# I How can the total orbital angular momentum be zero?

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1. Jun 15, 2017

### steve_a

I'm trying to understand the rotations of rigid diatomic molecules such as HCl. My understanding of the orbital angular momentum is that it is quantized with a total value equal to
$$E=\frac{\hbar^2}{2I}J(J+1)$$
where I is the rotational moment of inertia and J is the quantum number. Also, J can be 0, 1, 2, etc. According to this, the minimum rotational energy is 0, which is surprising but ok. Also, the total angular momentum is given by
$$L^2=\hbar^2 J(J+1)$$
This implies that the total angular momentum is 0 if J is 0. This is consistent with the rotational energy being 0.

This implies that I have complete knowledge of the angular momentum if J = 0 (i.e. it must be exactly 0). However, the Heisenberg uncertainty principle says that complete knowledge of the angular momentum would require complete lack of knowledge of the orientation. To the contrary, I know that the orientation is between 0 and 2π, which is not complete lack of knowledge and thus seems to violate the Heisenberg uncertainty principle.

If there were a zero-point energy for rotations, then everything would make sense, but it doesn't seem that there is one.

What am I doing wrong? Thanks for any replies!

-Steve

2. Jun 15, 2017

### Staff: Mentor

Which means you have absolutely no information about it: It can be oriented in any way.

Actually, this is a 3D problem, and you don't have any idea about the orientation in the full solid angle.

3. Jun 15, 2017

### steve_a

Thanks for your reply. I agree with your comment about the 3D aspect, so perhaps I should have been more careful and said that I know that the orientation is within 0 and π on the θ coordinate and within 0 and 2π on the Φ coordinate. However, that doesn't fundamentally change the problem.

I don't think that I agree that orientation in any direction is the same as complete lack of knowledge. For example, if I knew the linear momentum precisely, then my uncertainty in the linear position would range from -∞ to +∞. That's a vastly larger range than the possible orientations (2π radians, or 4π steradians). If I could keep track of the number of rotations that a particle made, then it's orientation could range from -∞ to +∞ (or the 3D version of this), which would be complete lack of knowledge; however, I can't, which means that I have some knowledge of the orientation.

Also, my understanding is that linear momentum is not quantized for a free particle because it has an infinite position interval. In contrast, angular momentum is quantized due to the finite interval for orientation. In effect, I think that the quantum treatment of angular momentum is similar to a particle in a box, but for a box with periodic boundary conditions. In both cases, the energies are similar (proportional to roughly the square of the quantum number) and the wavefunctions are similar. But, the particle in a box has a zero point energy whereas angular momentum does not, so I'm still puzzled.

4. Jun 15, 2017

### Staff: Mentor

How could you potentially know even less than "I have absolutely no information about the orientation and every orientation has the same probability"?

If you don't like the finite interval, transform the angle to some other function, then your parameter can go from minus infinity to plus infinity - that doesn't change the physics.

5. Jun 16, 2017

### Staff: Mentor

Look at the spherical harmonics. $Y_{0,0}$ is completely isotropic, meaning that a molecule in the rotational ground state can be found along any orientation with the same probability. In other words, it has the most uncertain orientation possible. (Note that this is the inverse of the classical result: in the classical rotational ground state, the molecule is not rotating, therefore has a very definite orientation.)

As for the comparison with the particle in a box, I would venture to say that it is because there is a uniform solution (in the case of rotation) that makes it possible to have a zero-energy solution. In the case of the particle in a box, because of the boundary conditions at the walls, the uniform solution is zero everywhere, but this is not the case for rotation.

6. Jun 16, 2017

### steve_a

Thank you to both of your for your replies. I'm starting to come around to your view, but am still confused.

First, mfb wrote that if I don't like the finite angle interval, then I could just transform it to some other function without changing the physics. I don't agree with this because the Heisenberg uncertainty relation applies to conjugate variables, such as x and p, so one can't just go and transform them as desired. Instead, one has to retain the conjugate nature.

Also, the reason that I am concerned about the finite interval is from the Heisenberg uncertainty relation. For angular momentum, it is often (more on this below) given as
$$\Delta L_z \Delta \phi \geq \frac{\hbar}{2}$$
where ΔLz is the angular momentum about the z axis and Δφ is the angular position about the z axis. If I know that L2 is equal to 0 (because I'm in the J = 0 quantum state), then this seems to imply that Lz must be 0 too. It also implies that Lx and Ly are also equal to 0, which is yet another problem because one isn't supposed to be able to measure Lx, Ly, and Lz at once. Anyhow, if Lz = 0, then ΔLz = 0, and ΔΦ must be equal to infinity just as is the case for the linear momentum problem. However, this isn't possible with a finite interval. Instead, the largest that Δφ can be is the case for uniform probability for all φ values, meaning P(φ) = 1/2π:
$$(\Delta \phi)^2 = \int_{0}^{2\pi} \frac{1}{2 \pi} \phi^2 d\phi = \frac{\pi^2}{3}$$
$$\Delta \phi = \frac{\pi}{\sqrt{3}}$$
This is clearly finite, showing that the finite angle interval leads to a finite angle uncertainty. In combination with the zero momentum uncertainty, this violates the Heisenberg uncertainty principle.

In response to DrClaude, I like your insight that the uniform solution for the wavefunction (for the angular momentum case but not for a particle in a finite box with reflective boundaries) would lead to zero energy. This follows from the time-independent Schrodinger equation.

I found two papers that are helpful. First "Quantum theory of rotation angles" by Barnett and Pegg (link: Phys Rev A, 1990) says that φ is not a good operator due to its periodic nature. I didn't fully understand the paper, but they propose a new operator that works better. In any case, this suggests that the Heisenberg uncertainty relation that I gave above may be incorrect. Secondly, "Uncertainty principle for angular position and angular momentum" by Franke-Arnold et al. (link: New Journal of Physics, 2004) gives the Heisenberg uncertainty relation as
$$\Delta \phi_\theta \Delta L_z \geq \frac{\hbar}{2} \mid 1-2\pi P(\theta) \mid$$
I don't understand where they got this equation from. However, if I stick in P(θ) = 1/2π, then it says that I'm allowed to know both φ and Lz exactly. In any case, both of these papers support the notion that I was using an incorrect form for the Heisenberg uncertainty relation due to the periodic nature of φ.

I still lack any sense of intuition, but I think there's progress.

7. Jun 16, 2017

### Staff: Mentor

I didn't say the uncertainty relation would hold for the transformed variables.

There is no meaningful mean angle. Your integral assumes $\phi=0$ would be special in some way. It is not.

8. Jun 16, 2017

### steve_a

You're correct that you didn't say that the uncertainty relation would hold for the transformed variables. But then, why would I want to transform them? My entire question was about the uncertainty relation, so there's no point in doing transformations that break that relationship.

Also, I made a typo in my equation for the variance of φ. I should have written the limits as -π to π, in which case φ=0 is special because it is the mean value within this interval. Or, I should have written the integrand as (φ-π)2 now using the fact that φ=π is the mean value with the limits of 0 to 2π. Either way, the integral evaluates to the same result that I gave, which was π2/3. Also either way, this is the variance in φ for a uniform probability distribution. It doesn't matter what domain is chosen for the angle. The point remains that the variance in φ is finite and the variance in L is 0, which appears to violate the Heisenberg uncertainty principle.

9. Jun 16, 2017

### Staff: Mentor

Forget the transformed variables. I hoped that they would help with understanding, but apparently they had the opposite effect.

Why -pi to pi? Why not 0 to 2 pi? Why not -pi/2 to 3pi/2? You see the problem? The interval choice is arbitrary, and the interval does not have a physically meaningful midpoint because the ends are identified with each other. Such a midpoint would be a preferred orientation, and we don't have that.

10. Jun 17, 2017

### steve_a

Thanks for sticking with this problem. Regarding the integral, it produces the same result independent of the domain of integration. The important thing is that the variance is the integral over the squared difference between the value and the mean of the value. The mean is necessarily going to be in the middle of the domain of integration for a uniform probability density. Using slightly better notation than before, a general version of the equation is:
$$\langle (\Delta \phi)^2 \rangle = \int_{\phi_0}^{\phi_0+2\pi} \frac{1}{2 \pi} (\phi-(\phi_0+\pi))^2 d\phi=\frac{\pi^2}{3}$$
In this case, the integral goes from φ0 to φ0+2π to show that the domain of integration (so long as it spans 2π) is irrelevant to the final solution.

11. Jun 17, 2017

### Staff: Mentor

You don't have a proper interval in the real numbers. The mean is meaningless.
Instead of an interval in the real numbers, think of it in terms of actual orientation. What is the mean orientation? What does that even mean?

Puns not intended, but unavoidable

12. Jun 17, 2017

### vanhees71

This is again a confusion due to not taking the math seriously enough. To represent an observable, an operator must be self-adjoint. Let's consider a particle on a circle (a rigid rotator). Then naively you could try to quantize the system by introducing usual polar coordinates and then $\varphi$ would be the "position coordinate" and the angular momentum $L$ generating translations of the angle. Then you'd assum(!) that there are two self-adjoint operators $\hat{\phi}$ and $\hat{L}$ with
$$[\hat{\phi},\hat{L}]=\mathrm{i} \hat{1}.$$
Here and in the following I use natural units with $\hbar=1$. Then you could introduce the "position representation", leading to "wave functions" $\psi(\phi)$ realizing the Hilbert space of the QT system, in which $\hat{\phi} \psi(\phi)=\phi \psi(\phi)$ and $\hat{L} \psi(\phi)=-\mathrm{i} \psi'(\phi)$.

So far everything looks as if we'd quantize simply position and momentum observables for a particle restricted to move on a line, but that's not true, because the angle variable (at least for the classical system) is defined modulo $2 \pi$. So we have to make sure that $\psi(\phi)$ and $\psi(\phi+2 \pi)$ for any $\phi \in \mathbb{R}$ represents the same state. To keep the argument simple we assume that the wave functions should be restricted to be periodic:
$$\psi(\phi+2 \pi)=\psi(\phi).$$
Now we look for the angular-momentum eigenfunctions:
$$\hat{L} u_{m}(\phi)=\ell u_{m}(\phi) \; \Rightarrow \; u_{m}(\phi)=\frac{1}{2 \pi} \exp(\mathrm{i} m \phi).$$
To be $2\pi$ periodic, we get $m \in \mathbb{Z}$, and everything seems to be fine, because also $\hat{L} u_m(\phi)$ is still periodic and thus in the Hilbert space of periodic functions $\mathrm{L}([0,2 \pi])_{\text{periodic}}$.

But now what about the angle operator? We have by definition
$$\hat{\phi} u_m(\phi)=\phi u_m(\phi),$$
and that's not anymore in the Hilbert space, just constructed. Of course, in general, with $\psi(\phi)$ the function $\phi \psi(\phi)$ is not periodic anymore. Indeed looking for a periodic $\psi$ such that $\phi \psi(\phi)$ is also periodic we get
$$(\phi+2 \pi) \psi(\phi)=\phi \psi(\phi) \; \Rightarrow 2 \pi \psi(\phi)=0 \;\Rightarrow \psi(\phi)=0.$$
This implies that $\hat{\phi}$ is not self-adjoint, i.e., there's no dense subspace of the Hilbert space, on which $\hat{\phi}$ 's codomain equals its domain.

Indeed the naive "canonical commutation relation" used above leads to a contradiction when applied to the angular-momentum eigenstate $|m \rangle$. You get for all $m \in \mathbb{Z}$
$$\langle m|[\hat{\phi},\hat{L}]|m \rangle=\langle \hat{\phi} m|\hat{L} m \rangle-\langle \hat{L} m|\hat{\phi}|m \rangle=m [\langle \hat{\phi} m|m - \langle m|\hat{\phi} m \rangle=0,$$
but on the other hand from the commutation relation we should get
$$\langle{m} |[\hat{\theta},\hat{L}]|m \rangle=\mathrm{i} \langle m|m \rangle=\mathrm{i}.$$
So the naive quantization here fails, because the naive assumption of a self-adjoint angle variable leads to a contradiction.

Here the solution is simple: The position on the circle is uniquely defined, and instead of using $\hat{\phi}$ we may use the unitary operator $\hat{U}=\exp(\mathrm{i} \hat{\phi})$. Indeed in the position representation
$$\hat{U} \psi(\phi)=\exp(\mathrm{i} \phi) \psi(\phi)$$
leads to a function in the here considered hilbert space since, if $\psi$ is square integrable and periodic, so is obviously $\hat{U}$. The commutation relation with $\hat{L}$ is
$$[\hat{U},\hat{L}] \psi=-\mathrm{i} \exp(\mathrm{i} \phi) \psi' + \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} \phi} [\exp(\mathrm{i} \phi) \psi]=-\hat{U} \psi.$$
This obviously doesn't lead to a contradiction when used in matrix elements with angular-momentum eigenstates. We obviously have
$$\hat{U} u_m(\phi)=u_{m+1}(\phi).$$
This means that $U$ is a kind of "ladder operator" for the eigenstates of $\hat{L}$.

13. Jun 19, 2017

### DrDu

Hendrik, I just followed a different line of thought:
What if instead of wrapping up the operator phi via exponentiation, we enlarge the definition of L to the whole real line? I think this is standard in solid state physics when you introduce Bloch functions.
To make clear, that phi is extended to the whole line, I use x instead of phi.
Namely we could split
$p=L+q$ where the eigenvalues k of q range from $-\pi/a$ to $+\pi/a$, where $a=2\pi r$ is the cirkumference of the cycle.
Namely
$p=\int_{-\pi/a}^{\pi/a} dk \sum_m (2\pi m/a +k) \exp(i(k+2\pi m/a))(x-x')$
and $L=\int_{-\pi/a}^{\pi/a} dk \sum_m (2\pi m/a ) \exp(i(k+2\pi m/a))(x-x')=-i\;\mathrm{sinc}(\pi/a (x-x') )\sum_n \delta'(x-x'-na)$.
Now the commutator becomes
$[L,x]=i\; \mathrm{sinc}(0)=i$.

Last edited: Jun 19, 2017
14. Jun 19, 2017

### steve_a

Thanks for these replies. However, at least to me, they don't appear to answer my original question. I'm wondering if there is an intuitive explanation for the fact that it's possible to know the angular momentum precisely (equal to 0) and also know the orientation with a finite variance? Offhand, this appears to violate the Heisenberg uncertainty principle. Thanks.

15. Jun 19, 2017

### Physics Footnotes

To resolve this apparent paradox you need to understand how to handle the standard QM formalism properly mathematically (i.e. as opposed to formally). In particular, you mustn't do what so many textbooks do and blindly accept formal claims about commutation relations and uncertainty relations justified by vague references to 'conjugate variables' and other so-called 'classical correspondence principles'.

Although that is certainly how Dirac did things, that was because he was in the discovery process. Thanks to von Neumann and others we can now be much more precise and convincing in our arguments, and when we do so problems like yours disappear.

Let's start with the fact that the (pure) states of your system are represented by elements $\phi,\psi,\chi,\cdots$ of some Hilbert Space $\mathcal H$. The observables you're interested in (e.g. in your case, angular momentum and orientation angle) are represented by self-adjoint operators $L,\Theta,\cdots$ on $\mathcal H$.

Now an important point (especially for your issue) is that self-adjoint operators do not always have a domain equal to the entire Hilbert Space. In fact it turns out that if we work in an assumed angle representation, $\mathcal H := L^2[0,2\pi)$, we have $\mathcal D (\Theta)=\mathcal H$ but $\mathcal D (L)\subset\mathcal H$. In other words, not all elements $\psi\in\mathcal H$ lie in the domain of $L$, and that is where your problems start.

Sidenote: Although I disagree with @vanhees71 on the details of angle representations, all that matters to this question really is that the mishandling (or total abandon) of operator domains is the real culprit in this puzzle. So let's save for another post the technical reasons why $\mathcal D (\Theta)=\mathcal H$ but $\mathcal D (L)\subset\mathcal H$.​

So how do domains cause the problem here?

Well, uncertainty relations are not to be assumed in advance, but rather proved from the definitions of the operators involved. The usual theorem we rely on (although it turns out we can do slightly better than this) is$$\Delta_{\psi} A \Delta_{\psi} B \geq \frac 1 2|\langle\left[A,B\right]\rangle_\psi|$$
Note, in particular, that this inequality is only defined on the domain of the commutator $\left[A,B\right]:=AB-BA$. If a state lies outside of this domain, there is no guarantee that the uncertainty relation has to hold in that state. And that is EXACTLY what goes awry in the case of rotation angles...

It turns out that the operator $\Theta$, which is defined on the entire Hilbert Space, projects most states out of the domain of the operator $L$, rendering the product $L\Theta$ (and hence the commutator) undefined on these states. Again, I'm happy to discuss the technicalities in a separate post, but all that matters to your question is this: the eigenstates of $L$ are precisely these sorts of states. In other words, the states of precise angular momentum are states for which the canonical uncertainty relation between $L$ and $\Theta$ is silent (because it turns out that $\Theta$ projects these states out of the domain of $L$).

That explains why the so-called canonical uncertainty relation does not cause contradictions, which was really your question, but many interesting questions remain. One of these is the question of whether some alternative uncertainty relation can be formulated for angle variables to capture their relationship with their conjugate partners, and the answer is absolutely yes. Despite popular belief to the contrary, we are not limited to the formal proclamations made by Dirac nearly a century ago and still propagated in textbooks to this day.

Again, I'm happy to discuss alternative formulations and technicalities in another post if you're interested. Here I just wanted to answer the question that is rightly bugging you. The answer, as I hope I've convinced you, lies in properly treating operator domains; something Dirac cannot be blamed for ignoring since they weren't properly elucidated until after his work had become the bible of quantum physics. We do not have the luxury of that excuse ;-)

Last edited: Jun 19, 2017
16. Jun 19, 2017

### dextercioby

@Physics Footnotes After reading through your post, I concluded: well, yet another guy who's dug deep into the Fr. Gieres' article on the mandatory technicalities (know-how of functional analysis) needed to properly understand QM in its textbook (which in this case better be Galindo & Pascual 1st volume) formulation.

There were a couple of recent discussions about the particle in a box problem which touched upon the mathematical finesse of standard QM. I mentioned there a section on the limitations of Heisenberg's "uncertainty principle" (well, in its Schroedinger-Robertson formulation) in B. Hall's book on QM for mathematicians. This applies mutatis mutandis to the problem discussed here, making x go to phi and p go to L-phi.

Last edited: Jun 19, 2017
17. Jun 20, 2017

### DrDu

I want to take up again mfb's line of thought, i.e. extending the range of $\phi$ to $(-\infty, \infty)$.
A general eigenstate (highly degenerate in this case) of L with eigenvalue m is then of the form $\exp(i m \phi)* \int_{-1/2}^{1/2} f(k) \exp(ik\phi) dk$
The uncertainty of the expectation value of $\phi$ then depends on the function f(k). If it is a delta function, then the variance of phi is infinite. The smallest variance arises for f(k)=const. Then $\Delta \phi= \sqrt{2\pi}$ (I hope I got this value right).
So yes, in this extended setting, angular momentum is not total momentum. Hence we can get some localization on phi even in an angular momentum eigenstate. Interestingly this $\Delta \phi$ is somewhat larger than what one would expect from a equal distribution of angles from -pi to pi for which $\Delta \phi= \sqrt{\pi^2/3}$.

Last edited: Jun 20, 2017
18. Jun 20, 2017

### vanhees71

I'd be interested to learn your opinion on this and particularly what's wrong with my argument that in the space of periodic square-integrable functions multiplication with $\theta$ leads out of this function space, because oviously if $\psi(\phi)$ is $2 \pi$ periodic, that's not the case for $\phi \psi(\phi)$.

19. Jun 20, 2017

### DrDu

As in the case with the Hamiltonian for a particle in the box, the (periodic) boundary conditions don't define the (Hilbert-) space but the domain of definition of the operator. L.

20. Jun 20, 2017

### vanhees71

Ok, so you take as Hilbert space simply $L^2[(0,2 \pi)]$ without constraints. Then, however the meaning of $\phi$ as an angle (i.e., a periodic quantity modulo $2 \pi$) is lost.

On the other hand, why then should I take the contraint for the domain of $L$ to be the periodic functions? You could impose any boundary conditions that make it a self-adjoint operator, or can you rule out by some argument any other possible boundary conditions somehow?

For me the most convincing argument for the commutation relations are that they occur from the Lie algebras of symmetry groups. For angular momentum you have the $\mathrm{su}(2)$ Lie algebra, defining the commutation relations between the $\hat{L}_k$ but never an "angle variable" occurs in this point of view at all.

21. Jun 20, 2017

### DrDu

A function can be discontinuous and yet periodic, namely a saw tooth. The angle is precisely of this type. We can also approximate a saw tooth arbitrarily well with continuous periodic functions, so if you want the Hilbert space to be closed, there are no special boundary conditions. The continuity and periodicity have to be encoded in the choice of the right self-adjoint extension of the operator $-id/d\phi$. All possible extensions can be seen to correspond to Aharonov Bohm terms $p+eA$ with $A = \Phi \delta(\phi)/(2\pi)$ where $\Phi$ is the magnetic flux enclosed.

22. Jun 20, 2017

### vanhees71

Ok, what's with the angle operator itself? Obviously it's not definable as a self-adjoint operator on the same domain as $-\mathrm{d}_{\phi}$, because then you'd run into the contradiction in my argument above.

23. Jun 20, 2017

### DrDu

I think this is what Physics Footnotes wanted to point out: The commutation relation between angle and L is almost useless as L is not defined $\phi |m>$.

24. Jun 20, 2017

### vanhees71

I don't understand. I thought, $\hat{L}$ is the angular-momentum operator, i.e., $-\mathrm{i} \mathrm{d}_\phi$. Now let's see if I can follow your previous posting. First I need a densely defined subspace where $\hat{L}$ is Hermitean. This means for all wave functions in this domain I should have
$$\langle \psi_1|\hat{L} \psi_2 \rangle=-\mathrm{i} \int_{0}^{2 \pi} \mathrm{d} \phi \psi_1^*(\phi) \psi_2'(\phi) \stackrel{!}{=} \langle \hat{L} \psi_1|\psi_2 \rangle = \mathrm{i} \int_0^{2 \pi} \mathrm{d} \phi \psi_1^{* \prime}(\phi) \psi_2(\phi).$$
Now Integrating by parts gives
$$\langle \psi_1 |\hat{L} \psi_2 \rangle=-\mathrm{i} [\psi_1^*(2 \pi) \psi_2(2 \pi) - \psi_1^*(0) \psi_2(0)] + \mathrm{i} \int_0^{2 \pi} \mathrm{d} \phi \psi_1^{*\prime}(\phi) \psi_2(\phi).$$
So to make $\hat{L}$ Hermitean, the non-integral term should vanish. This means
$$\frac{\psi_2(2 \pi)}{\psi_2(0)}=\frac{\psi_1^*(0)}{\psi_1^*(2 \pi)}. \qquad (1)$$
For $\psi_2=\psi_1$ we get
$$|\psi_1(2 \pi)|^2=|\psi_1(0)|^2,$$
which implies that there should be a real number $\Phi$ such that
$$\psi_1(2 \pi)=\exp(\mathrm{i} \Phi) \psi_1(0). \qquad (2)$$
The same should hold for $\psi_2$ of course, perhaps with another $\Phi'$. Plugging this, however, into (1) leads to $\Phi'=\Phi$.

Thus the functions should be integrable and with some fixed real number $\Phi$ obey the "quasi-periodic" boundary condition (2).

Now let's solve the eigenvalue problem
$$\hat{L} u_m=m u_m$$
giving
$$u_m(\phi)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} m \phi)$$
The boundary conditions give
$$2 \pi m =2 \pi n + \Phi \quad \text{with} \quad n \in \mathbb{Z}.$$
This means
$$u_m(\phi)=\frac{1}{\sqrt{2 \pi}} \exp \left [\mathrm{i} \phi \left (n + \frac{\Phi}{2 \pi} \right) \right].$$
These functions obviously build a complete set in $L^2([0,2 \pi])$, and thus the finite linear combinations of these functions build a dense subset, and also $\hat{L} \psi$ is then in this domain, i.e., we have a self-adjoint operator.

Now consider $\hat{\phi}$, defined as $\hat{\phi} \psi(\phi)=\phi \psi (\phi)$ is not self-adjoint with the same domain. It's of course Hermitean, but with $\psi$ in the domain of $\hat{L}$ just constructed above, $\hat{\phi} \psi$ is not, i.e., $\hat{\phi}$ is not self-adjoint. I guess, one can somehow extend it to an essentially self-adjoint operator, but the corresponding domain is disjunct from the domain of $\hat{L}$ constructed above, no matter how you choose $\Phi$ (different choices of $\Phi$ simply mean different self-adjoint extensions of $\hat{L}$). So you are not allowed to write down the simple commutation relation $[\hat{\phi},\hat{L}]=\mathrm{i} \hat{1}$, because there's no common domain, where this commutation relation makes sense.

I hope I got this at least approximately right, since I'm not so famliar with the formal mathematics.

Last edited: Jun 20, 2017
25. Jun 20, 2017

### DrDu

... or $\psi(2\pi)=\exp(i\lambda) \psi (0)$.