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This article concerns the rotation operator , as it appears in quantum mechanics.
With every physical rotation , we postulate a quantum mechanical rotation operator that is the rule that assigns to each vector in the space the vector that is also in . We will show that, in terms of the generators of rotation, where is the rotation axis, is angular momentum operator, and is the reduced Planck constant.
The rotation operator , with the first argument indicating the rotation axis and the second the rotation angle, can operate through the translation operator 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 according to Quantum Mechanics).
Translation of the particle at position to position :
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 is independent of the position. Because the translation operator can be written in terms of , and , we know that This result means that linear momentum for the system is conserved.
Classically we have for the angular momentum This is the same in quantum mechanics considering and as operators. Classically, an infinitesimal rotation of the vector about the -axis to leaving unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):
From that follows for states:
And consequently:
Using from above with and Taylor expansion we get: with the -component of the angular momentum according to the classical cross product.
To get a rotation for the angle , we construct the following differential equation using the condition :
Similar to the translation operator, if we are given a Hamiltonian which rotationally symmetric about the -axis, implies . This result means that angular momentum is conserved.
For the spin angular momentum about for example the -axis we just replace with (where is the Pauli Y matrix) and we get the spin rotation operator
Operators can be represented by matrices. From linear algebra one knows that a certain matrix can be represented in another basis through the transformation where is the basis transformation matrix. If the vectors respectively are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle between them. The spin operator in the first basis can then be transformed into the spin operator of the other basis through the following transformation:
From standard quantum mechanics we have the known results and where and are the top spins in their corresponding bases. So we have:
Comparison with yields .
This means that if the state is rotated about the -axis by an angle , it becomes the state , a result that can be generalized to arbitrary axes.
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