There are several analogies between diffraction patterns and Josephson junctions, especially between a double slit experiment and two Josephson junctions in a superconducting ring (like this):

  • Both are based on the interference of QM wavefunctions.
  • The intensity pattern observed in diffraction experiments corresponds to the maximum supercurrent.
  • The period of the pattern is determined by the ratio of the two slits compared to wavelength in optics whereas it depends on the ratio of the magnetic flux inside the ring to the flux quantum in superconductors.
  • the fininte size of the slits and the finite size of the Josephson junction yield a sinc-envelope of the observed pattern.

In optics it's pretty clear to me what happens when the light passes the aperture. The aperture is a modification in r-space and we observe the distribution of k-vectors on the screen in the far field. E.g. the typical double slit aperture described by two delta functions convoluted with a rectangular function in r-space results in a cosinus diffraction pattern (Fourier transform of two delta functions) multiplied (convolution theorem) with a sinc (FT of a rect) in k-space / on the screen.

Question: What is the analogy of Fourier optics when talking about Josephson junctions? What are the domains (like k and r) that are involved?

Thank you very much!

  • $\begingroup$ I guess it has something to to with the magnetic flux $\Phi$ and the elecitral charge $q$ which are kind of conjugate variables? Since $[\Phi]\cdot[q]=[\hbar]$... but this is just a guess... $\endgroup$
    – bernd
    Mar 28, 2015 at 10:59
  • 1
    $\begingroup$ I will try my best to answer you tomorrow, but I have been very busy recently. $\endgroup$
    – Xcheckr
    Apr 7, 2015 at 23:33

1 Answer 1


There is no "Fourier optics" in superconductors in general, only in Josephson junction, and it's due to the Aharonov-Bohm effect. In bulk superconductor, the Aharonov-Bohm gives the so-called Little-Parks effect : the oscillation of the critical temperature with respect to the magnetic flux enclosed in a non-connected superconductor (= a superconductor with a hole).

In short (and I confess, with a pretty bad explanation) the analogy is due to the kinetic-energy term $p-A$ in both cases, but the gauge-potential $A=0$ in optics whereas the momentum $p=0$ (so to say) in superconductors.

The final results in both cases are similar, provided the Fourier transform is generalised $px\rightarrow\left(p-A\right)x$. The position do not usually appear in the final result about Josephson junction since one usually averages over position. There are modern use of the position-dependency of the Josephson current, as it is pretty well introduced in the following articles

in particular you can find more details about your question in the supplementary materials of the article by Molenkamp et al.

There are nevertheless important differences:

  • the correlation length can be considered as infinite in superconductors. More precisely, the coherence length appears only with respect to the length of the Josephson junction. So you can think of making complicated circuits interrupted by short junctions, and the interferences will come from the loops in the circuit, like in SQUID. In contrast, the interference can appear due to large/long junctions, as you mentioned in your question

  • the interference in superconductors is due to the presence of flux quanta which make the superconductor topologically non-trivial. In short, inside the flux quantum, there is no superconductivity, and so the superconductor can be seen as a non-connected materials. It's because of this hole that the interference appear. In optics the hole is created by the slits, separated in space.

Note "Fourier optics" is a particular case of interferences, which requires the 4F correlator, see the associated Wikipedia page.

Well, I stop here, I find this answer pretty unclear, sorry. I may edit it later.


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