following Hamiltonian

$$H=\frac{1}{2}\mathbf{Q}^{T}\mathbf{C}^{-1}\mathbf{Q}+\frac{1}{2}\mathbf{\Phi}^{T}\mathbf{M}_{0}\mathbf{\Phi}-E_{J_{1}}\cos\left(\varphi_{1}\right)-E_{J_{2}}\cos\left(\varphi_{2}\right)$$

where

$$\begin{aligned}
\mathbf{C} & = & \left(\begin{array}{ccc}
C_{q}+C_{c} & 0 & -C_{c}\\
0 & C_{q}+C_{c} & -C_{c}\\
-C_{c} & -C_{c} & C_{r}+2C_{c}
\end{array}\right)\label{eq:C-2-Q-1-B}\end{aligned}$$

$\mathbf{M}_{0}$ diagonal with entries $\left(0,0,1/L_{r}\right)$ and the coordinate vector $\mathbf{\Phi}=(\Phi_{J_{1}},\Phi_{J_{2}},\Phi_{r})^{T}$ holds the fluxes across the inductive branches. Typically $C_{c}\ll C_{q}\ll C_{r}$ holds so that we can approximately write

$$\mathbf{C}^{-1}\cong\left(\begin{array}{ccc}
1/C_{q} & \frac{C_{c}^{2}}{C_{q}^{2}C_{r}} & \frac{C_{c}}{C_{q}C_{r}}\\
\frac{C_{c}^{2}}{C_{q}^{2}C_{r}} & 1/C_{q} & \frac{C_{c}}{C_{q}C_{r}}\\
\frac{C_{c}}{C_{q}C_{r}} & \frac{C_{c}}{C_{q}C_{r}} & 1/C_{r}
\end{array}\right)\label{eq:Cinv-2-Q-1-B}$$

so that we have

$$\begin{aligned}
g_{1} & = & \frac{1}{2\sqrt{Z_{1}Z_{r}}}\frac{C_{c}}{C_{q}C_{r}}\\
g_{2} & = & \frac{1}{2\sqrt{Z_{2}Z_{r}}}\frac{C_{c}}{C_{q}C_{r}}\end{aligned}$$

where $Z_{i}=\sqrt{L_{i}/C_{q}}$ and $Z_{r}=\sqrt{L_{r}/C_{r}}$. We note here that although there is no direct electrostatic dipole coupling between qubits in Eq. a mediated coupling $J_{0}$ appears in Eq. . As we will see below the magnitude of $J_{0}$ is non-negligible compared to $J$ in Eq. hence one should compute $J+J_{0}$ for the total exchange coupling rate as we did in Fig. . We note that

$$\begin{aligned}
J_{0} & = & \frac{1}{2\sqrt{Z_{1}Z_{2}}}\frac{C_{c}^{2}}{C_{q}^{2}C_{r}}\nonumber \\
 & = & \frac{2}{\omega_{r}}g_{1}g_{2}\label{eq:J0}\end{aligned}$$

We now apply the impedance formula for the $J$-coupling in Eq. to the circuit in Fig. . We need to first compute the two-port impedance matrix between the ports shunted by Josephson junctions. This can be done by an 