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

Fredrik

Option 2: This is the option that every text on the subject seems to prefer, and I don't see why. We define what it means for a sequence in [itex]\mathcal D[/itex] to be convergent by saying that [itex]\phi_n\rightarrow\phi[/itex] if there's a compact set K in [itex]\mathbb R^n[/itex] that contains the supports of all the [itex]\phi_n[/itex], and [itex]D^\alpha \phi_n[/itex] converges uniformly on K to [itex]D^\alpha\phi[/itex], for all [itex]\alpha[/itex]. Recall that [itex]D^\alpha[/itex] is defined by

[tex]|\alpha|=\alpha_1+\cdots+\alpha_n[/tex]

[tex]D^\alpha f(x)=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}f(x)[/tex]

Now we define "continuous" by saying that [itex]T:\mathcal D\rightarrow\mathbb C[/itex] is continuous at [itex]\phi\in\mathcal D[/itex] if

[tex]\lim_{n\rightarrow\infty}T(\phi_n)=T(\phi)[/tex]

for every sequence [itex]\{\phi_n\}[/itex] that converges to [itex]\phi[/itex]. Maybe these two options are actually equivalent, but I would expect that they're not, mostly because I think option 1 is more natural and intuitive. I don't see why these authors would all go for option 2 if they are equivalent.

Fredrik

Ah, that's actually very helpful. I just re-read a few statements from the part of Streater & Wightman where they define tempered distributions, keeping in mind what you just said, and suddenly what they're saying makes a lot more sense. Unfortunately I have to go to bed now, but I'll continue tomorrow.

Fredrik

The answer to my concerns in #34-35 (or at least a partial answer) is probably that there's nothing weird or unexpected about the fact that the concept of continuity depends on what topology you define on the set you're working with. The space of distributions isn't the dual space of the test functions. It's a dual space of the test functions. Or to be more accurate, the topological dual space of a set isn't defined. Only the topological dual space of a topological vector space is defined, so we have to choose a topology first, and then we get a dual space.

It still makes me wonder why their version of continuity is more desirable. It probably has something to do with the idea that derivatives of distributions should always exist, but I haven't thought that idea through yet.

Edit: I'm satisfied that I understand the definition of distributions and tempered distributions now, so we don't have to discuss them unless someone else wants to. I'm not saying that I understand every detail perfectly, but my understanding is good enough.

By they way, a thought occurred to me when I was replying to another thread. Isn't it weird to start the formulation of the simplest meaningful quantum theory (the non-relativistic QM of one spin-0 particle) by postulating that the states are rays in [itex]L^2(\mathbb R^3)[/itex] when the next postulate says that the time evolution is given by the Schrödinger equation? The Schrödinger equation only makes sense if the partial derivatives exist, and that implies continuity, but [itex]L^2(\mathbb R)[/itex] contains ridiculous functions like the one Hurkyl thought of in another thread to disprove a claim I made there:

strangerep

First, you can probably separate out the differentiability stuff from the topological
continuity stuff.

I think (in the absence of precise definitions of those two cases) that both notions
of convergence are equivalent. The first is just saying that you can construct a
Cauchy sequence and the second is saying that a sequence converges.
(Topologically, a sequence "converges to a point z" if, given an arbitrary open
set O containing z, all points of the sequence are eventually "in" O -- after some
integer n in the sequence.

The thing about restricting to infinitely differentiable functions is so that you
can do integral calculus, i.e., solve a DE given an initial condition, in such
a way that as to be compatible with the (rigged) Hilbert space structure.

Yes. This is a crucial point to understand in order to work with other
topologies (weak, weak*, and others). Generally speaking, convergence
is easier in weaker (coarser) topologies, but more theorems can be
proven with stronger (finer) topologies. Bit of a tradeoff depending on
exactly what one is trying to achieve.

Fredrik

I think I'm going to have to study some more functional analysis before I post a bunch of new questions. I'll read some more of Conway first, and maybe I'll read Maurin's lecture notes. Strangerep and Jostpuur, thanks for the answers so far. George, feel free to bump the thread when you're less busy. I'm interested in what you have to say about all of this.

dextercioby

Brilliant discussion on a very interesting topic, I must say. I was wondering, how would I build the RHS for the hydrogen atom ? The hamiltonian for the H atom has a mixed spectrum and I've never seen an application of the RHS formalism for an operator with mixed spectrum.

jostpuur

Equipping the [itex]\mathcal{D}[/itex] with a norm

[tex]
\|\psi\|^2 = \int\limits_{\mathbb{R}^n} dx\; |\psi(x)|^2
[/tex]

and defining [itex]\mathcal{D}^*[/itex] as a topological dual

[tex]
\mathcal{D}^* = \big\{ \phi^*\in \mathbb{C}^{\mathcal{D}} \;|\; \phi^*\;\textrm{is linear},\; \sup_{\|\psi\|\leq 1} |\phi^*(\psi)| < \infty\big\}
[/tex]

is probably not what anyone wants, because for example this collection of linear forms doesn't contain delta-functions [itex]\delta_x:\mathcal{D}\to\mathbb{C}[/itex], [itex]\delta_x(\psi)=\psi(x)[/itex].

Was that what you meant by your option 1?

I'm not 100% sure of this, but I've heard that it is possible prove the existence of such topology in [itex]\mathcal{D}[/itex], that the convergence [itex]\psi_n\to\psi[/itex] in that topology is equivalent with this definition that you described here. When [itex]\mathcal{D}[/itex] is equipped with such topology, then the standard collection of distributions [itex]\mathcal{D}^*[/itex] can be defined according to the definition of topological dual.

jostpuur

While I have understood this problem, it has not yet become clear to me how this problem is supposed to be solved in a more sophisticated formalism. The only solution that I have understood so far is this:

I'm not convinced that the distributions are very useful. The distributions are usually defined by using rather strange topologies or norms. The Swartz norm seems to be totally different from the original idea of a Hilbert space norm. However, the Hilbert space norm is highly relevant for the probability interpretation. How are we supposed to deal with quantum mechanical probabilities if the Hilbert space norm has been replaced with some of these distribution norms?

Fredrik

Ahh, excellent point. Yes, that's what I meant. I get it now. (I didn't until I read your post). Delta functions wouldn't be bounded (and therefore not continous) linear functionals if we go for option 1. For delta to be bounded, there must exist M>0 such that

[tex]|\delta f|\leq M\|f\|[/tex]

for all test functions f, but the inequality is equivalent to

[tex]\frac{|f(0)|}{\|f\|}\leq M[/tex]

and we don't have to look hard to find an f that violates this. Even some constant functions will do. For example, define f_r(x)=1 when |x|<r, and f_r(x)=0 outside that region. Now shrink that region (i.e. choose a smaller r) until the norm of f_r gets small enough to violate the inequality.

Fredrik

I don't think the construction of such a topology is very difficult. It would be harder to prove that the result is equivalent to what we already have, but even that looks doable. (I'm not sure I care enough to give it a try though ).

We can e.g. define a subset of [itex]\mathcal D[/itex] to be open if it can be expressed as [itex]T^{-1}(U)[/itex] where U is an open subset of the complex numbers, and T is continous in the sense defined above.

Alternatively, I think we can define a subset E of [itex]\mathcal D[/itex] to be open if for every sequence in [itex]\mathcal D[/itex] that converges to a a point in E (in the sense defined above), E contains all but a finite number of members of the sequence.

I'm guessing that these definitions are adequate (in the sense that they both define a topology on [itex]\mathcal D[/itex]), and equivalent (in the sense that those topologies are the same), but I haven't made any attempt to prove it or disprove it.

strangerep

Yes, that defines the "T-weak" topology on [itex]\mathcal D[/itex]. (Weak topologies are
defined implicitly/indirectly by a demand that a certain set of functions be continuous
under it.)

I don't follow this. If you haven't yet specifed a topology on [itex]\mathcal D[/itex]
then there's no meaning in the statement that a sequence of elements of
[itex]\mathcal D[/itex] "converges". But maybe I misunderstood and you
meant something else?

strangerep

The discrete part of the spectrum doesn't pose the same difficulties as encountered
with the continuous part. One has a resolution of unity as usual, consisting of an
integral over the continuous part, plus a sum over the discrete part. Rigged Hilbert
space and the nuclear spectral theorem give meaning to the continuous part.
(Not sure whether that's what you were asking, though.)

BTW, for anyone who's still a bit perplexed about the role of all this
rigged Hilbert space stuff, I noticed this pedagogical introductory paper:

Rafael de la Madrid,
"The role of the rigged Hilbert space in Quantum Mechanics",
Available as: quant-ph/0502053

It's written in a very physicist-friendly way, using the example of a rectangular
potential barrier to make everything concrete. It also explains the connections
between the bras and kets of rigged Hilbert space theory and distributions
quite clearly (while minimzing the heavy pure math that probably turns some
people off).

Fredrik

You understood me right. We define convergence of sequences first, and use that to define the topology. Convergence is defined as in post #35. I got that definition and the idea that it defines a topology on the test function space from the Wikipedia article on distributions. This is a direct link to the relevant section. Note in particular the sentence "It can be given a topology by defining the limit of a sequence of elements of D(U)." Unfortunately, Wikipedia doesn't say how the definition of convergence defines a topology. That's why I came up with those two guesses about how it can be done.

Fredrik

Thanks. I have read the first 9 pages now, and I'm very pleased with it so far. I'm definitely going to read the rest later.

Fredrik

I found a review of Maurin's books, and the reviewer is complaining a lot about how careless Maurin's presentations are. Link. I'm guessing it would be better to try to find the original articles than to read his books.

DarMM

I thought I'd add a bit to this since it's very interesting.

As you know the Hilbert Space for 1 particle spin-0 QM is [itex]L^2(\mathbb R^3, dx)[/itex]. Where dx is the Lesbesgue measure.

[itex]\mathbb R^3[/tex] coming from the fact that a particle can occupy any point in three dimensional space. So the set of all points [itex]\mathbb R^3[/tex] is the classical configuration space.

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)[/tex], the space of distributions. So the Hilbert space of QFT is:
[itex]L^2(\mathcal{D}^*(\mathbb R^3), d\nu)[/itex], the space of square integrable functions over the space of distributions with respect to some measure [itex]d\nu[/itex]. A free QFT and an interacting QFT differ by their choice of [itex]d\nu[/itex].
Quantum Fields [itex]\phi(x)[/itex] are then objects that when integrated against a function [itex]f(x)[/itex] give an object [itex]\int{\phi(x)f(x)}dx = \phi(f)[/itex], which is an unbounded operator on [itex]L^2(\mathcal{D}^*(\mathbb R^3), d\nu)[/itex].

Who said rigorous QFT was hard?

strangerep

OK, thanks. I think I get it now.
The key item is the use of uniform convergence. E.g., given two functions f,g it's
sensible to ask how "close" they are to each other, as follows: If we can say that
[itex]|f(x) - g(x)| < \epsilon[/itex] for all x, then [itex]\epsilon[/itex] is an expression
of the closeness of f and g (independently of x).

So one could define open sets in a manner reminiscent of open balls in a metric
topology: pick a function f and a "radius" r and then say that one particular
open set centered on f, of radius r, is all the other functions g s.t. [itex]|f(x) - g(x)| < r[/itex]
for all x. Then take unions, etc, to get a full topology.

Uniform convergence just extends this idea to sequences, the crucial bit
being how it's independent of x.

The fact that it's expressed in terms of arbitrary derivatives is just because we
want to deal with infinitely-differentiable functions only, and is a bit a red-herring
when one is focusing just on topological matters.