[math, computation] Derivation of BCH formula -part 2

In this post, following from the equations of last posting, we will wrap up the derivation of BCH formula

BCH formula

Summary of last posting

Derivation of BCH formula

Now using first equation we have

Then

by expanding this about epsilon, we will obtain G_1, G_2, and so on.

First take differenitation.

first and second equation depends on perspective. Here I choose second one

Since we will compute up to 1/24, we will collect up to cubic term. To obtain higher order terms we have to colect much higher order terms.

Using expression of G, we have

From

Now do the commutator computation up to order of 4

Now we are almost done. express same equation with different viewpoint.

And compare term by term

We can do more fancy way by using adjoint and exponential map which is in wiki.