BCH (Baker-Campbell-Hausdorff) formula for $[X,Y]=xY-yX$

Question

If some $X$ and $Y$ satisfy the commutation relation $[X,Y]=XY-YX=xY-yX$, where $x$ and $y$ are numbers (or commute mutually and with $X$ and $Y$), then what is the closed form of $\ln(\exp X \exp Y)$?

Quick calcualtion shows that $[X[X,Y]]=x[X,Y]$ and $[Y[X,Y]]=y[X,Y]$, from which follows that $\mathrm{ad}_X^n\mathrm{ad}_Y^m[X,Y] =x^ny^m[X,Y]$, so the result should be something like: \begin{equation} \ln(\exp X \exp Y)=X+Y+f(x,y)[X,Y] \end{equation}

Next step would be using this result in BCH formula, but it's rather complicated and readily available resources were useless because they mostly focus only on the first terms in expansion and examples with vanishing commutators after some order, so this is where I've got stuck.

Thanks for reply.

Answer 1

It turns out that a paper that answers this question has been uploaded recently on arXiv. The answer is immediately visible in abstract:

If \begin{equation} [X,Y]=xY-yX \end{equation} then \begin{equation} \ln(e^Xe^Y)=X+Y+ \frac {(ye^{-y}+xe^x)-(x+y)e^{x-y}} {xy(e^x-e^{-y})} (xY-yX) \end{equation} (if I didn't miss some sign while applying the formula from the abstract)

It seems that those kinds of identities are most easily obtained by finding some nice simple matrices with the same commutation relations and the result for a particular matrix representation will also hold generally for anything with the same commutation relations.

Answer 2

Yes, the nice simple matrices wished for in the OP's answer are, in fact, well-known 2x2 ones, leading to the answer directly without any machinery or technique.

If you wanted a name, the group is the 1-d affine group ($r\mapsto cr+d$), that of 2x2 upper triangular matrices with unit lower diagonal entry.

Observe first that the group structure is a familiar one of dilations, given the redefinitions $A\equiv X$ and $B\equiv Y -(y/x) X$, so that $$ [A,B]=xB, $$ so $A=z\partial_z$ and $B=z^x$, so $X=z\partial_z$ and $Y= z^x+(y/x)~z\partial_z$, so dilation maneuvers can be brought to bear.

However, the frontal, direct, approach is straightforward. Observe that $$ A=\begin{pmatrix}      x&0\\      0&0    \end{pmatrix} , \qquad B=\begin{pmatrix}      0&1\\      0&0    \end{pmatrix} , $$ satisfy the algebra, hence $$ X=\begin{pmatrix}      x&0\\      0&0    \end{pmatrix} , \qquad Y=\begin{pmatrix}      y&1\\      0&0    \end{pmatrix} , $$ so that $Y^n=y^{n-1} Y$, and consequently $$ e^X e^Y=\begin{pmatrix}      e^x&0\\      0&1    \end{pmatrix} \begin{pmatrix}      e^y&\frac{e^y-1}{y}\\      0&1    \end{pmatrix}= \begin{pmatrix}      e^{x+y}&\frac{e^x(e^y-1)}{y}\\      0&1    \end{pmatrix}~. $$

We just compare this to $$ e^{aX+bY}=e^Z=\exp \left ( \begin{matrix}      e^{ax+by}&b\\      0&0    \end{matrix}\right )= \begin{pmatrix}      e^{ax+by}&\frac{b(e^{ax+by}-1)}{ax+by}\\      0&1    \end{pmatrix}, $$ the last step by virtue of $Z^n=(ax+by)^{n-1} Z$.

Hence $$ x+y=ax+by, \qquad (ax+by) e^x(e^y-1)=by(e^{ax+by}-1), $$ which directly determine $$ a=\frac{(x+y)(e^x-1)}{x(e^{x+y}-1)}, \qquad b=\frac{(x+y)(e^y-1)e^x}{y(e^{x+y}-1)}, $$ as you determined.