Unbounded operators in nonrelativistic QM of one spin0 particleby Fredrik Tags: nonrelativistic, operators, particle, spin0, unbounded 

#37
Apr809, 04:48 AM

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The answer to my concerns in #3435 (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 nonrelativistic QM of one spin0 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: 



#38
Apr909, 12:14 AM

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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. 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. 



#39
Apr909, 04:46 PM

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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.




#40
Apr1209, 04:55 AM

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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.




#41
Apr1209, 10:57 AM

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[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 deltafunctions [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? 



#42
Apr1209, 11:14 AM

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#43
Apr1209, 12:25 PM

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[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. 



#44
Apr1209, 12:47 PM

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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. 



#45
Apr1209, 11:13 PM

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defined implicitly/indirectly by a demand that a certain set of functions be continuous under it.) 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? 



#46
Apr1309, 03:11 AM

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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: quantph/0502053 It's written in a very physicistfriendly 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). 



#47
Apr1309, 05:37 AM

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#48
Apr1309, 07:57 AM

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#49
Apr1309, 12:11 PM

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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.




#50
Apr1409, 02:03 PM

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As you know the Hilbert Space for 1 particle spin0 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? 



#51
Apr1409, 09:07 PM

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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 infinitelydifferentiable functions only, and is a bit a redherring when one is focusing just on topological matters. 



#52
Apr1409, 09:21 PM

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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 doublecheck 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. of knowledge. Books are supposed to be for lesser mortals. 



#53
Apr1409, 09:51 PM

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a couple of things from a direction I hadn't explored adequately... I hope you can say a bit more about that. Dirac braket notation, we have [tex] \phi(f) ~=~ \int{\phi(x)f(x)}dx ~=~ \langle\phi \int{x\rangle\langle x}dx ~ f\rangle ~=~ \langle \phi  f\rangle [/tex] with [tex] \langle xx'\rangle ~=~ \delta(x  x') ~,~~~\mbox{etc.} [/tex] 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 (higherorder?) functionals over these functions? 



#54
Apr1509, 04:29 AM

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As you know a distribution [itex]T(x)[/itex] is a continuous linear functional over the space of test functions [itex]\mathcal{D}[/itex]. However one can reverse this and consider a test function [itex]f[/itex] as a function on distributions. A given test function [itex]f[/itex] maps any distribution [itex]T[/itex] to [itex]T(f)[/itex], resulting in a map over the whole space of distributions. [itex]f:\mathcal{D}^*(\mathbb R^3) \rightarrow \mathbb R[/itex]. So one writes this function over distributions as [itex]f(T)[/itex]. One can then immediately arrive at more general functions for instance: [itex]f(T)g(T)[/itex] [itex]f(T)g(T)h(T)[/itex] or [itex]e^{f(T)}[/itex] 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: [itex]\phi(g) = \int{\phi(x)g(x)}dx[/itex] Then, [itex]\phi(g)f(T) = g(T)f(T)[/itex] The conjuagate momentum field [tex]\pi(x)[/tex] acts like the momentum operator from regular QM by being a derivative. In rough language: [itex]\pi(x)f(T) = i\frac{\delta}{\delta T}f(T)[/itex] The Hamiltonian is composed of [itex]\pi(x)[/itex] and [itex]\phi(x)[/itex] 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: [itex]\int{f(T)}dT[/itex]. 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 [itex]dT_{free}[/itex], 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 [itex]\phi(x)^4[/tex] theory for example, the Hamiltonian gains an additional term: [itex]\int{\phi(x)^4}dx[/itex] It is very difficult to know exactly how one should interpret this. [itex](\int{\phi(x)^4}dx)f(T) = ?[/itex] There are two problems here. [itex]\phi(x)[/itex] is a operator valued distribution. When integrated it gives operators on [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{free})[/itex]. Due to its distributional nature no sense can be made of [tex]\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 [itex](\int{\phi(x)^4}dx)[/itex] 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 [itex](\int{\phi(x)^4}dx)[/itex] as a quantum operator. This notion was actually already known to physicists as Wick ordering. So we replace: [itex](\int{\phi(x)^4}dx)[/itex] with [itex](\int{:\phi(x)^4:}dx)[/itex]. One can then prove that the full Hamiltonian is selfadjoint and semibounded (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 19681973. 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 [itex](\int{\phi(x)^4}dx)[/itex] with [itex](\int{:\phi(x)^4:}dx)[/itex]. However one finds that this does not result in the full Hamiltonian being a selfadjoint operator. If you are very clever you can prove that this problem can not be overcome, unless you leave [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{free})[/itex], that is leave Fock space and move to another space of square integrable functions over distributions. Let me call it: [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{Glimm})[/itex]. 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: [itex]\delta m^2(\int{:\phi(x)^2:}dx)[/itex] to the Hamiltonian. The [itex]\delta m^2[/itex] 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 [itex]L^2(\mathcal{D}^*(\mathbb R^3), dT_{Glimm})[/itex], 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 selfadjoint 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 [itex]\phi^4[/itex] 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 [itex](\int{\phi(x)^4}dx)[/itex] 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 YangMills 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. 


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