# A form of the Baker-Hausdorff equation

I wonder how many different ways are there of writing the Baker-Hausdorff equation! This is a form which I recently encountered and haven't been able to figure out how it comes,

$e^ae^Xe^b = e^{[X+L_X.(b-a)+(L_X \coth L_X).(b+a)+..]}$

where $L_X$ is an operator which acts on another matrix say $f$ as,

$L_X.f = \frac{1}{2}[X,f]$

I would be grateful if someone can help with the above.
(and motivate it)

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Help you with what, exactly? – Mariano Suárez-Alvarez Mar 1 '11 at 23:14
@Mariano If one can help derive this equality and hence show equivalence with the more usual expression. And if it is possible to explain under what situations does one chooses to think of it in the above way and not the usual expression. – Anirbit Mar 2 '11 at 6:46
$L_x$ seems to be an $\mathcal{ad}_X/2$ operator. But what is $(L_X \coth L_X)$? – draks ... Jul 16 '12 at 22:23
@draks Given usual notations I would thinks it is the product of the two operators $L_X$ and $coth L_X$. (...interesting that someone should discover this question after more than a year!..) – Anirbit Jul 18 '12 at 21:19
@Anirbit so $coth L_x$ maybe defined via a power series? (I was looking for some lost pearls...) – draks ... Jul 19 '12 at 17:32