exp X exp Y equals exp (X plus Y plus a half X brackets Y plus dot dot dot) where the dots indicate higher order terms that can all be written in terms of iterated brackets of X and Y.
The BakerCampbellHausdorff formula
The BakerCampbellHausdorff formula
In the last video, we found a local inverse for the matrix exponential exp from little g l n R to big G L n R. In other words, we found a neighbourhood U of the zero matrix in little g l n R and a neighbourhood V of the identity in big G L n R such that exp restricted to U from U to V is a bijection whose inverse we called log from V to U.
We also wrote down a power series expansion: log identity plus M, equals M minus a half M squared plus a third M cubed minus dot dot dot. valid for small M (as V is a neighbourhood of the identity, anything in V can be written as I plus M for some small M).
If X Y = Y X then we saw that exp X times exp Y equals exp (X plus Y). What happens if X Y is not equal to Y X? We can figure out a formula for log of exp X exp Y using our power series for \log.
Let's compute. log exp X exp Y equals log of (I plus X plus a half X squared plus dot dot dot) times (I plus Y plus a half Y squared plus dot dot dot). Multiplying the power series together, we get log of I plus X plus Y plus X Y plus a half X squared plus a half Y squared plus a half X Y squared plus a half X squared Y plus dot dot dot. Taking M to be X plus Y plus X Y plus a half X squared plus a half Y squared plus dot dot dot and using our power series for log I plus M we get M minus a half M squared plus a third M cubed + dot dot dot equals X plus Y plus X Y plus a half X squared plus a half Y squared plus dot dot dot, minus a half M squared which is X squared plus X Y plus Y X plus Y squared plus dot dot dot where the dots stand for terms which are at least cubic in X and Y. For now we'll ignore all cubic and higher order terms.
This simplifies to give log exp X exp Y equals X plus Y plus a half X Y minus Y X plus cubic and higher order terms
This quantity X Y minus Y X is going to turn out to be very important in this course. It's called the
We're now saying that log exp X exp Y is:

to first order just X plus Y,

has a second order correction term X brackets Y, which vanishes if X and Y commute,
and remarkably, all the cubic and higher order correction terms can be expressed as iterated commutators of X and Y. For example, the cubic term is one over 12 X brackets (X brackets Y) minus one over 12 Y brackets (X brackets Y). This fact that the higher order terms can be written in this way is genuinely miraculous and not at all obvious: proving it is one of the possible projects you could do.
You can write a general formula (due to Dynkin) for the nth order term, but this is not very illuminating. We will state the existence of this formula as a theorem:
The main point of the BakerCampbellHausdorff formula is that the group multiplication of the invertible matrices exp X and exp X in big G L n R is determined by the bracket operation (on little g l n R).
By keeping track of the cubic terms, check that the cubic term in the BakerCampbellHausdorff formula is one over 12 X brackets (X brackets Y) minus one over 12 Y brackets (X brackets Y).
Preclass exercise
Verify the BakerCampbellHausdorff formula for exp of the 3by3 matrix 0, a_1, c_1; 0, 0, b_1; 0, 0, 0 times exp of the 3by3 matrix 0, a_2, c_2; 0, 0, b_2; 0, 0, 0. [Hint: In this case, the formula is \exp\left(X+Y+\frac{1}{2}[X,Y]\right) with no further correction terms. Can you see why?]