Quantum Optics - Output field of Spontaneous Parametric Down-Conversion (SPDC)

Question

Following Ultrabright backward-wave biphoton source by Chih-Sung Chuu and S. E. Harris:

The output of a backward wave Spontaneous Parametric Down-Conversion (SPDC) is described by $$ a_s(t, z) = b_s(t) e^{-i\Omega_qt}\sin(\frac{q\pi z}{L}) $$

$$ a_i(t, z) = b_i(t) e^{-i\Omega_rt}\sin(\frac{r\pi z}{L}) $$

where the operator $b_i$ and $b_s$ are fields internal to the cavity and vary slowly with time. $\Omega_q$ and $\Omega_r$ are the cold cavity frequencies. The coupled equations for the slowly varying operators are:

$$ \frac{\partial b_s(t)}{\partial t} + \frac{\Gamma_s}{2}b_s(t) = -i\kappa_1 b^\dagger_i(t) + \sqrt{\gamma_s} b_s^{in}(t) $$ $$ \frac{\partial b_i^\dagger(t)}{\partial t} + \frac{\Gamma_i}{2}b_i^{\dagger}(t) = i\kappa_1 b^\dagger_s(t) + \sqrt{\gamma_i} b_i^{in\dagger}(t) $$

where $b_s^{in}(t)$ and $b_i^{in\dagger}(t)$ are the fields incident on the resonant cavity and $\Gamma_s$ and $\Gamma_i$ are the total cavity decay rates and $\gamma_s$ with $\gamma_i$ are the output coupling rates. The slowly varying output fields are:

$$b_s^{out}(t) = \sqrt{\gamma_s}b_s(t) - b_s^{in}(t)$$ $$b_i^{out\dagger}(t) = \sqrt{\gamma_i}b_i^\dagger(t) - b_i^{in\dagger}(t)$$

The final solution is in the form: $$a_s^{out}(\omega) = A_1(\omega)a_s^{in}(\omega) + B_1(\omega)a_i^{in\dagger}(-\omega_i)$$ $$a_i^{out\dagger}(-\omega_i) = C_1(\omega)a_s^{in}(\omega) + D_1(\omega)a_i^{in\dagger}(-\omega_i)$$

The Question

How was the solution for the output fields obtained? I understand that there is a Fourier Transform to the frequency regime and it looks like a Bogoliubov Transformation too, but I have no idea how to get it from the coupled differential equation...

Answer 1

Starting from the following equations $$ \begin{align} \frac{\partial b_{s}\left(t\right)}{\partial t}+\frac{\Gamma_{s}}{2} b_{s}\left(t\right) & =-i\kappa_{1}b_{i}^{\dagger}\left(t\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(t\right) \\ \frac{\partial b_{i}^{\dagger}\left(t\right)}{\partial t}+\frac{\Gamma_{i}}{2} b_{i}^{\dagger}\left(t\right) & =i\kappa_{1}b_{s}\left(t\right)+\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(t\right) \end{align} $$

perform the Fourier transform and note that $\frac{\partial}{\partial t} \rightarrow -i \omega$ (by properties of Fourier transforms).

$$ \begin{align} -i\omega b_{s}\left(\omega\right)+\frac{\Gamma_{s}}{2} b_{s}\left(\omega\right) & =-i\kappa_{1}b_{i}^{\dagger}\left(\omega\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) \\ -i\omega b_{i}^{\dagger}\left(\omega\right)+\frac{\Gamma_{i}}{2} b_{i}^{\dagger}\left(\omega\right) & =i\kappa_{1}b_{s}\left(\omega\right)+\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right) \end{align} $$

$$ \begin{align} \left(\frac{\Gamma_{s}}{2}-i\omega \right) b_{s}\left(\omega\right) & =-i\kappa_{1}b_{i}^{\dagger}\left(\omega\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) \\ \left(\frac{\Gamma_{i}}{2} -i\omega \right)b_{i}^{\dagger}\left(\omega\right) & =i\kappa_{1}b_{s}\left(\omega\right)+\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right) \end{align} $$ Now, solving for $b_{i}^{\dagger}\left(\omega\right)$ in the second equation and substituting into the first, one obtains

$$ \left(\frac{\Gamma_{s}}{2}-i\omega \right) b_{s}\left(\omega\right) =-\frac{i\kappa_{1}}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\left(i\kappa_{1}b_{s}\left(\omega\right)+\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) $$

$$ \left(\frac{\Gamma_{s}}{2}-i\omega \right) b_{s}\left(\omega\right)+\frac{i\kappa_{1}\left(i\kappa_{1} b_{s}\left(\omega\right)\right)}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)} =-\frac{i\kappa_{1}}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) $$

$$ \left(\left(\frac{\Gamma_{s}}{2}-i\omega \right) +\frac{i\kappa_{1}\left(i\kappa_{1} \right)}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\right) b_{s}\left(\omega\right)=-\frac{i\kappa_{1}}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) $$

$$ \left(\left(\frac{\Gamma_{s}}{2}-i\omega \right) -\frac{\kappa_{1}^{2}}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\right) b_{s}\left(\omega\right)=-\frac{i\kappa_{1}}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) $$

$$ \frac{1}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\left(\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right) -\kappa_{1}^{2}\right) b_{s}\left(\omega\right)=-\frac{i\kappa_{1}}{\left(\frac{\Gamma_{i}}{2} -i\omega \right)}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}b_{s}^{\text{in}}\left(\omega\right) $$

$$ \left(\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right) -\kappa_{1}^{2}\right) b_{s}\left(\omega\right)=-i\kappa_{1}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}\left(\frac{\Gamma_{i}}{2} -i\omega \right)b_{s}^{\text{in}}\left(\omega\right) $$

Neglecting $\kappa_{1}^{2}$ compared to $\kappa_{1}$ (I suspect this is related to the "small gain" assumption discussed in the paper, and since $\kappa_{1} \propto E_{p}$ this may be how this can be justified).

$$ \left(\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right)\right) b_{s}\left(\omega\right)=-i\kappa_{1}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}\left(\frac{\Gamma_{i}}{2} -i\omega \right)b_{s}^{\text{in}}\left(\omega\right) $$ $$ b_{s}\left(\omega\right)=\frac{-i\kappa_{1}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}\left(\frac{\Gamma_{i}}{2} -i\omega \right)b_{s}^{\text{in}}\left(\omega\right)}{\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right) } $$ Now, referring to the equations for the output fields

$$ \begin{align} b_{s}^{\text{out}}\left(t\right) & =\sqrt{\gamma_{s}}b_{s}\left(t\right)-b_{s}^{\text{in}}\left(t\right) \\ b_{i}^{\text{out}\dagger}\left(t\right) & =\sqrt{\gamma_{i}}b_{i}^{\dagger}\left(t\right)-b_{i}^{\text{in}\dagger}\left(t\right) \end{align} $$

and taking the Fourier transform one obtains

$$ \begin{align} b_{s}^{\text{out}}\left(\omega\right) & =\sqrt{\gamma_{s}}b_{s}\left(\omega\right)-b_{s}^{\text{in}}\left(\omega\right) \\ b_{i}^{\text{out}\dagger}\left(\omega\right) & =\sqrt{\gamma_{i}}b_{i}^{\dagger}\left(\omega\right)-b_{i}^{\text{in}\dagger}\left(\omega\right) \end{align} $$

Now, taking the first equation and substituting for $b_{s}\left(\omega\right)$ from the earlier computations

$$ b_{s}^{\text{out}}\left(\omega\right) =\sqrt{\gamma_{s}}\left(\frac{-i\kappa_{1}\left(\sqrt{\gamma_{i}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\sqrt{\gamma_{s}}\left(\frac{\Gamma_{i}}{2} -i\omega \right)b_{s}^{\text{in}}\left(\omega\right)}{\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right) }\right)-b_{s}^{\text{in}}\left(\omega\right) $$

$$ b_{s}^{\text{out}}\left(\omega\right) =\left(\frac{-i\kappa_{1}\left(\sqrt{\gamma_{i}\gamma_{s}}b_{i}^{\text{in}\dagger}\left(\omega\right)\right)+\gamma_{s}\left(\frac{\Gamma_{i}}{2} -i\omega \right)b_{s}^{\text{in}}\left(\omega\right)}{\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right) }\right)-b_{s}^{\text{in}}\left(\omega\right) $$

$$ b_{s}^{\text{out}}\left(\omega\right)=\frac{-i\kappa_{1}\sqrt{\gamma_{i}\gamma_{s}}}{\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right)}b_{i}^{\text{in}\dagger}\left(\omega\right)+\frac{\gamma_{s}}{\frac{\Gamma_{s}}{2}-i\omega}b_{s}^{\text{in}}\left(\omega\right)-b_{s}^{\text{in}}\left(\omega\right) $$

$$ b_{s}^{\text{out}}\left(\omega\right)=\frac{-i\kappa_{1}\sqrt{\gamma_{i}\gamma_{s}}}{\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right)}b_{i}^{\text{in}\dagger}\left(\omega\right)+\left(\frac{\gamma_{s}}{\frac{\Gamma_{s}}{2}-i\omega}-1\right)b_{s}^{\text{in}}\left(\omega\right) $$

$$ b_{s}^{\text{out}}\left(\omega\right)=\frac{-i\kappa_{1}\sqrt{\gamma_{i}\gamma_{s}}}{\left(\frac{\Gamma_{s}}{2}-i\omega \right)\left(\frac{\Gamma_{i}}{2} -i\omega \right)}b_{i}^{\text{in}\dagger}\left(\omega\right)+\left(\frac{\gamma_{s}-\frac{\Gamma_{s}}{2}+i\omega}{\frac{\Gamma_{s}}{2}-i\omega}\right)b_{s}^{\text{in}}\left(\omega\right) $$

Lastly, I believe the conversion to the $a$ functions involves changing $\omega \rightarrow \omega - \Omega$ in the appropriate places.

Getting the second equation, $a_{i}^{\text{out}\dagger}$, should follow a similar approach.