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

• A
Gold Member
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?

Delta2 and Demystifier

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

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

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

sysprog, Delta2, vanhees71 and 3 others
Gold Member
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...

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

vanhees71 and MichPod
hilbert2
Gold Member
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.

Gold Member
Not rigid space, it is called rigged Hilbert space (in the sense of "upgraded" Hilbert space).
oops. my poor English. sorry. fixing this.

Gold Member
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
Staff Emeritus
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.

PeterDonis
Mentor
2020 Award
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.

bhobba and Delta2
stevendaryl
Staff Emeritus
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
Mentor
2020 Award
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
Staff Emeritus
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.

bhobba and Demystifier
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."

dextercioby
PeterDonis
Mentor
2020 Award
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.

Gold Member
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
Staff Emeritus
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
Mentor
2020 Award
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
Gold Member
they're complete for the Hilbert space of all square integrable functions on the real numbers
Proof of completeness.

atyy
stevendaryl
Staff Emeritus
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
Gold Member
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.

atyy
stevendaryl
Staff Emeritus

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.

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vanhees71, mattt, MichPod and 2 others
Keith_McClary
Gold Member
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

atyy
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.

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vanhees71, strangerep and gentzen