Bivector grades of the squared angular momentum operator.
Posted by peeterjoot on September 6, 2009
The aim here is to extract the bivector grades of the squared angular momentum operator
I’d tried this before and believe gotten it wrong. Take it super slow and dumb and careful.
Suppose is a bivector, , the grade two product with a different unit bivector is
This same procedure will be used for the operator square, but we have the complexity of having the second angular momentum operator change the first bivector result.
In the first few lines of the bivector product expansion above, a blind replacement , and gives us
Using , eliminating the coordinate expansion we have an intermediate result that gets us partway to the desired result
An expansion of the first term should be easier than the second. Dropping back to coordinates we have
Okay, a bit closer. Backpedaling with the reinsertion of the complete vector quantities we have
Expanding out these two will be conceptually easier if the functional operation is made explicit. For the first
In operator form this is
Now consider the second half of (3). For that we expand
Since , and , we have
Putting things back together we have for (3)
This now completes a fair amount of the bivector selection, and a substitution back into (2) yields
The remaining task is to explicitly expand the last vector-trivector dot product. To do that we use the basic alternation expansion identity
To see how to apply this to the operator case lets write that explicitly but temporarily in coordinates
Considering this term by term starting with the second one we have
The curl of a gradient is zero, since summing over an product of antisymmetric and symmetric indexes is zero. Only one term remains to evaluate in the vector-trivector dot product now
Again, a completely dumb and brute force expansion of this is
With , the wedge in the first term is zero, leaving
In vector form we have finally
The final expansion of the vector-trivector dot product is now
This was the last piece we needed for the bivector grade selection. Incorporating this into (6), both the , and the terms cancel leaving the surprising simple result
The power of this result is that it allows us to write the scalar angular momentum operator from the Laplacian as
The complete Laplacian is
In particular in less than four dimensions the quad-vector term is necessarily zero. The 3D Laplacian becomes
So any eigenfunction of the bivector angular momentum operator is necessarily a simultaneous eigenfunction of the scalar operator.