Detailed derivation of the Input-Output theory

Basic definitions

Think of a system consisted of:

A long waveguide with length $d$ accommodating a photon field $\hat b_k$ .
A small resonator at $x=0$ accommodating an electric oscillation $\hat a$ .

that has the RWA Hamiltonian (5.60):

　 $\displaystyle H_{\rm RWA} = \omega\hat a^\dagger\hat a + \sum_k\left(g_k\hat a\hat b_k^\dagger + g_k^*\hat a^\dagger\hat b_k\right) + \sum_k\omega_k\hat b_k^\dagger\hat b_k$ --- (5.60)

where we use the unit system with $\hbar = 1$ . So, the momentum $p$ and the wavenumber $k$ are the same $p=k$ . (In general $p = \hbar k$ .)

The wavelength of the photon field is numbered as $\displaystyle \lambda_n \sim \frac{d}{n}$ . So, the wavenumber is counted as:

　 $\displaystyle k = p = \frac{2\pi}{\lambda_n} = \frac{2\pi}{d}n$

Hence, assuming that $d \gg 1$ , we can replace $\displaystyle \frac{2\pi}{d}\sum_k \sim \int_{-\infty}^\infty dk = 2\int_0^\infty \left|\frac{dk}{d\omega}\right|d\omega$ .

Applying it to the last term of $H_{\rm RWA}$ , we have:

　 $\displaystyle \sum_k\omega_k\hat b_k^\dagger\hat b_k \sim \frac{d}{\pi}\int_0^\infty \omega\hat b(\omega)^\dagger\hat b(\omega)\left|\frac{dk}{d\omega}\right|\,d\omega$

So the operator $\hat b_k$ has a scale dependency as $\displaystyle \hat b \sim \frac{1}{\sqrt{d}}$ . From the second term of $H_{\rm RWA}$ , the coupling constant $g_k$ has the same scale dependency $\displaystyle g_k \sim \frac{1}{\sqrt{d}}$

Now we define $\displaystyle J^{\rm QO}(\omega) := 2d|g(\omega)|^2\left|\frac{dk}{d\omega}\right|$ as a scale independent spectral function. Then, we have the relationship (5.62):

　 $\displaystyle \frac{1}{2\pi}\int_0^{\infty} J^{\rm QO}(\omega)e^{-i\omega t}\,d\omega \sim \sum_k |g_k|^2e^{-i\omega_k t} =: K(t)$ --- (5.62)

Also, in the following discussion, we use the Fourier transformation of the Heaviside step function $\Theta(t)$ :

　 $\displaystyle \int_0^\infty e^{i\omega u}\,du = \int_{-\infty}^\infty\Theta(u)e^{i\omega u}\,du = \pi\delta(\omega) + i\,{\mathcal PV}\left(\frac{1}{\omega}\right)$ --- (0)

Note that (0) is a relationship as a hyperfunction. So it's valid only when they are combined with the operation $\displaystyle \int_{-\infty}^\infty f(\omega)[*]\,d\omega$ .

Formal solution of the photon field $\hat b_k(t)$

From (5.60), we have the following Heisenberg equations (B.22):

　 $\displaystyle \frac{d}{dt}\hat a(t) = -i\omega\hat a(t) - i\sum_k g_k^*\hat b_k(t)$ --- (1)

　 $\displaystyle \frac{d}{dt}\hat b_k(t) = -i\omega_k\hat b_k(t) - ig_k\hat a(t)$ --- (2)

Define $\hat f(t)$ as $\displaystyle \hat b_k(t) = e^{-i\omega_k(t-t_0)}\left\{\hat b_k(t_0)+\hat f(t)\right\}$ with the initial condition $\hat f(t_0)=0$ . Then, from (2), we have:

　 $\displaystyle e^{-i\omega_k(t-t_0)}\frac{d}{dt}\hat f(t) = -ig_k\hat a(t)$

　 $\displaystyle\therefore\, \frac{d}{dt}\hat f(t) = -i e^{i\omega_k(t-t_0)}g_k\hat a(t)$

　 $\displaystyle\therefore\, \hat f(t) = -ig_ke^{-i\omega_kt_0}\int_{t_0}^te^{i\omega_k \tau}\hat a(\tau)\,d\tau$

　 $\displaystyle\therefore\, \hat b_k(t) = e^{-i\omega_k(t-t_0)}\hat b_k(t_0) -ig_k\int_{t_0}^t e^{-i\omega_k(t-\tau)}\hat a(\tau)\,d\tau$ --- (B.23)

A Langevin equation for the resonator

By substituting (B.23) into (1), we have:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -i\omega\hat a(t) - i\sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0) - \int_{t_0}^t\sum_k |g_k|^2e^{-i\omega_k(t-\tau)}\hat a(\tau)\,d\tau$

　 $\displaystyle \therefore\, \frac{d}{dt}\hat a(t) = -i\omega\hat a(t) - i \hat\xi(t) -\int_{t_0}^tK(t-\tau)\hat a(\tau)\,d\tau$ --- (5.61)

where $\displaystyle \hat\xi(t) := \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ .

Markovian approximation

We assume that the resonator resonates with the photon field with some frequency $\omega'$ with a slow modulation $\hat a_{\rm slow}(t)$ .

　 $\hat a(t) = \hat a_{\rm slow}(t)e^{-i\omega' t}$

and the contribution to the integration in (5.61) is dominated by the oscillation $e^{-i\omega' t}$ . So,

　 $\displaystyle \int_{t_0}^tK(t-\tau)\hat a(\tau)\,d\tau \simeq \hat a_{\rm slow}(t)\int_{t_0}^tK(t-\tau)e^{-i\omega'\tau}\,d\tau$

　　 $\displaystyle=\hat a_{\rm slow}(t)e^{-i\omega't}\int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du\ (u:=t-\tau)$

　　 $\displaystyle=\hat a(t)\int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du$

In the limit $t_0 \to -\infty$ , we have:

　 $\displaystyle \int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du = \frac{1}{2\pi}\int_0^\infty J^{\rm QO}(\omega)\left\{\int_0^{\infty}e^{i(\omega'-\omega)u}\,du\right\}d\omega$

　　 $\displaystyle = \frac{1}{2\pi}\int_0^\infty J^{\rm QO}(\omega)\left\{ \pi\delta(\omega'-\omega) + i\,{\mathcal PV}\left(\frac{1}{\omega'-\omega}\right)\right\}\,d\omega$

　　 $\displaystyle \simeq \frac{1}{2\pi}\int_{-\infty}^\infty J^{\rm QO}(\omega)\left\{ \pi\delta(\omega'-\omega) + i\,{\mathcal PV}\left(\frac{1}{\omega'-\omega}\right)\right\}\,d\omega$

　　　( $\because$ Contribution from around $\omega\sim\omega'$ is dominant.)

　　 $\displaystyle = \frac{1}{2}J^{\rm QO}(\omega') +i \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$

　　 $\displaystyle = \frac{\kappa}{2} + i\delta \omega$ --- (B.29)

where,

　 $\displaystyle\kappa := J^{\rm QO}(\omega')$

　 $\displaystyle\delta\omega := \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$ --- (B.30)

By combining these results, we have:

　 $\displaystyle \int_{t_0}^tK(t-\tau)\hat a(\tau)\,d\tau \simeq \left(i\delta\omega + \frac{\kappa}{2}\right)\hat a(t)$ --- (B.26)

And from (5.61), we have:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left\{i(\omega + \delta\omega)+\frac{\kappa}{2}\right\}\hat a(t)-i\hat\xi(t)$

From this result, you can see that $\hat a(t)$ oscillates as $\displaystyle\hat a(t) \sim \hat a(0)e^{-\frac{\kappa}{2}t} e^{-i(\omega+\delta\omega)t}$ . So,

　 $\omega' = \omega + \delta\omega$ .

This means that $\omega'$ is (implicitly) decided from the consistency condition with (B.30):

　 $\displaystyle\delta\omega = \omega' - \omega = \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$

Solution of the resonator $\hat a(t)$

Now we have the Langevin equation for the resonator:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left(i\omega'+\frac{\kappa}{2}\right)\hat a(t)-i\hat\xi(t)$ --- (3)

　 $\displaystyle \hat\xi(t) := \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ --- (4)

Define $\hat f(t)$ as $\displaystyle \hat a(t) = e^{-i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)}\hat f(t)$ . Then from (3), we have:

　 $\displaystyle e^{-i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)}\frac{d}{dt}\hat f(t) = -i\hat\xi(t)$

　 $\displaystyle\therefore\,\frac{d}{dt}\hat f(t) = -i e^{i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)}\hat\xi(t) =-i\sum_k g_k^* e^{i\left\{(\omega'-\omega_k)+\frac{\kappa}{2}\right\}(t-t_0)} \hat b_k(t_0)$

　 $\displaystyle\therefore\,\hat f(t) = -i \sum_k g_k^* \frac{e^{i\left\{(\omega'-\omega_k)+\frac{\kappa}{2}\right\}(t-t_0)}}{i(\omega'-\omega_k)+\frac{\kappa}{2}} \hat b_k(t_0) +C$

Hence, the solution $\hat a(t)$ is:

　 $\displaystyle \hat a(t) = Ce^{-i\left(\omega'+\frac{\kappa}{2}\right)(t-t_0)} -i \sum_k g_k^* \frac{e^{i\omega_k(t-t_0)}}{i(\omega'-\omega_k)+\frac{\kappa}{2}} \hat b_k(t_0)$ --- (B.32)

The constant $C$ is determined with the initial condition.

From this result, you can see that the contribution from coupling $g_k$ is dominant around the resonant frequency $\omega_k \sim \omega'$ . So changing the coupling strength $g_k$ outside $\omega_k \sim \omega'$ doesn't change the behavior of the system, and we can safely assume that $g_k$ is almost constant. In terms of $J^{\rm QO}(\omega)$ , this means that $J^{\rm QO}(\omega) \simeq J^{\rm QO}(\omega') = \kappa$

Then, compared to the free plane wave solution $\displaystyle \sum_k e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ , $\displaystyle \hat\xi(t) := \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ is rescaled with $g_k \propto \sqrt{\kappa}$ .

So, we define $\displaystyle\hat b^{\rm in}(t) := \frac{1}{\sqrt{\kappa}}\hat\xi(t)$ to have a renormalized operator, and (3) becomes:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left(i\omega'+\frac{\kappa}{2}\right)\hat a(t) -i\sqrt{\kappa}\hat b^{\rm in}(t)$ --- (5.64)

Note that in the discussions above, we used the initial condition at $t_0 \to -\infty$ . As a result, $\displaystyle\hat\xi(t)$ corresponds to the input photon generated at $t=t_0$ . So we named $\hat b^{\rm in}(t)$ to indicate the input photon.

Input-Output relations

It's also possible to consider the boundary condition at $t_0 \to \infty$ in the Markovian approximation. In this case, the calculation to get (B.29) becomes:

　 $\displaystyle \int_{0}^{t-t_0}K(u)e^{i\omega'u}\,du = \frac{1}{2\pi}\int_{0}^\infty J^{\rm QO}(\omega)\left\{\int_0^{-\infty}e^{i(\omega'-\omega)u}\,du\right\}d\omega$

　　 $\displaystyle = -\frac{1}{2\pi}\int_{0}^\infty J^{\rm QO}(\omega)\left\{\int_0^{\infty}e^{i(\omega-\omega')u}\,du\right\}d\omega$

　　 $\displaystyle \simeq - \frac{1}{2\pi}\int_{-\infty}^\infty J^{\rm QO}(\omega)\left\{ \pi\delta(\omega-\omega') + i\,{\mathcal PV}\left(\frac{1}{\omega-\omega'}\right)\right\}\,d\omega$

　　 $\displaystyle = -\frac{1}{2}J^{\rm QO}(\omega') +i \frac{1}{2\pi}\lim_{\epsilon\to 0+}\int_{|\omega-\omega'|>\epsilon}\frac{J^{\rm QO}(\omega)}{\omega'-\omega}\,d\omega$

　　 $\displaystyle = -\frac{\kappa}{2} + i\delta \omega$ --- (B.29)'

Note that we used the fact that $\delta(x)$ is an even function and $\displaystyle{\mathcal PV}\left(\frac{1}{x-x'}\right)$ is an odd function.

Compared to (B.29), (B.29)' has an opposite sign for $\kappa$ , and (5.64) becomes:

　 $\displaystyle \frac{d}{dt}\hat a(t) = -\left(i\omega'-\frac{\kappa}{2}\right)\hat a(t) -i\sqrt{\kappa}\hat b^{\rm out}(t)$ --- (5.64)'

where $\displaystyle\sqrt{\kappa}\hat b^{\rm out}(t) = \sum_kg_k^*e^{-i\omega_k(t-t_0)}\hat b_k(t_0)$ corresponds to the photon field pulling back to the current time $t$ from the final status at $t=t_0 \sim \infty$ .