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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.

Níckolas Alves
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Bard
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1 Answers1

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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.