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I read a paper on open quantum system, it's about non-Markovian process with memory effects. They describe a generic model of two qubits interacting with correlated multimode field.

They describe the interaction Hamiltonian as:

$H_i=\Sigma_k\sigma_z^i(g_kb{_k}^{i\dagger}+g_k^*b_k^{i})$,

and describe the unitary matrix of the time evolution as:

$U_i(t)=e^{\sigma_z^i\Sigma_k(b{_k}^{i\dagger}\xi_k(t_i)-b_k^{i}\xi_k^*(t_i))}$

and say that the U matrix work like this:

$U_i(t)|0\rangle\otimes|\eta\rangle=|0\rangle\otimes_kD(-\xi_k(t_i))|\eta\rangle$,

where $D(\xi_k)$ is the displacement operator.

It looks like the U matrix is a displacement operator, is there a reason for that? I see that the displacement operator is a change in the phase space, is this relevant for this?

Additionally, after this, they define the state with characteristics function and say that is a Gaussian state.

I want to ask why they use the displacement operator, and how it is related to characteristic function. Is there a connection between these two? I don't understand all the equations here, and why they want to describe the system like this.

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    $\begingroup$ I think people use the displacement operator because it is very common in / familiar to quantum optics. From the vacuum state, the displacement operator $D(\xi)$ prepares a coherent state with $\langle n \rangle = |\xi|^2$ photons on average. The variance of $n$ is also $|\xi|^2$. So this representation tells an expert a lot of information. It comes up in many settings. Also, is $\xi$ the "characteristic function" to which you refer? $\endgroup$
    – just a phase
    Commented Mar 26 at 0:25
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    $\begingroup$ @justaphase thanks! The characteristic function is defined with $\chi$ and not $\xi$. I updated the question, and added link to the paper. $\endgroup$
    – Yohay Halfon
    Commented Mar 26 at 9:27
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    $\begingroup$ Sorry for the late reply. I guess the charateristic function $\chi (\lambda)$ is like the expectation value of a displacement operator $D(\alpha)$ with $\alpha = i \lambda$. Basically, $\chi$ is defined to be $$\chi (\lambda) = \text{tr} \left[ \exp [i (\lambda a^\dagger + \lambda^* a) ] \rho \right] = \text{Tr} \left[ D ( i \xi ) \rho \right],$$since the displacement operator is $$D(\alpha) = \exp (\alpha a^\dagger - \alpha^* a )$$. $\endgroup$
    – just a phase
    Commented Mar 29 at 5:02

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