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Derivation of Input-Output relations for waveguide-QED

Target system

The system is similar to the following one.

enakai00.hatenablog.com

Here we restrict the states with only one excitation (either qubit or a propagating photon).

 \displaystyle \mid\psi(t)\rangle = c_1(t)\mid 1,{\rm vac}\rangle + \sum_k\psi_k(t)\mid 0,1_k\rangle --- (7.13)

The Heisenberg equations are:

 \displaystyle i\dot c_1(t) = \Delta c_1(t) + \sum_kg_k\psi_k(t),\ i\dot\psi_k(t) = \omega_k\psi_k(t) + g_k^* c_1(t) --- (7.14)

Using the same argument to derive (B.23) in the above article, we have the formal solution for \psi_k(t) as:

 \displaystyle\psi(t) = e^{-i\omega_k(t-t_0)}-ig_k^*\int_{t_0}^te^{-i\omega_k(t-\tau)}c(\tau)\,d\tau --- (1)

Position space

Wavefunction of the photon field is defined as:

 \displaystyle \Psi(x, t) := \frac{1}{\sqrt{d}}\langle x\mid\psi(t)\rangle =  \frac{1}{\sqrt{d}}\sum_k\psi_k(t)\langle x \mid 1_k\rangle =  \frac{1}{\sqrt{d}}\sum_k\psi_k(t)e^{ikx}

We included \displaystyle\frac{1}{\sqrt{d}} so that the wavefunction is scale-independent.

We split it into right-moving part \Psi_+(x, t) and left-moving part \Psi_-(x, t) as:

 \displaystyle \Psi_+(x, t) = \frac{1}{\sqrt{d}}\sum_{k > 0}\psi_k(t)e^{\pm ikx}

By substituting (1), we have:

 \displaystyle \Psi_\pm(x, t) = \Psi_\pm(x, t, t_0) -i \frac{1}{\sqrt{d}}\sum_{k > 0}g_k^*\int_{t_0}^t e^{\pm ikx-i\omega_k(t-\tau)}c_1(\tau)\,d\tau --- (7.17)

 \displaystyle \Psi_\pm(x,t,t_0) := \frac{1}{\sqrt{d}}\sum_{k>0}e^{\pm ikx-i\omega_k(t-t_0)}

where \displaystyle \Psi_\pm(x,t,t_0) represents a group of plane waves corresponds to input (t_0=-\infty) or output (t_0=\infty) photon beam. Note that in some locations, they are just formal expressions. For example, \Psi_+(x<0, t, \infty) corresponds to a right-moving output beam (before reaching the resonator) that doesn't exist in reality. It's a formal extension of the real output beam in x>0 to x<0.

Approximate calculation

We will apply approximation to the last term in (7.17) with the following assumptions:

 \displaystyle g_k = \frac{g}{\sqrt{d}},\ \omega = vk,\,\frac{2\pi}{d}\sum_{k>0} \sim \int_0^\infty \frac{dk}{d\omega}\,d\omega =  \frac{1}{v}\int_0^\infty \frac{dk}{d\omega}\,d\omega

The last one is the same as in this. Now,

 \displaystyle  -i \frac{1}{\sqrt{d}}\sum_{k > 0}g_k^*\int_{t_0}^t e^{\pm ikx-i\omega_k(t-\tau)}c_1(\tau)\,d\tau

  \displaystyle = -i\frac{g}{2\pi v} \int_{t_0}^t\left\{\int_0^\infty e^{\pm i\omega\frac{x}{v}-i\omega(t-\tau)}d\omega\right\}c_1(\tau)\,d\tau

  \displaystyle = -i\frac{g}{2\pi v} \int_{t_0}^t\left\{\int_0^\infty e^{i\omega(\tau-t\pm \frac{x}{v})}d\omega\right\}c_1(\tau)\,d\tau --- (2)

As the integration with \omega will be dominant around \displaystyle \tau \sim t\mp\frac{x}{v},

  • If the integral interval of \tau contains \displaystyle \tau \sim t\mp\frac{x}{v}, we replace c_1(\tau) with c_1\left(t\mp\frac{x}{v}\right), and extend the integral interval of \tau to -\infty to \infty.
  • If not, we approximate as (2) \sim 0.

The followings are the cases that result in (2) \sim 0. They corresponds to the following trivial cases.

  • \displaystyle \Psi_+(x<0, t) and t_0 = -\infty : In this case, \displaystyle \Psi_+(x<0, t) is an input beam before reaching to the resonator. So it's naturally the same as \displaystyle \Psi_+(x<0, t, -\infty).
  • \displaystyle \Psi_+(x>0, t) and t_0 = \infty : In this case, \displaystyle \Psi_+(x>0, t) is an output beam leaving the resonator. So it's naturally the same as \displaystyle \Psi_+(x>0, t, \infty).
  • \displaystyle \Psi_-(x>0, t) and t_0 = -\infty : In this case, \displaystyle \Psi_+(x>0, t) is an input beam before reaching to the resonator. So it's naturally the same as \displaystyle \Psi_-(x>0, t, -\infty).
  • \displaystyle \Psi_-(x<0, t) and t_0 = \infty : In this case, \displaystyle \Psi_+(x<0, t) is an output beam leaving the resonator. So it's naturally the same as \displaystyle \Psi_-(x<0, t, \infty).

Note: Strictly speaking, \displaystyle \Psi_+(x>0, t<\infty) is different from the asymptotic plane waves \displaystyle\Psi_+(x>0, t, t_0=\infty). In this rough approximation, we simply drop this information. Instead, we extract that information from \displaystyle \Psi_+(x>0, t<\infty) - \Psi_+(x>0, t, t_0=-\infty). In other words, we shouldn't use the trivial cases above to extract any information about the photon-resonator interaction.

In other cases, (2) represents the difference of the beam before and after interacting with the resonator. In the following calculation, we implicitly exclude the trivial cases above.

For the case t_0=-\infty,

 \displaystyle (2) = -i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty \int_0^\infty e^{i\omega(\tau-t\pm \frac{x}{v})}\,d\omega\,d\tau

  \displaystyle = -i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty\pi\delta\left(\tau-t\pm\frac{x}{v}\right) + i\,{\mathcal PV}\left(\frac{1}{\tau-t\pm\frac{x}{v}}\right)\,d\tau

  \displaystyle =-i\frac{g}{2 v}c_1\left(t\mp\frac{x}{v}\right)

For the case t_0=\infty,

 \displaystyle (2) = i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty \int_0^\infty e^{i\omega(\tau-t\pm \frac{x}{v})}\,d\omega\,d\tau

  \displaystyle = i\frac{g}{2\pi v}c_1\left(t\mp\frac{x}{v}\right)\int_{-\infty}^\infty\pi\delta\left(\tau-t\pm\frac{x}{v}\right) + i\,{\mathcal PV}\left(\frac{1}{\tau-t\pm\frac{x}{v}}\right)\,d\tau

  \displaystyle = i\frac{g}{2 v}c_1\left(t\mp\frac{x}{v}\right)

So we have the following non-trivial cases:

 \displaystyle\Psi_+(x>0,t) = \Psi^{\rm in}_+(x>0,t) - i\frac{g}{2v}c_1\left(t-\frac{x}{v}\right)

 \displaystyle\Psi_+(x<0,t) = \Psi^{\rm out}_+(x<0,t) + i\frac{g}{2v}c_1\left(t-\frac{x}{v}\right)

 \displaystyle\Psi_-(x<0,t) = \Psi^{\rm in}_+(x<0,t) - i\frac{g}{2v}c_1\left(t+\frac{x}{v}\right)

 \displaystyle\Psi_-(x>0,t) = \Psi^{\rm out}_-(x>0,t) + i\frac{g}{2v}c_1\left(t+\frac{x}{v}\right)

where \displaystyle\psi^{\rm in}_{\pm}(x, t) := \psi_{\pm}(x, t, -\infty),\, \psi^{\rm out}_{\pm}(x, t) := \psi_{\pm}(x, t, \infty)

Input-output relations

From the equation:

 \displaystyle \lim_{x\to 0} \Psi_+(x>0,t) = \lim_{x\to 0} \Psi_+(x<0,t)

 \displaystyle \lim_{x\to 0} \Psi_-(x>0,t) = \lim_{x\to 0} \Psi_-(x<0,t)

we have:

 \displaystyle\Psi^{\rm out}_\pm(0, t) = \Psi^{\rm in}_\pm(0,t) - i\frac{g}{v}c_1(t)

Using \displaystyle\gamma = \frac{2|g|^2}{v}, this can be written as:

 \displaystyle\Psi^{\rm out}_\pm(0, t) = \Psi^{\rm in}_\pm(0,t) - i\sqrt{\frac{\gamma}{2v}}c_1(t) --- (7.20)