StanislavDolgopolov

OK, consider a sample. An impurity state covers up to 1000 sites, and all these atoms vibrate at zero temperature, so impurity state boundaries are not fixed in space. Two conduction electrons are bound with the impurity state and form a coherent BEC with electron pairs of other impurity states. Coherent wave functions of BEC bosons are strongly overlapped in space, physical boundaries of electronic states are flexible in space; there are no obstacles for a paired charge transport between neighboring impurity states. If we consider Anderson localization, there are no fixed impurities at all, so every local state can drift due to real-space-vibrations of atom potentials. Electron pairs drift together with their states, the BEC provides that the pair drift is dissipationless.

The special circumstance is the degenerative wave function of bosonic pairs. The second circumstance is the fact that the local state band is not identical to a normal valence band. In a valence band the barriers are pinned to each atom, whereas boundaries of every local state can fluctuate, so in an external potential the local state can be flexible in space.

Yes. The electrons don't jump the energy gap. Every electron pair is a combined particle with zero spin, that is a boson. The boson population at low temperature can condense into one ground state (Bose-Einstein-Condensate), then all states of bosons are energetically identical, so wave functions of bosons may be overlapped in real space and every boson can tunnel in the common space without energy loss (since it is in ground state). Nothing new. Superfluidity works this way.

Yes, the electrons in a local well are no longer usual conduction electrons. They form a new band - band of local states, which is usually much narrower than the conduction band.

Yes, I would discuss the topic with community, since it seems to be important for general understanding of superconductivity. The conduction electrons are extended throughout the crystal before they fall into a local state. Inside the local state two electrons must be a singlet pair.

Crystals may contain electronic real-space-eigenstates as ground states, which are spatially much larger than one unit cell, such as impurity states, standing waves at Brillouin zone edges, Anderson localization, etc. Every eigenstate is usually occupied by two conduction electrons with opposite spins, forming a singlet pair. Notably: if the eigenstate is limited in real space, then the excitation energy of each singlet electron is not necessarily negligible, so below a certain temperature the singlet pair can be lasting. Isn't this a long-debated pairing mechanism in superconductors ?

Thank you for the detailed analysis of the experiment. For us is most important that the temperature of the ring remains below Tc. Will the persistent supercurrent (in the ring with T<Tc) disappear due to the remote piece of wire with T>Tc?

This particle drift occurs because the wave length of electron is longer than the distance between atoms of lattice. So every electron itinerates (tunnels) from one site to another. The electron pairs can also tunnel like single electrons.

The motion of electron pairs is rather a drift than a thermal motion, like the drift of conduction electrons between atoms of crystal lattice, because the pair drift and single electron drift are possible at T=0.

We don't you need an electric field, because the charge drift is not aligned into a certain direction, it is rather chaotic like electron motion in metals.

The answer to your question - the total current in the wire is zero because the thermal gradient may cause a charge density gradient, so currents in both directions are compensated. Note, we can eliminate the thermal gradient if we create the non-SC zone by a magnetic field H>Hc and not by T>Tc. Then we don't need consider any thermal currents. How about my (initially posed) question: does the supercurrent in the ring vanish due to the compensated charge drifts in the wire?

The non-zero supercurrent flows in the ring (and not in the wire). Toward the wire and back to the ring there is a charge drift (like a charge drift of electrons in a metal, but the electrons are paired in the superconductor). The drifts from the ring to the wire and back to the ring are equal, so the total current in the wire is zero. The question is: does the drift disturb the persistent supercurrent in the ring? (the persistent supercurrent was induced in the ring before we connect the ring to the wire).

Will a persistent supercurrent in a superconducting (SC) aluminum ring decay, if we connect the SC aluminum ring to an aluminum wire and the remote end of the wire is located in a separate chamber with T > Tc (or H > Hc)? This question is more complicated than it seems, most physicists cannot answer it unambiguously and there is no experiments to the issue. Imagine, a persistent supercurrent flows in a SC aluminum ring. Then we connect the SC aluminum ring (without solder) to an aluminum wire, the second end of the wire is in a separate chamber with T > Tc (or H > Hc) and is not SC. The temperature of the SC ring is stable below Tc. Thus the SC ring is directly connected to a non-SC zone where electron pairs dissipate their supercurrent momenta on atom lattice. Will the remote non-SC zone suppress the persistent supercurrent in the SC ring? Experimental setup to the question is shown in Figure 1 in https://vixra.org/pdf/2310.0103v1.pdf Quote Edit Share Pin Topic

I also was not sure that the article is about that question. However, the experiment shows that superconducting electrons stay permanently in paired states, so we can assert that a paired electron never was normal (at least during the measurement on the device). This fact has interesting consequences. For example, now we may not put all electrons and states into one Slater determinant; but we must form two Slater determinants (a normal and a superconducting ones). And electrons may not hop from one determinant to another one.

An interesting result in Nature "A superconductor free of quasiparticles for seconds" https://www.nature.com/articles/s41567-021-01433-7 . It shows that superconducting (paired) electrons don't hop into normal (unpaired) states for seconds. The measurement device detects single pair-breaking-events for a large pair population, so the average life time of each pair is much longer than a few seconds (up to many hours). Thus, the superconducting and normal electrons are not interchangeable in the momentum space during the measurement. More intriguing: all modern theories of superconductivity assume that all conduction electrons (superconducting + normal) are interchangeable within one shared momentum space, i.e. every electron may be awhile superconducting and awhile normal. So it's unclear are superconducting and normal electrons interchangeable or not?