Register to reply 
Unbounded operators in nonrelativistic QM of one spin0 particle 
Share this thread: 
#1
Apr309, 02:15 PM

Emeritus
Sci Advisor
PF Gold
P: 9,224

What exactly are the axioms of nonrelativistic QM of one spin0 particle? The mathematical model we're working with is the Hilbert space [itex]L^2(\mathbb R^3)[/itex] (at least in one formulation of the theory). But then what? Do we postulate that observables are represented by selfadjoint operators? Do we say that a measurement of an operator [itex]A[/itex] on a system prepared in state [itex]\psi\rangle[/itex] yields result a_{n} and leaves the system in the eigenstate [itex]n\rangle[/itex] with probability [itex]\langle n\psi\rangle^2[/itex]? Then how do we handle e.g. the position and momentum operators, which don't have eigenvectors?
Can the problem of unbounded operators be solved without the concept of a "rigged Hilbert space"? Is it easy to solve when we do use a rigged Hilbert space? What is a rigged Hilbert space anyway? I think I brought this up a few years ago, but apparently I wasn't able to understand it even after discussing it. I think I will this time, because of what I've learned since then. Don't hold back on technical details. I want a complete answer, or the pieces that will help me figure it out for myself. 


#2
Apr309, 03:20 PM

P: 102

I too would like a clarification on the subject of "rigged Hilbert space". Sometimes it seems like it is just a word people throw in to justify introducing nonnormalizable eigenstates and treating them in a similar way as other eigenstates with the substitution [itex]\langle nm\rangle =\delta_{nm}\rightarrow\langle xx'\rangle=\delta(xx')[/itex]. Is this just some trick or is there more to it.
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 ..? 


#3
Apr409, 01:31 AM

Sci Advisor
P: 1,881

of operators (a Banach algebra), satisfying some extra axioms that make it into a C* algebra. Then construct a Hilbert space on which the elements of the algebra act as operators. This is called the "GNS" construction. The algebraic approach avoids some of the operator ambiguities that can arise with the "Hilbert space first" approach. (Personally I dislike both names, and prefer the more explicit "Gelfand Triple Space", though I think I'm alone in that usage.) operators. The general spectral theorem for s.a. operators on infdim Hilbert space is littered with domain stuff. But the whole point of the RHS idea is to avoid that stuff and provide a rigorous mathematical underpinning of Dirac's original braket stuff that uses such improper eigenvectors. Do you have a copy of Ballentine's QM textbook? It's one of the few that explain and emphasize how all the Diracstyle QM we know and love is really all being done in an RHS. (Ballentine also shows how some of the operators in nonrel QM arise by considering unitary representations of the Galilei group, which was another part of your question.) I just looked at the RHS Wiki page but it's very brief and doesn't tell you much. Although Ballentine describes RHS, it's only at an introductory level. There's an old book by Bohm & Gadella, "Dirac Kets, Gamow Vectors, and Gel'fand Triplets" which explains a bit more, but they too don't get into the mathematical guts. Forums post, but maybe I can get you started... The basic idea is to start with a Hilbert space "H" and then construct a family of subspaces. To do this, take the formula for your Hilbert space norm, and then modify it to make it harder for all states to have a finite norm. E.g., change the usual norm from [tex] \int dx \psi^*(x) \psi(x) [/tex] to something like [tex] \int dx x^n \psi^*(x) \psi(x) [/tex] Clearly, for n>0, only a subset of the original [itex]\psi[/itex] functions still have finite norm. It is therefore a "seminorm" (meaning that it's defined only a subset of H). This family of seminorms, indexed by n, define a family of progressing smaller and smaller subsets of the original Hilbert space H. It turns out that each such subspace is a linear space, and is dense in the next larger one. More generally, this construction comes under the heading of "Nuclear Space", with a corresponding family of "seminorms". The Wiki page for Nuclear Space has some more info. Now, to proceed further, you need to know a couple of things about infdim vector spaces and their duals. First a Hilbert space H is isomorphic to its dual (i.e., isomorphic to the set of linear mappings from H to C). Then, if you restrict to a linear subspace of H, (let's call it [itex]\Omega_1[/itex], corresponding the case n=1 above), the dual of [itex]\Omega_1[/itex], which I'll denote as [itex]\Omega^*_1[/itex], is generally larger than H. I.e., we have [itex]\Omega_1 \subset H \subset \Omega^*_1[/itex]. Note that the usual norm and inner product are illdefined between vectors belonging to the dual space [itex]\Omega^*_n[/itex], but we still have well defined dualpairing between a vector from [itex]\Omega^*_n[/itex] and a vector from [itex]\Omega_n[/itex]. This is enough for Diracstyle quantum theory. Actually, I'm getting a bit ahead of myself. First, we should take an inductive limit [itex]n\to\infty[/itex] of the [itex]\Omega_n[/itex] spaces, which I'll denote simply as plain [itex]\Omega[/itex] without the subscript. This is the subspace of functions from H which vanish faster than any power of x. The "Rigged Hilbert Space", or "Gel'fand Triple", is the name given to the triplet of densely nested spaces: [tex] \Omega \subset H \subset \Omega^* [/tex] The word "rigged" should be understood to mean "equipped and ready for action". (Even with this explanation I personally still think it's a poor name.) [Continued in next post because of "Database error"...] 


#4
Apr409, 01:35 AM

Sci Advisor
P: 1,881

Unbounded operators in nonrelativistic QM of one spin0 particle
[Continuation of previous post...]
The master stroke now comes in that the socalled improper states of position, momentum, etc, in Dirac's braket formalism correspond to vectors in [itex]\Omega^*[/itex]. It is possible to take a s.a. operator on [itex]\Omega[/itex], and extend to an operator on [itex]\Omega^*[/itex], where the extension is defined in terms of its action on elements of [itex]\Omega[/itex] via the dualpairing. Taking this further, there is a generalization of the usual spectral theorem, called the GelfandMaurin Nuclear Spectral Theorem which shows that eigenvectors of A in the dual space [itex]\Omega^*[/itex] are "complete" in a generalized sense, even though they're not normalizable. So although people "throw around" the phrase "rigged Hilbert space" it's actually very important to the mathematical underpinnings of QM, though perhaps less so if you just want to do Diracstyle basic calculations. The RHS *is* the arena for modern QM, rather than the simpler Hilbert space as widely believed. The RHS, with the GM Nuclear Spectral Theorem, is a far more general mathematical foundation than the trick of "finite boxes", etc, that jensa asked about. There's also an old textbook by Maurin "General Eigenfunction Expansions..." which gives the rigorous proof (though not very clearly, imho). But I think I'll stop here and see what followup questions arise. 


#5
Apr409, 05:18 AM

P: 2,022

This connection between unboundedness of operators and the nonnormalizable eigenvectors is nonexisting, and indicates some confusion.
Example 1: If you consider the system defined by a Hamiltonian (this is the infinitely deep potential well) [tex] H = \frac{\hbar^2}{2m}\partial_x^2 + \infty\;\chi_{]\infty,0[\cup ]L,\infty[}(x) [/tex] and solve it's eigenstates, you get a sequence of normalizable eigenvectors [itex]\psi_1\rangle,\psi_2\rangle,\ldots[/itex], and in this basis the Hamiltonian is [tex] H = \frac{\hbar^2\pi^2}{2mL^2}\left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & \cdots \\ 0 & 4 & 0 & 0 & \cdots \\ 0 & 0 & 9 & 0 & \cdots \\ 0 & 0 & 0 & 16 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}\right) [/tex] This is an unbounded operator, but still it can be diagonalized in the Hilbert space in the standard sense. Example 2: If you regularize the differential operator [itex]\partial^2_x[/itex] by making some cutoff in the Fourierspace, you obtain some pseudodifferential operator which will be approximately the same as the [itex]\partial_x^2[/itex] for wave packets containing only large wave lengths. So fix some large [itex]R[/itex] and set [tex] (H_R \hat{\psi})(p) = \frac{p^2}{2m} \chi_{[R,R]}(p)\hat{\psi}(p), [/tex] which is the same thing as [tex] (H_R \psi)(x) = \frac{1}{2m} \frac{1}{2\pi\hbar} \int\limits_{R}^R\Big( \int\limits_{\infty}^{\infty} p^2 \psi(x') e^{i(xx')p/\hbar} dx'\Big) dp. [/tex] This operator is bounded and [itex]\H_R\ = \frac{R^2}{2m} < \infty[/itex]. However, it's eigenvectors are outside the Hilbert space [itex]L^2(\mathbb{R})[/itex]. Conclusion: So it is possible to have an unbounded operator so that its eigenvectors are inside the Hilbert space, and it is possible to have a bounded operator so that its eigenvectors are outside the Hilbert space. 


#6
Apr409, 06:10 AM

P: 2,022

If we ask a question that what is the probability for a momentum to be in an interval [itex][p_0\Delta, p_0+\Delta][/itex], we get the answer from the expression
[tex] \frac{1}{2\pi\hbar} \int\limits_{p_0\Delta}^{p_0+\Delta} \hat{\psi}(p)^2 dp. [/tex] I'm not convinced that it is useful to insist on being able to deal with probabilities of precise eigenstates. Experimentalists cannot measure such probabilities either. 


#7
Apr409, 10:24 AM

Emeritus
Sci Advisor
PF Gold
P: 9,224

Good answers, both of you. I appreciate that you are taking the time to explain these things to me. I need some time to think about the technically advanced parts of Strangerep's posts, so for now I'll just reply to Jostpuur. I'll reply to Strangerep later today, or tomorrow.
Jostpuurs post #6 brings up one of the things I was thinking about when wrote the OP and when I was reading Strangerep's reply. The RHS stuff is interesting, and I definitely want to learn it, but I feel that as long as we're just talking about the nonrelativistic quantum theory of one spinless particle, it should be possible to avoid the complications by stating the axioms of the theory of physics carefully, instead of changing the mathematical model (by replacing the Hilbert space by a Gelfand triple). What I mean by the "axioms of the theory of physics" are the statements that tell us how to interpret the mathematics as predictions of probabilities of possible results of experiments. Am I right about this, or do we absolutely need something like a RHS just to state the simplest possible quantum theory in a logically consistent way? Joostpur, I agree that your example 1 proves that it's possible for an unbounded operator on a Hilbert space to have eigenvectors. I didn't expect that. The Hilbert space in your example is [itex]L^2([0,L])[/itex], not [itex]L^2(\mathbb R^3)[/itex], but those two spaces are isomorphic (unless I have misunderstood that too), so this should mean that there's an unbounded operator on [itex]L^2(\mathbb R^3)[/itex] that has an eigenvector. I don't understand 100% of example 2, but I accept it as a convincing argument that it's possible for a bounded operator to fail to have eigenvectors. The part that's confusing me is that I don't see what the "eigenvectors" of H_{R} are. I assume that they are some sort of distributions. 


#8
Apr409, 07:36 PM

Sci Advisor
P: 1,881

Expressed more precisely, let me quote the HellingerToeplitz theorem (from Lax, p377): "An operator M that is defined everywhere on a Hilbert space H and is its own adjoint, (Mx,y) = (x,My), is necessarily bounded." The proof is only a few lines. There follows a corollary (further down on p377): "It follows from this [...] that unbounded operators that are their own adjoints can be defined only on a subspace of the Hilbert space". simplify your example... Your eigenvectors can be written as infinitelength vectors: [tex] e_1 := (1,0,0,0,....)~~ e_2 := (0,1,0,0,....)~~ \dots~ e_k := (0,0,...,0,1,0,0,...)~~ [/tex] Forgetting the constants, your Hamiltonian can be written as [tex] H_{nm} ~=~ n^2 ~ \delta_{nm} [/tex] Now, I can construct a particular linear combination of the [itex]e_k[/itex]: [tex] \psi := \sum_{k=1}^\infty \frac{1}{k} ~ e_k [/tex] whose squared norm is [tex] (\psi,\psi) ~=~ \sum_{k=1}^\infty \frac{1}{k^2} ~<~ \infty [/tex] So [itex]\psi[/itex] is in the Hilbert space. Now consider [tex] \phi := H \psi ~=~ \sum_{k=1}^\infty k^2 ~ \frac{1}{k} ~ e_k ~=~ \sum_{k=1}^\infty k ~ e_k [/tex] whose squared norm is [tex] (\phi,\phi) ~=~ \sum_{k=1}^\infty k^2 ~\to~ \infty [/tex] and therefore [itex]\phi[/itex] is not in the Hilbert space. Hence, it's incorrect to say that this Hamiltonian is an operator on the entire Hilbert space. It's only a welldefined operator on a subspace of the Hilbert space. The rigged Hilbert space formalism was developed to make sense of this. The Hamiltonian *can* be diagonalized in a sense, but must be done in terms of generalized eigenvectors in a larger space of tempered distributions (the [itex]\Omega^*[/itex] from my earlier post). is not welldefined on all of [itex]L^2(\mathbb{R})[/itex]. I.e., it's not an operator on all of [itex]L^2(\mathbb{R})[/itex]. You seem to be defining the eigenvectors on a subset of [itex]L^2(\mathbb{R})[/itex] (with finite "R") and then assuming they remain welldefined when you take [itex]R\to\infty[/itex]. But the limit [tex] \lim_{R\to\infty} \H_R\ ~=~ \lim_{R\to\infty} \frac{R^2}{2m} [/tex] does not exist. these subtleties. Dirac was one of them. He just knew intuitively that it was ok, somehow. Later, some mathematicians came along and made it more rigorous and respectable, using rigged Hilbert space and related concepts. And Fredrick's question was clearly asking about the mathematically precise stuff. 


#9
Apr409, 11:27 PM

P: 2,022

I was already aware of the fact that unbounded operators are often[itex]{}^*[/itex] defined on some subsets of the original Hilbert space, although I thought it would not be necessary to get into that matter in my post. For example the domain of [itex]H=\frac{\hbar^2}{2m}\partial_x^2[/itex] is
[tex] D(\partial_x^2) = \big\{\psi\in L^2(\mathbb{R})\;\; \int\limits_{\infty}^{\infty} p^4 \hat{\psi}(p)^2 dp < \infty \big\}, [/tex] but this doesn't usually get mentioned in every post that is concerned with this Hamiltonian. ([itex]{}^*[/itex]: Or let's say that I was under belief that this "often" the case, while not being aware of the fact that it is always the case, with selfadjoint operators.) [tex] \psi\mapsto M_f\psi,\quad (M_f\psi)(x) = f(x)\psi(x) [/tex] defines a bounded operator [itex]M_f:L^2(X)\to L^2(X)[/itex], and [itex]\M_f\\leq \f\_{\infty}[/itex]. My example belongs to this class of operators. 


#10
Apr409, 11:52 PM

P: 2,022

The harmonic oscillator is an example of operator which is defined on some subspace of [itex]L^2(\mathbb{R}^n)[/itex], and which has a sequence of eigenvectors whose span is dense in [itex]L^2(\mathbb{R}^n)[/itex]. [tex] f(x)\psi(x) =\lambda \psi(x) [/tex] must be true for a.e. [itex]x\in X[/itex]. This cannot happen unless [itex]f(\overline{x})=\lambda[/itex] with some [itex]\overline{x}\in X[/itex], and unless [itex]\psi(x)=0[/itex] for a.e. [itex]x\neq \overline{x}[/itex]. So the eigenvectors would have to be [itex]\psi(x) = \chi_{\{\overline{x}\}}(x)[/itex]. If the measure [itex]\mu[/itex] is such measure that [itex]\mu(\{\overline{x}\})=0[/itex], then the eigenvector doesn't exist because it is zero. What happens in the nonrigorous formalism is that this kind of eigenvector is multiplied with an infinite constant so that it becomes nonzero. If the Hamiltonian is defined in the Fourierspace by a multiplication [tex] (H\hat{\psi})(p) = \frac{p^2}{2m}\hat{\psi}(p) [/tex] then in the nonrigorous formalism the eigenvectors are deltafunctions [itex]\delta(pp')[/itex], and in the spatial representation they are the plane waves [itex]e^{ixp/\hbar}[/itex]. If the Hamiltonian is made bounded by force by multiplying the operated function with [itex]\chi_{[R,R]}(p)[/itex], then the same eigenvectors still work, but the eigenvalues are different. The eigenvalues are the same for [itex]R\leq p\leq R[/itex], but go to zero for other [itex]p[/itex]. 


#11
Apr509, 09:11 AM

Emeritus
Sci Advisor
PF Gold
P: 9,224

Edit: I do now, thanks to Jostpuur. See my next post. [tex]H_R\psi(x)=\frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}dp\ e^{ipx}\frac{p^2}{2m}\tilde\psi(x)\chi_{[R,R]}(p)=\frac{1}{2\pi}\int_{R}^{R}dp\ e^{ipx}\frac{p^2}{2m}\int_{\infty}^{\infty}dx'\ e^{ipx'}\psi(x')[/tex] We can take the expression on the righthand side as the definition of H_{R}, and if we do I think it's clear that this operator is welldefined. The middle expression implies that it's not injective. The integral doesn't depend on what values the Fourier transform [itex]\tilde\psi[/itex] has outside of the interval [R,R]. (LOL, the tilde is invisible in itex mode). 


#12
Apr509, 09:31 AM

P: 2,022

Recall that if an operator [itex]T:V\to V[/itex] between two norm spaces is unbounded, then it does not mean that [itex]\T\psi\=\infty[/itex] for some [itex]\psi\in V[/itex]. This would be contradictory with the implicit assumption [itex]T(V)\subset V[/itex]. Instead it means that there is a sequence of vectors [itex]\psi_1,\psi_2,\psi_3,\ldots\in V[/itex] such that [itex]\\psi_n\\leq 1[/itex] for all n, and [tex] \sup_{n\in\mathbb{N}} \T\psi_n\ = \infty. [/tex] When [itex]V[/itex] is a subset of some larger vector space in which the norms of vectors are infinite, and the image point [itex]T\psi[/itex] is defined with some formula in this larger vector space, it can happen that [itex]\T\psi\=\infty[/itex] for some [itex]\psi\in V[/itex]. In this case we don't obtain an operator [itex]T:V\to V[/itex], but instead an operator [itex]T:D(T)\to V[/itex] where [tex] D(T) = \{\psi\in V\;\; \T\psi\<\infty\} [/tex] is the domain of the operator. This is what happens in the example of the infinite potential well. The Hamiltonian is not defined on the entire Hilbert space [itex]L^2([0,L])[/itex], but only on some subspace. However, this subspace is dense in the Hilbert space. I don't think that I said anything wrong in my example 1 though. Despite the fact that the Hamiltonian is not defined in the entire Hilbert space, the Hamiltonian has a sequence of orthogonal eigenvectors, whose span is dense in the Hilbert space, so the Hamiltonian is pretty diagonalizable there. 


#13
Apr509, 10:31 AM

Emeritus
Sci Advisor
PF Gold
P: 9,224

Assume that [itex]H=(\hbar/2m)d^2/dx^2[/itex] is a linear operator from [itex]L^2([0,L])[/itex] into [itex]L^2([0,L])[/itex]. Then it's defined on the specific [itex]\psi[/itex] that Strangrep defined, but that [itex]\psi[/itex] satisfies [itex]\H\psi\=\infty[/itex], and that contradicts the assumption that the range of H is a subset of [itex]L^2([0,L])[/itex]. 


#14
Apr509, 06:23 PM

Sci Advisor
P: 1,881

I've now edited my earlier post to fix this. Sorry for my lack of care. 


#15
Apr509, 06:23 PM

Emeritus
Sci Advisor
PF Gold
P: 9,224

Edit: I think I see the point you were going for. I was asking about how to introduce observables into the theory, and you just meant that this is one way to define them. Is it a better way than to define the observables as selfadjoint operators on a separable Hilbert space? Does the C*algebra/GNS approach have anything to do with the RHS concept, or are the two unrelated? (They seem unrelated to me). I chose to ask specifically about nonrelativistic QM of one spin0 particle because it's the simplest of all relevant quantum theories, and I felt that it should be possible to define it in a pretty simple way. The traditional way (which is kind of sloppy) is to postulate among other things that states are represented by the rays of a (separable) Hilbert space (or specifically [itex]L^2(\mathbb R^3)[/itex]) and that the time evolution of a state is given by the Schrödinger equation. I think I would prefer to drop the explicit stuff about the Schrödinger equation, and instead postulate something about inertial observers and unitary representations of the Galilei group. 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'm also interested in how the axioms must be changed when we go from nonrelativistic to special relativistic quantum mechanics, and finally to general relativistic quantum mechanics. (But we can ignore that last one in this thread ). Is the bottom line here that the members of H* are distributions with H (square integrable functions) as the test function space, and that the members of [itex]\Omega^*[/itex] are tempered distributions? Hm, what you said to Jostpuur in #8 looks like a "yes" to that question. I just realized that there's one small difference. The members of [itex]L^2(\mathbb R^3)[/itex] are not all infinitely differentiable, and test functions are usually assumed to be. It seems a bit strange and complicated to define a sequence [itex]\Omega_n[/itex] instead of defining [itex]\Omega[/itex] right away, but then I didn't understand S & W on a first read, and they seem to go straight for [itex]\Omega[/itex] (if I remember correctly). Maybe that's why I didn't understand them. 


#16
Apr509, 07:58 PM

Sci Advisor
P: 1,881

relativity and causality, but more about constructing a quantum theory starting from an algebra of observables, instead of starting from a Hilbert space. You can read the axioms for C*algebras for yourself, but briefly, they're a subclass of Banach *algebras, which in turn are *algebras with a norm defined on every element satisfying certain extra axioms (e.g., the norm is submultiplicative  See the top of Wiki Banach algebra page for what that means). One then considers linear functionals in the dual space of this normed algebra to arrive at a way of mapping observables to numbers, thus getting a quantum theory. 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). GelfandMaurin generalization. (Looking at the index, I couldn't even find "nuclear spaces" mentioned.) Reed & Simon Vol1 talk about nuclear stuff, but in the context of tempered distributions. That's why I asked a question here a while ago about proofs of the GM theorem. Eventually I got hold of Maurin's book, difficult though it is. (respectively) the Schwartz space of test functions, and its dual space of tempered distributions. The important thing about functional analysis is that abstracts away from specific spaces to general properties of infdim linear spaces, abstracting a lot of messy integration stuff by expressing things as linear operators instead. Although I too found functional analysis quite challenging and bewildering initially, I've now come to prefer it immensely and I only drop back to explicit integral stuff when considering specific examples. Functional analysis is an essential tool for the mathematical physicist, imho. their universal enveloping algebra), and (2) find all unitary irreducible representations of this algebra. (3) Construct tensorproduct spaces thereof. The details fill many books of course. Weinberg takes this approach (moreorless) in his volumes. too bad for the Galilei case (Ballentine covers it), but constructing a relativistic position operator is still controversial and problematic. example of my [itex]\Omega[/itex] space (i.e., a dense subspace of H). originally) are just a rigorous way to define and generalize the notion of "...functions vanishing faster than any power of x...". The Wiki page on "nuclear space" has a bit more, though it doesn't mention the GM theorem. Try to find Maurin's book if you can. (or maybe vol4 in the series by Gelfand & Vilenkin  I couldn't obtain the latter, but many authors reference it). 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 It covers a lot of the theorems, but skips the lengthy proofs. 


#17
Apr509, 11:33 PM

Sci Advisor
P: 1,190

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
[tex]P_{x\in V} \equiv \int_V d^3\!x\,x\rangle\langle x[/tex] 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. 


#18
Apr609, 07:12 AM

Mentor
P: 6,219

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.
This is an example of something much more general. If two selfadjoint operators satisfy a canonical commutation relation, then it is easy to show that at least one of the operators must be unbounded. "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 wellknown 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." 


Register to reply 
Related Discussions  
Raising and lowering operators for spin  Quantum Physics  3  
Spin angular momentum operators  Advanced Physics Homework  1  
Spin angular momentum operators  Quantum Physics  5  
Spin operators  Advanced Physics Homework  2  
Help me (Spin Operators)  Introductory Physics Homework  6 