It is widely stated that when a Josephson junction is placed in a superconducting circuit, the cooper pairs can tunnel through and create a net current without a bias voltage. However, given that the junction is symmetric, wouldn't there be equal probability that the cooper pairs tunnel in both directions, cancelling out any net current? Is there some mechanism that destroys this symmetry? External flux or something?
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2 Answers
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The direction and magnitude of the current depends on the difference in phase of the order parameter on the two sides of the junction. In the simplest case the current from the left to right right of a junction is gven by $I=I_{\rm max} \sin (\theta_{\rm right}-\theta_{\rm left})$ where $\langle \psi\psi\rangle \propto e^{i\theta}$ is the superconducing order parameter.
answered Aug 5, 2020 at 19:13
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EDIT: My previous answer was wrong, I think. The theory is explained here. It seems like in a perfectly symmetric system there is no a priori way to say in which direction the current will flow. Instead it is some kind of symmetry breaking. So think of it as "turning the voltage down to zero" (i.e. $V\to0^+$), but observing that a current remains.
EDIT2: For more info, you can refer to this excellent post.
