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

[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.

 

[A. Neumaier]

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 nuclear space of the rigged Hilbert space.

But one can use elements of the nuclear space, which is a space of integrable and highly differentiable states. Hence there is no need to have the Hilbert space.

 

[martinbn]

What are examples of rigged Hilbert spaces used in QM? The example with the Sobolev spaces $H^s$, $L^2$, $H^{-s}$, has only separable spaces involved.

 

[A. Neumaier]

What are examples of rigged Hilbert spaces used in QM? The example with the Sobolev spaces $H^s$, $L^2$, $H^{-s}$, has only separable spaces involved.

Typical examples are spaces of Schwartz functions (bottom level) and Schwartz distributions (top level). See https://en.wikipedia.org/wiki/Spaces_of_test_functions_and_distributions

Sobolev spaces are not good enough since not all powers of $p$ are defined on them.

 

[vanhees71]

But one can use elements of the nuclear space, which is a space of integrable and highly differentiable states. Hence there is no need to have the Hilbert space.

Sure, the nuclear space is a proper, dense subspace of the Hilbert space. I guess, if you look at the mathematical details you even have to use elements of the nuclear space, i.e., where the momentum operators of the particles are well-defined, but that you have to decide as a mathematician :-).

 

[A. Neumaier]

you even have to use elements of the nuclear space, i.e., where the momentum operators of the particles are well-defined,

The momentum operator is already defined on a (much bigger) Sobolev space. The nuclear space is needed to have all powers of the momentum operators well-defined.

 

[vanhees71]

Fine. All you need is that there's a dense subset of the Hilbert space, where the momentum eigenstates are defined.

 

[A. Neumaier]

Fine. All you need is that there's a dense subset of the Hilbert space, where the momentum eigenstates are defined.

Yes.

 

[martinbn]

Typical examples are spaces of Schwartz functions (bottom level) and Schwartz distributions (top level). See https://en.wikipedia.org/wiki/Spaces_of_test_functions_and_distributions

Sobolev spaces are not good enough since not all powers of $p$ are defined on them.

The space of tempered distributions is weak* separable.

 

[A. Neumaier]

The space of tempered distributions is weak* separable.

so what?

 

[martinbn]

so what?

I was asking for an example of non sebarable spaces in a Gelfand triple used in QM? Just curious.