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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.

glS
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bkphysics
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1 Answers1

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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.

Cosmas Zachos
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