The iterated commutators are
$$
[-i\mathbf P\cdot\mathbf y,a_{\mathbf q}^\dagger]
=-i\mathbf q\cdot\mathbf y\,a_{\mathbf q}^\dagger,
$$
so the <Baker--Campbell--Hausdorff formula> gives
$$
\boxed{e^{-i\mathbf P\cdot\mathbf y}a_{\mathbf q}^\dagger
 e^{i\mathbf P\cdot\mathbf y}
=e^{-i\mathbf q\cdot\mathbf y}a_{\mathbf q}^\dagger.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 1 / ii / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301/1/ii/c/solution