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In Quantum Mechanics one often deals with wavefunctions of particles. In that case, it is natural to consider as the space of states the space $L^2(\mathbb{R}^3)$. On the other hand, on the book I'm reading, there's a construction which it's quite elegant and general, however it is not rigorous. For those interested in seeing the book, it's "Quantum Mechanics" by Cohen-Tannoudji.

The book proceeds as follows: the first postulate of Quantum Mechanics states that for every quantum system there is one Hilbert space $\mathcal{H}$ whose elements describe the possible states of the system. The idea then is that $\mathcal{H}$ doesn't necessarily is a space of functions.

Indeed, Cohen defines (or doesn't define) $\mathcal{H}$ as the space of kets $|\psi\rangle\in \mathcal{H}$, being the kets just vectors encoding the states of the system.

The second postulate states that for each physically observable quantity there is associated one hermitian operator $A$ such that the only possible values to be measured are the eigenvalues of $A$ and such that

  1. If $A$ has discrete spectrum $\{|\psi_n\rangle : n \in \mathbb{N}\}$ then the probability of measuring the eigenvalue $a_n$ on the state $|\psi\rangle$ is $\langle \psi_n | \psi\rangle$ considering that $|\psi\rangle$ is normalized.

  2. If $A$ has continuous spectrum $\{|\psi_{\lambda}\rangle : \lambda \in \Lambda\}$ then the probability density on state $|\psi\rangle$ for the possible eigenvalues is $\lambda \mapsto \langle \psi_\lambda | \psi\rangle$

If, for example, the position operator $X$ for particle in one-dimension, exists, and if its eigenvectors are $|x\rangle$ with eigenvalues $x$, for each $x\in \mathbb{R}$, the probability density of position is $\langle x |\psi\rangle$ which is a function $\mathbb{R}\to \mathbb{C}$ and we recover the wavefunction.

This formulation, though, seems to be more general. In that case, wavefunction is just the information about one possible kind of measurement which we can obtain from the postulates. There is nothing special with it.

Now, although quite elegant and simple, this is not even a little rigorous. For example: the position operator hasn't been defined! It is just "the operator associated to position with continuous spectrum", but this doesn't define the operator. On the book, it is defined on the basis $\{|x\rangle\}$, but this set is defined in terms of it, so we get circular.

Another problem is that usually we are dealing with unbounded operators which are not defined on the whole of $\mathcal{H}$. And an even greater problem is that $\mathcal{H}$ was never defined!

I've been looking forward to find out how to make this rigorous, but couldn't find anything useful. Many people simply say that the right way is to consider always $L^2(\mathbb{R}^3)$, so that all of this talk is nonsense. But I disagree, I find it quite natural to consider this generalized version.

The only thing I've found was the idea of rigged Hilbert spaces, known also as Gel'fand triple. I've found not much material about it, but anyway, I didn't understand how it can be used to make this rigorous.

In that case, how does one make this idea of space of states, or space of kets, fully rigorous, overcoming the problems I found out, and possibly any others that may exist? Is it through the Gel'fand triple? If so, how is it done?

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  • $\begingroup$ You may find something useful here: science.unitn.it/%7Emoretti/Zaragoza-Lectures.pdf $\endgroup$ – Aretino Sep 20 '15 at 15:44
  • $\begingroup$ Another couple of comments, even if I'm quite rusty on such things. There does not exist a single $\mathcal{H}$, that's why you didn't find its definition. You can decide what Hilbert space to employ according to what you need. An example: if you choose $\mathcal{H}=L^2(\mathbb{R}^3)$ with coordinates representing the position of a particle, then the position operator simply corresponds to multiplication by $x$ while the momentum operator is given by $\partial/\partial x$ (times a constant). But you can choose instead another space where momentum is a multiplication and position a derivative. $\endgroup$ – Aretino Sep 20 '15 at 16:32
  • $\begingroup$ The different views are actually essentially one and the same. Whatever the Hilbert space is, you want the "wavefunction" (a vector in this space) to be square-integrable. This because the total probability must be finite (in fact one). For a finte dimensional Hilbert space this does not impose any constraint: any vector is "square-integrable" (the sum of the square moduli of each components is finite). For a function space on e.g, $\mathbb{R}^3$, this defines $L^2(\mathbb{R}^3)$. The probability of finding the particle anywhere in the space is one. $\endgroup$ – lcv Dec 11 '15 at 8:45
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First remark. QM postulates state that the Hilbert space is separable. Recall that Riesz-Fischer theorem ensures that separable Hilber spaces are all isometrically isomorphic, so it does not really matter which one you choose until you are speaking about the general theory. In concrete realizations, it is obviously useful pick the Hilbert that best reflects the properties of the system.

Second remark. QM postulates don't and can't say how operators are made in specific situations. This is an experimental fact. They only say that these must be linear operators (generally unbounded) on $\mathcal H$, self-adjoint (not simply Hermitean!) if representing observables (i.e., quantities that can be actually measured on the system). It is a consequence of the noncommutative structure of the observables $C^*$-algebra that there exist couples of incompatible observables. (This leads to the Heisenberg principle.) Who says how operators are made? This is a prerogative of the quantization procedure. Such a procedure establishes a correspondence between classical and quantum observables, making precise the intuitive idea by Dirac. In a precise sense (Groenewold theorem), does not exists a "universal" quantization procedure (again, observables are experimental). One requires that $X$ and $P$ must be implemented one as a multiplication operator and the other as a differential operator (which one is a matter of taste, leading to the so-called x-representation and p-representation), continuous and essentially self-adjoint when considered on rapidly decreasing functions, self-adjoint (not both bounded) when considered on a maximal domain of $L^2$, satisfaying canonical commutation rules. [For quantities that have not a classical counterpart, such as spin, the actual form is induced extrapolating by experimental data algebraic properties such as commutation rules, spectra and so on.]

Third remark. Wave functions are a very peculiar kind of state. $\mathcal H$ has a lot of more other elements. Precisely, let $\mathcal A$ be the $C^*$-algebra of the physical system. Gleason theorem establishes a 1-1 correspondence between trace-class operators in $\mathcal A$ and the rays of a projective Hilbert space $\mathcal H$. The so-called wave functions are associated to the projectors of the form $(\psi, \, ) \psi$ (round brackets stands for scalar product in $\mathcal H$, $\psi \in \mathcal H$.). Physicists refer to $\psi$ as the wave function and this is misleading and, strictly speaking, uncorrect.

Gel'fand triple and Gel'fand-Naimark-Segal (GNS) construction constitute a way of automatically get the correct Hilbert space for the system. However, it is not something you can hope to find in a QM textbook for physicists, since the mathematical apparatus of QM is so strong that usually one can forget technical details, "canonically" realize the Hilbert space and the observables and perform actual calculations, which are the only important things in physics.

If you are interested in the last part, you can see

[1] Bogolioubov (et alii), Axiomatic quantum field theory

[2] Lansdmann, Mathematical topics between classical and quantum mechanics

[3] Dixmier, $C^*$-algebras (without applications to QM)

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  • $\begingroup$ "Physicists refer to $\psi$ as the wave function and this is misleading and, strictly speaking, uncorrect." How can a definition be incorrect? $\endgroup$ – lcv Dec 11 '15 at 8:50
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There are two mathematically rigorous (and quite general) Hilbert Space formalisms you might be looking for. Both can be seen as attempts to salvage the engine of Dirac's original bra-ket algorithm, while avoiding its mathematical embarrassments.

The first - created by von Neumann - replaces Dirac's kets by vectors in an abstract Hilbert space $\mathscr{H}$, and replaces his transformations with linear operators on this space. To avoid the need of Dirac's (in)famous delta functions, von Neumann rejects the fundamental status of eigenvectors in Quantum Mechanics. Instead, the central player in von Neumann's game is spectral decomposition (specifically, of self-adjoint operators, although this can be generalized). This approach reduces to eigenvectors and eigenvalues when the operators happen to be finite dimensional (or, in the infinite dimensional case, compact), but handles the continuous case without batting an eyelid.

In the end, however, physicists were reluctant to abandon the highly intuitive eigenvector-based formalism of Dirac, and definitely did not want to muck around with all the subtleties of operator-domains inherent in von Neumann's approach.

And that is why the next fellow entered the fray...

The second - created by Gelfand et al - attempts to preserve more of Dirac's legacy, while straying as little as possible from von Neumann's Hilbert Space. Specifically, they proceed by starting with a Hilbert Space $\mathscr{H}$ and then constructing two auxiliary spaces from it. First a dense inner-space $\Phi$, and then an outer-space $\Phi^*$ (being the dual of the inner-space). Taken together, this framework is referred to as a Rigged Hilbert Space or Gelfand Triple: $$\Phi\subset\mathscr{H}\subset\Phi^*$$

Very briefly, the reasons they do this are:

  • The inner-space $\Phi$ supports operator algebra without all the hassle of domains inherent with von Neumann's approach.
  • The continuous spectrum can be handled in a rigorous eigenvector-eigenvalue fashion because the Dirackian delta 'functions', which don't live in the Hilbert Space, can be found in the dual $\Phi^*$ (technically, as distributions rather than functions).

Where does this leave the poor mathematical physicist with a desire for rigor?

Well, the truth is, nobody really uses Gelfand triples. Just to construct one triple you need to construct an infinite sequence of topologies! Not only that, but the whole construction depends on the specific form of the Hamiltonian, so it is nowhere near as general as von Neumann's approach. (Indeed, finding a Gelfand triple for a specific quantum system would easily qualify you for a PhD!)

To me, the Rigged Hilbert Space approach is reminiscent of the non-standard analysis discovered in the 1960s. Nobody actually uses it, but it makes them feel better about cancelling the $du$ in: $$\frac{dy}{du}\frac{du}{dx}$$ But let's be honest, you were going to do that anyway right?

Bottom line?

Learn von Neumann's rigorous Hilbert Space quantum mechanics.

There are plenty of excellent books out there. Even von Neumann's seminal Mathematical Foundations of Quantum Mechanics (1932) is a fantastic resource (although the old-fashioned pre-LaTeX notation is a bit intimidating). There were lots of excellent modern books published in the 1970s and 80s on the subject (Prugovecki's Quantum Mechanics in Hilbert Space comes to mind). For a superb up-to-the-minute treatment of Quantum Mechanics in the thorough style von Neumann would salute is Valter Moretti's Spectral Theory and Quantum Mechanics (which I happen to be reading at the moment!).

Note on the issue of generality...

Much of your question concerns the issue of generality. Von Neumann's approach has the virtue of not committing to any specific $L^2$ space, and as a result produces theorems that hold for all quantum mechanical systems. But that is not always advantageous. Sometimes you want to find results about specific (types of) quantum mechanical systems. In this case, you should choose a specific Hilbert Space (often, but not always, an $L^2$ space). But even then, you know you can rely on all the tools of the trade you learned/developed in a general Hilbert Space.

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