Exponentiation of linear combination of commuting Vector fields

Question

I have to prove the formula: $$e^{a\partial/ \partial\lambda +b \partial / \partial\mu}=e^{a\partial/ \partial\lambda}e^{b\partial/ \partial\mu}$$ if $\partial/ \partial\lambda$ and $\partial/ \partial\mu$ commute This is I think is a result of the Baker-Campbell-Hausdorff formula. But how can I prove this? I thought of doing this with the taylor expansion of the exponential. I worked it out till second order, but isn't there a more elegant way? Is it possible to do this by using the binomial coefficient? The problem when using that is that the operators don't commute in general and the order matters.

Answer 1

BCH is overkill here. Since $\partial_\lambda$ commutes with $\partial_\mu$, any arbitrary product of $\partial_\lambda$s and $\partial_\mu$s can be rearranged to place the former on the left, and$$\begin{align}e^{a\partial_\lambda+b\partial_\mu}&=\sum_{n\ge0}\frac{1}{n!}(a\partial_\lambda+b\partial_\mu)^n\\&=\sum_n\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}(a\partial_\lambda)^k(b\partial_\mu)^{n-l}\\&=\sum_{k,\,\ell\ge0}\frac{(a\partial_\lambda)^k(b\partial_\mu)^\ell}{k!\ell!}\\&=\sum_k\frac{(a\partial_\lambda)^k}{k!}\sum_\ell\frac{(b\partial_\mu)^\ell}{\ell!}\\&=e^{a\partial_\lambda}e^{b\partial_\mu}.\end{align}$$