The vacuum has zero momentum and is invariant under translations. Hence
$$
e^{-i\mathbf P\cdot\mathbf y}|\mathbf q\rangle
=e^{-i\mathbf P\cdot\mathbf y}a_{\mathbf q}^\dagger
 e^{i\mathbf P\cdot\mathbf y}|0\rangle
=\boxed{e^{-i\mathbf q\cdot\mathbf y}|\mathbf q\rangle}.
$$
Thus a one-particle <momentum eigenstate> remains the same ray and acquires the translation phase appropriate to its momentum.

Solution

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

 Solution

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