Maths - "Baker-Campbell-Hausdorff formula

There is a very useful identity, which is,

e^(a+b) = e^a * e^b

However this only applies when 'a' and 'b' commute, so it applies when a or b is a scalar for instance.

The more general case where 'a' and 'b' don't necessarily commute is given by:

e^c = e^a * e^b

where:

c = a + b + a×b + 1/3(a×(a×b)+b×(b×a)) + ...

where:

• × = vector cross product

This is a series known as the Baker-Campbell-Hausdorff formula.

This shows that if when a and b become close to becoming parallel then a×b approaches zero and c approaches a + b so the rotation algebra approaches vector algebra.

Derivation

We will use the formula for an infinite series of an exponent:

e(q) =

∞ ∑ n=0

(q)n n!

So multiplying two of these together we get:

e(a) *e(b) =

∞ ∑ n=0

(a)n n!

∞ ∑ m=0

(b)m m!

Now we want to move the second summation outside the first part, but since the first part has already been summed 'n' times we need to substitute n with n-m to correct for that:

e(a) *e(b) =

∞ ∑ n=0

∞ ∑ m=0

(a)(n-m) (n-m)!

(b)m m!

I don't think what we have done so far has implied any commutation because the operands are still in the same order.

 

Prerequisites

On this page we discussed how to represent and evaluate functions where the domain and codomain are complex numbers.