Past exam of the condensed matter physics major option course of the University of Oxford / 2015 / 8 / d / Solution

The AC Josephson effect is $V=({\Phi }_0/2\pi )\dot{\delta }$. Since the imposed flux is held fixed when defining a small-signal inductance, differentiate c to obtainOn the stable small-current branch with $\cos \varphi >0$, $\delta \simeq 0$ and the result reduces to the printed expression,The stable small-signal inductance of a symmetric DC SQUID requires $|I|\ll 2I_C|\cos \varphi |$, stronger than $I\ll I_C$ near a half-integer flux quantum. If $\cos \varphi <0$, the stable zero-current phase is $\delta \simeq \pi$ and $\cos \delta \simeq -1$. Its inductance is positive:For example, $\Phi ={\Phi }_0$ makes the printed signed formula negative when expanded about $\delta =0$, but that phase is an energy maximum. Expanding about the stable $\delta =\pi$ corrects the sign. At ideal half-integer flux the critical current vanishes and the linear inductance formally diverges; a finite transport current cannot satisfy the small-current approximation there. Geometric inductance and other nonidealities then have to be retained instead of extending the simple formula through its pole.