More generally, there are also "extended" canonical transformations which involve the time parameter. For more detail, try Goldstein, or Jose & Saletan.

 

[lugita15]

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?

 

[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:

<br /> U(t_1,t_2)= T\{exp{ {\large (} -\frac{i}{\hslash} \int_{t_1}^{t_2} \textbf{d}t \, H(t) {\large)} }\} \equiv <br /> \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)} \}<br />

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.