How to entangle two superconducting qubits?

Question

I do understand qubits entanglement, and their spins but I do not understand that at circuit level how can we entangle 2 qubits? Is it related to 2 qubits gates?

Answer 1

As discussed in the paper by Vool and Devoret, there is a relatively simple way to write the Hamiltonian for an electrical circuit. For example, the Hamiltonian of a transmon qubit (a Josephson junction in parallel with a capacitor) is $$ H = \frac{Q^2}{2C} - E_J \, \cos(2 \pi \Phi / \Phi_0)$$ where $Q$ is the charge on the capacitor, $C$ is the capacitance, $\Phi$ is the flux in the junction, $\Phi_0 = h / (2e)$ is the flux quantum, and $E_J$ is the junction energy scale. Note that the flux and charge have the same commutator as position and momentum, $[\Phi, Q] = i \hbar$.

Now suppose we have two transmons with capacitances $C_a$ and $C_b$ and junction energies $E_{J,a}$ and $E_{J,b}$. If we connect them through a capacitor with capacitance $C_g \ll C_a, C_b$, then the Hamiltonian is approximately$^{[1]}$ \begin{align} H &= \frac{Q_a^2}{2 C_a} - E_{J,a} \cos(2 \pi \Phi_a / \Phi_0) \\ &+\frac{Q_b^2}{2 C_b} - E_{J, b} \cos(2 \pi \Phi_b / \Phi_0) \\ &\underbrace{+ \frac{Q_a Q_b}{C_a C_b / C_g}}_\text{coupling} \, . \end{align} The final term couples the two transmons together and can therefore create entnaglement. If we restrict the analysis to only the two lowest levels of each transmon, the coupling term can be rewritten as $$g \left( \sigma_{y,a} \otimes \sigma_{y, b} \right)$$ where $$g/\hbar = \frac{1}{2} \frac{C_g}{\sqrt{(C_a + C_g)(C_b + C_g)}} \sqrt{\omega_a \omega_b}$$ is called the "coupling strength" and $\omega_{a,b}$ refers to the resonance frequencies of the transmons. Detailed calculations showing these relationships with more attention to details can be found in my theory GitHub repository.

[1]: The form of the Hamiltonian is exactly as written, but the precise definitions of $C_a$, $C_b$, and $C_g$ change slightly in the coupled system.