Solution
= Solution
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.