Completeness of Eigenfunctions of Hermitian Operators

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• gaiussheh
gaiussheh
TL;DR Summary
What is the theorem that clearly states the completeness of eigenfunctions?
I know that completeness (roughly) implies that (almost) all functions can be decomposed into a sum of eigenfunctions of a Hermitian operator. ##\psi=\sum_n \alpha_n \psi_n##. Clearly, there have to be some restrictions on the function itself, and the operator as well. But what is that?

My question arises from the fact that the eigenfunctions of a particle in a hard box model (i.e., infinite potential well) can not be used to express wavefunction that is non-zero outside of the potential well. Surely, it is because the Hamiltonian is only defined within the well, and so are the eigenfunctions. Still, if we extend the definitions to ##(-\infty,+\infty)##, like what textbooks do (i.e. define ##V[x]=+\infty ## for ##x## outside the box), it seems to be a contradiction of the completeness theorem, as the function is defined on the whole axis, and the boundary conditions are met as ##\psi -\rightarrow 0## as ##x\rightarrow \infty##, so there must be some condition that is not met. (Is it because you can't really define an operator that "equals to" ##\infty##?) Nevertheless, can you expand an arbitrary wave function using the basis of hydrogen atom orbitals? There seems to be nothing wrong with the Hamiltonian itself apart from ##x=0## is not continuous.

I think your question fits under Sturm-Liouville theory. From a mathematical point of view, problems on a compact domain are much simpler than cases where some aspect of the domain extends to infinity.

Alternatively, stated in algebraic terms, you'd want to look at spectral theorems .

Demystifier
Haborix said:
I think your question fits under Sturm-Liouville theory. From a mathematical point of view, problems on a compact domain are much simpler than cases where some aspect of the domain extends to infinity.

Alternatively, stated in algebraic terms, you'd want to look at spectral theorems .
Oh, I think I have no issue with Sturm-Liouville Theory. I believe the real matter is that in any real quantum mechanical problem, the wave function is defined in ##\mathbb{R}^3##, not a compact subset of it. I wish to find the theorem that clearly states the conditions (i.e., what condition the self-adjoint operator must meet to provide a set of eigenvalues and eigenfunctions, and what kind of functions can be expressed under this eigenfunction basis?) Sturm-Liouville is good as long as you are dealing with potential well, of course.

gaiussheh said:
Still, if we extend the definitions to ##(-\infty,+\infty)##, like what textbooks do (i.e. define ##V[x]=+\infty ## for ##x## outside the box), it seems to be a contradiction of the completeness theorem, as the function is defined on the whole axis, and the boundary conditions are met as ##\psi -\rightarrow 0## as ##x\rightarrow \infty##, so there must be some condition that is not met.
If you accept that "(almost) all functions can be decomposed" must be read as "all square integrable functions can be decomposed", then my impression is that the remaining part of your problem is related to the recently discussed issue that square integrable functions go to zero at infinity if suitably interpreted, but not if interpreted via point evaluation:
gentzen said:
In which sense(s) do square integrable functions go to zero at infinity?
Of course, they cannot go to zero at infinity in the sense of point evaluation, because point evaluation is not the appropriate concept for square integrable functions. There was a recent discussion in the Quantum Physics Forum, which focused on 1D functions:
gentzen said:
The problem in Demystifier's example can be fixed by modifying the function on a set of measure zero. This no longer works for your example. For your example, for any ##\epsilon >0##, one can modify the function on a set of measure ##<\epsilon##.

gaiussheh said:
(Is it because you can't really define an operator that "equals to" ##\infty##?) Nevertheless, can you expand an arbitrary wave function using the basis of hydrogen atom orbitals? There seems to be nothing wrong with the Hamiltonian itself apart from ##x=0## is not continuous.
I think the Hamiltonian itself is just fine, and that "you can't really define an operator that "equals to" ##\infty##" is also unrelated.

May be you want to look at the spectral theorem.

gaiussheh said:
TL;DR Summary: What is the theorem that clearly states the completeness of eigenfunctions?

My question arises from the fact that the eigenfunctions of a particle in a hard box model (i.e., infinite potential well) can not be used to express wavefunction that is non-zero outside of the potential well.
For an infinite potential well, unbounding states which are non zero ourside the well have infinite and continuous energy which would be controvertial. Finite potential system might be more convenient for your purpose.

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anuttarasammyak said:
For an infinite potential well, unbounding states which are non zero ourside the well have infinite and continuous energy which would be controvertial. Finite potential system might be more convenient for your purpose.
Is it possible to construct the full hilbert space using the solutions of finite potential well? i.e. do they form a complete basis?

gaiussheh said:
Oh, I think I have no issue with Sturm-Liouville Theory. I believe the real matter is that in any real quantum mechanical problem, the wave function is defined in ##\mathbb{R}^3##, not a compact subset of it. I wish to find the theorem that clearly states the conditions (i.e., what condition the self-adjoint operator must meet to provide a set of eigenvalues and eigenfunctions, and what kind of functions can be expressed under this eigenfunction basis?) Sturm-Liouville is good as long as you are dealing with potential well, of course.

Look at the Spectral Theorem for Unbounded Self-Adjoint Operators on Hilbert Space.

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gaiussheh said:
Is it possible to construct the full hilbert space using the solutions of finite potential well? i.e. do they form a complete basis?
Yes, it is, not from mathematical proof but from physical insights for me. If it were not, there should be states energy of which we cannot measure in the system. Degeneracy is another issue here.

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The case where both the absolutely continuous and the singular continuous spectrum are the empty set (and so the spectrum is just the pure-point spectrum), is very similar to the finite dimensional case.

The Theorem is a generalization of this result to any (densely defined) Self-Adjoint Operator (bounded or unbounded).

dextercioby
mattt said:
Look at the Spectral Theorem for Unbounded Self-Adjoint Operators on Hilbert Space.
Thanks - before going through the proof I just want to ask what's the problem with the Hamiltonian in the infinite potential well - what is the condition that is unmet in the theorem that makes it incomplete in ##L^2##? Is this fixed if I change it to, say, Hamiltonian of a finite potential well or a hydrogen atom?

gaiussheh said:
Thanks - before going through the proof I just want to ask what's the problem with the Hamiltonian in the infinite potential well - what is the condition that is unmet in the theorem that makes it incomplete in L2? Is this fixed if I change it to, say, Hamiltonian of a finite potential well or a hydrogen atom?
As for square well potential, rectangular shape of edge cause discontinuity in derivatives of wave function. <p^4> and higher order ones diverge in infinite well potential energy eigenstates. <p^6> and higher order ones diverge for finite well potential energy eigenstates. Thus we know that energy eigenstates are incomplete because momentum eigenstate is not expanded by them. I think it is due to not physical or natural, toy property of square well potential which are multi-valued at the edges.

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anuttarasammyak said:
<p^4> and higher order ones diverge in infinite well potential energy eigenstates. <p^6> and higher order ones diverge for finite well potential energy eigenstates. Thus we know that energy eigenstates are incomplete because momentum eigenstate is not expanded by them.
This argument does not convince me. Even if <p^6> and higher order ones diverge, why should that imply incompleteness of the eigenstates for L^2? In fact, I see no reason why the eigenstates for finite well potential (discrete and continuous together) should not be complete.

For infinite square well ##\frac{\partial^2 \phi}{\partial x^2}## diverges or has delta functions peaked at x=-w/2,w/2
For finite square well ##\frac{\partial^3 \phi}{\partial x^3}## diverges or has delta functions peaked at x=-w/2,w/2
where w is the width of well which is centered at x=0 and ##\phi## is bounded or unbounded energy eigenstates. p is x differential operatior in mind, these behavior at well edges cause divergence of <p^6> and higher order.
We may say that in square well cases any wave function of ##C^\infty## cannot be expresed as superposition of ##\phi##.

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anuttarasammyak said:
For infinite square well ##\frac{\partial^2 \phi}{\partial x^2}## diverges or has delta functions peaked at x=-w/2,w/2
For finite square well ##\frac{\partial^3 \phi}{\partial x^3}## diverges or has delta functions peaked at x=-w/2,w/2
where w is the width of well which is centered at x=0 and ##\phi## is bounded or unbounded energy eigenstates. p is x differential operatior in mind, these behavior at well edges cause divergence of <p^6> and higher order.
We may say that in square well cases any wave function of ##C^\infty## cannot be expresed as superposition of ##\phi##.
I remember that QM does not even require the wave function to be ##\mathbb{C}^1##. Physics usually requires only ##\mathbb{C}^0##, and ##\frac{{\rm d} \psi}{{\rm d}x}## can be discontinuous at where ##V[x]=\infty##. Even if I say that one can not really define a function to be infinity, The strictest requirement is ##\mathbb{C}^1##

Your concern is completeness. 　There exists ##C^\infty## wave function, e.g. Gaussian wave function, which is a member of state vectors. I think it would be much difficult if not impossible to prove that ##C^\infty## Gaussian can be expressed as superpositon of ##C^0,C^1,C^2## wave functions.

anuttarasammyak said:
We may say that in square well cases any wave function of ##C^\infty## cannot be expresed as superposition of ##\phi##.
OK, now I see where you are coming from. But your conclusion is not justified. The "weights" of the delta peaks in the higher order derivatives can converge to zero in the limit of an infinite sum of eigenstates, even if the individual eigenstates have non-zero weights for those peaks. (Of course, you could already cancel the weight of one specific delta peak in the linear combination of two eigenstates. I just guess that this exact cancelation in a finite sum is not what will actually happen most of the time.)

anuttarasammyak said:
Your concern is completeness. 　There exists ##C^\infty## wave function, e.g. Gaussian wave function, which is a member of state vectors. I think it would be much difficult if not impossible to prove that ##C^\infty## Gaussian can be expressed as superpositon of ##C^0,C^1,C^2## wave functions.
I see. I think you can't build a ## C^\infty ## function by only using up to ## C^1 ## functions. (I don't quite know if this is true). In that sense, of course, it is incomplete. Does that imply ##V[x]## should be ## C^\infty ## as well?

gaiussheh said:
Does that imply V[x] should be C∞ as well?
Yes, I guess so.
gaiussheh said:
Is this fixed if I change it to, say, Hamiltonian of a finite potential well or a hydrogen atom?
I think a hydrogen atom hamiltonian satisfies this condition except the origin.

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anuttarasammyak said:
Yes, I guess so.

I think a hydrogen atom hamiltonian satisfies this condition except the origin.
I just found a counterexample to your theory. define step functions ##\mathbb{1}_{(-\infty,0]}[x]=1~({\rm iff.} x\leq0)## and ##\mathbb{1}_{(0,\infty)}[x]=1~({\rm iff.} x>0)##. Clearly ##\mathbb{1}_{(-\infty,0]}[x]+\mathbb{1}_{(0,\infty)}[x]=\mathbb{1}[x]=1## is ##C^{\infty}##.

gentzen
Making use of your counter example, let me express my concern using Heaviside function.
$$H_a(-x)+H_a(x)=2a$$ for x=0 and 1 otherwise. Corresponding to ambiguity of square well potential at edges we are not sure a = 1/2 or not.
Now I admit that it is too much mathematical concern on toy model and any wave function would be expressed as sum of energy eigenfunction except around two edge points of infinitely small areas.

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anuttarasammyak said:
For infinite square well ∂2ϕ∂x2 diverges or has delta functions peaked at x=-w/2,w/2
For finite square well ∂3ϕ∂x3 diverges or has delta functions peaked at x=-w/2,w/2
In other words in momentum representation
$$\phi(p) \sim p^{-2}$$ for infinite square well and
$$\phi(p) \sim p^{-3}$$ for finite square well for large p.
Any wavefunction of momentum representation with any kind of damping for large p should be sum of these energy eigenfunctions. I know it should be so but still feel it amazing.

anuttarasammyak said:
Making use of your counter example, let me express my concern using Heaviside function.
$$H_a(-x)+H_a(x)=2a$$ for x=0 and 1 otherwise. Corresponding to ambiguity of square well potential at edges we are not sure a = 1/2 or not.
Now I admit that it is too much mathematical concern on toy model and any wave function would be expressed as sum of energy eigenfunction except around two edge points of infinitely small areas.
That's strange. Even if it converges almost everywhere except only a few points, the derivatives will diverge, and your Hamiltonian will broken. I'll open a new thread to discuss this.

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