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A Kraus Operator in Fock basis

  1. Dec 28, 2016 #1
    The Kraus operator is defined as,
    $$A_{k}(t)={\sum_{\{k_i\}}^{k}}'\langle\{k_i\}|U(t)|\{0\}\rangle$$

    is given in eqn(5) in the [Arxiv link](https://arxiv.org/pdf/quant-ph/0407263.pdf)

    the matrix representation of $A_k(t)$ is given in eqn (7) as

    $$A_k(t)=\sum_{m,n}A_{m,n}^{k}(t)|m\rangle\langle{n}|$$

    with

    $$A_{m,n}^{k}(t)={\sum_{\{k_i\}}^{k}}' \langle{m}|\langle{\{k_i\}}|U(t)|\{0\}\rangle|n\rangle$$ given in eqn(8)

    Using Wigner-Weisskopf approximation defined in eqn(9),
    $$b^{\dagger}(t)=u(t)b^{\dagger}(0)+\sum_{i=1}^{\infty}v_{i}(t){b_{i}}^{\dagger}(0)$$ It is said that $|U(t)|\{0\}\rangle|n\rangle$ can be written as $\frac{[b^{\dagger}(-t)]^n}{\sqrt{n!}}|\{0\}\rangle|0\rangle$, But I don't know how to proceed from here to get eqn (10), I tried to expand $b^{\dagger}(t)$ binomially and acted each term on $|\{0\}\rangle|0\rangle$ but it becomes clumsy. I don't know whether approach is correct or not? I need a hint to proceed with the problem.
     
  2. jcsd
  3. Jan 2, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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