exponential and Lie bracket

Question

I've been reading that one can compute a form of the Baker-Campbell formula with direct computations through the power series of $\exp(X)$, $\exp(Y)$ and $\log(1+W)$, where $X$, $Y$ are non commuting variables and $W=\exp(X)\exp(Y)-1$. During the calculations for the first three terms i came across the following $$X^2+2XY+Y^2=(X+Y)^2+[X,Y]$$ $$X^3+3X^2Y+3XY^2+Y^3=(X+Y)^3+2X[X,Y]+2[X,Y]Y+[X,Y]X+Y[X,Y]$$ where [X,Y] denotes the Lie Bracket. So, i was wandering if there is a way to express $X^4+4X^3Y+6X^2Y^2+4XY^3+Y^3$ in a similar way (with brackets), or generally every expression of the form $\sum_{n=0}^{m}\binom{m}{n}X^{m-n}Y^{n}$.

Answer 1

Sure. One way I like to think about commutators (Lie brackets) is

"if you have $YX$, and you'd rather have $XY$, you have to add $[Y,X]$", i.e.

$$ YX = XY + [Y,X]. $$

So, when you expand a binomial of the form $(X + Y)^n$, you're going to get a lot of terms that aren't 'in the right order', and you have to add commutators if you want to swap things around. That is,

$$ (X + Y)^3 = X^3 + XXY + YXX + XYX + \dots $$

and so if you want an expression of the form $$ X^3 + 3XXY + \dots $$

you're going to have to add $[Y,X]X$ and $X[Y,X]$ to 'fix' the first term, and another $X[Y,X]$ to fix the second.

So, in general, when you expand $(X+Y)^n$, you're going to get all the terms in the usual binomial expansion, just with the $X$'s and $Y$'s in a different order, which you can straighten out by introducing enough brackets. I can't tell you off the top of my head what the closed form would be, but there's probably a straightforward combinatorial approach.