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:

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

  • 3
    $\begingroup$ If $\alpha=1$ is the momentum basis, if $\alpha=i$ is the position basis $\endgroup$
    – Mauricio
    Mar 10, 2023 at 17:05

1 Answer 1


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.

It is easy to geometrically construct other answers but harder to show that they correspond to pure states. For example, a quasiprobability distribution that is a series of parallel lines would suffice. But one cannot attain this by simply taking a superposition of infinitely squeezed states, because that would imply that any superposition of position eigenstates will be unchanged by displacements in momentum, which must be false. The quasiprobability distribution that is a series of parallel lines is really a mixture of infinitely squeezed states, not a superposition thereof.

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 $$\sum_{s,t=-\infty}^\infty \exp(-i s \hat{p}\alpha)\exp(2\pi i t \hat{x}/\alpha)|\mathrm{vac}\rangle.$$

The common theme is that infinities are necessary: the quasiprobability distribution cannot be localized; only a state whose quasiprobability distribution has infinite support can be an eigenstate of a displacement operator.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.