qubit coupled to a readout resonator which in turn coupled to external drive line of characteristic impedance $Z_{0}$. $C_{d}$ is the shunt capacitance of the drive port.](figs/Q-coupled-to-Z0-Cd-Vd)

To see that Purcell rates $\frac{1}{T_{1}^{i,d}}$’s are independent of the drive port shunt capacitances $C_{p(d)}$’s we workout $Z_{i,p(d)}(\omega_{i})$ for the example circuit shown in Fig. . Assuming $C_{r}\gg C_{\kappa}$, $C_{q}\gg C_{c}$ and $C_{d}\gg C_{\kappa}$ one can show that

$$\mathrm{Im}\left[Z_{12}(\omega)\right]\simeq\frac{\left(\frac{C_{c}C_{\kappa}}{C_{q}C_{d}}\right)L_{r}\omega}{1-(\omega/\omega_{r})^{2}}\label{eq:Z12-example}$$

with $\omega_{r}=1/\sqrt{L_{r}C_{r}}$ and port $1$ being defined across the Josephson junction and port $2$ across $C_{d}$. So that the Purcell rate $\frac{1}{T_{1}^{i,j}}$ derived in Eq. gives

$$\begin{aligned}
1/T_{1} & =\frac{2}{L_{q}}\mathrm{Im}\left[Z_{12}\left(\omega_{q}\right)\right]^{2}\frac{\omega_{q}^{2}Z_{0}C_{d}^{2}}{1+\omega_{q}^{2}Z_{0}^{2}C_{d}^{2}}\nonumber \\
 & =\frac{2}{L_{q}}\left(\frac{C_{c}C_{\kappa}}{C_{q}C_{r}}\right)^{2}\frac{(\omega_{q}/\omega_{r})^{4}}{\left[1-(\omega_{q}/\omega_{r})^{2}\right]^{2}}\left(\frac{Z_{0}}{1+\omega_{q}^{2}Z_{0}^{2}C_{d}^{2}}\right)\label{eq:T1-Cd}\end{aligned}$$

where $\omega_{q}$ is the qubit frequency and $L_{q}$ qubit inductance. The expression in the Eq. above will be independent of $C_{d}$, the total shunt capacitance of the drive port, in the limit of $\omega_{d}^{2}Z_{0}^{2}C_{d}^{2}\ll1$ which holds for typical parameter values in the actual experiments.

One can similarly calculate coupling of the qubit to the voltage source $V_{d}$ using Eqs. and to get

$$\begin{aligned}
\varepsilon_{12} & =e^{i\theta_{2}}\sqrt{\frac{\omega_{q}}{2\hbar L_{q}}}\mathrm{Im}\left[Z_{12}(\omega_{q})\right]C_{d}\nonumber \\
 & =\frac{1}{\sqrt{2\hbar Z_{q}}}\left(\frac{C_{c}C_{\kappa}}{C_{q}C_{r}}\right)\frac{(\omega_{q}/\omega_{r})^{2}}{1-(\omega_{q}/\omega_{r})^{2}}\left(\frac{e^{i\theta_{2}}}{\sqrt{1+\omega_{d}^{2}Z_{0}^{2}C_{d}^{2}}}\right)\end{aligned}$$

where $\theta_{2}=\frac{\pi}{2}-\arctan(\omega_{d}Z_{0}C_{d})$ and $Z_{q}=\sqrt{L_{q}/C_{q}}$. Above expression for the coupling $\varepsilon_{12}$ of the qubit to its voltage source $V_{d}$ will be again independent of $C_{d}$ in the limit of $\omega_{d}^{2}Z_{0}^{2}C_{d}^{2}\ll1$ which 