Unbounded operators in non-relativistic QM of one spin-0 particle

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The discussion revolves around the axioms of non-relativistic quantum mechanics (QM) for a spin-0 particle, focusing on the mathematical framework of Hilbert space L^2(ℝ³). Key points include the representation of observables by self-adjoint operators and the challenges posed by position and momentum operators, which lack eigenvectors. The concept of rigged Hilbert spaces (RHS) is introduced as a solution to the issues with unbounded operators, providing a rigorous foundation for Dirac's bra-ket formalism. Participants express a desire for clarity on the necessity of RHS in formulating quantum theory and its implications for understanding eigenstates and probabilities. Overall, the conversation emphasizes the importance of a robust mathematical structure in quantum mechanics.
  • #61
Hm, I might as well post what I have right now, even though I still haven't solved all the problems. It should make it easier for others to suggest improvements.

1. The possible states of a physical system are represented by the unit rays of an infinite-dimensional separable Hilbert space H over the complex numbers.

2. Let O and O' be two inertial observers, and let S and S' be the inertial frames naturally associated with their motion. Let g be the Galilei transformation that represents a coordinate change from S to S'. If O describes an isolated system as being in state F, then O' must describe the system as being in the state that contains the vector f'=U(g)f, where f is any vector in F and U:G→GL(H) is an irreducible unitary representation of the covering group of the Galilei group.

3. The physical states are represented by the unit rays of the nuclear space N generated by the U(g) operators (i.e. the space N that's invariant under any product of a finite number of U(g) operators). Measurable quantities are represented by self-adjoint operators on N. These operators are therefore called "observables". If a system is in state F and a measurement of an observable B is performed, the state of the system will change according to ρ → ∑b Pb ρ Pb

Axiom 2 is a bit awkward, but at least it defines momentum and spin as well, in addition to the Hamiltionian. Maybe position too, but I still don't quite get that part. I also realize now that I don't quite understand the significance of using the covering group instead of the Galilei group itself. I said that g is a Gallilei transformation, so what do we make of the expression U(g)? I mean, there are two members of G corresponding to each g, so which one of them do we use? It probably doesn't matter, but something should be said about this in the axiom.

Axiom 3 still isn't able to deal with continuous spectra, so it needs to be changed.
 
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  • #62
Fredrik said:
1. The possible states of a physical system are represented by the unit rays of an infinite-dimensional separable Hilbert space H over the complex numbers.

2. Let O and O' be two inertial observers, and let S and S' be the inertial frames naturally associated with their motion. Let g be the Galilei transformation that represents a coordinate change from S to S'. If O describes an isolated system as being in state F, then O' must describe the system as being in the state that contains the vector f'=U(g)f, where f is any vector in F and U:G→GL(H) is an irreducible unitary representation of the covering group of the Galilei group.
It's probably cleaner to start with the dynamical group as an axiom --
in your case Galilei. But when one tries to add a position operator, so
that one has a "Heisenberg-Galilei" group, one finds that the position
operator has to be a multiple of the Galilean boost operator. (See
Ballentine pp66-84). (The relativistic case is more problematic, of
course.)

But let's assume you start with the Heisenberg group, and require a
unitary irreducible representation thereof. A short proof then shows
that a finite-dimensional Hilbert space can't work, so you're forced to
inf-dim. That pretty much pushes you to L^2(R^3) (assuming you want 3
configuration degrees of freedom), or something that generalizes it. So
I don't think you need to postulate a specific Hilbert space. Simply
choosing a dynamical group and requiring unitary irreducible
representation(s) thereof is enough. Since some of the operators are
unbounded, that forces us to rigged Hilbert space (at least). I.e., we
want a rigged Hilbert space that carries a faithful irrep of a chosen
dynamical group.

Then, maths alone (spectral theorem) tells us that the space (and
indeed arbitrary operators on the space) can be decomposed in terms of
eigenspaces of a self-adjoint operator. Each such eigenspace
corresponds (via the associated eigenvalue) to a possible measurement
outcome. (This is why we chose self-adjoint operators: they have real
eigenvalues. We could have used more general operators if complex
eigenvalues were ok.) If we can find another (different) self-operator
that commutes with the first, we can further decompose each eigenspace.
And if we can find a *complete* set of commuting self-adjoint
operators, we can decompose right down to eigenvectors. All this pretty
much follows from having a (rigged) Hilbert space, hence need not be
explicit in the axioms.

So you don't really need axiom 3 in its current form (since the
spectral theorem also decomposes any operator in terms of projection
operators). You just need to say something about how states correspond
to density operators (and particular measurements to projection
operators).

Axiom 2 is a bit awkward, but at least it defines momentum and spin as well, in addition to the Hamiltionian. Maybe position too, but I still don't quite get that part.
Position is easy enough in Galilei. (Do you have a copy of
Ballentine yet? :-)

But I've never much liked the way position tends to be grafted
on as an afterthought.

More generally, one needs to start from a symplectic formulation (i.e.,
symplectic group on phase space), since that brings in non-trivial
dynamics, interactions, etc. But then one must use something like
Weyl-Wigner-Groenewold-Moyal quantization to establish a satisfactory
correspondence between phase space functions and quantum operators.
People continue to research this.

I also realize now that I don't quite understand the significance of
using the covering group instead of the Galilei group itself.
It helps you construct representations corresponding to
half-integral spin.

I said that g is a Gallilei transformation, so what do we make of the expression U(g)? I mean, there are two members of G corresponding to each g, so which one of them do we use? It probably doesn't matter, but something should be said about this in the axiom.
You could perhaps say that g is a member of covering Galilei.
 
  • #63
Thank you for another good answer. You've been more helpful to me already than my advisor for my M.Sc. thesis ever was. :smile:
strangerep said:
It's probably cleaner to start with the dynamical group as an axiom --
in your case Galilei.
I'm not sure what you mean. If you meant that I should use the Galilei group instead of its covering group, we wouldn't be able to let U be a representation. I'm sure you know about the annoying minus sign that we can't get rid of: U(g)U(h)=±U(gh).

If you meant that I should drop that stuff about coordinate systems, I don't think that's a good idea. The physical intepretation of the U(g) operators must be included in the axioms in one way or another, and it might as well be this way.

strangerep said:
But when one tries to add a position operator, so
that one has a "Heisenberg-Galilei" group, one finds that the position
operator has to be a multiple of the Galilean boost operator. (See
Ballentine pp66-84). (The relativistic case is more problematic, of
course.)
I guess I'm going to have to start reading Ballentine at Google Books. Amazon.com says "Shipped on April 22, 2009" and "Delivery estimate: May 19, 2009". Unbelievable. I'll probably start tomorrow.

strangerep said:
So you don't really need axiom 3 in its current form (since the
spectral theorem also decomposes any operator in terms of projection
operators). You just need to say something about how states correspond
to density operators (and particular measurements to projection
operators).
Maybe I'm saying too much, but some version of the rule ρ → ∑b Pb ρ Pb must be included, and it should be a version of it that can handle both discrete and continuous spectra.

Unfortunately I don't know those spectral theorems. I still have a lot of functional analysis to read.

strangerep said:
More generally, one needs to start from a symplectic formulation (i.e.,
symplectic group on phase space), since that brings in non-trivial
dynamics, interactions, etc. But then one must use something like
Weyl-Wigner-Groenewold-Moyal quantization to establish a satisfactory
correspondence between phase space functions and quantum operators.
This something I've never heard of, but it sounds interesting.

strangerep said:
It helps you construct representations corresponding to
half-integral spin.
Yes, I understand that part. I was referring to some of the finer details, in particular, if h and h' are the two members of the covering group that correspond to the Galilei transformation g, are U(h)f and U(h')f in the same ray? I'm guessing that the answer is yes, and that U(h')f=±U(h)f.

strangerep said:
You could perhaps say that g is a member of covering Galilei.
Yes, but I also have to say which member it is, or rather which two members it is. Something like this:

2. Let O and O' be two inertial observers, and let S and S' be the inertial frames naturally associated with their motion. Let g be the Galilei transformation that represents a coordinate change from S to S'. If O describes an isolated system as being in state F, then O' must describe the system as being in the state represented by the ray F' that contains the vector f'=U(h)f, where f is any vector in F, U:G→GL(H) is an irreducible unitary representation of the covering group of the Galilei group, and h is either of the two members of G that the covering homomorphism takes to g.
 
  • #64
Fredrik said:
It's probably cleaner to start with the dynamical group as an axiom --
in your case Galilei.
I'm not sure what you mean. If you meant that I should use the Galilei group instead of its covering group, we wouldn't be able to let U be a representation. I'm sure you know about the annoying minus sign that we can't get rid of: U(g)U(h)=±U(gh).

Hmmm. Now I not sure what's actually bugging you about this (and also the related
stuff near the end of your post). You're talking about the double-valuedness that arises
from the presence of SO(3) inside Galilei, right? If so, then I don't think this is an issue
if you're restricting to spin-0. But maybe it's better to have a mega-space containing
all spins, and then restrict attention to spin-0. The sign ambiguity arising from 2\pi
rotations is then handled more cleanly: one just finds all unitary irreducible representations
of SO(3), then discovers superselection rules between states of integer and half-integer
spin. (But maybe I've misunderstood what's bugging you?)

I guess I'm going to have to start reading Ballentine at Google Books. Amazon.com says "Shipped on April 22, 2009" and "Delivery estimate: May 19, 2009". Unbelievable. I'll probably start tomorrow.
Standard shipping, huh? Maybe it will arrive sooner.

So you don't really need axiom 3 in its current form (since the
spectral theorem also decomposes any operator in terms of projection
operators). You just need to say something about how states correspond
to density operators (and particular measurements to projection
operators).
Maybe I'm saying too much, but some version of the rule ρ → ∑b Pb ρ Pb must be included,
That's part of what I meant by correspondence between measurements and
projection operators.

and it should be a version of it that can handle both discrete and continuous spectra.
Unfortunately I don't know those spectral theorems. I still have a lot of functional
analysis to read.
The spectral theorem allows you to replace the sum by an
integral. The main content of the theorem is proving that a "projection-valued
Lebesgue measure" exists, making the integral mathematically meaningful.
For mixed spectra you have two terms, one with an integral plus one with
a sum.

About the rest of your post,... maybe we should wait until you have a copy
of Ballentine in front of you.
 
  • #65
strangerep said:
You're talking about the double-valuedness that arises
from the presence of SO(3) inside Galilei, right?
Yes.

strangerep said:
If so, then I don't think this is an issue
if you're restricting to spin-0.
I realize now that I didn't make myself clear earlier. What I eventually decided to do was to try to write down the axioms of the non-relativistic quantum theory of one particle with arbitrary spin, in such a way that all we have to do to get the special relativistic theory later is to replace "Galilei" with "Poincaré".

strangerep said:
one just finds all unitary irreducible representations
of SO(3), then discovers superselection rules between states of integer and half-integer
spin. (But maybe I've misunderstood what's bugging you?)
It's just that if we have U(g)U(h)=±U(gh), then U isn't a representation. I'd rather talk about a "representation" than a "projective representation" in the axioms. That's why I picked the covering group instead of the group itself.

strangerep said:
Standard shipping, huh? Maybe it will arrive sooner.
Expedited shipping actually. That's why it's so ridiculous.

strangerep said:
The spectral theorem allows you to replace the sum by an
integral. The main content of the theorem is proving that a "projection-valued
Lebesgue measure" exists, making the integral mathematically meaningful.
Is that the spectral theorem that can only be found in two out of print books, one of which is badly written? :smile:

strangerep said:
About the rest of your post,... maybe we should wait until you have a copy
of Ballentine in front of you.
That's fine with me.
 
  • #66
Fredrik said:
What I eventually decided to do was to try to write down the axioms of the non-relativistic quantum theory of one particle with arbitrary spin,
OK, good idea. In that case, the eigenspaces of J^2 corresponding to different total spin
don't mix under the group action, so it's cleaner when talking about integer vs half-integer
spins. It's also an excellent way to show how superselection rules can arise from one's
original choice of group.

in such a way that all we have to do to get the special relativistic theory
later is to replace "Galilei" with "Poincaré".
OK.

It's just that if we have U(g)U(h)=±U(gh), then U isn't a representation. I'd rather talk
about a "representation" than a "projective representation" in the axioms. [...]
Ah, now wait a minute... that phase ambiguity which enters in a projective representation
is a distinct issue from the double-valuedness of half-integer spin reps. In a projective
unitary rep, we can have U(g)U(h) = e^{ia} U(gh). At the Lie algebra level, this is the
same as adding a multiple of the identity to the right hand side of all the commutation
relations. However, in some cases it can be shown that these multiples must be 0 (using
consistency arguments from the Jacobi identity, etc). In other cases, it can be shown that
the phase has no physical effect, so the multiple can be chosen as 0. For Galilei, however, it
turns out there's one case where it can't be removed in either of these ways. Comparison
with classical mechanics then shows that this remaining central element should be
interpreted as mass.

For Lorentz/Poincare, it turns out that the phase can always be set harmlessly to 1.


Is that the spectral theorem that can only be found in two out of print books,
one of which is badly written? :smile:
No. To get the idea, any respectable book on Functional Analysis should have it.
Look for a spectral theorem concerning (noncompact) bounded Hermitian-symmetric
operators on an (ordinary) inf-dim Hilbert space.

I'm halfway through writing a summary paper (originally intended for personal
use only) which states and proves all the spectral theorems, in increasing order
of difficulty. Now I'm thinking I should eventually post some version of it in the
tutorial forum. I wonder if I'll ever finish it sufficiently.
 
  • #67
strangerep said:
It's also an excellent way to show how superselection rules can arise from one's original choice of group.
I agree. (I assume that you're talking about using the Galilei group instead of its covering group, because as I understand it, the superselection rules have to be imposed as an axiom if we don't. Hm, that is a reason to use the Galilei group instead of the covering group).

strangerep said:
Ah, now wait a minute... that phase ambiguity which enters in a projective representation
is a distinct issue from the double-valuedness of half-integer spin reps.
...
For Lorentz/Poincare, it turns out that the phase can always be set harmlessly to 1.
I don't think so. But maybe I just don't understand what you mean by double-valuedness. This is what I'm thinking: SO(3) is isomorphic to SU(2)/Z2. SU(2) can be mapped continuously and bijectively onto S3 (a 3-sphere). So SU(2) can be interpreted as a 3-sphere, and SO(3) can be interpreted as a 3-sphere with opposite points identified, or equivalently, as the set of straight lines through the origin in R4. This means that SO(3) isn't simply connected, but SU(2) is.

Now Wigner's theorem about symmetries says that given a simply connected symmetry group G (i.e. a group of functions that map the set of rays of the Hilbert space H bijectively onto itself), we can find a map U:G→GL(H) such that U(g) is either linear and unitary for all g in G or antilinear and antiunitary for all g in G, and U(g)U(h)=U(gh) for all g,h in G.

As I understand it, when G=SO(3), there's no U:G→GL(H) with these properties, and it can be traced back to SO(3) not being simply connected. The best we can do is to find a set of operators that satisfy an identity that is sometimes written as U(g)U(h)=±U(gh), but even that is being really sloppy with the notation. The truth is more complicated:

I'll use the notation h→g to represent a continuous curve from h to g. The best we can do when G=SO(3) is to find a set

{U(h→g) in GL(H) | h,g in G}

such that each U(h→g) is linear and unitary and

U(h→gh)U(e→h)=(-1)f(e→h,h→gh,e→gh)U(e→gh)

where f(e→h,h→gh,e→gh)=0 when the curve constructed by joining the three curves e→h, h→gh and the "reverse" of e→gh to each other can be continuously deformed to a point, and f(e→h,h→gh,e→gh)=0 otherwise. (What I mean by the "reverse" of a curve C:[0,1]→G, is the curve B:[0,1]→G defined by B(t)=C(1-t) ).

My main concern here is that the notation "U(g)" is inappropriate when G=SO(3), because the domain of U isn't G. It's a set of curves in G.

So the bottom line is that if we take the symmetry group to include SU(2), we have to impose superselection rules manually, and if we take it to include SO(3), we have to use a very awkward notation, or be intentionally sloppy with the notation. Both of these things are undesirable, but the first is easier to deal with, I think.

I got this idea from Weinberg, who seems to have gotten it from Streater & Wightman, or maybe directly from Wigner.

strangerep said:
In a projective
unitary rep, we can have U(g)U(h) = e^{ia} U(gh). At the Lie algebra level, this is the
same as adding a multiple of the identity to the right hand side of all the commutation
relations. However, in some cases it can be shown that these multiples must be 0 (using
consistency arguments from the Jacobi identity, etc). In other cases, it can be shown that
the phase has no physical effect, so the multiple can be chosen as 0. For Galilei, however, it
turns out there's one case where it can't be removed in either of these ways. Comparison
with classical mechanics then shows that this remaining central element should be
interpreted as mass.
I'm definitely going to learn this stuff, but I'm going to be a bit busy the next few days.

strangerep said:
I'm halfway through writing a summary paper (originally intended for personal
use only) which states and proves all the spectral theorems, in increasing order
of difficulty. Now I'm thinking I should eventually post some version of it in the
tutorial forum. I wonder if I'll ever finish it sufficiently.
If you do, you could probably publish it in some sort of review journal. I'm sure it would be interesting to other people as well. I would definitely read it.
 
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  • #68
Fredrik said:
[...] phase ambiguity which enters in a projective representation
is a distinct issue from the double-valuedness of half-integer spin reps.
...
For Lorentz/Poincare, it turns out that the phase can always be set harmlessly to 1.
I don't think so. ...
I'll try to clarify. (If you have a copy of Weinberg vol-1, this is in section 2.7 on
Projective Representations.) In fact, it's probably simpler if I just quote Weinberg (p83-84).

The phase of any representation U(T) of a given group can be chosen so that \phi=0 [...]
if two conditions are met:

(a) The generators of the group in this representation can be redefined [...] so as to
eliminate all central charges from the Lie algebra.

(b) The group is simply connected [...]


I was talking about the redefinition of generators (which can always be done in
Poincare, but not in Galilei), whereas you were talking about simple-connectedness
(which must be considered in both cases).

... but I'm going to be a bit busy the next few days.
Yeah, I should try and get some more actual work done too. :-)
 
  • #69
Thanks, I had completely forgotten about (a). (And yes, I have a copy of Weinberg, vol. 1).
 
  • #70
Fredrik said:
Amazon.com says "Shipped on April 22, 2009" and "Delivery estimate: May 19, 2009". Unbelievable.
Thank FSM that Amazon sucks at estimates. I picked up my Ballentine at the post office today. I actually got it two days ago, but the slip ended up in the middle of a pile of mail that I didn't look at right away. I have started to read it, and I can already tell it was a good buy. Lots of good stuff in there.
 
  • #71
Fredrik said:
Now Wigner's theorem about symmetries says that given a simply connected symmetry group G (i.e. a group of functions that map the set of rays of the Hilbert space H bijectively onto itself), we can find a map U:G→GL(H) such that U(g) is either linear and unitary for all g in G or antilinear and antiunitary for all g in G, and U(g)U(h)=U(gh) for all g,h in G.

<Note for moderator> I revive this thread because I currently have in interest in group theory and its applications to QM.

Wigner's theorem does not refer to group theory and has nothing to do with <simple connectedness>. For the original form, please, see Wigner's book in the English translation of 1959, Section 20 and especially its appendix (pp. 220 until 236 - English 1959 translation from original German).

The modern form appears in several sources. I pick the version from Thaller's book which is in the attachment to this post.

Now, how can we formulate quantum mechanics from a symmetry-based approach ?
I may call this the Wigner-Weyl formulation of quantum mechanics. The basic postulate is this:

If the system under quantization is described at a classical level by an irreducible set of observables and whose classical equations of motion are invariant under a set of continuous transformations called symmetries which form a Lie group G, then the set of all possible representatives of quantum states is given by an infinite-dimensional, complex , separable Hilbert space on which the operators of a linear unitary* irreducbile representation of a group G' act.

Notes:
1. According to the laws of classical mechanics in the Newtonian formulation, a possible group of symmetries for them is given by the full Galilei group (the specially-relativistic correspondent of it is the full Poincare group). Apparently, this is not the largest symmetry group of Newton's equations.
2. For G a connected Lie-group, G' is the universal covering group of a non-trivial central extension of G. (example: the set of all proper classical rotations form a group isomorphic to SO(3) which is connected and path-connected).
3. If G is a non-connected Lie-group, then G' is the universal covering group of a non-trivial central extension of the connected component of G (example: the set of all Lorentz transformations forms a group isomorphic to O(1,3). The group O(1,3) has 4 distinct components, thus being disconnected).
4. G can admit no non-trivial central extensions. For G simply-connected, the classes of inequivalent non-trivial central extensions of G are bijectively related to the classes of inequivalent non-trivial central extensions of Lie(G). (an example of group with non-trivial central extensions is the full Galilei group of classical mechanics. An example of group with only trivial central extensions: the restricted Poincare group (= the component of the full Poincare group connected to identity).
5. If G is simply-connected, its universal covering group is isomorphic to G. (example, SU(2) and SL(2,C)).
6. If Psi is a unit ray of the Hilbert space, any modulus one vector in Psi is called a representative of a ray Psi (representative of a quantum state Psi).
7.*: Discrete symmetry transformations such as temporal inversion are handled based on Wigner's theorem using antiunitary operators.
8. The fundamental mathematical theorem underlying the postulate is the main result of Bargmann's famous 1954 article (transparent or not in the article): <Let G be a connected Lie group. Then its (projective) representations on a projective Hilbert space are in bijective correspondence with the vector unitary representations of G' (described above in notes 2-->5 and mentioned in the postulate)>.
9. For an exact symmetry, the Hamilton operator must be a central element in Lie(G') (actually a representation of Lie(G') through essentially self-adjoint operators on the Hilbert space of the quantum system).

Another comment to which I don't have a reference right now (but hope to have in the future) is this:
10. The postulate can be extended for rigged Hilbert spaces.

Without the sources for 10, I can only speculate on the identification of the G@rding domain of Lie(G') and the <small space> in the Gelfand triple.

I'm in search of mathematical proofs of the statements in notes 1--> 10 and any other theorem of group theory (harmonic analysis) which can be used in a symmetry-approach to QM. For this I've opened a thread on March 19th in the <Algebra> subforum on a particular theorem which I quote here as well (No answer, no group theorist there, I guess).

<Let G be a connected and simply-connected Lie group. Then all the linear irreducible representations of G on a TVS are faithful (= single-valued = true representations = representation morphism is injective)>. [/size]

In the meantime, I found a proof, but for it to make sense I first need to find out a rigorous definition of <multi-valued> representation of a Lie group. I've searched the literature and came up with this one: "A multi-valued linear representation of a multiple-connected and connected Lie group G on a TVS is a linear representation of the universal covering group of G". But I'm dissatisfied, for it seems to assume the theorem already: it doesn't define a multi-valued representation for a simply-connected group (which doesn't exist as specified by the theorem). So I can't use it to prove the theorem.

Another definition would be then: A multi-valued linear representation of a connected Lie group is defined through the following property: there exist at least two distinct elements of G, a and b, such as U(a)=U(b).

The proof I was seeking is given below for SU(2) (adaptation from a text by Elie Cartan) but applies to all G's in the hypothesis (if G is non-compact, then the matrices mentioned in the text are infinite dimensional, if one insists on unitarity).

<SU(2) is compact, then all its irreducible linear representations are finite-dimensional and equivalent to unitary ones. Let's assume SU(2) had multi-valued representations {U} (whatever that means, see the issue with the definition above) and for g(t) a continuous curve in SU(2), U(g(t)) is a continuous curve in Aut(V) (V is the TVS of the multi-valued representation). Even in the absence of unitarity of U, the set of all U's in Aut(V) is imposed to have a structure of a topological space. (If U are unitary operators, then the set of all U's in Aut(V) is a Banach space in the strong topology). Following the continuous variation of the representing matrix U_ij as the point in the group space describes a suitable closed contour starting and finishing at some origin, the matrix would start as the unit matrix and end as a different matrix. Continuously deforming the contour in Aut(V), the final matrix will stay the same, i.e. different than the unit matrix. But SU(2) is simply-connected, so any closed contour in it can be continuously deformed to a point. Using the continuity of the representation morphism, the representation matrices would coincide, thus a contradiction>.

This proof is crystal clear to me and leads me to think that multi-valued representations of connected Lie-groups indeed exist iff the manifold of a connected Lie-group is multiple-connected when seen as a topological space.

So one can go further and issue a corollary:
<Let G be a connected, m-connected Lie group. Then all its irreducible linear representations in a TVS are at most m-valued>.
 

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  • #72
I can't edit on my post, so I'll have to quote it.

dextercioby said:
1. According to the laws of classical mechanics in the Newtonian formulation, a possible group of symmetries for them is given by the full Galilei group (the specially-relativistic correspondent of it is the full Poincare group). Apparently, this is not the largest symmetry group of Newton's equations.

Indeed, the symmetry group of Newton's equations is larger than the full Galilei group, it is a 13-parameter Lie group (connected or not, I will investigate, definitely the connected component of the identity is non simply-connected as it would necessarily contain a subgroup isomorphic to SO(3)).

dextercioby said:
Another comment to which I don't have a reference right now (but hope to have in the future) is this:
10. The postulate can be extended for rigged Hilbert spaces.

Without the sources for 10, I can only speculate on the identification of the G@rding domain of Lie(G') and the <small space> in the Gelfand triple.

I promise to come back on this issue, as I consider it to be of upmost importance.

For the interested reader, if any, I will also make some comments on previous posts in this thread, hopefully to enchance its content, so that it would turn out in some sort of learning material.
 
  • #73
dextercioby said:
Indeed, the symmetry group of Newton's equations is larger than the full Galilei group, it is a 13-parameter Lie group (connected or not, I will investigate, definitely the connected component of the identity is non simply-connected as it would necessarily contain a subgroup isomorphic to SO(3)).

Is there any particular reason you start with the Newtonian formulation rather than
a more modern Hamiltonian formulation? For the latter, there's a large body of work
on symmetries, starting from the basic inhomogeneous symplectic group, and
progressing to more sophisticated stuff. Have you tried the following book?

Marsden & Ratiu,
"Introduction to Mechanics & Symmetry,
Springer 1991, (2nd Ed), ISBN 0-387-98643-X)

Armed with such insights, the relation between classical and quantum
becomes much more transparent in this setting. (Quantum commutators
as deformations of classical brackets.)

It's more powerful to try and quantize in terms of a full dynamical group
rather than just a symmetry group.

Arnold Neumaier treats some of this in his book, but I'm sure he'll have
more to say on this when he has time... :-)

Another comment to which I don't have a reference right now (but hope to
have in the future) is this:
10. The postulate can be extended for rigged Hilbert spaces.

Try this:

S. Wickramasekara & A. Bohm,
"Symmetry Representations in the Rigged
Hilbert Space Formulation of Quantum Mechanics",
(Available as math-ph/0302018)

Wickramasekara & Bohm said:
Abstract:

We discuss some basic properties of Lie group representations in
rigged Hilbert spaces. In particular, we show that a differentiable
representation in a rigged Hilbert space may be obtained as the
projective limit of a family of continuous representations in a nested scale
of Hilbert spaces. We also construct a couple of examples illustrative
of the key features of group representations in rigged Hilbert spaces.
Finally, we establish a simple criterion for the integrability of an
operator Lie algebra in a rigged Hilbert space.


dextercioby said:
Without the sources for 10, I can only speculate on the identification of the
G@rding domain of Lie(G') and the <small space> in the Gelfand triple.

That's the way RHS usually works; -- find the largest space for which all quantities
of interest are defined everywhere (i.e., the <small space> in the Gelfand triple),
then take its closure and dual to get the other spaces in the triple. More sophisticated
treatments involve scales of Hilbert spaces, or partial-inner product (PIP) spaces,
but I'm not yet convinced that the boil down to much more than good old RHS (except
of course for the greater rigor obtained thereby).
 
  • #74
strangerep said:
Marsden & Ratiu,
"Introduction to Mechanics & Symmetry,
Springer 1991, (2nd Ed), ISBN 0-387-98643-X)

Armed with such insights, the relation between classical and quantum
becomes much more transparent in this setting. (Quantum commutators
as deformations of classical brackets.)

It's more powerful to try and quantize in terms of a full dynamical group
rather than just a symmetry group.

Thanks for the input. I will try to find the book and see if one indeed obtains more things from such a perspective than the symmetry-based approach to QM offers.

strangerep said:
S. Wickramasekara & A. Bohm,
"Symmetry Representations in the Rigged
Hilbert Space Formulation of Quantum Mechanics",
(Available as math-ph/0302018)

I knew about this article, but I had some older articles in mind.
 
  • #75
dextercioby said:
[...] the symmetry group of Newton's equations is larger than the full Galilei group, it is a 13-parameter Lie group (connected or not, I will investigate, definitely the connected component of the identity is non simply-connected as it would necessarily contain a subgroup isomorphic to SO(3)).

Since you recently mentioned these posts elsewhere, I'll re-revive this unfinished thread...

Did you finish investigating the larger group? If so, what did you find?
 
  • #76
strangerep said:
Since you recently mentioned these posts elsewhere, I'll re-revive this unfinished thread...

Did you finish investigating the larger group? If so, what did you find?

Yes for the first question, for the second: arxiv.org article under <math-ph/0102011v2> says it better than me. (I don't have access, however, to Niederer's article in HPA. It would be nice to see the original argument, too.)
 
  • #77
dextercioby said:
[...] arxiv.org article under <math-ph/0102011v2>[...]

Thanks. I remembered the group vaguely in the context of the
Schrodinger equation, but I'd forgotten the name "Niederer".
 

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