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The energy spectrum of a BCS superconductor contains an energy gap. How does the existence of this energy gap explain that there is no dissipation of energy in a superconductor? There are also energy gaps in semiconductors. But they do have resistance and energy dissipation happens.

Solidification
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  • The energy gap does not explain the non-dissipative properties of superconductors, Cooper pairing does. There are plenty of superconductors without a full gap. See this question https://physics.stackexchange.com/questions/339747/what-is-the-link-between-the-bcs-ground-state-and-superconductivity/339794?r=SearchResults#339794 – KF Gauss May 02 '20 at 23:31

2 Answers2

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These are two very different types of gaps. Also, the nature of electronic transport in superconductors and semiconductors is very different.

Superconductors

The carriers of electric current in superconductors are Cooper pairs – bound states of pairs of electrons, with a charge $-2 e$. Cooper pairs are bosons and, as such, they are allowed to bunch together in unlimited quantities in a single quantum state. This is what happens below the temperature of superconducting transition: All Cooper pairs end up in the quantum state with the lowest energy – the ground state – separated by an energy gap from the lowest excited state. In this state, if a Cooper pair was to scatter it would have to lose some of its energy and go to a quantum state with lower energy. But it cannot do this because it is already in the ground state, which is the state with lowest energy possible.

One might argue that a Cooper pair could scatter by breaking apart into two electrons. However, to break a Cooper pair one would first have to spend an energy equal to the gap size, which is not small. This can actually be done if one either places the superconductor in a sufficiently strong magnetic field (above the critical field) or one lets a too strong current (above the critical current) through the material.

TL;DR: In superconductors carriers of electric current are bosonic quasiparticles residing in their lowest-energy state. Therefore, they cannot lose energy hence no energy dissipation.

Semiconductors

Semiconductor is another word for an insulator with a relatively small gap. And indeed, a pure semiconductor does not conduct current at zero temperature, i.e., its resistivity is infinite. The reason for that is that electrons in the valence band (the one that forms the bottom of the gap) cannot move because the valence band is completely filled and electrons are fermions, hence they are not allowed to transition to quantum states occupied by other electrons.

Nevertheless, the electrons could propagate if they receive enough energy to overcome the energy gap and to reach the empty conduction band, where they can move more or less freely. This is what happens at finite temperatures. Theoretically, any insulator can conduct some current at elevated temperatures. However, the carrier density, hence the conductivity, is proportional to $exp(-\Delta / T)$, where $\Delta$ is the gap size. So, if the gap is large (e.g., 1 eV $\approx 11600$ K or higher) the current will be negligibly small.

Another mechanism to make semiconductors actually conduct is to dope them, i.e., to alloy them with impurities that change the number of electrons per atom. This process either creates positively charged carriers (holes, $p$-type) in the valence band or negatively charge carriers (electrons, $n$-type) in the conduction band. These carriers can sustain some current but the resistivity is usually high because of the relatively low carrier density (compared to metals). Also, the dopants themselves get charged and they serve as very efficient carrier scatterers.

TL;DR: In contrast to superconductors, the gap in semiconductors hinders the flow of the current. Semiconductors have finite conductivity despite the gap, thanks to thermal excitations and doping.

This is a rather rough picture but I hope it serves as an answer.

Riddler
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  • The seems to be a range of energies to which cooper pairs condense, not a single lowest state. See last figure http://musr.ca/theses/Sonier/MSc/node15.html Does this complicate/change your answer? – Solidification May 02 '20 at 17:19
  • Note that the figures in your link show only the DOS of fermionic quasiparticles. Cooper pairs condense to a single (in the simplest cases) quantum state characterized by one macroscopic wave function. – Riddler May 03 '20 at 05:59
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It does not. In particular, notice that there are also gapless superconductors. The crucial characteristic is not the energy gap but the fact that a superconductor is described by a macroscopic wave function. In Bardeen-Cooper-Schrieffer theory (BCS), this is given by a coherent superposition of Cooper-pair states. This macroscopic wave function gives the superconductor a certain rigidity to some excitations, and as a result gives the Meissner effect and the zero d.c. resistance.

fra_pero
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