Closed form of Baker Campbell Hausdorff theorem with cyclic bracket structure I would like to know if there exists a closed form of the Baker Campbell Hausdorff theorem subject to the conditions that $[x,[x,y]] \sim x$ and $[y,[x,y]] \sim y$. 
The simple cases that I know a closed form exists are when $[x,y]$ is a scalar, then the expansion truncates as 
$\log(e^x e^y) = x + y + \frac{1}{2}[x,y]$ 
and if $[x,y] = sy$ for some constant $s$, then we have
$\log(e^x e^y) = x + \frac{sy}{1-e^{-s}}$.
These lead me to believe that there should be some closed form since I know the higher order terms in the BCH expansion are proportional to $x$ and $y$.
My specific problem is trying to apply these with $x = a \frac{d^2}{dp^2} $ and $y = b p^2$ for some (possibly complex) constants $a$ and $b$. 
The commutator is 
$[x,y] = ab(4 p \frac{d}{dp} + 2)$
The nested commutators are 
$[x,[x,y]] = 8a^2b\frac{d^2}{dp^2}$ and
$[y,[x,y]] = -8ab^2p^2 $.
Indeed these are proportional to the original x and y. Is there a known closed form of the BCH theorem for this example? 
Thanks in advance!
 A: You are dealing with a realization of su(2), standard in CFT and stringery. 
Define 
$$
\sigma_x=i(x+y)\\
\sigma_y=x-y\\
\sigma_z=-[x,y]/(4ab),
$$
so that 
$$
    \left[\sigma_x, \sigma_y\right] = 2i\sigma_z \, \\
    \left[\sigma_y, \sigma_z\right] = 2i\sigma_x \, \\
    \left[\sigma_z, \sigma_x\right] = 2i\sigma_y  ~,
$$ 
recognizable as the commutation relations of su(2), realized by the faithful representation of the Pauli 2x2 matrices,
$$
\begin{align}
   \sigma_x &=
    \begin{pmatrix}
      0&1\\
      1&0
    \end{pmatrix} \\
 \sigma_y &=
    \begin{pmatrix}
      0&-i\\
      i&0
    \end{pmatrix} \\
  \sigma_z &=
    \begin{pmatrix}
      1&0\\
      0&-1
    \end{pmatrix} \,.
\end{align}
$$
Because they are so easy to manipulate and exponentiate, they serve to instantly produce the generic composition law of group elements of the group SU(2),
$$
\begin{align}
  e^{i a(\hat{n} \cdot \vec{\sigma})} e^{i b(\hat{m} \cdot \vec{\sigma})}
    =& I(\cos a \cos b - \hat{n} \cdot\hat{m} \sin a \sin b) \\
&+ i(\hat{n} \sin a \cos b + \hat{m}  \sin b \cos a - \hat{n}\times\hat{m} ~ \sin a \sin b  )\cdot  \vec{\sigma } \\
    =& I\cos{c} + i (\hat{k} \cdot \vec{\sigma}) \sin{c} \\
    =& e^{i c \left(\hat{k} \cdot \vec{\sigma}\right)},
\end{align}
$$
where the basic spherical law of cosines, $\cos c = \cos a \cos b - \hat{n} \cdot\hat{m} \sin a \sin b~$ yields c, so that 
$$\hat{k} = \frac{1}{\sin c}\left(\hat{n} \sin a \cos b + \hat{m} \sin b \cos a - \hat{n}\times\hat{m}  \sin a \sin b\right) ~. $$
You only need carefully normalize the coefficients (angles) and unit vectors involved in arbitrary expressions you'd like to CBH-compose.
Since these are group identities, the combinatorics is identical for any faithful representation, so you obtained the answer in the 2x2 Pauli matrix one, but it will also, then, mutatis mutandis, hold for your differential operators. 
