Green–Kubo formula for electrical conductivity of a driven $0$–$\pi$ qubit

Abstract

One of the most important paradigms of quantum computation rests on employing the Cooper pair condensate states in a Josephson junction. In using these, the configuration of great current interest is the $0$–$\pi$ qubit. We present the linear response of this to an external drive by solving the Liouville equation for the phase-space distribution function. Thus, we obtain “Ohm’s law” or the expression for electrical conductivity for this system in terms of novel correlation functions. This general result has been tested for the $0$–$\pi$ qubit parameters that are used in recent experiments.

Appendix: Kinetic energy part of Hamiltonian

The gauge invariant kinetic energy term of the circuit of the $0$–$\pi$ qubit with the effect of a time-dependent external flux can be calculated as [22]

$$\begin{aligned} \, \mathcal L_{\mathrm K}^{}(\vec{\boldsymbol\Phi},\dot{\vec{\boldsymbol\Phi}},t)= \frac{1}{4C_\Sigma}\bigl[4C_\Sigma^{} \bigl(C_\Sigma^{}\dot\Phi_1^2+2\dot\Phi_1^{}\dot\Phi_2^{}+2\dot\Phi_2^2&-2C_{\mathrm J}^{}\dot\Phi_2^{}\dot\Phi_3^{}+C_{\mathrm J}^{}\dot\Phi_3^2+C_C^{}C_{\mathrm J}^{}\dot\Phi_{ \mathrm{ext} ,1}^{}(t)+\dot\Phi_{ \mathrm{ext} ,3}^{}(t)\bigr)^2\bigr], \end{aligned}$$

(A.1)

where $\vec{\boldsymbol\Phi}$ and $\dot{\vec{\boldsymbol\Phi}}$ include all three components of $\boldsymbol\Phi$ and $\dot{\boldsymbol\Phi}$, and $C_\Sigma=C_C+ C_{\mathrm J}$. The Hamiltonian for the circuit can be calculated using Legendre transformation. In the kinetic energy part of the Hamiltonian $\mathcal H^{(0)}_{\mathrm K}$, the constants $A,B,\ldots,F$ can be calculated as

$$ \begin{aligned} \, &A=\frac{1}{32C_C^{}C_{\mathrm J}^{}C_\Sigma^3} \bigl[2C_{\mathrm J}^{}(16C_C^3+39C_C^2C_{\mathrm J}^{}+29C_C^{}C_{\mathrm J}^2+7C_{\mathrm J}^3)\bigr], \\ &B=\frac{1}{32C_C^{}C_{\mathrm J}^{}C_\Sigma^3} \bigl[2C_{\mathrm J}^{}C_\Sigma^2(C_C^{}+7C_\Sigma^{})\bigr], \\ &C=\frac{1}{32C_C^{}C_{\mathrm J}^{}C_\Sigma^3} \bigl[C_\Sigma^2(14C_C^2+27C_C^{}C_{\mathrm J}^{}+14C_{\mathrm J}^2)\bigr], \\ &D=\frac{1}{32C_C^{}C_{\mathrm J}^{}C_\Sigma^3}\bigl[-4C_{\mathrm J}^{}C_\Sigma^2(C_C+7C_\Sigma^{})\bigr], \\ &E=\frac{28C_{\mathrm J}^{}C_\Sigma^3}{32 C_C^{}C_{\mathrm J}^{}C_\Sigma^3}, \\ &F=\frac{1}{32C_C^{}C_{\mathrm J}^{}C_\Sigma^3}[-2C_{\mathrm J}^{}C_\Sigma^{}(14C_C^2+29C_C^{}C_{\mathrm J}^{}+14C_{\mathrm J}^2)]. \end{aligned}$$

(A.2)