# What are the eigenstates of the Displacement operator?

I know that the displacement operator:

$$\hat{D}(\alpha)=e^{\alpha \hat{a}^{\dagger}-\alpha^*\hat{a}}$$

acts on the vacuum as: $$\hat{D}(\alpha) \vert 0\rangle =\vert \alpha\rangle$$

But what are the eigenstates $$\vert \Psi \rangle$$ of $$\hat{D}(\alpha)$$, such that:

$$\hat{D}(\alpha)\vert \Psi \rangle = \lambda (\alpha)\vert \Psi \rangle \ ?$$

• If $\alpha=1$ is the momentum basis, if $\alpha=i$ is the position basis Mar 10, 2023 at 17:05

Displacements shift a state's quasiprobability distribution by $$\alpha$$ in phase space. This question is thus equivalent to asking what phase-space distributions are unchanged by shifts $$\alpha$$.
One answer is a state whose quasiprobability distribution is a straight line pointing in the same direction as $$\alpha$$. Such a state is an infinitely squeezed state, with the squeezing in direction perpendicular to $$\alpha$$. For example, a position eigenstate is infinitely squeezed in terms of its position, so it's momentum distribution is a flat line, thus being unchanged by displacements in momentum.
Another answer is a state whose phase-space structure is repeated periodically in $$\alpha$$. These are things like GKP states, which are explicitly constructed as superpositions of coherent states spaced on an infinite grid $$\sum_{s,t=-\infty}^\infty \exp(-i s \hat{p}\alpha)\exp(2\pi i t \hat{x}/\alpha)|\mathrm{vac}\rangle.$$