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.

## Best Answer

Consider first the classical theory. Turning points are these points $x$ where the total energy is completely stored in the potential energy, $E = V(x)$. Hence the kinetic energy is zero at these points. For example consider the potential of the Harmonic Oscillator $V(x) = x^2 / 2$. Then the turning points are $x_t = \pm \sqrt{2 E}$. The pendulum oscillates between $x_{-} = - \sqrt{2 E}$ and $x_{+} = - \sqrt{2 E}$, and hence makes a turn at these maximal points.

Now turn to the quantum mechanic picture, or to be more exact to the semiclassical WKB method. One tries to approximate the quantum mechanic solution by some other functions. In the case of the Harmonic Oscillator these approximating functions are exactly the Airy functions (with some prefactors). So looking at the solution from wikipedia, one sees that the behavior of the wavefunction changes at the turning points (the potential is drawn in blue, so the turning points are approximately where the wavefunctions intersects the blue graph). For the region between the $x_t$ one has an oscillation whereas for greater $x$ the wavefunction is exponentially damped. This is just as for the Airy function.

http://upload.wikimedia.org/wikipedia/commons/9/9e/HarmOsziFunktionen.png