This question consists of the basic calculations which are often referred to as "trivial" in most references in circuit-QED. However, are not so trivial for anyone who starts fresh into the quantum optics involved for studying the input/output (I/O), and therefore the measurements in actual architecture of QC. (With some quality answer this can become a resource for starters to refer to). The starting point is to study the Quantum Langevin Equations from some 'rare' quantum optics texts. The notation is standard:
$\Omega:$ Cavity Frequency
$\Gamma$: Dissipation/decay rate
$b_{\text{in/out}}$: Input/Output field operators
$a$ & $a^\dagger$: Cavity modes
The solution to the Langevin equation reads:
$a(t)=a(0)\text{e}^{-(i\Omega +\Gamma/2)}-\sqrt{\Gamma}\int_0^tds$ $\text{e}^{-(i\Omega+\Gamma/2)(t-s)}b_{\text{in}}(s)$.
(1) Prove the commutators:
$[a^\dagger(t),b_{\text{in}}(t')]=0 \quad$ for $t<t'$
$[a^\dagger(t),b_{\text{in}}(t)]=-\frac{1}{2}\sqrt{\Gamma} \quad$
$[a^\dagger(t'),b_{\text{in}}(t')]=-\sqrt{\Gamma}\text{e}^{-(i\Omega+\Gamma/2)(t-t')} \quad$ for $t>t'$
$[a^\dagger(t),a^\dagger(s)]=\text{e}^{-\frac{1}{2}\Gamma|t-s|}$
(2) Given the time reversed Langevin equation
$\dot{a}(t)=(-i\Omega+\Gamma/2)a(t)-\sqrt{\Gamma}b_{\text{out}}(t)$
find the solution for $a(t)$ in terms of the output field at a future time $T$.
(3) Prove the relation
$\tilde{a}(\omega)=\frac{\sqrt{\Gamma}}{i(\omega-\Omega)-\Gamma/2}\tilde{b}_{\text{in}}(\omega)$
where the standard Fourier transform is defined as: $\tilde{f}(\omega)=\frac{1}{2\pi} \int_{-\infty}^{\infty}\text{e}^{i\omega t}f(t)dt$
(4) Given the I/O relation and boundary condition:
$b_{\text{out}}(t)=b_{\text{in}}(t)+\sqrt{\Gamma}a(t)$
Prove the simpler relation:
$\tilde{b}_{\text{out}}(\omega)=\text{e}^{i\delta(\omega)}\tilde{b}_{\text{in}}(\omega)$
where $\delta(\omega)$ is a phase shift which preserves the commutation relations for the I/O operators under the Fourier transform.