Writing normally ordered operator $a^{\dagger n}a^m$ in terms of anti-normal order I am looking for a formula to write a normally ordered operator ${a}^{\dagger n} a^m$ as a function of  anti-normally ordered operators $a^i {a}^{\dagger j}$. Any help with this would be much appreciated. 
 A: There are far too many expressions around to know which one you'd find the most useful. It may be that your starting expression is just a generic term in an expansion of something easier, in which case you'd end up reduplicating steps. 
The standard generic technique is relying on generating functions:
$$
a^{\dagger ~ n } a^m=\left. \partial_\sigma ^n \partial_\tau ^m \left ( e^{\sigma a^\dagger}  e^{\tau a}\right ) ~~  \right|_{\sigma=\tau=0} .
$$
But you know from the lowest order (degenerate) CBH identity that 
$$
e^{\sigma a^\dagger}  e^{\tau a} =    e^{\sigma \tau [a^\dagger, a]}  e^{\tau a}  e^{\sigma a^\dagger} =  e^{-\sigma \tau }  e^{\tau a}  e^{\sigma a^\dagger} , 
$$
so that your original expression is anti normal ordered by 
$$
a^{\dagger ~ n } a^m= \left.\partial_\sigma ^n \partial_\tau ^m \left ( e^{-\sigma \tau  }  e^{\tau a}  e^{\sigma a^\dagger}  \right ) ~~ \right|_{\sigma=\tau=0} .
$$


*

*Confirm to your satisfaction how this works for, e.g., n=1, m=3, $$ 
 \left.\partial_\sigma  \partial_\tau ^3 \left ( e^{-\sigma \tau  }  e^{\tau a}  e^{\sigma a^\dagger}  \right ) ~~ \right|_{\sigma=\tau=0} = \left.  \partial_\tau ^3 \left (    e^{\tau a}   ( a^\dagger-\tau)  \right ) ~~ \right|_{\tau=0}= a^3 a^\dagger -3a^2.$$ This never involves expanding exponentials, as you fear--merely the eigenfunction property of exponentials  w.r.t. derivation.


For an arbitrary normally ordered operator, substitute $a^\dagger \mapsto \partial_\sigma, \quad a \mapsto \partial_\tau$, and repeat the maneuver. Experiment with simple functions, not mere monomials, to see how it works.
