The Baker-Campbell-Hausdorff formula

The Baker-Campbell-Hausdorff formula

In the last video, we found a local inverse for the matrix exponential exp : 𝔤 𝔩 ( n , 𝐑 ) G L ( n , 𝐑 ) . In other words, we found a neighbourhood U of 0 𝔤 𝔩 ( n , 𝐑 ) and a neighbourhood V of I G L ( n , 𝐑 ) such that exp | U : U V is a bijection whose inverse we called log : V U .

We also wrote down a power series expansion: log ( I + M ) = M - 1 2 M 2 + 1 3 M 3 - valid for small M (as V is a neighbourhood of the identity, anything in V can be written as I + M for some small M ).

If X Y = Y X then we saw that exp ( X ) exp ( Y ) = exp ( X + Y ) . What happens if X Y Y X ? We can figure out a formula for log exp ( X ) exp ( Y ) using our power series for log .

Let's compute. log ( exp ( X ) exp ( Y ) ) = log ( ( I + X + X 2 2 + ) ( I + Y + Y 2 2 + ) ) . Multiplying the power series together, we get log ( I + X + Y + X Y + X 2 2 + Y 2 2 + X Y 2 2 + X 2 Y 2 + ) . Taking M = X + Y + X Y + X 2 2 + Y 2 2 + and using our power series for log ( I + M ) we get M - 1 2 M 2 + 1 3 M 3 + = X + Y + X Y + 1 2 X 2 + 1 2 Y 2 + - 1 2 ( X 2 + X Y + Y X + Y 2 + ) + 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 ) = X + Y + 1 2 ( X Y - Y X ) +

This quantity X Y - Y X is going to turn out to be very important in this course. It's called the commutator bracket or Lie bracket of X and Y and is written [ X , Y ] . This vanishes if X and Y commute.

We're now saying that log ( exp ( X ) exp ( Y ) ) is:

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 1 12 [ X , [ X , Y ] ] - 1 12 [ Y , [ X , 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 n th order term, but this is not very illuminating. We will state the existence of this formula as a theorem:

Theorem (Baker-Campbell-Hausdorff formula):

exp ( X ) exp ( Y ) = exp ( X + Y + 1 2 [ X , Y ] + ) where the dots indicate higher order terms that can all be written in terms of iterated brackets of X and Y .

The main point of the Baker-Campbell-Hausdorff formula is that the group multiplication of the invertible matrices exp ( X ) and exp ( Y ) in G L ( n , 𝐑 ) is determined by the bracket operation (on 𝔤 𝔩 ( n , 𝐑 ) ).

Exercise:

By keeping track of the cubic terms, check that the cubic term in the Baker-Campbell-Hausdorff formula is 1 12 [ X , [ X , Y ] ] - 1 12 [ Y , [ X , Y ] ] .

Pre-class exercise

Exercise:

Verify the Baker-Campbell-Hausdorff formula for exp ( 0 a 1 c 1 0 0 b 1 0 0 0 ) exp ( 0 a 2 c 2 0 0 b 2 0 0 0 ) . [Hint: In this case, the formula is exp ( X + Y + 1 2 [ X , Y ] ) with no further correction terms. Can you see why?]