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

[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.

 

[A. Neumaier]

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

Ah. You can take a direct integral of uncountably many Schwartz spaces to get something nonseparable. I guess that a rigorous treatment of the infrared problem in quantum field theory might need such a space.

 

[Nullstein]

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.

I checked his book "Methods of Hilbert spaces" and he also requires the nuclearity property.

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.

You don't need nuclearity to define what a generalized eigenvector is, but you need it to prove that the set of generalized eigenvectors is complete. And this is essential for most applications of the theory, such as inserting $\int\left|x\right>\left<x\right|\mathrm d x$.

 

[A. Neumaier]

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?

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.

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.

I checked his book "Methods of Hilbert spaces" and he also requires the nuclearity property.

You don't need nuclearity to define what a generalized eigenvector is, but you need it to prove that the set of generalized eigenvectors is complete. And this is essential for most applications of the theory, such as inserting $\int\left|x\right>\left<x\right|\mathrm d x$.

Sorry, I didn't read the exact statement you had claimed. Yes, nuclearity is needed for this.

My claim should have been that self-adjoinedness (hence the full strength of the spectral theorem) is not needed for quantum theory. All operators actually used in quantum mechanics always have a common domain of definition, hence the extra condition in Gelfand and Vilenkin is trivially satisfied.

One can build quantum mechanics in a very elementary way on an inner product space (not necessarily nuclear) carrying a unitary representation of a symmetry group. The unitary representation is usually given explicitly, and its infinitesimal generators are the observables of interest. These have a spectral resolution which is easy to construct explicitly in terms of representation theory. Thus no functional analysis is needed. (If desired, you can get it via B^* algebras, without assuming nuclearity.)

The only nontrivial issue is whether a given Hamiltonian generates a representation of U(1), which is usually unchecked in quantum mechanics books, hence can be assumed without proof in a first course. (To prove it, if desired, semiboundedness is sufficient.)

In fact, one can set up quantum mechanics even without the notion of eigenvalues and eigenvectors. For the basics of quantum mechanics, the spectral resolution is only needed in two places:

1. To justify the resolution of unity, needed only for position and measurement, where it can be proved directly, avoiding spectra.
2. To state Born's rule in its idealized form, where it can be replaced by a more widely applicable rule involving POVMs.

For the latter, see, e.g., my paper

• A. Neumaier, Born's rule and measurement, 2019.

Of course one needs eigenvalues and eigenvectors later, for angular momentum (where existence can be settled directly) and for the Hamiltonian to describe observable spectra. But as I had already mentioned, almost no textbook on quantum mechanics treats the latte in a rigorous way, hence an informal presentation of this is fully adequate.

 

[Nullstein]

My claim should have been that self-adjoinedness (...) is not needed for quantum theory.

But if an operator is symmetric, but not self-adjoint, its spectrum necessarily contains at least a half-plane of complex numbers, so measurements could yield complex results and the corresponding translation operator wouldn't be unitary and hence not probability-preserving.

All operators actually used in quantum mechanics always have a common domain of definition, hence the extra condition in Gelfand and Vilenkin is trivially satisfied.

That's not enough for nuclearity though. A nuclear space must satisfy a couple of rather convoluted axioms. Just being a common domain of some operators is not enough. For example, every Hilbert space is the common domain of the set of bounded operators on that Hilbert space. Yet, Hilbert spaces are not nuclear.

Of course one needs eigenvalues and eigenvectors later, for angular momentum (where existence can be settled directly) and for the Hamiltonian to describe observable spectra. But as I had already mentioned, almost no textbook on quantum mechanics treats the latte in a rigorous way, hence an informal presentation of this is fully adequate.

Well, most non-relativistics QM relies heavily on the position and momentum operators, which are essentially self-adjoint on the Schwartz space, which is the basic example of a nuclear space. That's why everything works and nobody needs to care about these issues, because they have been worked out already. But already in the case of a free scalar quantum field, working with RHS becomes much more non-trivial.

It's alright if quantum physicists use non-rigorous math, because most of the time, the details had already been worked out by mathematicians before. But this thread was specifically about a mathematical detail, so I think it's important to be as explicit as possible.

 

[vanhees71]

One of my math professors once said that the (separable) is almost as good-natured as finite-dimensional vector spaces, and that's why physicists get away with their sloppy math almost all the time ;-).

 

[A. Neumaier]

But if an operator is symmetric, but not self-adjoint, its spectrum necessarily contains at least a half-plane of complex numbers, so measurements could yield complex results and the corresponding translation operator wouldn't be unitary and hence not probability-preserving.

That's why I required that observables of interest are infinitesimal generators of (dynamical) symmetries. This is satisfied in practice and produces self-adjointness without having to talk about it.

That's not enough for nuclearity though. A nuclear space must satisfy a couple of rather convoluted axioms.

Yes. My claim was only that one does not need nuclearity in most of quantum theory.

One just needs a common domain on which the dynamical symmetry group generated by the observables of interest has a unitary representation. These are elementary concepts and suffice for all conceptual issues in quantum mechanics. The rigged Hilbert spaces I am talking about consist of such a space and its dual, without any assumption of nuclearity. The Hilbert space in the middle is not even needed, though it should be introduced to maintain continuity with tradition.

That later one may need more advanced mathematical tools to prove existence results such as the unique solvability of the Schrödinger equation or the existence of ground states is natural. This is already the case in classical mechanics, where the solvability of the Navier-Stokes equations is an unsolved problem, and such questions are left to mathematicians.

But already in the case of a free scalar quantum field, working with RHS becomes much more non-trivial.

The RHS setting Without requiring nuclearity is easy to set up and use for free quantum fields. All creation and annihilation operators act on a common domain.

It's alright if quantum physicists use non-rigorous math, because most of the time, the details had already been worked out by mathematicians before. But this thread was specifically about a mathematical detail, so I think it's important to be as explicit as possible.

My point is that all conceptual definitions needed for the development of quantum physics can be made rigorous without using advanced concepts (not even needing Hilbert spaces), and the solution of certain problems can be precisely stated without proof - while the proofs themselves and the tools needed are left to the mathematical physicists.

Just as in classical physics, physicists are not required to know Sobolev spaces or weak^* convergence. Most phycisists cannot work with these, although they are important in the analysis of the differential equations the physicists deal with.

 

[Demystifier]

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.

I also used to think so, until very recently I encountered a physical problem in which such subtleties turned out to be very important. So I joined with a mathematical physicist to do it correctly. https://arxiv.org/abs/2107.08777

 

[vanhees71]

Well, yes, indeed theoretical physicists tend not to think very much about such subtleties, but from time to time you start nevertheless thinking.

Sometimes you even need to argue with the domain and co-domain of operators and whether an operator is self-adjoint or only Hermitian. It's surprising how simple these questions can be. E.g., when I prepared my lectures about orbital angular momentum for quantum mechanics, I started in the usual way using Dirac's representation free formalism. From the Lie algebra (angular-momentum algebra) you find of course that angular momentum $\vec{L}^2$ takes the values $\ell(\ell+1)$ with $\ell \in \{0,1/2,1,\ldots \}$. On the other hand "we all know" that orbital angular momenta only have integer values for $\ell$.

Now, why the heck is this so? Of course, first I started to look in some textbooks. There you find the argument within Schrödinger wave mechanics that when separating the Laplace operator in spherical coordinates you get the eigenfunctions of $\hat{L}_z=-\mathrm{i} \partial_{\varphi}$ to be $\exp(\mathrm{i} m \varphi)$, and then "$m$ must be in $\mathbb{Z}$, because the wave function must be a unique function of $\vec{x}$. Thus as function of $\varphi$ ist must be $2 \pi$ periodic and thus $m \in \mathbb{Z}$." Well, hm, sounds convincing, but than before I carefully told my students that pure states are represented by normalized wave functions and that an overall phase doesn't change the state. The if I make $m$ half-integer, all that happens when checking what happens when changing $\varphi$ to $\varphi+2 \pi$ is that I get a factor $(-1)$, which is just an unimportant phase factor! So this argument is hand-waving and a bit dishonest towards the students, and of course you always hope that they'll catch you on such a hand-waving "demonstration".

Now fortunately there's a nice way to write $\hat{L}_z$ in terms of annihilation and creation operators of the 2D harmonic oscillator in the $xy$-plane, and this leads indeed to $m \in \mathbb{Z}$.

On the other hand, there should also be an argument within Schrödinger's wave mechanics, and there the argument that $\hat{L}_j$ should be self-adjoint and not only Hermitian plays an important role! Indeed you can just bravely trying to find spherical harmonics with $\ell$ and $m$ half-integer. Checking $\ell=1/2$ is already sufficient. You easily find the solutions for $m=\pm 1/2$, looking innocent enough, being even square integrable on the unit sphere! But then check that applying $\hat{L}_-$ to the $m=1/2$ solution gives the solution for $m=-1/2$, and this indeed fails! You also don't get $0$ when applying $\hat{L}_-$ to the solution for $m=-1/2$, but according to the abstract Dirac formalism this should be so! Now indeed, the solution of this apparent paradox is that the solutions for $m = \pm 1/2$ are not in the domain of $\vec{\hat{L}}$ as self-adjoint operators, i.e., their repeated application to this apparent solutions lead out of this domain, and thus you cannot realize the case $\ell=1/2$ as square-integrable functions on the unit sphere, and thus you have to abandon the idea that there were an orbital angular momentum with $\ell=1/2$. That really shows that spin 1/2 can only be unerstood within QT, and it's not simply some "canonical quantization" of classical physics.

Another related question is, what is the representation of an angle in quantum theory. Naively you'd expect it's the canonical conjugate variable to an angular-momentum component (as in classical mechanics with the analogy between commutators in QT and Poisson brackets in CT in mind). Then you immediately also run into a contradiction, because then from the treatment of linear momentum and position operators you know that then both should have a continuous spectrum, and it's entire $\mathbb{R}$, but that contradicts the angular-momentum algebra proof that $\hat{L}_z$ has only $\mathbb{Z}$, which we had just shown before with all the quibbles to throw away the false half-integer eigenvalues. Here the solution is simple: Knowing from the analysis leading to this result, that it is indeed true, that a rotation by $2 \pi$ indeed should be represented by a phase factor $+1$ when applied to the wave function, we should have the wave functions $2 \pi$-periodic wrt. the azimuthal angle $\varphi$. The naive realization as the canonical conjugate variable would be of course to multiply the wave function with $\varphi$, but starting with a $2 \pi$-periodic function and then applying $\hat{\varphi}$ leads out of this function space, since $\varphi \psi$ wan't be $2\pi$ periodic anymore, and thus $\hat{\varphi}$ is not self-adjoint, because the domain and the codomain should coincide for this.

So finally, what can be used as a representation of an angle are the functions $\cos \hat{\varphi}$ and $\sin \hat{\varphi}$ (defining uniquely a point on the unit circle) or even the unitary operator $\exp(\mathrm{i} \hat{\varphi})$ with $\hat{\varphi}$ defined in the "naive sense" as multiplying the position or momentum wave function by $\varphi$ or rather the functions thereof.