Meaning of the Angular Momentum in Quantum Mechanics There was a discussion about the angular momentum in QM in an online course that I am attending now. The discussion and the answers did not satisfy me so I wanted to ask it on Physics SE. 
In the classical picture we think of hydrogen atom as an electron circling around a nucleus with plus charge, which has obvious flaws such as the instability of the atom because of the EM radiation.
In QM we have the operator $\mathbf{L = p \times r}$ and we know that the the electron does not radiate EM-Waves because it is not actually circling around the nucleus. It is sometimes here and there. However the wave function of the electron may have an eigenvalue of $\mathbf{L}^2$ ie. it can have a definite angular momentum. Since we know  that the angular momentum is also conserved in QM ($[ \mathbf{L}^2,H ]=0$) I would assume (classically) that the point at which we measure the angular momentum should be the nucleus since only then the angular momentum can be conserved (correct me if I am wrong). However that would mean that the electron is actually circling around the nucleus.
How can it have an angular momentum without circling around the nucleus and its angular momentum being conserved? If it is circling, we are back to square one, why does it not radiate EM-Waves? Alongside these sub-questions that beg for an answer, my real question is this: How correct it is to associate the classical angular momentum with the angular momentum algebra in QM, which arise actually from a rotational symmetry. I don't know much about the experimental side of the story but if we are able to measure the angular momentum of an electron bound to the nucleus, then how does it fit in to all of this?
 A: It is a misconception to say that

In QM [...] we know that the the electron does not radiate EM-Waves because it is not actually circling around the nucleus. It is sometimes here and there.

In QM the notion of "circling round the nucleus" does indeed fail to make sense, but this is not why electrons don't radiate EM waves. Instead, an atom in its ground state will not radiate because its state space simply does not contain any states with lower energy. 
In essence, this is due to the uncertainty principle. If you have a hydrogen atom and you try to compress the 'electron orbit' any further, then the electron's position uncertainty will decrease, and this will drive a corresponding increase in the momentum uncertainty. This momentum uncertainty is proportional to $p^2$ and thus to the kinetic energy. This means that a decrease in the mean radius will decrease the potential energy but increase the minimum allowed kinetic energy, so there is a trade-off involved. For the ground state, this trade-off is optimal, and you cannot increase or decrease the orbit's spatial extent without increasing the energy.
On another track, you ask

How can it have an angular momentum without circling around the nucleus and its angular momentum being conserved?

The answer to that is that you need to stop thinking about the electron "circling" anything. The only relevant physical quantity is the electron's wavefunction. The electron is said to have orbital angular momentum if its wavefunction has significant changes in phase over paths which go around the nucleus. That's it.
Finally, in terms of 

How correct it is to associate the classical angular momentum with the angular momentum algebra in QM, which arise actually from a rotational symmetry.

you are severely underestimating the extent to which angular momentum arises as the algebra associated with rotational symmetry within classical physics. Indeed, the algebra is exactly the same; you only replace commutators $i[·,·]$ with Poisson brackets $\{·,·\}$. Angular momentum is as related to rotational symmetry in classical physics as it is in quantum mechanics.
A: The 'quantum' angular momentum, in the classical limit, does reduce to the 'classical' angular momentum. In this sense, they are the same thing. Where the classical angular momentum occurs in the classical Hamiltonian, we replace that with the quantum angular momentum in the quantum Hamiltonian.
Of course, 'QM being QM', brings with it some new aspects. In this case, there is an uncertainty relation between angular momentum and angular position:
$$\Delta L_z\  \Delta \phi \sim \hbar$$
The angular momentum is then well-defined only when the angular position is completely indeterminate - and this is why you cannot picture something rotating, even though angular momentum continues to 'measure' rotation. This is just like being unable to associate a constantly changing position to a state of definite linear momentum, once again because of the uncertainty relation.
The interpretation of the electron being 'sometimes here' and 'sometimes there' in a state of fixed $L$ is at best misleading. You only force it into a state of definite angular position by a position measurement, until which point you have to live with the indeterminacy. However, in the position state, it has no determinate angular momentum, and is not stationary; it therefore evolves by "spreading out".
Regarding EM waves, the reason none of these states radiate is because we don't include the electromagnetic field in the Hamiltonian $H$ of the hydrogen atom (except for the Coulomb force). When we do, we see that there are transitions between various states of different energies (i.e. different eigenstates of $H$)- this is what leads to the emission spectra, for example. The difference between this and classical models is that transitions can only occur between these discrete states, and not to arbitrary "orbits". This means that transitions cannot occur to energies below the ground state and the system won't indefinitely lose energy.
Measurements of angular momentum may be carried out through the motion of the atom in a magnetic field gradient (see Stern-Gerlach experiment and this question).
A: While there is an origin for the position operator $\vec r$, and therefore the angular momentum $\vec L = \vec r \times \vec p$ has an origin, it is a little glib to say that there is an angular momentum operator.  What you can is three operators $\hat L_x,$ $\hat L_y,$ and $\hat L_z$ whose forms are a inspired by the forms of the three components of the classical vector $\vec L = \vec r \times \vec p.$
And really you could make lots more operators too like $n_x\hat L_x+n_y\hat L_y+n_z\hat L_z$ for any three real numbers $n_x,$ $n_y,$ $n_z$ or $\hat L_x\hat L_x+\hat L_y\hat L_y+\hat L_z\hat L_z.$
Now it might even be the case that you want to consider the system of the proton and electron, then for your generalized coordinates select the center of mass for three of the coordinates and the position of the electron relative to the center of mass as the final three.  The Hamiltonian then emerges as a regular Schrödinger equation kinetic energy term (with a reduced mass) in the relative position and a kinetic energy term for the center of mass, with the potential all being in terms of the relative coordinates.  When the Hamiltonian is expressed as a sum of terms that depend on just some coordinates, this leads to a natural separation of variables solution.  This is just to say that if you are considering the proton then the position vector for your angular momentum might be for the center of mass of the proton-electron system, as is the origin of the standard hydrogen wavefunction.  So $(\hat x,\hat y, \hat z)$ (which unlike the triple $\hat L_x,$ $\hat L_y,$ $\hat L_z$, is an observable since the components commute) has for an origin, the center of mass, not the location of the proton.
Now you can see another issue.  The various angular momentum operators are not the vector components of some single vector observable called angular momentum.  You cannot measure "the" angular momentum vector.  And when you are in an eigenstate of $\hat{L^2}=\hat L_x\hat L_x+\hat L_y\hat L_y+\hat L_z\hat L_z$ you do not have a definite value of some angular momentum vector, there no observable called the angular momentum vector.  So you don't measure it, and the particles or even the system doesn't have one.  There are many operators, but there is no vector observable for angular momentum like there is with position or momentum.
So now when you say that $[\hat{L^2},\hat H]=0$ you are not saying that there is some vector that is conserved.  All you are saying is that there are common eigenvectors to both the operators, and since one of them is the generator of time translations, the common eigenvectors only change their phase in time.  You can also argue that $[\hat{L^2},\hat L_z]=0$ and $[\hat{L}_z,\hat H]=0$ so there are eigenvectors common to all three operators.  But still there is no vector.
There isn't a vector to be conserved. This can sound strange, since $[\hat{L}_x,\hat H]=[\hat{L}_y,\hat H]=[\hat{L}_z,\hat H]=0$ but since the observables $\hat L_x,$ $\hat L_y,$ and $\hat L_z$ do not commute with each other, they do not have common eigenvectors so cannot all be observed.  There isn't a vector to be conserved, even though any one component, if it existed as an eigenvalue or eigenvector would be conserved.  There simply isn't an angular momentum vector.
And this is above and beyond how the position and momentum are operators.  At least all three components of the position can be measured.  Or instead all three components of the momentum could be observed.  So the whole vector operator $(\hat x, \hat y, \hat z)$ has common eigenvectors, so can be observed.  And thus you can ask if it is conserved (it isn't).  You can't even ask about angular momentum.
So in classical physics you had a position and momentum, in quantum theory you replace each with operators, but at least the whole vector (one vector or the whole other vector) can be observed, so you can discuss whether you are an eignvector of position or momentum (but can't be eigen to both).  Now with angular momentum, all you have are things like $n_x\hat L_x+n_y\hat L_y+n_z\hat L_z$ for any three real numbers $n_x,$ $n_y,$ $n_z$ or $\hat L_x\hat L_x+\hat L_y\hat L_y+\hat L_z\hat L_z.$  But you can't have more than one component because they don't commute with each other. The trade off between position and momentum has turned into a trade off between different components of angular momentum.
So you don't have three components of a vector, so if someone asked you what plane your alleged particle was orbiting you'd be completely at a loss.  because the idea of it orbiting makes it seem like it has three components of an angular momentum vector.  It does not.
I think many textbooks do not make a good clear distinction between having three observables such as $\hat L_x,$ $\hat L_y,$ and $\hat L_z,$ and having a single observable vector.  if position and momentum are introduced first, you might get wrong ideas deeply and implicitly in your brain before you the time you finally get to angular momentum.  Best to fix it.
And thinking classically too can get you into trouble unless you use a version of quantum mechanics that is compatible with whatever particular classical baggage you want to use (which is possible).
For your final questions, you cannot measure the angular momentum vector, not even theoretically, let alone experimentally.  And it has the same relationship to symmetric as in classical physics, and the algebra is an algebra of operators, and the fact that they do not commute is exactly the whole point of there not being an angular momentum vector observable.
