Applying the vector derivative commutator (or not).
Let’s express the and unit vectors explicitly in terms of the standard basis. Starting with we have
Explicitly in vector form, eliminating the exponential, this is , but it is more convenient to keep the exponential as is.
For we have
Again, explicitly this is , but we’ll use the exponential form above. Last we want
Summarizing we have
Or without exponentials
Now, having worked out the cool commutator result, it appears that it will actually be harder to use it, then to just calculate the derivatives directly (at least for the derivatives). For those we have
This multiplication takes a vector in the plane and rotates it 90 degrees, leaving an inwards facing radial unit vector in the x,y plane.
Now, having worked out the commutator method, lets at least verify that it works.
Much harder this way compared to taking the derivative directly, but we at least get the right answer. For the derivative using the commutator we have
Good, also consistent with direct calculation. How about our derivatives? Lets just calculate these directly without bothering at all with the commutator. This is
Finally, last we have the derivatives of . Those are
Summarizing, all the derivatives we need to evaluate the square of the angular momentum operator are
Bugger. We actually want the derivatives of the bivectors and so we aren’t ready to evaluate the squared angular momentum. There’s three choices, one is to use these results and apply the chain rule, or start over and directly take the derivatives of these bivectors, or use the commutator result (which didn’t actually assume vectors and we can apply it to bivectors too if we really wanted to).
An attempt to use the chain rule get messy, but it looks like the bivectors reduce nicely, making it pointless to even think about the commutator method. Introducing some notational conveniences, first write . We’ll have to be a bit careful with this since it commutes with , but anticommutes with or (and therefore ). As usual we also write for the Euclidean pseudoscalar (which commutes with all vectors and bivectors).
This gives us just
and calculation of the bivector partials will follow exclusively from the partials tabulated above.
Our other bivector doesn’t reduce quite as cleanly. We have
So for this one we have
Tabulating all the bivector derivatives (details omitted) we have
Okay, we are now armed to do the squaring of the angular momentum, but once again it’s late to start now. To be continued.