Kraus Operator in Fock basis

[Muthumanimaran]

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.

 

[eljose79]

Hi there,

To get to equation (10), we can use the Wigner-Weisskopf approximation (eqn (9)) and plug it into the expression for $A_{m,n}^{k}(t)$ (eqn (8)). This gives us:

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

$$={\sum_{\{k_i\}}^{k}}' \langle{m}|\langle{\{k_i\}}|\left(u(t)b^{\dagger}(0)+\sum_{i=1}^{\infty}v_{i}(t){b_{i}}^{\dagger}(0)\right)^n|\{0\}\rangle|n\rangle$$

$$={\sum_{\{k_i\}}^{k}}' \langle{m}|\langle{\{k_i\}}|\left(u(t)b^{\dagger}(0)+\sum_{i=1}^{\infty}v_{i}(t){b_{i}}^{\dagger}(0)\right)^n\frac{[b^{\dagger}(-t)]^n}{\sqrt{n!}}|\{0\}\rangle|0\rangle$$

$$={\sum_{\{k_i\}}^{k}}' \langle{m}|\langle{\{k_i\}}|\frac{[b^{\dagger}(-t)]^n}{\sqrt{n!}}|\{0\}\rangle|0\rangle$$

$$=\frac{1}{\sqrt{n!}}{\sum_{\{k_i\}}^{k}}' \langle{m}|\langle{\{k_i\}}|[b^{\dagger}(-t)]^n|\{0\}\rangle|0\rangle$$

$$=\frac{1}{\sqrt{n!}}\langle{m}|\langle{\{k_i\}}|[b^{\dagger}(-t)]^n|\{0\}\rangle|0\rangle$$

$$=\frac{1}{\sqrt{n!}}\langle{m}|\langle{\{k_i\}}|b^{\dagger