A crystal at any finite temperature has phonons, as longer wavelength phonons require less energy to be excited. It seems that the electron (or cooper pair composite particle) wave would be scattered by phonon-induced inhomogeneities in the potential, much like a bumpy road where the bumps are continuously changing. Scattering would create resistance, but in superconductors (below the critical temperature) there is none. Why is said scattering completely suppressed at finite temperature?
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asked Apr 5, 2017 at 13:52
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$\begingroup$ Because the phonon does not have enough energy to disrupt the Cooper pairing. Why doesn't a mosquito hitting your windshield force your car off the road? $\endgroup$– Jon CusterCommented Apr 5, 2017 at 15:02
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1$\begingroup$ @Jon Custer: Why can't the cooper pair itself get scattered by the potential undulations? That still would create resistance. $\endgroup$– Kevin KostlanCommented Apr 5, 2017 at 17:26
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1$\begingroup$ @KevinKostlan I'm not an expert but I'm pretty sure that what you said there is incorrect. I think Cooper pairs do scatter but that there's no resistance because of how the superconducting ground state works. $\endgroup$– DanielSankCommented Apr 5, 2017 at 17:46
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1$\begingroup$ @DanielSank: How would scattering not reduce the forward component of the current (the momentum operator applied to the wave-function), and thus act as a sort of damping? Scattering scrambles velocities and therefore drives the average velocity toward zero. $\endgroup$– Kevin KostlanCommented Apr 5, 2017 at 23:12
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1$\begingroup$ @KevinKostlan All I can now is that I think that's not how it works in superconductivity. In a superconductor, the charge carriers aren't independent little balls bouncing around off of scattering centers. It's a little more like you shine a single coherent laser through a cluster of reflectors. There's interference, but the amount of light power coming out is exactly the same as what you put in. $\endgroup$– DanielSankCommented Apr 5, 2017 at 23:15
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2 Answers
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To be in the superconducting state, the temperature must be below the superconducting gap $2\Delta$ (modulo constants of order 1).
This automatically means that thermal energy $k_B T$ is lower than the superconducting gap. Thus, the only thermally populated excitations, be they phonons or anything else, do not have enough energy to break a Cooper pair.
But what if we don't break a Cooper pair, but just change its total momentum slightly? This is definitely possible through scattering via acoustic phonons for example, which can have very low momentum. However, the issue here is Pauli exclusion, all the other momentum states for the Cooper pair are occupied by other Cooper pairs. Because Cooper pairs are built from two fermions, Pauli exclusion still holds even though they are composite bosons. So small angle scattering is also suppressed.
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$\begingroup$ "Pauli exclusion still holds even though they are composite bosons" — this sounds contradictory. What does it mean then to be a boson (even if composite)? Shouldn't the behavior of the center of mass of this composite obey Bose-Einstein statistics? $\endgroup$– RuslanCommented Sep 11, 2020 at 6:56
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1$\begingroup$ The center of mass is still "bosonic" in the sense that many different Cooper pairs can have the same total momentum. This happens by having different individual momenta for the internal fermions that make up each pair. This is the way you reconcile Pauli exclusion for composite bosons made of fermions. However, when you talk about the phase space for a Cooper pair to scatter into, most states are Pauli-blocked for the underlaying fermions by fermions in other Cooper pairs. The only way around this blocking is to form a supercurrent where all pairs coherently have a nonzero COM momentum. $\endgroup$– KF GaussCommented Sep 11, 2020 at 8:40
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Typical phonon energies are in the meV range. Typical Cooper pair binding energies are also in that energy ball park (or even lower). So indeed a phonon with ~meV energy can break a Cooper pair apart. However, to have sufficiently many phonons excited with enough energy to break so many Cooper pairs apart the superconducting phase is destroyed, the temperature must be high enough. At room temperature, $k_{\rm B}T \approx 25$meV, which is a lot here. At the boiling point of He, e.g., $k_{\rm B}\cdot4.2{\rm K}\approx 0.36$meV and there won't be many phonons with sufficient energies to break up a Cooper pair with say $1$meV binding energy.
Breaking Cooper pairs apart is necessary to destroy the superconducting phase; as long as Cooper pairs are bound, they are Bosonic particles obeying Bose-Einstein statistics with macroscopic ground state occupation. That ground state will only be destroyed if a significant amount of Cooper pairs is broken up.
