Assume, without the loss of generality, that the angular momentum operator has only one component, $\hat{L_z}$. It can be shown that the following eigenvalue equation holds true.
$$ L_{z} |l, m \rangle = m \hbar |l, m \rangle \quad \text{such that} \quad m \in \mathbb{Z}. $$
This is to say that the eigenvalues of the orbital angular momentum operator are integer multiples of $\hbar$. You can show this is two ways:
In the co-ordinate (polar) representation, $\hat{L_z} = -i \hbar \frac{\partial}{\partial \phi}$. The eigenvalue equation,
$$ \hat{L_z} |l, m \rangle = l_z |l, m\rangle, $$
implies that the (unnormalized) eigenfunction of $\hat{L_z}$ is:
$$ e^{\frac{il_z\phi}{\hbar}}. $$
We can prove the desired result by:
- Imposing the condition that the aforementioned eigenfunction has to be single-valued; and
- $\hat{L_z}$ is a hermitian operator. That is:
$$ \langle \psi_1 | \hat{L_z} | \psi_2 \rangle = \langle \psi_2 | \hat{L_z} | \psi_1 \rangle ^{*} $$
We can also have angular momentum whose eigenvalues are half integer multiplies of $\hbar$. This is referred to as spin angular momentum. One can show that in the matrix mechanical formulation of quantum mechanics, the eigenvalues of angular momentum can in general be both half or full integer multiples of $\hbar$.
One shouldn't think of the spin angular momentum as having a classical counterpart. Let me give you an example:
Apply the rotation operator (whatever its functional form) on a quantum state with spin pointing in the positive $z$ direction. The result is the following:
$$ \hat{R}(2 \pi) | + z \rangle = -| + z \rangle, $$
where I have chosen to rotate the spin state by 360 degrees. Defying classical intution, the state of, say electron, isn't what we expect: classically, we expect the result of the application of the rotation operator to be the same as the result of our applying the identity operator, which'll yield the initial state as the final state. It is only when one rotates the electrons by $4 \pi$ radians that the final and initial states are the same. The result is both counter-intuitive and against classical intuition.
Note: To be pedantic, the quantum states on both the right and left sides are equivalent/same. (why?) So please choose to ignore the terminology, in its correct QM interpretation, since in the previous paragraph what I'm tying to convert is that the expressions on the right and the left hand sides don't match.
Therefore, one shouldn't be making a correspondence between the spin angular momentum and classical intuition/results. For example, the spin of an electron can be fully specified on a Bloch Sphere and one usually uses the term that one can perform "rotations" on the Bloch sphere, which can be thought of (for example) as a result of time evolution on a Bloch sphere. Yes, we can but one shouldn't think of these "rotations" as operations that have a classical counterpart in all cases, as for example in the previous example. Therefore, the quantum operator of interest in this case is usually not a rotation operator bur rather a unitary operator, which can be thought of as generalization of a rotations operator in either a complex vector space or a Hilbert space.
