We know that for a position variable $x$ and momentum $p$, the uncertainties of the two quantities are bounded by
$$\Delta x \Delta p \gtrsim \hbar$$
Now, this is usually first explained with $x$ being a simple linearly measured position and $p$ being linear momentum. But it should apply to any good coordinate and its conjugate momentum. It should, for instance, apply to angle $\phi$ about the $z$ axis, and angular momentum $L_z$:
$$\Delta \phi \Delta L_z \gtrsim \hbar$$
The thing is, $\Delta \phi$ can never be greater than $2\pi$. I mean, you have to have some value of $\phi$ and $\phi$ only runs from 0 to $2\pi$. Therefore
$$\Delta L_z \gtrsim \hbar/\Delta \phi \geq \hbar/2\pi$$
But, uh-oh! This means it is impossible for $\Delta L_z$ to be zero, and we should never be able to have angular momentum states with definite $L_z$ values.
Of course, it doesn't mean that. But I have never figured out how this is not in contradiction with the Schroedinger eqn. calculations that give us states with definite values of $L_z$. Can anyone help me out?
One answer I anticipate is that $\phi$ is sort of "abstract" in that if you chose your origin at some other point you will get completely different values of $\phi$ and $L_z$, and ipso facto, usual considerations don't apply. I don't think this will work, though. Consider a "quantum bead" sliding around on a rigid circular ring and you get the exact same problem with no ambiguity in $\phi$ or $L_z$. (Well, there will be some limited ambiguity in $\phi$, but still, there won't be in $L_z$.)
Answer
The problem here is there is at this time still no "legitimate" self-adjoint phase operator. As you phrase the problem, you assume that $\hat \phi$ and $\hat L_z$ would have the same commutation relations as $\hat x$ and $\hat p$, and in particular given that $\hat L_z\mapsto -i\hbar d/d\phi$ the $\hat \phi$ operator would be multiplication of an arbitrary function $f(\phi)$ by $\phi$, i.e. $$ \hat L_zf(\phi)=-i\hbar \frac{df}{d\phi}\, ,\qquad \hat \phi f(\phi)= \phi f(\phi) $$ Thus far everything is fine except that, when it comes to boundary condition, we must have $f(\phi+2\pi)=f(\phi)$. However, the function $\phi f(\phi)$ does not satisfy this. As a result, the action of a putative $\hat \phi$ as defined above takes a "legal" function $f(\phi)$ that satisfies the boundary conditions to an "illegal" one $\phi f(\phi)$, and make $\hat \phi$ NOT self-adjoint (which means trouble).
The uncertainty relation assumes that the operators involved as self-adjoint. Since there is (thus far) no known definition of $\hat \phi$ that makes it self-adjoint, the quantity $\Delta \phi$ cannot be computed in the usual way and indeed is not necessarily well defined for arbitrary states. In other words, there is no mathematical reason to believe that $\Delta \phi\Delta L_z\ge \hbar /2$.
Indeed an obvious "problem" with your expression is obtained by taking $f(\phi)$ to be an eigenstate of $\hat L_z$. Then clearly $\Delta L_z=0$ so the putative variance $\Delta \phi$ would have to be arbitrarily large, which is impossible given that $\phi$ physically ranges from $0$ to $2\pi$.
The problem of constructing a self-adjoint phase operator is an old one. It has been the subject of several questions on this site, including this one. Finding a good definition of a phase operator remains an open research problem.
Edit: added some clarifications after a query.
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