3
$\begingroup$

In superconducting quantum computing resonators are typically an LC circuit. If L and C are fixed the resonator will have a fixed resonance frequency. However, I heard that in the group of Prof. John Martinis they are using a different type of resonators that their frequency is tunable. The tunability is somehow possible by applying a DC flux bias. Can anyone elaborate on the subject? How does it work? Usages?

$\endgroup$

1 Answer 1

3
$\begingroup$

There's a superconducting circuit element called Josephson junction, which is roughly a nonlinear inductor. The inductance of a Josephson junction depends on current via the relation $$L(I) = \frac{L_0}{\sqrt{1 - (I/I_c)^2}}$$ where $L_0$ is the inductance of the junction with no bias current and $I_c$ is the so-called "critical current" which is the maximum current that can pass through the junction before it stops being superconducting. The resonance frequency of an LC resonator is $\omega_0 = 1 / \sqrt{LC}$, so if you change $L$ you change the resonance frequency. Therefore, if you can include a Josephson junction into the resonator, you have a current-dependent resonance frequency.

Often, we use a dc SQUID, which is two junctions connected in parallel, instead of a single junction. When you put some magnetic flux through the SQUID loop, it puts current through the two junctions and changes their inductance, therefore shifting the resonance frequency.

That's pretty much how most modern superconducting qubits are frequency tuned.

$\endgroup$
2
  • $\begingroup$ But if I understand correctly adding a junction will result in a non-linear inductance $L_0$ , and because the energy of the inductance term of the circuit is $\Phi^2/2L$ then the circuit will not be an harmonic oscillator anymore. Is it in a regime where the non-linearity is negligible? $\endgroup$
    – Qexp
    Mar 6, 2019 at 19:15
  • 1
    $\begingroup$ @Qexp that's correct. Josephson junctions are not linear so adding one to a circuit adds some nonlinearity. However, especially at low power, the nonlinearity can be rather weak. $\endgroup$
    – DanielSank
    Mar 7, 2019 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.