Azimuthal coordinate operator: Hermition or not? Self-adjoint or not?

Question

I am told that the azimuthal coordinate operator $\hat{\phi}$ is not self-adjoint. I am told this by people who I am sure know much more about this stuff than I do. To my unsophisticated mind, "non self-adjoint" means "not Hermitian". So I think I will demonstrate this to myself by calculating some matrix elements. So,

$$\langle m|\phi|n\rangle = \frac{1}{2\pi}\int_0^{2\pi}d\phi \text{ e}^{-im\phi}\phi\text{ e}^{in\phi} $$

For $m=n$, this is just $\pi$, which is real. I was hoping for complex or imaginary. But all is not lost, so I continue for $m\neq n$:

$$\frac{1}{2\pi}\int_0^{2\pi}d\phi\text{ } \phi\text{ e}^{i(n-m)\phi} $$

Let $\mu \equiv n-m$, then

$$\frac{1}{2\pi}\int_0^{2\pi}d\phi\text{ } \phi\text{ e}^{i(n-m)\phi}= \frac{1}{2\pi}\int_0^{2\pi}d\phi\text{ }\phi \frac{1}{i\mu}\frac{d}{d\phi}\text{ e}^{i\mu\phi} = \frac{-i}{2\pi\mu}\int_0^{2\pi}d\phi\text{ }\frac{d}{d\phi}(\phi \text{ e}^{i\mu\phi})-\text{ e}^{i\mu\phi} = $$ $$\frac{-i}{2\pi\mu}\left[ \phi \text{ e}^{i\mu\phi} - \frac{\text{ e}^{i\mu\phi}}{i\mu} \right]_0^{2\pi}=\frac{-i}{\mu}=\frac{-i}{n-m} $$

Now I try to calculate $\langle n|\phi|m\rangle$. It will obviously be (well, if I am doing this right)

$$ \langle n|\phi|m\rangle=\frac{-i}{m-n} $$

But this not at all what I expected because the complex conjugate gives me my original result for $\langle m|\phi|n \rangle$. That is $\langle m|\phi|n\rangle = \langle n|\phi|m\rangle^*$. This to my mind, is pretty much the definition of Hermitian. So how is $\hat{\phi}$ not Hermitian?

So where have I gone wrong? Do I not understand the difference between Hermitian and self-adjoint? I'm not mathematically sophisticated enough to know what "essentially self-adjoint" means. Does that come into play somehow? Is my calculation just off-base from the get-go?

Answer 1

This is an attempt to rephrase @ACuriousMind's answer linked in the comments using a less mathematically sophisticated language as requested by the OP -- this is just an attempt at translation, not an original explanation (any errors are mine). As anticipated in the comments, as a physicist-not-a-mathematician, I'm not going to be careful about distinguishing Hermiticity from self-adjointness or essential self-adjointness, possibly to my peril.

Finite dimensional intuition

Let's start with finite dimensional spaces, and remind ourselves what Hermitian means in this context. Of course, we know an operator $A$ is Hermitian if $A=A^\dagger$. For finite dimensional operators, we can check this condition by choosing some basis and writing out all the matrix elements $A_{mn}$ in that basis, and then checking explicitly that $A_{mn}=\bar{A}_{nm}$, where overbar denotes the complex conjugate.

A more sophisticated way to say what we're doing is to recall what it means to evaluate the matrix elements. It means we are evaluating the inner products $\langle n | A | m \rangle$ for some complete and orthonormal basis $|m\rangle$. In order to be a complete basis, the states $|m\rangle$ must span the space of all states. In other words, we must be able to construct any possible state as a linear superposition of the basis states $|m\rangle$. This is why, when answering a question about whether an operator is Hermitian or self-adjoint (or other questions about an operator), it is important to understand the space of states over which the operator is acting. These subtleties only become more important when dealing with infinite dimensional operators.

Now, often, we start off working with some "canonical" space, where we get used to various operations being well-defined, but when we move to a variant of the original space, certain operations may become ill-defined.

Again, let's just start with a finite-dimensional example. We can start with $R^3$, or 3-dimensional vectors. Then an operation that makes total sense is to rescale a vector by a constant $a$. This operation is Hermitian. However, now let's restrict ourselves to the subspace vectors in $R^3$ that have unit norm. Then it doesn't make sense to rescale vectors by any constant other than $1$ -- to do so, would take us out of the space we started in.

Operators and wavefunctions on a circle

OK now let's go to the example you're interested in.

In the "canonical" example of an infinite straight line, we can define Hermitian operators $x$ and $p$. Not only that, but their products $xp$ and $px$, and their commutator $[x,p]=xp-px$, are well-defined and Hermitian as well. Of the various properties we have to check, one of them is that all of these operators ($x$, $p$, $xp$, $px$, $[x,p]$) take wavefunctions (normalizable complex-valued functions of the real line) and map them to wavefunctions.

Now, the case you want to study is a particle moving on a circle. We will model this by saying that the infinite straight line is periodic, where we identify $x$ with $x+L$. This means that the wavefunction $\psi$ must obey $\psi(x)=\psi(x+L)$. (This is mathematically equivalent to your example, if you replace $x$ with $\phi$, $p$ with $L_\phi$, and $L$ with $2\pi$.)

Operator $p$

Now, consider the operator $p$. Does $p$ map wavefunctions from the space of periodic functions, to the space of periodic functions? Sure, because $\psi'(0) = \psi'(L)$, and as naive physicists we'll just assume that the wavefunction is as smooth as we need it to be to take as many derivatives as we want.

Operator $x$

What about the operator $x$? Well, it's perfectly possible to multiply $\psi$ by $x$ at any point. Furthermore, we can still normalize $x\psi$. So, we can call it a Hermitian operator. However, there is a subtlety, because $x$ will not map periodic functions to periodic functions; $x \psi$ will not be periodic in general. This is because the periodicity condition, applied to $x\psi$, becomes $0 = L \psi(L)$, which is only true if $\psi(L)=0$ (and, of course, periodicity also implies that $\psi(0)=0$). This only works for a special subspace of wavefunctions. However, it does't affect the ability to compute a value for $x\psi$, or to evaluate the integral of $|x\psi|^2$, so at this level we are ok.

Operator $px$

Now imagine we want to compute the action of the operator $px$ on our circle space. There was no problem with computing the action of this operator on the infinite line. However, now we run into a problem on the circle. The problem is that the input to $p$ should be a function that is periodic, so that we can differentiate it. However, we've just seen that $x\psi$ does not generally produce a periodic function. Therefore, $px$ is not defined for all wavefunctions on the circle. It is only defined for the special subspace of wavefunctions that also obey the stronger condition $\psi(0)=\psi(L)=0$. Note that this subspace does not include the functions $e^{i k x}$, or in azimuthal coordinates $e^{i m\phi}$.

Uncertainty principle

In @ACuriousMind's answer, they point out that there is a trick for defining $\langle \psi| [p, x] | \psi \rangle$ for wavefunctions on the circle, to make sense of the uncertainty principle. Instead of evaluating $[p, x] | \psi\rangle$, which would involve evaluating $px \psi$, which we have just seen is not always defined, we can instead evaluate $p \psi$ and $x \psi$, then compute the integral $\int( \bar{(p \psi)}(x \psi) - \bar{(x \psi)} (p \psi))$. In this example, we are able to sidestep the issue of evaluating $px\psi$, so we don't run into the problem of differentiating a discontinuous function. Then, everything is well defined, and the apparent issue with the uncertainty principle brought up in that question is resolved.

Summary

Anyway, to summarize, there is a subtlety with the $x$ operator when working on a circle, because $x\psi$ is not periodic. This doesn't necessarily cause a major problem on its own, but you can quickly run into issues if you start mixing $x$ and $p$, as @ACuriousMind stated in the comments.