Mentor

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

This is a very interesting thread in which I would like to participate actively, and from which I would like to learn, but I'm too busy with work for the next week or two to do the necessary reading and thinking.
 Quote by jensa Sometimes people introduce boxes with periodic boundary conditions and let the size of those boxes go to infinity at the end.... is this more rigorous? Probably not ..?
Even for a non-relativistic particle in a box, there is a lot a technical "grit". The position operator doesn't have any eigenstates that live in the Hilbert space of states, it only has distributional eigenstates. Also, the momentum operator is unbounded, and thus, by the Hellinger-Toeplitz theorem (as already posted by strangerep), the momentum operator cannot act on all the states in the Hilbert space of states.

This is an example of something much more general. If two self-adjoint operators satisfy a canonical commutation relation, then it is easy to show that at least one of the operators must be unbounded.
 Quote by Fredrik This would give us both the Schrödinger equation and a definition of the Hamiltonian, the momentum operators and the spin operators (and probably the position operator too, but I haven't fully understood that part...something about central charges of the Lie algebra).
I think you're referring to the fact that (unlike the case for the Poincare group) non-relativistic quantum mechanics deals with representations of a central extension of the Galilean group, not with representations of the Galilean group. This is related to mass in non-relativistic quantum mechanics. Ballentine never uses the term "central extension," but, unlike most (all?) standard quantum mechanics texts, he does give a non-rigorous version. See: page 73, Multiples of identity (c); page 76; pages 80-81.
 Quote by Fredrik Hm, this part sounds familiar. I read the part about tempered distributions in Streater and Wightman recently, but I didn't try to understand every word. They defined a space of test functions that vanish faster than any power of x, and defined a tempered distribution to be a member of its dual space. The part I didn't understand was the exact definition of "vanish faster than any power of x". I'm going to read that part again, and see if I can understand it.
I think that you would like chapter 9, Generalized Functions, form the book Fourier Analysis and Its Applications by Gerald B. Folland.
 Quote by Avodyne I don't see why it's necessary to go beyond Hilbert space.
I think that it's just a matter of taste whether one uses Hilbert spaces or rigged Hilbert spaces as a rigourous basis for quantum mechanics. For example, Reed and Simon write (in v1 of their infamous work):

"We must emphasize that we regard the spectral theorem as sufficient for any argument where a nonrigorous approach might rely on Dirac notation; thus, we only recommend the abstract rigged space approach to readers with a strong emotional attachment to the Dirac formalism."

I also think that the reason for the popularity of the Hilbert space approach is historical.

In the early 1930s, before the work of Schwartz and Gelfand on distributions and Gelfand triples, von Neumann came up a rigorous Hilbert space formalism for quantum theory.

I think if a rigorous rigged Hilbert space version of quantum theory had come along before the rigorous Hilbert space version of quantum theory, then the Hilbert space version might today be even less well-known than the rigged Hilbert space version actually is. Students would now be hearing vague mutterings about "making things rigourous with Gelfand triples," instead of hearing vague mutterings about "making things rigourous with Hilbert spaces."

 In my understanding, the reason why standard Hilbert space formalism is not suitable for QM is rather simple. Let's say I want to define an eigenfunction of the momentum operator. In the position space such an eigenfunction has the form (I work in 1D for simplicity) $$\psi(x) = N \exp(ipx)$$ where $$N$$ is a normalization factor. This wavefunction must be normalized to unity, which gives $$1 =\int \limits_V |\psi(x)|^2 dx = N^2V$$ where $$V$$ is the "volume of space", which is, of course, infinite. This means that the normalization factor is virtually zero $$N = 1/\sqrt{V}$$ So, the value of the wavefunction at each space point is virtually zero too, and it can't belong to the Hilbert space. But the wavefunction is not EXACTLY zero, because its normalization integral is equal to 1. So, here we are dealing with resolving uncertain expressions like "zero x infinity". As far as I know, there is a branch of mathematics called "non-standard analysis", which tries to assign a definite meaning to such "virtually zero" or "virtually infinite" quantities and to define mathematical operations with them. I guess that using methods of non-standard analysis in quantum mechanics could be an alternative solution for the "improper states" in QM (instead of the rigged Hilbert space formalism). Did anyone hear about applying non-standard analysis to QM?
 Mentor I'm still a bit confused by distributions and tempered distributions. Let's see if we can sort this out. We define D to be the set of all $C^\infty$ functions from $\mathbb R^n$ to $\mathbb C$ with compact support. (This D isn't used in the construction of a rigged Hilbert space. I'm defining it just for completeness). We say that $\phi_n\rightarrow\phi$ if there's a compact set K that contains the supports of all the $\phi_n$, and every $D^\alpha\phi_n$ converges uniformly on $\mathbb R^n$ to $D^\alpha\phi$. Here we're using the notation $$|\alpha|=\alpha_1+\cdots+\alpha_n$$ $$D^\alpha f(x)=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}f(x)$$ The members of D are called test functions. Now we define a distribution as a linear function $T:D\rightarrow\mathbb C$, which is continuous in the following sense: $$\lim_{n\rightarrow\infty}T(\phi_n)=T(\lim_{n\rightarrow\infty}\phi_n)$$ I think I would prefer to do it a bit differently (if the following is in fact equivalent to the above, but it might not be). We define an inner product and the associated norm by $$\langle f,g\rangle=\int_D f g\ d\mu$$ where $\mu$ is the Lebesgue measure on $\mathbb R^n$. Now we define the space of distributions to be the dual space of D. Is this definition equivalent to the first?
 Mentor Now let's consider tempered distributions. We define S as the set of all $C^\infty$ functions from $\mathbb R^n$ to $\mathbb C$ that satisfy $$\sup_x\{|x^\alpha D^\beta\phi(x)|}<\infty$$ for all $\alpha$ and $\beta$. I'm using the notation $$x^\alpha=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$$ The set S is called the Schwarz class (at least by Folland). Now, I would like to say that a member of its dual space is called a tempered distribution, but that doesn't seem to be how it's done. Instead of using the standard norm, we define an infinite number of new norms: $$\|\phi\|_{r,s}=\sum_{\alpha\leq r}\sum_{\beta\leq s}\sup_x\{x^\alpha D^\beta \phi(x)\}$$ and define a tempered distribution as a linear functional that's bounded with respect to at least one of these norms. I'm pretty confused by this. Is the set of tempered distributions the dual space of the Schwarz class or not?

Recognitions:
 Quote by meopemuk So, the value of the wavefunction at each space point is virtually zero too, and it can't belong to the Hilbert space. But the wavefunction is not EXACTLY zero, because its normalization integral is equal to 1. So, here we are dealing with resolving uncertain expressions like "zero x infinity".
Yes, but that's why the more general versions of the spectral theorem are formulated
in terms of "projection-valued measures" to which Lebesgue integration theory is applicable.

 Did anyone hear about applying non-standard analysis to QM?
I remember a fairly recent paper:

 math-ph/0612082 Title: An approach to nonstandard quantum mechanics Authors: Andreas Raab Abstract: We use nonstandard analysis to formulate quantum mechanics in hyperfinite-dimensional spaces. Self-adjoint operators on hyperfinite-dimensional spaces have complete eigensets, and bound states and continuum states of a Hamiltonian can thus be treated on an equal footing. We show that the formalism extends the standard formulation of quantum mechanics. To this end we develop the Loeb-function calculus in nonstandard hulls. The idea is to perform calculations in a hyperfinite-dimensional space, but to interpret expectation values in the corresponding nonstandard hull. We further apply the framework to non-relativistic quantum scattering theory. For time-dependent scattering theory, we identify the starting time and the finishing time of a scattering experiment, and we obtain a natural separation of time scales on which the preparation process, the interaction process, and the detection process take place. For time-independent scattering theory, we derive rigorously explicit formulas for the M{\o}ller wave operators and the S-Matrix.
I only skimmed it at the time. I should take a closer look.

Mentor
 Quote by George Jones This is an example of something much more general. If two self-adjoint operators satisfy a canonical commutation relation, then it is easy to show that at least one of the operators must be unbounded.

 Quote by George Jones I think you're referring to the fact that (unlike the case for the Poincare group) non-relativistic quantum mechanics deals with representations of a central extension of the Galilean group, not with representations of the Galilean group. This is related to mass in non-relativistic quantum mechanics. Ballentine never uses the term "central extension," but, unlike most (all?) standard quantum mechanics texts, he does give a non-rigorous version. See: page 73, Multiples of identity (c); page 76; pages 80-81.
I think that settles it...I'm going to have buy Ballentine's book.

 Quote by George Jones I think that you would like chapter 9, Generalized Functions, form the book Fourier Analysis and Its Applications by Gerald B. Folland.
Thanks for the tip. It looks good. Link.

Mentor
 Quote by Avodyne I don't see why it's necessary to go beyond Hilbert space. Rather than defining a position operator, we could define projection operators with eigenvalue 1 if the particle is in some particular volume V, and 0 otherwise; heuristically, these would be $$P_{x\in V} \equiv \int_V d^3\!x\,|x\rangle\langle x|$$ Similarly for a volume in momentum space. Then, instead of defining a hamiltonian whose action could take a state out of the Hilbert space, we could define a unitary time evolution operator.
Yes, this makes a lot of sense. Ballentine seems to agree. Quote from page 28:

We now have two mathemathcally sound solutions to the problem that a self-adjoint operator need not possess a complete set of eigenvectors in the Hilbert space of vectors with finite norms. The first, based on the spectral theorem (Theorem 4 of Sec. 1.3), is to restate our equations in terms of projection operators which are well defined in Hilbert space, even if they cannot be expressed as sums of outer products of eigenvectors in Hilberbert space. The second, based on the generalized spectral theorem, is to enlarge our mathematical framework from Hilbert space to rigged Hilbert space, in which a complete set of eigenvectors (of possibly infinite norm) is guaranteed to exist.

Mentor
 Quote by strangerep ...more about constructing a quantum theory starting from an algebra of observables, instead of starting from a Hilbert space.
OK, but why would we want to? If QM can be made rigorous using either a Hilbert space and the spectral theorem, or a rigged Hilbert space and the generalized spectral theorem (see the Ballentine quote in my previous post), then we don't need C*-algebras for mathematical rigor, and in the end, it gives us a Hilbert space anyway.

 Quote by strangerep When starting from a Heisenberg algebra, one usually employs the regularized (Weyl) form of the canonical commutation relations to banish the pathological behaviour that caused the need for RHS in the other formalism. The GNS construction basically means being given a vacuum vector (and its linear multiples, i.e., a 1D linear space), and multiplying it by all the elements of the algebra to generate a full Hilbert space (of course, I'm skipping lots of technicalities here).
I don't really understand any of this, but you don't have to break your back trying to explain it all to me. I'm probably going to have to postpone a serious attempt to understand some of the things we're discussing in this thread until I have studied some more functional analysis. I will of course try to understand as much as possible now, but I might not always be successful.

 Quote by strangerep Of which book? (I don't have Conway.)
Yes, Conway. I only mentioned the page number to indicate how much I would have to read to really understand it.

 Quote by strangerep Functional analysis is an essential tool for the mathematical physicist, imho.
I don't doubt that. It kind of bugs me though when I think about how things were handled at my university. We weren't really encouraged to take more math classes. Most of us didn't even study complex analysis, and probably only about one or two physics students per year bothered to study the analysis class based on "Principles of mathematical analysis" by Rudin. I'm glad I was one of them.

 Quote by strangerep Edit: I just remembered... there's some old ICTP lecture notes by Maurin on this stuff, available as: streaming.ictp.trieste.it/preprints/P/66/012.pdf
Cool, I'll check it out tomorrow.

Recognitions:
 Quote by Fredrik Is this definition equivalent to the first?
There's a distinction to be made be the "algebraic dual", which seems
to be what your 2nd defn is (by default), and the "topological dual"
which corresponds to your 1st defn.

The linear functionals in the topological dual space are required to be
continuous. Therefore, the topological dual is (in general) a subspace
of the algebraic dual (since there's an extra restriction is imposed
upon the former).

FAPP, you might as well just concentrate on the topological dual.

(I'll respond to the tempered distribution question in a later post.)

Recognitions:
 Quote by Fredrik Is the set of tempered distributions the dual space of the Schwarz class or not?
I don't have Folland unfortunately, but my understanding is that the
space of tempered distributions is indeed the (topological) dual of the
Schwartz space.

This coincides with Lax (p559) and also the Wiki page
http://en.wikipedia.org/wiki/Distribution_(mathematics) .
See the section "Tempered distributions and Fourier transform".
Warning: read the paragraph starting "The space of tempered
distributions is defined ...." very carefully since the wording
is a little confusing if read too quickly.

Recognitions:
 Quote by Fredrik Yes, this makes a lot of sense. Ballentine seems to agree. Quote from page 28: We now have two mathemathcally sound solutions to the problem that a self-adjoint operator need not possess a complete set of eigenvectors in the Hilbert space of vectors with finite norms. The first, based on the spectral theorem (Theorem 4 of Sec. 1.3), is to restate our equations in terms of projection operators which are well defined in Hilbert space, even if they cannot be expressed as sums of outer products of eigenvectors in Hilberbert space. The second, based on the generalized spectral theorem, is to enlarge our mathematical framework from Hilbert space to rigged Hilbert space, in which a complete set of eigenvectors (of possibly infinite norm) is guaranteed to exist.
I think Ballentine's spectral theorem (Theorem 4 of Sec. 1.3) is distinctly different
from what Avodyne was saying (which involves a restriction to finite domains in both
position and momentum space, and is thus only a subspace of the unrestricted Hilbert space
that Ballentine refers to.

Recognitions:
Quote by Fredrik
 ...more about constructing a quantum theory starting from an algebra of observables, instead of starting from a Hilbert space.
OK, but why would we want to? If QM can be made rigorous using either a Hilbert space and the spectral theorem, or a rigged Hilbert space and the generalized spectral theorem (see the Ballentine quote in my previous post), then we don't need C*-algebras for mathematical rigor, and in the end, it gives us a Hilbert space anyway.
For QFT in infinite-dimensions, even the rigged Hilbert space is not big enough.
The singular behaviours arising from the interaction terms in the Hamiltonian
become far more pathological than mere delta distributions.

Secondly, there are general theorems about how not all operators are sensible
observables. There's a little more on this at the end of a previous thread:
http://www.physicsforums.com/showthr...=262821&page=2

Thirdly, there are so-called Bogoliubov transformations of the a/c ops in QFT
which take you from one Hilbert space into another unitarily inequivalent one.
This stuff is important in condensed matter physics, and other stuff like
Hawking-Unruh.

Hence, plenty of people believe that it's better to start with the algebra of
relevant physical observables, and then construct states as linear mappings
from the algebra to C. In that formalism, issues like those above are more
out in the open.

Mentor
 Quote by strangerep There's a distinction to be made be the "algebraic dual", which seems to be what your 2nd defn is (by default), and the "topological dual" which corresponds to your 1st defn.
I wasn't aware of the distinction. (I checked the Wikipedia article, so I am now). The type of "dual" I had in mind was the one that consists of bounded linear functionals, with the word "bounded" meaning that there exists an M>0 such that $|Tx|\leq M\|x\|$ for all x. It's fairly easy to show that a functional is bounded in this sense if and only if it's continuous with respect to the metric induced by the norm.

So what I had in mind is the topological dual of a vector space with the topology induced by the inner product. What's confusing me is the definition of continuity. It seems natural to me to define the standard inner product, and use the concept of continuity induced by it. But instead, we choose to define convergence of a sequence in D, and use that to define continuity. These definitions may be equivalent, but it's not obvious that they are.

It gets even more weird in the case of tempered distributions. We have defined an infinite number of norms, which should give us infinitely many kinds of continuity. Maybe I'm just overlooking something simple, but it seems weird that none of the texts (Wikipedia, Streater & Wightman, Folland) say anything about it.

Mentor
 Quote by strangerep For QFT in infinite-dimensions, even the rigged Hilbert space is not big enough. The singular behaviours arising from the interaction terms in the Hamiltonian become far more pathological than mere delta distributions.
Interesting...and strange. The strange part is that we can actually make predictions in spite of all of this.

 Quote by strangerep Secondly, there are general theorems about how not all operators are sensible observables. There's a little more on this at the end of a previous thread: http://www.physicsforums.com/showthr...=262821&page=2
I don't see anything about it in that thread. but I recognize the claim from here. Their refererence is this, but I don't really understand it. I think I understood one thing though. A self-adjoint operator is only an observable if it preserves each superselection sector. For example, it can't take a bosonic state to a fermionic state.

Recognitions:
 Quote by Fredrik I don't see anything about it in that thread.
OK. There's some papers by Gotay, e.g., paper math-ph/9809011,
main subject of the current thread, so you probably won't want
to wade through 50-pages of nontrivial math. Maybe just look
at p18: the paragraph starting with
 In particular, one can see at the outset that it is impossible for a prequantization to satisfy the “product → anti-commutator” rule.
and the manipulations that follow. And maybe look through the
early sections to get the context.

EDIT: A better reference might be math-ph/9809015.

But like I said, this goes off on an tangent from the current thread.

Recognitions:
 Quote by Fredrik So what I had in mind is the topological dual of a vector space with the topology induced by the inner product. What's confusing me is the definition of continuity. It seems natural to me to define the standard inner product, and use the concept of continuity induced by it. But instead, we choose to define convergence of a sequence in D, and use that to define continuity. These definitions may be equivalent, but it's not obvious that they are.
I'm starting to lose track of what you're referring to, hence unable to offer a pinpoint

 It gets even more weird in the case of tempered distributions. We have defined an infinite number of norms, which should give us infinitely many kinds of continuity.
Note that each of these norms defines a subspace. I.e., the sequence of norms defines
a sequence of spaces, each densely nested in the previous (larger) one. So "continuity"
applies in the context of the norm topology on each subspace. The tempered distribution
case applies to the (dual of) the "inductive limit" of this sequence of spaces (afaiu).

Mentor
 Quote by strangerep I'm starting to lose track of what you're referring to, hence unable to offer a pinpoint answer. Maybe you should re-summarize?
OK, let's focus on "distributions" first, not "tempered distributions". We define the test function space (I'll call it $\mathcal D$) as the set of $C^\infty$ functions from $\mathbb R^n$ into $\mathbb C$ that have compact support. I would like to define a "distribution" as a member of the dual space $\mathcal D^*$, defined as the set of continuous linear functions $T:\mathcal D\rightarrow\mathbb C$. But to do that, we need to define what "continuous" means. There are at least two ways to do that.

Option 1: Define the usual inner product. The inner product gives us a norm, and the norm gives us a metric. Now we can use the definition of continuity that applies to all metric spaces.

$T:\mathcal D\rightarrow\mathbb C$ is continuous at $g\in\mathcal D$ if for each $\epsilon>0$ there's a $\delta>0$ such that $\|f-g\|<\delta\implies|T(f)-T(g)|<\epsilon$. T is continous on a set $U\subset\mathcal D$ if it's continuous at each point in U. Alternatively, and equivalently, T is continuous on $U\subset\mathcal D$ if $T^{-1}(E)$ is open for every open $E\subset\mathbb C$.

Weird, it takes a lot less to cause a database error now than a couple of weeks ago. I'll continue in the next post.