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A superconductor has zero resistance. What about an electron in a vacuum? Could this simple system be considered superconducting?

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  • $\begingroup$ Alex, any particular reason why you say in a vacuum? I am no expert by any means, but I can't see how this enters into it. Hope you get a good answer anyway, it's certainly not a superconductor using current theories imo , it's charge is stable over a huge range of temperatures. $\endgroup$ – user81619 Jul 31 '15 at 21:18
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    $\begingroup$ I used a vacuum in the example in hopes of simplifying. I didn't want anyone to assume I was talking about an electron in a piece of metal. I think what I was hoping to get out of it is a very good and in retrospect obvious reason why it's not. $\endgroup$ – Alex Jul 31 '15 at 21:24
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    $\begingroup$ Nice question Alex. :) $\endgroup$ – Harshal Gajjar Aug 1 '15 at 2:54
  • $\begingroup$ You should change the title of the question,it doesn't really match with question. $\endgroup$ – Paul Aug 1 '15 at 16:23
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    $\begingroup$ I wonder whether this question makes any sense. A simple electron (in vacuum or not) has no resistance because the genuine notion of resistance does not exist for a single particle. Superconductivity and resistance are many-body problems. The only common notion between a superconductor and an electron is that you may describe both of them using a wave-function. Still, the wave-function of a single object has no real interest, since most of the quantum properties come from many-body problems (interference, interactions, quantization ...) $\endgroup$ – FraSchelle Aug 5 '15 at 8:53
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You are right, electrons in vacuum can carry a current without resistivity. However, superconductivity is not only zero resistance, but something more.

Superconductivity is defined by zero resistivity and by the presence of the Meissner effect, i.e., the expulsion of magnetic fields from the system.

The zero resistivity of electrons in vacuum is usually called ballistic conductivity, which means absence of scattering and, consequently, absence of resistivity. Ballistic conductivity is realized not only in vacuum, but also in other systems, for examples carbon nanotubes. In this systems the electrons can travel without undergoing scattering, since an ideal nanotube has no impurity and it is hollow at the center.

You may think that this is only matter of definitions, since both ballistic systems (electrons in vacuum, nanotubes) and superconductors exhibit zero resistance. However, the Meissner effect (expulsion of the magnetic field) is what differentiates superconductivity from ballistic conductance.

Also, the physical origin of zero resistance is completely different in the two systems. In ballistic conductors (electrons in vacuum, nanotubes), zero resistance arises from the fact that electron scattering is negligible in absence of impurities. Electrons simply move undisturbed. In superconductors, scattering is still present (superconductors can have impurities and disorder), but the weak attraction between electrons (mediated by the ions in the crystal) overcomes that, and therefore electrons move coherently without any dissipation.

Another difference between ballistic conductivity and superconductivity is the presence of a phase transition at a critical temperature $T_c$. Superconductors exhibit superconductivity only under a certain temperature $T_c$, and the transition between the normal state (usually metallic) and superconductivity is very, very sharp. In vacuum, the presence of ballistic conductivity does not depend on temperature. In nanotubes, perfect ballistic conductivity is obtained at low temperatures, but there is no sharp transition between zero and non-zero resistivity. Actually the dependence on the temperature is very smooth and in fact carbon nanotubes are nearly ballistic even at room temperatures. The increasing resistivity at higher temperatures is due to the fact that the system becomes more and more "disordered" as temperature increases.

Addendum: I should stress that the Meissner effect is not a synonym of perfect diamagnetism. Electron in vacuum (like plasma) can exhibit perfect diamagnetism, which is a direct consequence of zero resistance. Zero resistance in fact implies that current loops (eddy currents) are generated as a response to a variation of the external magnetic field, and will exactly cancel this variation. In a perfect diamagnetic system (as the electron-in-vacuum system) the magnetic field can be still non-zero (for example, the electron moving in vacuum generates a finite magnetic field). Inside a superconductor, the magnetic field is always zero, at least under a certain critical field, where the superconductor has a phase transition to a normal system (metal) or to a more complicated mixed state (type II superconductors). The difference between Meissner effect and diamagnetism is kind of subtle to understand but is physically well defined. I think a good introduction to this is here

Superconductivity: Meissner effect + zero resistance Perfect diamagnetism: zero resistance

2nd Addendum: Another difference between ballistic conductivity and superconductivity is the presence of a superconducting order parameter $\Delta e^{\imath \varphi}$. This is not only mathematical object to describe the pairing between electrons, but has direct physical consequences. One example is the Josephson effect. Clearly no Josephson junction can be realized between two ballistic conductors. (although it is possible to realize Josephson junctions between a superconductor and a ballistic conductor).

3rd Addendum: The weak attraction between electrons in conventional, low-temperature superconductors is described in the Bardeen-Cooper-Schrieffer (BCS) theory as an effective interaction between electrons mediated by the lattice distortions (phonons). In BCS, the weak attraction is described in the mean field approximation as an order parameter $\Delta e^{\imath \varphi}$. This mean field describes the phase transition between superconducting state and normal state (this is similar to the case of ferromagnets, which are described by a different order parameter, namely, the magnetization). No interaction plays a role in a system made of a single electron in vacuum. This is also the case in a system made of many electrons in vacuum (as long as the density is low) or in a ballistic system (nanotubes).

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    $\begingroup$ Type 2 superconductors do exhibit the Meissner effect up to a critical magnetic field $H_1$. Above this critical field, they exhibit a mixed phase where the magnetic field partially penetrates into the system. This mixed phase can be described as a coexistence of a normal phase and a superconducting phase (with Meissner effect), and survives up to a critical field $H_2$ where superconductivity is completely destroyed. $\endgroup$ – sintetico Jul 31 '15 at 22:48
  • $\begingroup$ This is an irrelevant answer. You are only summarizing superconductivity, which anyone can read on Wikipedia. To me, you haven't addressed the question. For instance, the "critical temperature" might be infinite, i.e. the electron could always be superconducting. Since there is no "bulk" to the material here, it's vacuously true that there is no B-field "inside it" (which is what expulsion of the field means here). To me, what really makes an electron obviously not a superconductor, is that there is only one electron, whereas superconductor requires two electrons to form a Cooper pair. $\endgroup$ – Chris Gerig Aug 3 '15 at 15:47
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    $\begingroup$ @ChrisGerig The main difference is not the critical temperature (one can have a critical temperature with a transition to zero resistance but no superconductivity). The difference between superconductivity and other systems with zero-resistance is the Meissner effect. This is the main point. I fully agree with you that Cooper pairs and BCS theory are the microscopic explanation of superconductivity, which as far as we know works extremely well with conventional superconductors. However one can have 2 or many electrons in vacuum which have zero resistance and no superconductivity. $\endgroup$ – sintetico Aug 3 '15 at 16:57
  • $\begingroup$ @ChrisGerig I was not digging in the microscopic theory (BCS and its generalization) because I wanted to stress the physical differences rather than the theory. And also because not everybody agrees that a generalized BCS can explain high temperature superconductors. $\endgroup$ – sintetico Aug 3 '15 at 17:03
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    $\begingroup$ Ok I see now what is your point, you define the bulk as the inside of the electrons. Following your definition all electrons, whether in vacuum or not, exhibit Meissner effect... Moreover, following your definition, the bulk has spatial dimension equal to zero (electrons are point-like). However, I don't see how this definition can be physical or meaningful. $\endgroup$ – sintetico Aug 3 '15 at 20:04
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This is a well advanced discussion but I'll give my two cents here. There is a major quantum mechanics difference between a superconductor and electrons in vacuum.

Electons in vacuum have each its own wave function. In a superconductor, Cooper pairs condensate into a single wave function.

So what?

Well, it becomes very easy to scatter electrons in vacuum and, hence dissipate the current. We know this very well. It is called a synchrotron storage ring. You find them almost anywhere nowadays. In a synchrotron, electrons circulate in a closed orbit inside a vacuum chamber. The amount of electrons is even measured as a current circulating in the ring. However, as the vacuum is not perfect, electrons eventually scatter and the current dissipates. And you have to fill the ring with more electrons. So, any small impurity in the vacuum dissipates the current.

Because Cooper pairs share the same quantum state, it is very hard to dissipate them. Actually, the best conventional superconductors (Pb, Nb, NbTi, NbN) are pretty bad conductors. They have a lot of scattering. But in the condensate state, it is very hard to scatter a single pair and dissipate the current. Hence the zero resistance state. Conventional superconductors can tolerate a large amount of (non magnetic) impurity and disorder (hence normal state scattering) without loosing their superconducting properties.

In summary, although electrons in perfect vacuum may conduct electricity with no dissipation, it is easy to scatter them. Cooper pairs share a single quantum state and require a lot of energy (the gap) to be scattered.

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  • $\begingroup$ +1 One can say that the superconducting state is "robust" against impurities and disorder, and this is a direct consequence of the Cooper condensation, as you said, of the electrons into a single wavefunction. This shares similarities with other phenomena, for example, the Bose-Einstein condensation. $\endgroup$ – sintetico Aug 11 '15 at 20:58
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    $\begingroup$ @sintetico ...and superfluid Helium as well. $\endgroup$ – rpsml Aug 11 '15 at 22:10
  • $\begingroup$ Also one can say for this reasons that superconductivity is a pure quantum phenomenon, which takes place at a macroscopic scale. The physics of electrons or in general charged particles in vacuum can be well described classically (of course in certain limits/approximations). $\endgroup$ – sintetico Aug 12 '15 at 18:21

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