Bivector form of quantum angular momentum operator (derivatives of the unit vectors)
Posted by peeterjoot on August 26, 2009
Derivatives of the unit vectors
To properly evaluate the angular momentum square we’ll need to examine the and variation of the unit vectors , , and . Some part of this question can be evaluated without reference to the specific vector or even which derivative is being evaluated. Writing for one of , , or , and for the mapping of this vector under rotation, and for the desired or partial derivative, we have
Since , we have
So substitution of , back into (18) supplies
Writing the bivector term as
The change in the rotated vector is seen to be entirely described by the commutator of that vectors image under rotation with . That is
Our spherical polar rotor was given by
Lets calculate the bivector for each of the and partials. For we have
Explicitly, this is the bivector , a wedge product of a vectors in direction with one in the perpendicular plane (curiously a vector in the plane rotated by polar angle , not the equatorial angle ).
FIXME: picture. Draw this plane cutting through the sphere.
For the partial variation of any of our unit vectors our bivector rotation generator is
This one has no variation at all with angle whatsoever. If this is all correct so far perhaps that is not surprising given the fact that we expect an extra in the angular momentum operator square, so a lack of dependence in the rotation generator likely means that any additional dependence will cancel out. Next step (for a different day) is to take these rotation generator bivectors, apply them via commutator products to the , , and vectors, and see what we get.