Say that an operator is given by $$D(\alpha) = \exp(\alpha a^{\dagger} - \alpha^{*}a) = \sum_{n=0} ^ {\infty} (n!)^{-1}(\alpha a^{\dagger} - \alpha^{*}a)^n.$$

Define $\{(a^{\dagger})^m a^n\}$ as the average of the ${n+m \choose m} = {n+m \choose n}$ differently ordered terms.

For example, $$\{a^{\dagger}a^2\} = \frac{a^{\dagger}a^2 + a a^{\dagger} a + a^2 a^{\dagger}}{3}.$$

How do I express $D(\alpha)$ in a neat, symmetric form (where $a$ and $a^{\dagger}$ are on equal footing)? For example $$D(\alpha) = \sum_{n, m=0} ^ \infty \frac{\alpha^n(-\alpha^*) ^{m}}{n!m!} \{(a^{\dagger})^n a^m\}.$$

I am not sure how to do a binomial expansion of the first equation. There are posts which explain how to do it, for example here, but I am not sure how to use on of the answers to proceed.

  • $\begingroup$ You are probably talking about the analog of Weyl ordering for creation and annihilation operators. There are powerful ways to do this, and I'd be shocked if no quantum optics text had that, but a tour-de-force of combinatorics is not a pathway to that goal. $\endgroup$ Jan 26, 2022 at 16:22
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    $\begingroup$ It is assumed you have read up. $\endgroup$ Jan 26, 2022 at 16:29

1 Answer 1


The best way to treat them on equal footing, IMO, is to split the displacement operator in the following way $$D(\alpha) = \exp(\alpha a^{\dagger} - \alpha^* a) = \frac{e^{-|\alpha|^2/2}e^{\alpha a^{\dagger}}e^{-\alpha^* a} +e^{|\alpha|^2/2}e^{-\alpha^* a}e^{\alpha a^{\dagger}}}{2}$$ where one uses Baker-Campbell-Hausdorff formula. Now you can expand in a power series and match terms with identical powers of $a$ and $a^{\dagger}$ from the two contributions.


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