A Does the quantum space of states have countable or uncountable basis?

[MichPod]

TL;DR Summary

Hard to express, but smth like "an apparent problem in choosing between a "big" Hilbert space of quantum states with non-countable basis and a "small" Hilbert space generated by a countable set of basis vectors".

It's probably more kind of math question.

I consider a wave function of a harmonic oscillator, i.e. a particle in a parabolic well of potential. We know that the Hamiltonian is a Hermitian operator, and so its eigenstates constitute a full basis in the Hilbert space of the wave function states. We also know that this basis is countable.

On the other side, the arbitrary state may also be considered as a weighted integral of delta-functions, and such delta-functions are obviously a non-countable basis which creates a "bigger" space than what the eigenstates of the original Hamiltonian could generate.

I then wonder what should be called a "true" Hilbert space for such a case? A continuous spectrum of eigenfunctions is common in QM, it is generated for instance by a "free" (no potential part) Hamiltonian, so it's not a completely weird idea that the space of the function should have uncountable basis for practically ANY potential, including parabolic - after all, the wave function at a single moment may be arbitrary. Also a question may be asked, how a delta-wave-function would evolve in a parabolic well potential, but such a function obviously lies out of the subspace generated by the countable set of eigenstates of the Hamiltonian of the parabolic potential, so how such a problem should be approached, while we are used to a countable basis for the Harmonic oscillator?

 

[atyy]

There is the rigged Hilbert space formalism, which allows delta functions to be considered "generalized" basis functions (although these are not permitted in the Hilbert space formalism). However, we still do not consider a delta function to be an allowed quantum state (which must be square integrable), although we allow it to be a basis vector for mathematical operations.

http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf
Quantum Mechanics in Rigged Hilbert Space Language
Rafael de la Madrid Modino

https://arxiv.org/abs/quant-ph/0502053
The role of the rigged Hilbert space in Quantum Mechanics
R. de la Madrid

https://arxiv.org/abs/1411.3263
Quantum Physics and Signal Processing in Rigged Hilbert Spaces by means of Special Functions, Lie Algebras and Fourier and Fourier-like Transforms
Enrico Celeghini, Mariano A. del Olmo

 

[MichPod]

There is the rigged Hilbert space formalism, which allows delta functions to be considered "generalized" basis functions (although these are not permitted in the Hilbert space formalism). However, we still do not consider a delta function to be an allowed quantum state (which must be square integrable), although we allow it to be a basis vector for mathematical operations.

Thanks, atyy. Having said that, could you clarify what space should be considered as "full", "true" etc. for the case of harmonic oscillator, and if it's a rigged space, what does it practically/theoretically mean that this space cannot be generated by the harmonic oscillator Hamiltonian? Like, in the end, does the Hamiltonian produce a full basis via its eigenstates or not? And whether some state from the rigged space (a nontrivial one i.e. laying out of the space with countable basis) may be used as an initial condition for a harmonic oscillator Schrodinger equation? I think the questions I am asking may be somehow incorrect as well, so some clarifications of the validity of such questions may be helpful also. I should read the books you provided by 'immediate help' may be of value now...

 

[gentzen]

Not rigid space, it is called rigged Hilbert space (in the sense of "upgraded" Hilbert space).

 

[hilbert2]

The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.

 

[MichPod]

Not rigid space, it is called rigged Hilbert space (in the sense of "upgraded" Hilbert space).

oops. my poor English. sorry. fixing this.

 

[MichPod]

The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.

But, speaking formally, why would the Hamiltonian define what is the Hilbert space for a particular arrangement, but not, say, a coordinate or momentum operator for which the spectrum is (always) continuous?

 

[stevendaryl]

The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.

I would clarify that statement a little bit. What you really mean is that space of all eigenstates of the Hamiltonian. There might be countably many independent eigenstates, or there might be continuum many. But the Hilbert space is not forced on you by the Hamiltonian. The Hilbert space is (as I understand it) the collection of all square-integrable functions of 3D space. The Hilbert space is the same whether the Hamiltonian is the harmonic oscillator or the Coulomb potential.

Maybe somebody can correct me if I'm wrong about this...

 

[PeterDonis]

The Hilbert space is (as I understand it) the collection of all square-integrable functions of 3D space. The Hilbert space is the same whether the Hamiltonian is the harmonic oscillator or the Coulomb potential.

Not quite. The Hilbert space for a particular system is the set of square-integrable functions that are solutions of the Schrodinger Equation with the applicable Hamiltonian. This is generally a proper subset of all square integrable functions (i.e., some functions are not possible solutions and are excluded from the Hilbert space). As long as that proper subset also satisfies all the axioms of a Hilbert space, there is no issue.

 

[stevendaryl]

Not quite. The Hilbert space for a particular system is the set of square-integrable functions that are solutions of the Schrodinger Equation with the applicable Hamiltonian.

That's a little bit apples-to-oranges, because the Hilbert space contains functions of space, while solutions to Schrodinger's equation are functions of space and time. Do you mean the space of all functions $f(x)$ such that there exists a solution $\psi(x,t)$ to the Schrodinger equation, and a time $t_0$ such that $f(x) = \psi(x,t_0)$?

 

[PeterDonis]

Do you mean the space of all functions $f(x)$ such that there exists a solution $\psi(x,t)$ to the Schrodinger equation, and a time $t_0$ such that $f(x) = \psi(x,t_0)$?

Yes.

 

[stevendaryl]

Yes.

Okay, so what's an example of a square-integrable function that can't be a solution to Schrodinger's equation?

Oh, I guess an infinite square well gives an example: No function that is nonzero outside the well is a possible solution.

 

[atyy]

Thanks, atyy. Having said that, could you clarify what space should be considered as "full", "true" etc. for the case of harmonic oscillator, and if it's a rigged space, what does it practically/theoretically mean that this space cannot be generated by the harmonic oscillator Hamiltonian? Like, in the end, does the Hamiltonian produce a full basis via its eigenstates or not? And whether some state from the rigged space (a nontrivial one i.e. laying out of the space with countable basis) may be used as an initial condition for a harmonic oscillator Schrodinger equation? I think the questions I am asking may be somehow incorrect as well, so some clarifications of the validity of such questions may be helpful also. I should read the books you provided by 'immediate help' may be of value now...

The eigenstates of the Hamiltonian is a basis for all the physical states. This is a countably infinite basis. Each Hamiltonian eigenstate is an allowed physical state.

The eigenstates of the position and momentum operators are not bases for all physical states, but they are generalized bases (normally, we aren't so particular, and we don't bother to use the term "generalized" and we simply call them "bases"). They are uncountably infinite bases. In the Hilbert space formulation, all bases are countable (the Hilbert spaces of quantum mechanics are separable Hilbert spaces). In the rigged Hilbert space formulation, when we allow generalized bases, those can be countable (eg. Hamiltonian eigenstates) or uncountable (position eigenstates). While each Hamiltonian eigenstates is an allowed physical state, each position eigenstate is not an allowed physical state.

The rigged Hilbert space formalism roughly corresponds to what physicists do with Dirac notation. However, the first rigorous mathematical formulation of quantum mechanics was the Hilbert space formalism, which meant that the Dirac notation was somewhat "quick and dirty". Later it was found that many of the "quick and dirty" practices could be included in a rigorous formulation, if one extended the Hilbert space formalism to rigged Hilbert spaces.

 

[atyy]

The reference by Celeghini and del Olmo says:
"lt seems trivial, but $\{|n \rangle \}$ has the cardinality of the natural numbers $\aleph_{0}$ and, as all bases in a Hilbert space have the same cardinality, the structure we have constructed (the quantum space on the line $\mathbb{R}$) is not an Hilbert space but a Rigged Hilbert space."

 

[PeterDonis]

Oh, I guess an infinite square well gives an example: No function that is nonzero outside the well is a possible solution.

More generally, any Hamiltonian that includes a potential which is only finite in a bounded region will have a restricted set of possible solutions.

Even more generally, any Hamiltonian that includes a potential which is not bounded at infinity (e.g., the harmonic oscillator) will, I think, have a restricted set of possible solutions.

 

[MichPod]

In the rigged Hilbert space formulation, when we allow generalized bases, those can be countable (eg. Hamiltonian eigenstates) or uncountable (position eigenstates). While each Hamiltonian eigenstates is an allowed physical state, each position eigenstate is not an allowed physical state.

Do I understand you correctly that the same rigged Hilbert space is somehow considered as having both countable and uncountable basis?
Also, it looks like a plain wave solution of a free Hamiltonian cannot be considered as an "allowed physical state" with this approach. Not arguing, just thinking...

 

[stevendaryl]

Even more generally, any Hamiltonian that includes a potential which is not bounded at infinity (e.g., the harmonic oscillator) will, I think, have a restricted set of possible solutions.

So the basis states for Harmonic Oscillator states are not complete (in the sense that there are square-integrable functions that are not expressible as linear combinations?)

 

[PeterDonis]

So the basis states for Harmonic Oscillator states are not complete

They're complete for the Hilbert Space of possible solutions of the harmonic oscillator Schrodinger Equation. I don't think they're complete for the Hilbert space of all square integrable functions on the real numbers, but that's not the relevant Hilbert space for the harmonic oscillator.

 

[Keith_McClary]

they're complete for the Hilbert space of all square integrable functions on the real numbers

Proof of completeness.

 

[stevendaryl]

They're complete for the Hilbert Space of possible solutions of the harmonic oscillator Schrodinger Equation. I don't think they're complete for the Hilbert space of all square integrable functions on the real numbers, but that's not the relevant Hilbert space for the harmonic oscillator.

At this point, I'm just curious about the mathematical issue:

Some definitions:

If $f(x)$ and $g(x)$ are square-integrable functions, say that $f \approx g$ if
$\int_{-\infty}^{+\infty} |f(x) - g(x)|^2 dx = 0$.

Let $\psi_n(x)$ be the $n^{th}$ Harmonic oscillator basis state. If $f(x)$ is a complex-valued function on $(-\infty, +\infty)$, let's say that $f(x)$ is representable by HO basis states if there is a sequence of complex numbers $C_n$ such that $\sum_n C_n \psi_n(x)$ converges to some function $g(x)$ where $f \approx g$.

So is every square-integrable function $f(x)$ representable by HO basis states? If not, what is a counter-example?

 

[Keith_McClary]

So is every square-integrable function representable by HO basis states? If not, what is a counter-example?

Yes. You probably didn't see my post about completeness a minute before.

 

[stevendaryl]

Proof of completeness.

There's a step that is omitted in the proof. Maybe it's because it's obvious? There are two different claims: (1) that $f(x)$ has a nonzero overlap with at least one basis state.
(2) that $f(x)$ can be represented as a (possibly infinite) sum of basis states.

The article you pointed to proved the first statement. Does the second follow obviously? I suppose you could just keep taking out the part of $f(x)$ that is not in the overlap, and then apply the theorem again. If that process converges, then you get a representation?

A second observation is that the proof is for functions that are square-integrable when multiplied by $e^{-x^2}$.

 

[atyy]

Do I understand you correctly that the same rigged Hilbert space is somehow considered as having both countable and uncountable basis?
Also, it looks like a plain wave solution of a free Hamiltonian cannot be considered as an "allowed physical state" with this approach. Not arguing, just thinking...

In the Hilbert space formalism, the Hilbert space of quantum mechanics is always separable and has a countable basis.

In the rigged Hilbert space formalism, the generalized formalism allows both countable and uncountable bases in a generalized sense.

Yes, a plain wave solution cannot be considered an allowed physical state in both the Hilbert space and the rigged Hilbert space formalisms.

Both Hilbert space and rigged Hilbert space formalisms are the same with respect to physical content, so the Hilbert space formalism is enough. The rigged Hilbert space formalism makes the elegant but apparently mathematically dirty Dirac formalism closer to something mathematically respectable.

 

[Keith_McClary]

A second observation is that the proof is for functions that are square-integrable when multiplied by .

That's a bit confusing:

An equivalent formulation of the fact that Hermite polynomials are an orthogonal basis for L2(R, w(x) dx) consists in introducing Hermite functions (see below), and in saying that the Hermite functions are an orthonormal basis for L2(R). - Wikipedia

 

[A. Neumaier]

The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.

All Hilbert spaces used in quantum mechanics (and most in QFT) are separable, i.e., have a countable basis. Most are also infinite-dimensional - then they also have continuous representations.

For example, the state space of a single particle can be expressed in the position or momentum representation, where each state is a superposition of uncountably many improper states, or in an angular momentum representation, where the basis is countable, given by the spherical harmonics (n,s,p,d,f,...)

This is independent of whether particles can escape to infinity. The latter is a property of the particular Hamiltonian used, not of the Hilbert space.

Okay, so what's an example of a square-integrable function that can't be a solution to Schrodinger's equation?

Oh, I guess an infinite square well gives an example: No function that is nonzero outside the well is a possible solution.

The Hamiltonian of the infinite square well does not act on the space of square integrable functions on $R$, only on the space of square integrable functions on the support of the well. Thus the latter is the correct Hilbert space for this potential, and this is not a counterexample. Indeed, there are no counterexamples.

They're complete for the Hilbert Space of possible solutions of the harmonic oscillator Schrodinger Equation. I don't think they're complete for the Hilbert space of all square integrable functions on the real numbers, but that's not the relevant Hilbert space for the harmonic oscillator.

No. The harmonic oscillator basis is complete for that space. More generally, the collection of orthonormal eigenstates of any self-adjoint Hamiltonian on $:^2(R)$ with discrete spectrum is complete in this space. Eigenstates of self-adjoint Hamiltoniand with continuous spectrum need the rigged Hilbert space extension, which allows for superpositions of uncountably many basis states.

a planr wave solution of a free Hamiltonian cannot be considered as an "allowed physical state" with this approach.

No, because it is not square integrable.

 

[hilbert2]

No, because it is not square integrable.

What I meant with my own comment was that in the case of a plane wave solutions the $\langle E\rangle$ is infinite in a confining potential, making them even more unphysical. So there are no scattering solutions that would be useful in any situation when it's the harmonic oscillator or a particle in a box.

 

[martinbn]

Are there any examples of spaces with uncoutble bases used in quantum theory?

 

[A. Neumaier]

Are there any examples of spaces with uncountable bases used in quantum theory?

Yes, in quantum field theory, whenever there are infinitely many superselection sectors. For example, the massless free field in 2 dimensions has a naturally associated nonseparable Hilbert space in which all superselection sectors are represented.

 

[stevendaryl]

Are there any examples of spaces with uncoutble bases used in quantum theory?

A toy example:

Consider an infinite lattice of harmonic oscillators. A natural basis for this system would be the states $|\vec{n} \langle$, where $\vec{n}$ is an infinite array of nonnegative integers, where $n_j$ is the state of harmonic oscillator number $j$.

This is an uncountable basis.

Confronted with such a situation, physicists would make the ansatz that the total energy is finite, in which case, we can come up with a countable basis for the states satisfying that criterion, which is the same as the original basis, but with the restriction that $\vec{n}$ has at most finitely many nonzero elements. Note, this ansatz doesn't say that there is an upper bound on the number of oscillators that are not in the ground state. It just has to be that the amplitude for there being $k$ oscillators that are not in the ground state goes to zero as $k \rightarrow \infty$.

This toy example is actually related to quantum field theory, where there is a similar assumption that we need only to consider the basis in which there are finitely many particles of each type. (Once again, this doesn't impose an upper bound on the number of particles, but makes it vanishingly unlikely to find states with $k$ particles in the limit as $k \rightarrow \infty$)

If you actually allowed infinitely many particles, each with positive energy, then your system would have infinite total energy and would have an uncountable basis. (I think)

 

[MichPod]

I still have my rather basic questions on basic QM.

In the Hilbert space formalism, the Hilbert space of quantum mechanics is always separable and has a countable basis.
[...]
Yes, a plain wave solution cannot be considered an allowed physical state in both the Hilbert space and the rigged Hilbert space formalisms.

Then how can we meaningfully speak in QM about any Hamiltonians with a continuous spectrum? Like how can we speak about a free Hamiltonian $-\frac{1}{2m}\frac{\partial ^2}{\partial x^2}$ and am I not correct that it provides a set of eigenstates with uncountable basis? Has it any stationary physical states at all in QM if a plain wave is not considered an allowed physical state?

 

[A. Neumaier]

I still have my rather basic questions on basic QM.Then how can we meaningfully speak in QM about any Hamiltonians with a continuous spectrum? Like how can we speak about a free Hamiltonian $-\frac{1}{2m}\frac{\partial ^2}{\partial x^2}$ and am I not correct that it provides a set of eigenstates with uncountable basis? Has it any stationary physical states at all in QM if a plain wave is not considered an allowed physical state?

These eigenstates are not in the Hilbert space, but in the larger upper (dual) space of the rigged version of it. Thus the eigenstates do not form a basis of the Hilbert space in the sense of functional analysis.

 

[pellis]

If you're only considering spin states, then the basis should be finite, I think.

 

[vanhees71]

The Hilbert space is countable if the system is "confining" in the sense that it's impossible to give the particle an energy kick that makes it fly to infinity. A particle-in-box system and the harmonic oscillator are confining in that way, but a hydrogen atom or a finite square well aren't. Neither is the Morse potential.

The Hilbert space of non-relativistic quantum mechanics with a fixed number of particles is $\mathrm{L}^2(\mathbb{R}^{3N})$, which is separable, i.e., it has a countable set of orthonormalized vectors. You can take, e.g., the harmonic-oscillator energy eigenbasis (a $3N$-dimensional harmonic oscillator).

In the physicists' sloppy slang they call also a generalzed basis like the "position eigenbasis" a "basis", but it's rather a generalized basis. To get as proper mathematical formulation, which is closest to the convenient (and usually amazingly well working) sloppy notions of the physicists, have a look at the formalism of "Rigged Hilbert space", as already mentioned several times in this thread.

 

[pellis]

The Hilbert space of non-relativistic quantum mechanics with a fixed number of particles is L2(R3N), which is separable, i.e., it has a countable set of orthonormalized vectors.

I get L2(R^3N), but as a math-poor chemist I‘m still not completely clear on exactly how that works for, say, the continuum of states of an ionised hydrogen atom.

According to Rigged Hilbert Spaces (Wikipedia):

“They bring together the 'bound state' (eigenvector) and 'continuous spectrum', in one place”.

OK.

Two questions, if I may:

1) How are the countable set of orthonormalised (basis – OK?) vectors used to characterise an unbound state in the continuum? Is it just a case of some linear combination of the orthonormal vectors approximating a continuum state?

2) Countable as in Aleph-Null?

 

[A. Neumaier]

1) How are the countable set of orthonormalised (basis – OK?) vectors used to characterise an unbound state in the continuum? Is it just a case of some linear combination of the orthonormal vectors approximating a continuum state?

A continuum eigenstate is a kind of limit of linear combinations of the countable basis, but the limit moves outside the Hilbert space into the top of the rigged Hilbert space. This is like a limit of continuous functions to approximate a delta function that is outside the space of continuous functions.

2) Countable as in Aleph-Null?

Yes. A basis that can be labelled 1,2,3,...

 

[pellis]

Many thanks, I can grok that.

 

[bhobba]

If you want the gory technical detail see:
http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf

It however does not include the proof of the important Nuclear Spectral Theorem. I did find a proof not of the full theorem, but of what is mostly used in QM. See Attachment.

Thanks
Bill

 

[Nullstein]

There is a lot of confusion in this thread. The Hilbert space of a particle (with spin $s$) in non-relativistic quantum mechanics is always $L^2(\mathbb{R}^2)\otimes\mathbb{C}^{2s+1}$, independent of the Hamiltonian and it has a countable basis. Even though the Hamiltonian usually is not defined on all states in the Hilbert space, its corresponding time-evolution operator $U(t)=e^{-i t H}$ is a bounded, unitary operator and thus defined on the whole Hilbert space, so every state is the solution to the Schrödinger equation at some time $t$, because it can always act as initial value. (The Hamiltonian must of course be self-adjoint for this to be true, but this is a general requirement anyway.) And by the way, the rigged Hilbert space formalism is nice, because it makes the Dirac formalism rigorous, but it's completely unnecessary. All computations can be done and are usually even easier without it. (If we want to stay rigorous. Otherwise, one could just apply the sloppy Dirac formalism directly. It's not like one can just say "rigged Hilbert space" and the calculation magically becomes rigorous. One really needs to care about all the subtleties about nuclear spaces if one wants to apply the rigged Hilbert space formalism.)

 

[bhobba]

And by the way, the rigged Hilbert space formalism is nice, because it makes the Dirac formalism rigorous, but it's completely unnecessary.

I would not say completely unnecessary, but I get the gist. I learned QM from Dirac and Von-Neumann. Von-Neumann was a breeze - it was just a review/extension of Hilbert Spaces I learned at Uni. Good assignment while studying; if your teacher is into that learning style (mine was - but I did mine on applications to numerical analysis). Dirac was another matter. That damnable Dirac Delta function confused the bejesus out of me. I did a long sojourn into RHS's. I came out the other end with my issues resolved. But then I did what I should have done - read Ballentine - QM A Modern Introduction. Chapter 2 explains all you really need to know about it. Lest anyone think it is completely useless - resonances are best done using RHS's. But that is advanced.

Thanks
Bill

 

[Nullstein]

Well, "unnecessary" might have been a bit harsh. I really wanted to clear up some confusion. Rigged Hilbert spaces don't replace Hilbert spaces. We are always dealing with Hilbert spaces in QM. Rigged Hilbert spaces are a mathematical tool do do certain calculations on Hilbert spaces in a different way that is syntactically closer to Dirac's formalism than ordinary functional analysis. So the answer to the question, whether the Hilbert space has a countable basis, doesn't change by introducing rigged Hilbert spaces into the discussion and it might even cause more confusion. I didn't want to say that rigged Hilbert spaces can't be useful in general.

 

[A. Neumaier]

We are always dealing with Hilbert spaces in QM.

This may be true in your own community ('we') but it is not true if one replaces your subjective 'we' by 'the community of quantum physicists'.

They routinely use bras and kets that do not belong to a Hilbert space but to the upper floor of a rigged Hilbert space. Only the state vectors (wave functions) composed of these must lie in the Hilbert space since they must be normalized to have a probabilistic meaning.

the answer to the question, whether the Hilbert space has a countable basis, doesn't change by introducing rigged Hilbert spaces

It seems to be your confusion that the opposite was asserted. Asserted was only that the uncountable basis used in a continuous representation was not a Hilbert space basis but a basis of the RHS.

 

[Nullstein]

This may be true in your own community ('we') but it is not true if one replaces your subjective 'we' by 'the community of quantum physicists'.

They routinely use bras and kets that do not belong to a Hilbert space but to the upper floor of a rigged Hilbert space. Only the state vectors (wave functions) composed of these must lie in the Hilbert space since they must be normalized to have a probabilistic meaning.

That's not in disagreement with what I said. Rigged Hilbert spaces are a tool. We can use distributional states as intermediate objects, but in the end, we always have to end up in the Hilbert space again. And I also agree that it may be useful to do it this way, but it's not strictly necessary. There are other ways and if we care about rigor, one has to acknowledge that rigged Hilbert spaces require much higher level mathematics.

It seems to be your confusion that the opposite was asserted. Asserted was only that the uncountable basis used in a continuous representation was not a Hilbert space basis but a basis of the RHS.

I'm not sure of that. When we talk about a basis in QM, we really mean the functional analytict definition of a basis (Schauder basis), which requires the underlying space to be a Banach space. The space of tempered distributions is not a Banach space, so the relevant notion of basis can only be the algebraic one (Hamel basis), where only finite linear combinations are allowed, and this one of course must be uncountable. However, the set of generalized eigenvectors of e.g. the position operator (delta functions) hardly suffices to span the space of tempered distributions with only finite linear combinations. The generalized eigenfunctions are complete in the sense that any vector in the Hilbert space can be represented using an integral that involves the generalized eigenfunctions, but that's a different notion than the generalized eigenfunctions forming a basis of a vector space (neither a Schauder basis nor a Hamel basis).

 

[A. Neumaier]

Rigged Hilbert spaces are a tool.

The Hilbert space is no less a tool than the rigged Hilbert space.

We can use distributional states as intermediate objects, but in the end, we always have to end up in the Hilbert space again.

No. Scattering states are also in the continuum, and not in the Hilbert space. In the end we end up not in the Hilbert space but with cross sections and probabilities.

 

[Nullstein]

The Hilbert space is no less a tool than the rigged Hilbert space.

Right, but Hilbert spaces can't be avoided, since they are required axiomatically by the axioms of QM. Rigged Hilbert spaces may or may not be used in QM. It's largely a matter of personal preferences.

No. Scattering states are also in the continuum, and not in the Hilbert space. In the end we end up not in the Hilbert space but with cross sections and probabilities.

Well, in order to get probabilities rather than probability densities, one needs to integrate in the end. Again, I don't argue that one shouldn't use distributions in intermediate calculations. (That would be stupid.)

 

[A. Neumaier]

but Hilbert spaces can't be avoided, since they are required axiomatically by the axioms of QM.

No, only conventionally. Most quantum physics only pays lip service to the Hilbert space and its requirements. What is treated as basic is a historical accident.

For example, outside mathematical physics, nobody cares about domains of definition of operators (no unbounded operator acts on the physical Hilbert space) , and nobody cares about checking whether a Hamiltonian used is self-adjoint. Indeed, this is often a difficult analytic question beyond the capabilities of typical quantum theorists. But the notion of self-adjoint operator is needed in the axioms for Born's rule.

One can rewrite everything in terms of operators acing on the nuclear space (as observables) and kets in the dual nuclear space (as states). One can even relax the notion of the rigged Hilbert space without impact on the level of rigor typical for quantum theory by dropping the somwhat technical nuclear property.

Working with operators (as observables) and their exponentials defined on an inner product space and its dual (for states) is just what one needs in quantum mechanics, and has none of the technical difficulties that Hilbert space has.

 

[Nullstein]

No, only conventionally. Most quantum physics only pays lip service to the Hilbert space and its requirements. What is treated as basic is a historical accident.

For example, outside mathematical physics, nobody cares about domains of definition of operators (no unbounded operator acts on the physical Hilbert space) , and nobody cares about checking whether a Hamiltonian used is self-adjoint. Indeed, this is often a difficult analytic question beyond the capabilities of typical quantum theorists. But the notion of self-adjoint operator is needed in the axioms for Born's rule.

One can rewrite everything in terms of operators acing on the nuclear space (as observables) and kets in the dual nuclear space (as states). One can even relax the notion of the rigged Hilbert space without impact on the level of rigor typical for quantum theory by dropping the somwhat technical nuclear property.

Working with operators (as observables) and their exponentials defined on an inner product space and its dual (for states) is just what one needs in quantum mechanics, and has none of the technical difficulties that Hilbert space has.

It may be possible to relax the axioms of QM, I don't know about that. I certainly haven't a clear exposition of that, though, that derives all the standard results of QM from such a relaxed set of axioms. What about simple results like conservation of probability? Don't you need self-adjointness for that? I suspect that a potential analogous criterion in the language of nuclear spaces will certainly be much more technical. Generally, I would say that Hilbert spaces offer much more structure and thus have fewer technical difficulties than nuclear spaces. E.g., as far as I'm concerned, not every self-adjoint operator even has a set of generalized eigenvectors. There is a subtle technical requirement. On the other hand, the spectral theorem in the Hilbert space setting works like a charm.

 

[A. Neumaier]

E.g., as far as I'm concerned, not every self-adjoint operator even has a set of generalized eigenvectors.

You are not concerned enough. Maurin's spectral theorem guarantees precisely what is needed for Dirac's formalism.

It may be possible to relax the axioms of QM, I don't know about that.

It is not a relaxation but fully equivalent, by completion. Thus it is just a matter of preference. Having to assume nontrivial concepts and results from functional analysis (such as selfadjointness and the spectral theorem) to even formulate an axiom (Born's rule) is very strange...

What about simple results like conservation of probability? Don't you need self-adjointness for that?

In a *-algebra of linear operators on an inner product space, $H^+=H$ and $i\hbar\dot\psi=H\psi$ imply with a 1-line proof that #~psi^*\psi## has zero derivative.

 

[Nullstein]

You are not concerned enough. Maurin's spectral theorem guarantees precisely what is needed for Dirac's formalism.

In Gelfand, Vilenkin, Generalized Functions IV, they additionally require that the operator $A$ can be restricted to a dense, nuclear space $\Phi$ such that $A:\Phi\rightarrow\Phi$ maps from $\Phi$ to $\Phi$ and $\Phi \subseteq H \subseteq \Phi^*$ forms a rigged Hilbert space. It's certainly not trivial that such a set $\Phi$ exists for an arbitrary self-adjoint operator $A$. Can you point me to a proof of your theorem that doesn't make such assumptions?

It is not a relaxation but fully equivalent, by completion. Thus it is just a matter of preference. Having to assume nontrivial concepts and results from functional analysis (such as selfadjointness and the spectral theorem) to even formulate an axiom (Born's rule) is very strange...

I'd like to see the proof for your claimed equivalence. You may find it strange, but the math of rigged Hilbert spaces is way more complicated than the spectral theorem, so you should find that strange as well.

In a *-algebra of linear operators on an inner product space, $H^+=H$ and $i\hbar\dot\psi=H\psi$ imply with a 1-line proof that #~psi^*\psi## has zero derivative.

Again, the condition $H:\Phi\rightarrow\Phi$ from above slipped in that may not be satisfiable by a general self-adjoint operator.

 

[A. Neumaier]

Can you point me to a proof of your theorem that doesn't make such assumptions?

For Maurin's spectral theorem see the books by Maurin. One gets from the commutative B^*-algebra generated by a self-adjoint operator a unitary representation on the spectrum, which is precisely what Dirac's notation is about.

the math of rigged Hilbert spaces is way more complicated than the spectral theorem,

Nuclearity is not needed.

For the purposes of quantum physics, all one needs is an inner product space, the common domain of all operators considered, and its dual. This is assumed axiomatically. Then one can already do everything quantum physicists commonly do in their semiformal way, without any functional analytic baggage. The operators actually needed have explicit spectral resolutions, in particular those whose measurement is discussed (position, momentum, angular momentum and spin) since they are infinitesimal generators of symmetry groups with a unitary representations.

The only operator where one needs some analysis is the Hamiltonian, where one has to assume that it is Hermitian and its range of values is bounded below. Then one can state without proof that this implies a spectral resolution. The proof of this can be left to the mathematical physicist - it follows by constructing the Hilbert space and applying standard results on semibounded Hermitian forms.

 

[vanhees71]

The Hilbert space is no less a tool than the rigged Hilbert space.

No. Scattering states are also in the continuum, and not in the Hilbert space. In the end we end up not in the Hilbert space but with cross sections and probabilities.

Exactly! Note that to define cross sections from the S-matrix elements you need to use "proper" states, i.e., members of the Hilbert space and not from the dual of the nucleus space of the rigged Hilbert space. For a detailed treatment of this physical approach (in contradistinction to the more pragmatic approach of introducing a finite space-time volume by Fermi) in the context of relativistic QFT (but equally well applicable in non-relativistic QM scattering theory) see Peskin&Schroeder, Introduction to QFT.