# Displacement operator action on exponential of ladder operators

How do I evaluate the following: $$D_{\alpha}^{-1} e^{-i\Omega aa^{+}t} D_{\alpha},$$ where $D_a$ is the usual displacement operator for coherent states $$D_{\alpha} = e^{\alpha a^{+} - \alpha^* a}.$$

I have calculated that $$D_{\alpha}^{-1} aa^{+} D_{\alpha} = (a+\alpha)(a^+ + \alpha^*)$$ but I confused how to take the next step to the exponential. Any help would be appreciated, thanks!

• The Baker-Campbell-Hausdorff identity might be useful here: en.wikipedia.org/wiki/… – probably_someone Jun 28 '18 at 3:23
• Is there a direct way to go from the formula for the product of ladder operators to that of the exponent of the product? – onknc Jun 28 '18 at 3:39

The exponential is just $e^{x} = \sum_{n} \frac{1}{n!} x^{n}$ therefore $De^{x} D^{-1} = \sum_{n} \frac{1}{n!} Dx^{n}D^{-1} = \sum_{n} \frac{1}{n!} (DxD^{-1})^{n}$ because $D$ has an inverse.
Since you know $D^{-1}_{\alpha} a a^{\dagger} D_{\alpha} = (a + \alpha)(a^{\dagger}+\alpha^{*})$ and $D_{\alpha} = e^{\alpha a^{\dagger}-\alpha^{*} a}$ always has an inverse then $D^{-1}_{\alpha} e^{-i\Omega a a^{\dagger} t} D_{\alpha} = e^{-i\Omega D^{-1}_{\alpha} a a^{\dagger} D_{\alpha}}$ and you just substitute in the solution.
In fact if you'd worked out $D^{-1}_{\alpha} a D_{\alpha}$ and $D^{-1}_{\alpha} a^{\dagger} D_{\alpha}$ then you could have just substituted these in place of $a$ and $a^{\dagger}$. For this problem you don't need to do this, however if you'd been looking at a term like $a a a^{\dagger} a^{\dagger}$ then you could just make the same substitutions and solve it the same way as the original problem.
• I don't see how that is relevant, I never made the assumption that $e^{A+B} = e^{A}e^{B}$ instead what I said was $e^{A} e^{B} e^{-A} = e^{e^{A} B e^{-A}}$ which is why I pointed out that $D$ needed to have an inverse (to be precise a right inverse). The only place where I can see the Baker-Campbell-Hausdorff formula as having relevance is in computing $D^{-1}_{\alpha} a a^{\dagger} D_{\alpha}$ – N A McMahon Jun 28 '18 at 4:16