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Why isn't momentum a function of position?

by lugita15
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Ilmrak
#19
Mar6-12, 04:09 PM
P: 97
Quote Quote by lugita15 View Post
Well, what I'm trying to understand is why the Hamiltonian operator can have a dependence on time, the parameter of the time translation group, but the spin angular momentum operator cannot have a dependence on angle, the parameter of the (intrinsic) rotation group.
In a time translational invariant theory, total energy can't depend on time because it's the Noether charge related to that symmetry.

Ilm
lugita15
#20
Mar6-12, 04:17 PM
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Quote Quote by Ilmrak View Post
In a time translational invariant theory, total energy can't depend on time because it's the Noether charge related to that symmetry.
Yes, I know that, but you can have a system with a time-dependent external force, and you'll get a time-dependent Haimiltonian operator. Is the reason why you can't have analogous situations for momentum or angular momentum something rigorous in the quantum mechanical formalism, or is it just due to our classical knowledge of how things are supposed to work?

I suppose my question could be phrased in purely mathematical terms: under what conditions do the generators of a Lie algebra inherit the parametric dependence of the associated Lie group?
kith
#21
Mar6-12, 05:45 PM
P: 729
Quote Quote by lugita15 View Post
Is the reason why you can't have analogous situations for momentum or angular momentum something rigorous in the quantum mechanical formalism, or is it just due to our classical knowledge of how things are supposed to work?
Interesting question.

I don't think the answer is something special to quantum mechanics. Also in the classical Lagrangian/Hamiltonian formalism, conjugate momenta don't depend on their generalized coordinates. Maybe the question is similar to "why do we have second order differential equations?".

Time is special, because it is not a conjugate quantity to anything or a function of such quantities. I.e. it's just a parameter and not an observable in Hamiltonian mechanics.
lugita15
#22
Mar6-12, 05:50 PM
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Quote Quote by kith View Post
Interesting question.

I don't think the answer is something special to quantum mechanics. Also in the classical Lagrangian/Hamiltonian formalism, conjugate momenta don't depend on their generalized coordinates. So the answer seems to be related to the fact, that a classical state is characterized by two independent values for every degree of freedom.

Time is special, because it is not a conjugate quantity to anything or a function of such quantities. I.e. it's just a parameter and not an observable in Hamiltonian mechanics.
But as I told Tom, for the spin angular momentum operator there is no "intrinsic angle operator" conjugate to it, so why can't the spin angular momeentum operator depend on an angle parameter?
kith
#23
Mar6-12, 06:18 PM
P: 729
Quote Quote by lugita15 View Post
But as I told Tom, for the spin angular momentum operator there is no "intrinsic angle operator" conjugate to it, so why can't the spin angular momeentum operator depend on an angle parameter?
Well, spin is a very special case because in NRQM, it is introduced ad hoc. If we want to think about your question in terms of the Hamiltonian formalism, we have to use a theory which describes the relevant observables in this framework. For spin, this would be QFT.
lugita15
#24
Mar6-12, 06:26 PM
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Quote Quote by kith View Post
Well, spin is a very special case because in NRQM, it is introduced ad hoc.
That's actually not true. If you construct Hilbert space operators out of the representation theory of the Galilei group, just as you do the analogous thing with the Poincare group in QFT, you will naturally get spin angular momentum; see Ballentine for details. What is genuinely relativistic is the spin-statistics theorem.
kith
#25
Mar6-12, 06:46 PM
P: 729
Quote Quote by lugita15 View Post
That's actually not true. If you construct Hilbert space operators out of the representation theory of the Galilei group, just as you do the analogous thing with the Poincare group in QFT, you will naturally get spin angular momentum; see Ballentine for details. What is genuinely relativistic is the spin-statistics theorem.
Which section in Ballentine do you have in mind? Substituting QxP by QxP+S (as he does in chapter 3) seems pretty ad hoc to me.

For relativistic QM, there are physical reasons for the existence of spin. I really wonder what such reasons could be in the non-relativistic case.
strangerep
#26
Mar6-12, 07:35 PM
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Quote Quote by lugita15
What is genuinely relativistic is the spin-statistics theorem.
Actually, the boson-fermion superselection rule can be derived from the properties
of the rotation group alone.

Hegerfeldt, Kraus, Wigner,
"Proof of the Fermion Superselection Rule without the
Assumption of Time-Reversal Invariance",
J. Math. Phys., vol 9, no 12, (1968), p2029.

Abstract: The superselection rule which separates states with integer
angular momentum from those with half-integer angular momentum is proved using
only rotational invariance.


Their argument is essentially a more rigorous version of the one in Ballentine's
section 7.6 about rotations by ##2\pi##. This sort of thing can be developed to
reveal a spin-statistics theorem for non-relativistic QM.

Quote Quote by lugita15
I suppose my question could be phrased in purely mathematical terms: under
what conditions do the generators of a Lie algebra inherit the parametric
dependence of the associated Lie group?
Never. The Lie algebra is derived from the Lie group by differentiating wrt
the parameters and then setting the parameters to 0.

The confusion about time in dissipative Hamiltonian systems is a different
issue. Let's go back to your original question:
Quote Quote by lugita15
[...] the Hamiltonian 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.
Cases where the Hamiltonian depends on time involve (at least) a second
subsystem in some way. E.g., a dissipative system can gain or lose energy
from/to another system, a system under the influence of a time-dependent
external force presumes the existence of another system responsible for that
force, etc.

So in general we have a composite system whose total Hamiltonian is
time-independent. But for the component subsystems, their individual evolution
parameters might not coincide with a global time parameter associated with the
total Hamiltonian. One chooses the component-specific evolution parameters to
make the maths as convenient as possible, and (presumably) to coincide with
some notion of local clock associated with that subsystem.

Herein lies a deep question about the distinction between kinematics and
dynamics. There is a (no-acceleration) theorem of Currie-Jordan-Sudarshan
which shows that assuming a common evolution parameter associated with
interacting particles is not viable in general: their respective worldlines in a
common Minkowski space fail to transform in a way which is compatible with
the interacting versions of the Hamiltonian and Lorentz boost operators.

The usual way to construct dynamics is the so-called "instant form" in which
we add an interaction term to the Hamiltonian (and to the Lorentz boost
generators in the relativistic case). This is motivated by our familiarity
with our everyday picture of Euclidean 3D space and our imagined reference
frame coordinatized implicitly in a way which is compatible with free
dynamics. Ballentine describes this briefly on p83 where he justfies modifying
only the Hamiltonian in the Galilean algebra to accommodate external fields.

But this is not the only possible way that we can try to make a "split" between
kinematics and dynamics. There's also the "point form" and "front form" which
modify other generators, but I won't delve into the details here.

It may even be the case that none of these relatively simple approaches are
truly adequate for all purposes. Sudarshan and collaborators also experimented
with more general alternatives in which the evolution parameter is determined
dynamically rather than via a once-and-for-all split between kinematics and
dynamics (which is what's done in the other forms of dynamics I listed above).

I can probably dig out more references for the above if necessary, but that's
probably enough for now.
dextercioby
#27
Mar7-12, 02:30 PM
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Quote Quote by kith View Post
Which section in Ballentine do you have in mind? Substituting QxP by QxP+S (as he does in chapter 3) seems pretty ad hoc to me.

For relativistic QM, there are physical reasons for the existence of spin. I really wonder what such reasons could be in the non-relativistic case.
Spin occurs naturally in the theory of vector representations of a central extension of the universal covering group of the (proper) Galilei group in the same way it emerges from the theory of vector representations of the universal cover of the restricted Poincaré group, i.e. through one of its Casimirs. This is a result known from the work of Bargmann and especially Levy-Leblond way back from the 1960's.

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).
lugita15
#28
Mar7-12, 02:36 PM
P: 1,583
Quote Quote by dextercioby View Post
Spin occurs naturally in the theory of vector representations of a central extension of the universal covering group of the (proper) Galilei group in the same way it emerges from the theory of vector representations of the universal cover of the restricted Poincaré group, i.e. through one of its Casimirs. This is a result known from the work of Bargmann and especially Levy-Leblond way back from the 1960's.
OK, this seems pretty convincing.
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 seems less convincing. Why in the world would you choose to linearize the Schrodinger equation, other than the fact that you know it somehow gives you spin?
dextercioby
#29
Mar7-12, 02:48 PM
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Quote Quote by lugita15 View Post
[...] This seems less convincing. Why in the world would you choose to linearize the Schrodinger equation, other than the fact that you know it somehow gives you spin?
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> ?
lugita15
#30
Mar7-12, 03:37 PM
P: 1,583
Quote Quote by strangerep View Post
Actually, the boson-fermion superselection rule can be derived from the properties of the rotation group alone.

Hegerfeldt, Kraus, Wigner,
"Proof of the Fermion Superselection Rule without the
Assumption of Time-Reversal Invariance",
J. Math. Phys., vol 9, no 12, (1968), p2029.

Abstract: The superselection rule which separates states with integer
angular momentum from those with half-integer angular momentum is proved using
only rotational invariance.


Their argument is essentially a more rigorous version of the one in Ballentine's
section 7.6 about rotations by ##2\pi##. This sort of thing can be developed to
reveal a spin-statistics theorem for non-relativistic QM.
Are you talking about the Jabs paper discussed in this thread? You said this in that thread:
Quote Quote by strangerep View Post
The thing that still leaves me a little perplexed is this: although demanding a consistent sense for the rotation transformations sounds asthetically pleasing, I have trouble seeing why it's essential (a priori) from a physical perspective. But hey, the double-valuedness of rotations is tricky at the best of times -- needing the "Dirac belt trick" or similar devices to illiustrate it.
Have your qualms been resolved since then?
Never. The Lie algebra is derived from the Lie group by differentiating wrt
the parameters and then setting the parameters to 0.

The confusion about time in dissipative Hamiltonian systems is a different
issue.
So are you saying that for a dissipative Hamiltonian system, the time-dependent operator does not generate the associated time evolution operators of the system? How can that possibly be?
Cases where the Hamiltonian depends on time involve (at least) a second
subsystem in some way. E.g., a dissipative system can gain or lose energy
from/to another system, a system under the influence of a time-dependent
external force presumes the existence of another system responsible for that
force, etc.

So in general we have a composite system whose total Hamiltonian is
time-independent. But for the component subsystems, their individual evolution
parameters might not coincide with a global time parameter associated with the
total Hamiltonian. One chooses the component-specific evolution parameters to
make the maths as convenient as possible, and (presumably) to coincide with
some notion of local clock associated with that subsystem.
OK, 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.
Herein lies a deep question about the distinction between kinematics and
dynamics. There is a (no-acceleration) theorem of Currie-Jordan-Sudarshan
which shows that assuming a common evolution parameter associated with
interacting particles is not viable in general: their respective worldlines in a
common Minkowski space fail to transform in a way which is compatible with
the interacting versions of the Hamiltonian and Lorentz boost operators.
Does this limitative theorem extend even to nonrelativistic QM? That would be surprising.
The usual way to construct dynamics is the so-called "instant form" in which
we add an interaction term to the Hamiltonian (and to the Lorentz boost
generators in the relativistic case). This is motivated by our familiarity
with our everyday picture of Euclidean 3D space and our imagined reference
frame coordinatized implicitly in a way which is compatible with free
dynamics. Ballentine describes this briefly on p83 where he justfies modifying
only the Hamiltonian in the Galilean algebra to accommodate external fields.
Here is what Ballentine has to say on the subect:
One may ask why only the time displacement generator H should be changed by the interactions, while the space displacement generators P are unchanged. If the system under consideration were a self-propelled machine, we could imagine it displacing itself through space under its own power, consuming fuel, expelling exhaust, and dropping worn-out parts along the way. If P generated that kind of displacement, then the form of the operators P certainly would be altered by the interactions that were responsible for the displacement. But that is not what we mean by the operation of space displacement. Rather, we mean the purely geometric operation of displacing the system self-congruently to another location. This is the reason why P and the other generators of symmetry operations are not changed by dynamical interactions. However, H is redefined to be the generator of dynamic evolution in time, rather than merely a geometric displacement along the time axis.
This 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?
lugita15
#31
Mar7-12, 03:40 PM
P: 1,583
Quote Quote by dextercioby View Post
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
#32
Mar7-12, 04:21 PM
P: 353
Klein-Gordon and Schrodinger's equations are linear. Surely you don't mean linearization but finding degree one equation.
dextercioby
#33
Mar7-12, 06:05 PM
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Yes, linearization = linear dependence of first derivatives.
strangerep
#34
Mar7-12, 07:53 PM
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Quote Quote by lugita15 View Post
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
#35
Mar8-12, 01:18 AM
P: 1,583
Quote Quote by strangerep View Post
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
#36
Mar8-12, 03:24 AM
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Quote Quote by lugita15 View Post
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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