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Given the characteristic function defined as: $$\chi(\beta)=\text{tr}[\rho D(\beta)],$$ with $D(\alpha)=e^{\alpha a^\dagger-\bar\alpha a}$ the displacement operator. Is it possible that for some $\rho$ we have $$\chi(\beta)=\textbf{1}_A,$$ for $\textbf{1}_A$ the indicator function of some area $A$ in phase-space?

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    $\begingroup$ I'd guess not? Consider that the Wigner at $\alpha=0$ reads $W(0)=\frac1{\pi^2}\int d^2\eta\chi(\eta)=|A|/\pi^2$, and so for large enough $A$ the Wigner wouldn't be normalised $\endgroup$
    – glS
    Commented May 10 at 7:06
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    $\begingroup$ Actually it is normalized given that the integral of the Wigner in all its domain is exactly the value of the characteristic function at 0, so as far as the area contains the 0 is normalized. $\endgroup$
    – Nicolas Medina Sanchez
    Commented May 10 at 11:38
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    $\begingroup$ There would be a limit to the size of $A$, because that maximum amplitude of a Wigner function is bounded. $\endgroup$
    – flippiefanus
    Commented Jun 8 at 4:16

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