Rotation operator (quantum mechanics)

This article concerns the rotation operator, as it appears in quantum mechanics.
Quantum mechanical rotations.
With every physical rotation formula_1, we postulate a quantum mechanical rotation operator formula_2 which rotates quantum mechanical states.
In terms of the generators of rotation,
where formula_5 is rotation axis, and formula_6 is angular momentum.
The translation operator.
The rotation operator formula_7, with the first argument formula_8 indicating the rotation axis and the second formula_9 the rotation angle, can operate through the translation operator formula_10 for infinitesimal rotations as explained below. This is why, it is first shown how the translation operator is acting on a particle at position x (the particle is then in the state formula_11 according to Quantum Mechanics).
Translation of the particle at position formula_12 to position formula_13: formula_14 
Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):
Taylor development gives:
with
From that follows:
This is a differential equation with the solution 
Additionally, suppose a Hamiltonian formula_21 is independent of the formula_12 position. Because the translation operator can be written in terms of formula_23, and formula_24, we know that formula_25 This result means that linear momentum for the system is conserved.
In relation to the orbital angular momentum.
Classically we have for the angular momentum formula_26 This is the same in quantum mechanics considering formula_27 and formula_28 as operators. Classically, an infinitesimal rotation formula_29 of the vector formula_30 about the formula_8-axis to formula_32 leaving formula_8 unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):
From that follows for states:
And consequently:
Using 
from above with formula_43 and Taylor expansion we get:
with formula_45 the formula_8-component of the angular momentum according to the classical cross product.
To get a rotation for the angle formula_47, we construct the following differential equation using the condition formula_48:
Similar to the translation operator, if we are given a Hamiltonian formula_21 which rotationally symmetric about the formula_8-axis, formula_54 implies formula_55. This result means that angular momentum is conserved.
For the spin angular momentum about the formula_8-axis we just replace formula_57 with formula_58 and we get the spin rotation operator 
Effect on the spin operator and quantum states.
Operators can be represented by matrices. From linear algebra one knows that a certain matrix formula_60 can be represented in another basis through the transformation
where formula_62 is the basis transformation matrix. If the vectors formula_63 respectively formula_64 are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle formula_47 between them. The spin operator formula_66 in the first basis can then be transformed into the spin operator formula_67 of the other basis through the following transformation:
From standard quantum mechanics we have the known results formula_69 and formula_70 where formula_71 and formula_72 are the top spins in their corresponding bases. So we have:
Comparison with formula_69 yields formula_76.
This means that if the state formula_72 is rotated about the formula_78-axis by an angle formula_47, it becomes the state formula_71, a result that can be generalized to arbitrary axes.