Recognitions:

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

 Quote by Fredrik I found a review of Maurin's books, and the reviewer is complaining a lot about how careless Maurin's presentations are. Link.
Hmmm, that reviewer uses rather savage language, but I have to say I agree with
his remarks about lack of care. I noticed the same thing as soon as I started to
try and follow Maurin's proof of the nuclear spectral theorem. I had to keep flipping
backwards tediously to double-check notations, often finding things that I felt
sure were typos. Worse, even though the book came with an errata list at the
back, I had trouble matching it with what was in the text(!) (sigh).

Look's like I'll have to try harder to obtain a copy of Gelfand & Vilenkin and
see whether their presentations is any better.

TBH, I'm quite surprised and disappointed that I couldn't find a clear pedagogical
proof in a more modern text, given it's importance in modern quantum theory.

 I'm guessing it would be better to try to find the original articles than to read his books.
Dunno. Original papers tend to be written for people with a greater level
of knowledge. Books are supposed to be for lesser mortals.

Recognitions:
 Quote by DarMM I thought I'd add a bit [...]

 For Quantum Field Theory the classical configurations are the set of all possible field configurations. This set turns out to be $$\mathcal{D}^*(\mathbb R^3)$$, the space of distributions. So the Hilbert space of QFT is: $L^2(\mathcal{D}^*(\mathbb R^3), d\nu)$, the space of square integrable functions over the space of distributions with respect to some measure $d\nu$.
I guess this gets into the problematic area of "multiplication of distributions"?
I hope you can say a bit more about that.

 A free QFT and an interacting QFT differ by their choice of $d\nu$. Quantum Fields $\phi(x)$ are then objects that when integrated against a function $f(x)$ give an object $\int{\phi(x)f(x)}dx = \phi(f)$, which is an unbounded operator on $L^2(\mathcal{D}^*(\mathbb R^3), d\nu)$.
Let me see... in the simpler case of rigged Hilbert space, translating into
Dirac bra-ket notation, we have
$$\phi(f) ~=~ \int{\phi(x)f(x)}dx ~=~ \langle\phi| \int{|x\rangle\langle x|}dx ~ |f\rangle ~=~ \langle \phi | f\rangle$$
with
$$\langle x|x'\rangle ~=~ \delta(x - x') ~,~~~\mbox{etc.}$$

So in the more general case we're dealing with functions over distributions (which
involving products of distributions in general), and then trying to figure out
how to construct (higher-order?) functionals over these functions?

Recognitions:
 Quote by strangerep I guess this gets into the problematic area of "multiplication of distributions"? I hope you can say a bit more about that.
Yes, in fact it's one of the big issues in rigorous QFT. Your question below will naturally lead to it.

 Quote by strangerep So in the more general case we're dealing with functions over distributions (which involving products of distributions in general), and then trying to figure out how to construct (higher-order?) functionals over these functions?
Let me give a simple example of a function over distributions.
As you know a distribution $T(x)$ is a continuous linear functional over the space of test functions $\mathcal{D}$. However one can reverse this and consider a test function $f$ as a function on distributions. A given test function $f$ maps any distribution $T$ to $T(f)$, resulting in a map over the whole space of distributions.
$f:\mathcal{D}^*(\mathbb R^3) \rightarrow \mathbb R$.

So one writes this function over distributions as $f(T)$. One can then immediately arrive at more general functions for instance:
$f(T)g(T)$
$f(T)g(T)h(T)$
or $e^{f(T)}$

Since one can choose a any test function, this results in an enormous space of functions over Distributions.

In the case of a free quantum field, the quantum field operator acts as a "multiplication operator". For example:
$\phi(g) = \int{\phi(x)g(x)}dx$
Then,
$\phi(g)f(T) = g(T)f(T)$

The conjuagate momentum field $$\pi(x)$$ acts like the momentum operator from regular QM by being a derivative. In rough language:
$\pi(x)f(T) = -i\frac{\delta}{\delta T}f(T)$

The Hamiltonian is composed of $\pi(x)$ and $\phi(x)$ and so it acts in a natural way.

Like all quantum mechanical theories though, we don't allow just any functions, but only square integrable ones. So one requires a notion of integration for this functions, formally something like:
$\int{f(T)}dT$. This can be done, although the details are a bit involved. For the purpose of the remainder of this post let's assume we have done this for the free theory. We have the correct measure $dT_{free}$, the Hilbert space of square integrable functions with respect to it and the Hamiltonian, field and conjugate momentum operators acting in the way I described above. This space is basically Fock space.

However when one comes to $\phi(x)^4[/tex] theory for example, the Hamiltonian gains an additional term: [itex]\int{\phi(x)^4}dx$

It is very difficult to know exactly how one should interpret this.
$(\int{\phi(x)^4}dx)f(T) = ?$

There are two problems here. $\phi(x)$ is a operator valued distribution. When integrated it gives operators on $L^2(\mathcal{D}^*(\mathbb R^3), dT_{free})$. Due to its distributional nature no sense can be made of $$\phi(x)^4[/itex]. One could attempt to come up with some way of powering distributions, however this powering system must respect the fact that $(\int{\phi(x)^4}dx)$ is meant to be a linear operator and respect certain commutation relations. Let's approach this problem first in two dimensions then in three dimensions. Two dimensions It turned out (again avoiding details) that there was one unique notion of powering which respects the structure of $(\int{\phi(x)^4}dx)$ as a quantum operator. This notion was actually already known to physicists as Wick ordering. So we replace: $(\int{\phi(x)^4}dx)$ with $(\int{:\phi(x)^4:}dx)$. One can then prove that the full Hamiltonian is self-adjoint and semi-bounded (stable, no negative energy), the time evolution operator is unitary, e.t.c. Everything one wants from a QFT basically. This was done by James Glimm and Arthur Jaffe in the period 1968-1973. Three dimensions Here things get significantly harder. Where as one can put the proof of the two dimensional case into a semester long graduate course, the proof of the three dimensional case is a monster. However let me paint a picture. Basically we try what we did before and replace $(\int{\phi(x)^4}dx)$ with $(\int{:\phi(x)^4:}dx)$. However one finds that this does not result in the full Hamiltonian being a self-adjoint operator. If you are very clever you can prove that this problem can not be overcome, unless you leave $L^2(\mathcal{D}^*(\mathbb R^3), dT_{free})$, that is leave Fock space and move to another space of square integrable functions over distributions. Let me call it: $L^2(\mathcal{D}^*(\mathbb R^3), dT_{Glimm})$. However finding this space is incredibly diffcult, so one uses the condition that on this correct space the physical mass of a particle should be finite. This condition show you how to get to this new space. Basically, cutoff the fields so that you do the analysis in a more clear cut way, then use the condition of finite mass. This condition tells you that you should add the term: $\delta m^2(\int{:\phi(x)^2:}dx)$ to the Hamiltonian. The $\delta m^2$ is a constant that will go infinite when the cutoff is removed, whose explicit form is given by the finite mass condition. As the cutoff is removed, this extra term does the job of "pushing" us into the correct space $L^2(\mathcal{D}^*(\mathbb R^3), dT_{Glimm})$, which is alot easier than finding it abstractly. Physicists will know this procedure as mass renormalization. We then take the remove the cutoff. James Glimm proved that this moves us into the correct space and that the full Hamiltonian is self-adjoint there. Glimm and Jaffe later proved that the Hamiltonian is also semibounded (in what is considered the most difficult paper in mathematical physics, lots of heavy analysis). Shortly after came all the the other properties one wants, Lorentz invariance, e.t.c. Thus proving the rigorous existence of $\phi^4$ in three dimensions. As you can imagine the four dimensional case is extremely difficult. For five dimensions and greater, Jürg Fröhlich has shown that there is no space where the $(\int{\phi(x)^4}dx)$ term makes sense, unless you set the interaction to zero, that is make it free. People have also shown the existence of Yukawa theories in two and three dimensions, as well as gauge and Higgs theories in two and three dimensions. Tadeusz Balaban has gotten pretty far although with Yang-Mills theory in two and three dimensions. He has even gotten some results for the four dimensional case. One final thing I wish to emphasize is that renormalization is not the difficulty in rigorous quantum field theory. Renormalization is a necessary technique, as it has been known for a long time the most interacting theories simply cannot live in the same space as the free theory (Haag's theorem). The actual difficulty is with standard analytic stuff, proving convergence and obtaining estimates which tell you something about a limit besides the fact that it exists. I hope this helps. Mentor  Quote by DarMM For Quantum Field Theory the classical configurations are the set of all possible field configurations. This set turns out to be [tex]\mathcal{D}^*(\mathbb R^3)$$, the space of distributions. So the Hilbert space of QFT is: $L^2(\mathcal{D}^*(\mathbb R^3), d\nu)$, the space of square integrable functions over the space of distributions with respect to some measure $d\nu$. A free QFT and an interacting QFT differ by their choice of $d\nu$.
I'm really confused by this. I thought the Hilbert space of (say) Klein-Gordon theory was supposed to be the Fock space constructed from the Hilbert space H1 of one-particle states, which is defined as the subset of $L^2(\mathbb R^3)$ that consists of square-integrable positive-frequency solutions of the Klein-Gordon equation. Is this wrong, or just another way of saying the same thing? Note btw that a square-integrable solution of the field equation must be continuous, because its partial derivatives must exist, so this H1 isn't $$L^2(\mathbb R^3)$$. It's just a "small" subset of it.

I don't understand why the space of distributions is the classical configuration space. Can you elaborate on that?

Also, if it's possible to explicitly write down the measure of the free theory and explain why it's the right one, I'd appreciate it.

 Quote by DarMM Who said rigorous QFT was hard?
Some moron probably

Seriously, I think it's hard just to find out where you can learn this stuff. I took three classes on quantum mechanics and two on quantum field theory at my university, and they didn't say anything about these things. (They mentioned the word "distribution" and said that you can use that concept to make sense of the delta function, but that's where they stopped).

 Quote by DarMM $\mathcal{D}$. However one can reverse this and consider a test function $f$ as a function on distributions. A given test function $f$ maps any distribution $T$ to $T(f)$, resulting in a map over the whole space of distributions. $f:\mathcal{D}^*(\mathbb R^3) \rightarrow \mathbb R$. So one writes this function over distributions as $f(T)$. One can then immediately arrive at more general functions for instance: $f(T)g(T)$ $f(T)g(T)h(T)$ or $e^{f(T)}$ Since one can choose a any test function, this results in an enormous space of functions over Distributions.
I don't understand what you're saying about these "more general" functions. Isn't f(T)g(T) defined as (fg)(T) and exp(f(T)) as $\exp\circ f(T)$? In that case, how are they more general? The product of two test functions is another test function. Same thing with compositions.

I'll stop here because it seems dumb to ask about advanced stuff when I'm still struggling with the "easy" stuff.

Recognitions:
 Quote by Fredrik I'm really confused by this. I thought the Hilbert space of (say) Klein-Gordon theory was supposed to be the Fock space constructed from the Hilbert space H1 of one-particle states, which is defined as the subset of $L^2(\mathbb R^3)$ that consists of square-integrable positive-frequency solutions of the Klein-Gordon equation. Is this wrong, or just another way of saying the same thing?
It's just another way of saying the same thing. $L^2(\mathcal{D}^*(\mathbb R^3), d\nu)$ can be decomposed quite easily into direct sums of tensor products of square-integrable positive-frequency solutions of the Klein-Gordon equation. However in rigorous QFT this is sometimes avoided because for interacting theories the Hilbert space $L^2(\mathcal{D}^*(\mathbb R^3), d\nu)$ is not a Fock space over positive-frequency solutions.

 Quote by Fredrik I don't understand why the space of distributions is the classical configuration space. Can you elaborate on that?
Basically you know from QM that the Hilbert space is the square integrable functions over the classical configuration space. It is of course difficult to know what this space should be, when does a given function count as a sensible classical configuration. However it turns out that if you start with any set of functions, in order to define a measure on this space you'll have to make the space larger to the point where it encompesses distributions. If you choose a larger space of functions it turns out that alot of the space has measure zero, except the subspace of distributions. So either way no matter how big or small a space you pick it has to be the space of distributions.

 Quote by Fredrik I don't understand what you're saying about these "more general" functions. Isn't $f(T)g(T)$ defined as $(fg)(T)$ and $exp(f(T))$ as $\exp\circ f(T)$ In that case, how are they more general? The product of two test functions is another test function. Same thing with compositions.
I was unclear before, $f(T)g(T) \neq (fg)(T)$. It's actually simpler, you just evaluate $f(T)$ and $g(T)$ and then multiply them.
By more general functions I just meant I can define more functions than only the linear ones.

 Quote by Fredrik Also, if it's possible to explicitly write down the measure of the free theory and explain why it's the right one, I'd appreciate it.
Why it's the right one is easy. Basically instead of defining the free theory on continuous spacetime, put it on a lattice. On the lattice you can find the correct measure easily, since you're now back in a system of finite degrees of freedom. It's basically just $L^2(\mathbb R^n)$ with $n$ a very large number equal to the number of lattice points. If you take the limit as the lattice returns to a continuum you'll get the free theory measure I'm talking about.
You can also solve the Klien-Gordon equation to obtain the free field and work out the generating functional for the theory. If you have the generating functional then by a theorem called Minlos theorem, you know there exists a measure on the space of distributions which reproduces this generating functional and this the correct measure.

I'll define the measure in my next post, however you actually already know the measure and how to integrate with it.
 Mentor Thank you DarMM. This is good stuff, and very interesting. Maybe you can also tell me what I would have to read to learn this stuff?
 Mentor Strangerep, I skimmed the rest of the RHS article that you recommended. The interesting stuff was on the pages I had already read, so I didn't study the details of his specific example. One thing in particular that I had failed to understand before I read his paper is that the RHS definition of a ket is as an anti-linear functional on the nuclear space. (Is that the correct term for $\Omega$?) I liked his motivation for the definition of $\Omega$. Short version: If we define a position operator $Q:D(Q)\rightarrow H$ by Qf(x)=xf(x) for all x, then its domain can't be all of H, because Qf is only square integrable if $$\int|xf(x)|^2\ dx < \infty$$ So the idea is to find a subspace of H on which the entire algebra of operators generated by Q,P and H (including operators of the form $Q^iP^jH^k$ where i,j,k are arbitrary positive integers) is well-defined. One thing that still seems weird to me is that people keep claiming that you need the RHS approach specifically to motivate bra-ket notation. I mean, all the problems that the RHS stuff is supposed to solve are there even if you don't use bra-ket notation. The first QM book I studied didn't use bra-ket notation, and it still pretended that Q and P had eigenvectors. It just ignored the fact that plane waves aren't square integrable and that delta functions aren't even functions. If you ignore those problems, the Riesz representation theorem is all you need to justify the notation. We can define a ket to be a member of H, and a bra to be a member of H*. The bra written as $\langle f|$ is the linear functional $|g\rangle\mapsto(|f\rangle,|g\rangle)$.

Recognitions:
 Quote by Fredrik [...] the RHS definition of a ket is as an anti-linear functional on the nuclear space. (Is that the correct term for $\Omega$?)
Yes, what I wrote earlier as $\Omega$ is a nuclear space. In the paper
quant-ph/0502053, the author denotes it as $\Phi$ instead.

 So the idea is to find a subspace of H on which the entire algebra of operators generated by Q,P and H (including operators of the form $Q^iP^jH^k$ where i,j,k are arbitrary positive integers) is well-defined.
Precisely. We want a representation of the enveloping algebra of observables,
and/or the full Lie group (not just the basic Lie algebra).

 One thing that still seems weird to me is that people keep claiming that you need the RHS approach specifically to motivate bra-ket notation. I mean, all the problems that the RHS stuff is supposed to solve are there even if you don't use bra-ket notation. [...]
Yes. It would be clearer to say "...motivate the use of Dirac's improper eigenvectors of
position and momentum...". I.e., to be able to write (eg)
$$1 ~=~ \int\!\! dx ~ |x\rangle\langle x|$$
and have it actually mean something.

 If you ignore those problems, the Riesz representation theorem is all you need to justify the notation. We can define a ket to be a member of H, and a bra to be a member of H*. [...]
Yes, up to a point, but note that H is self-dual, i.e., H is isomorphic to H*.
So we don't really see the problems until unbounded operators are in play.

P.S., I finally got hold of the books by Gelfand and Vilenkin: "Generalized Functions
vol 4 -- Applications of Harmonic Analysis." Also vol 5. After some cursory reading
it looks better than Maurin's book. (Possibly because the G-V book was originally
Russian and then translated by a technically competent person. I.e., more people
have been over it with a fine tooth comb.)

P.P.S., Looks like you started are really good thread. Several days later, I'm still
trying to relate the stuff DarMM said with what I already know about QFT. :-)
 Mentor Yes, this has been a good discussion. I have learned a lot from it. I'm still confused though, and I'd like to go back to the simplest possible QM theory for a while. (See the thread title). How can we state its axioms in a way that makes sense? They are usually stated in a way that ignores all the complications. Something like this: 1. The possible states of a physical system are represented by the unit rays of a separable Hilbert space over the complex numbers, on which there exists a self-adjoint operator H called the Hamiltonian. 2. The time evolution of an isolated system which is initially in state F is given by f(t)=exp(-iHt)f, where f is any vector in the ray F. 3. Measurable quantities are represented by self-adjoint operators. Self-adjoint operators are therefore called "observables". If a system is in state F when a measurement of an observable B is performed, the state of the system will change to state |b> (an eigenvector of B with eigenvalue b) with probability ||2, where f is any vector in F. The result of a measurement of B that leaves the system in state |b> is b. This is actually a less sloppy formulation than the ones that were shown to me when I first studied QM, but there are many flaws here. These are some of my thoughts: * If we're dealing with the QM of one spin-0 particle, we can be more specific in axiom 1 and say that the Hilbert space is L2(R3) rather than just some Hilbert space. On the other hand, since all separable infinite-dimensional Hilbert spaces are isomorphic to each other, it doesn't really matter. Maybe we should at least add the requirement that the Hilbert space is infinite-dimensional. * Instead of postulating the existence of the Hamiltonian directly, we should be postulating that we're dealing with an irreducible representation of the covering group of the Galilei group. Actually, now that I think about it, once we start talking about representations, the axioms of non-relativistic QM of one spin-0 particle aren't significantly more complicated than the the axioms of special relativistic QM of one particle with arbitrary spin, so we might as well go for a slightly more general set of axioms (but stick with one-particle theories for now). * Axiom 2 can be dropped from the list, since it's implicit in the definition of the Hamiltonian as the generator of translations in time. That doesn't mean that we delete axiom 2. We just reinterpret it as the definition of an "isolated system". * Axiom 3 is still giving me a headache. It ignores several important issues, including: a) An observable can have several eigenvectors with the same eigenvalues, b) Position and momentum aren't even observables if we define them as self-adjoint operators on the Hilbert space, c) If we generalize the definition of observables to densely defined operators, some observables don't have eigenvectors, d) A generalized observable may have a continuous spectrum, a discrete spectrum, or a combination of both. * The first of those issues seems easy enough to deal with. Axiom 3 above is equivalent to saying that a density operator ρ changes to ∑b Pb ρ Pb when we measure B. (Sorry, LaTeX doesn't work). Here Pb=|b>
 Mentor 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.

Recognitions:
 Quote by Fredrik 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.

Mentor
 Quote by strangerep 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.

 Quote by strangerep 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.

 Quote by strangerep 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.

 Quote by strangerep 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.

 Quote by strangerep 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.

 Quote by strangerep 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.

Recognitions:
Quote by Fredrik
 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).
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.

Mentor
 Quote by strangerep You're talking about the double-valuedness that arises from the presence of SO(3) inside Galilei, right?
Yes.

 Quote by strangerep 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".

 Quote by strangerep 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.

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

 Quote by strangerep 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?

 Quote by strangerep 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.

Recognitions:
 Quote by Fredrik 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?
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.

Mentor
 Quote by strangerep 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).

 Quote by strangerep 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.

 Quote by strangerep 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.

 Quote by strangerep 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.

Recognitions:
Quote by Fredrik
 [...] 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. :-)

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