quantum information - How can one get the density operator from the characteristic function?

To solve analytically the master equation of two qubits interacting with a cavity mode through their environment we use the charactristic function,
$$\chi (\beta)=\operatorname{tr}[\rho D(\beta)],$$ with $D(\beta)$ being the displacement operator.

How do we get the $\rho$ from the $\chi$ expression?

Answer

The fundamental reconstruction theorem in phase-space quantum mechanics inverts the Wigner transform of an arbitrary operator $\hat G$ through the Weyl transform, $$ \hat G= \frac {\hbar}{2\pi} \int\!\! d\tau d\sigma~~ e^{i(\tau \hat p + \sigma \hat x)} ~~\operatorname{tr} [\hat G~ e^{-i(\tau \hat p + \sigma \hat x) } ] . $$ (If you blink in a right-brain hemisphere vision, your might dream of a collapsing 2d operator δ-function there, but don't worry if this does not evoke anything. The trivial formal proof is in our Concise Treatise of Quantum Mechanics in Phase Space, Ch 0.18. It first appeared in Groenewold's 1946 breath-taking dissertation.)

Mindful of the connection to optical phase space, $$ \hat D(\beta)= e^{\beta a^\dagger -\beta ^* a}= e^{\hat x (\beta -\beta^*)/\sqrt{2}-i\hat p (\beta + \beta^*)/\sqrt{2}}, $$ and using $\hat ρ$ as your operator, you translate this to a linear combination of translation operators, your building blocks, $$ \hat\rho= \int \frac{d^2\beta }{\pi} ~~ \hat D(-\beta) ~~ \operatorname{tr} [ \hat\rho \hat D(\beta) ]= \int \frac{d^2\beta }{\pi} ~~ \hat D(-\beta) ~~ \chi(\beta) . $$