# Why don't low frequency phonons scatter electrons in a superconductor?

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?

• 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? – Jon Custer Apr 5 '17 at 15:02
• @Jon Custer: Why can't the cooper pair itself get scattered by the potential undulations? That still would create resistance. – Kevin Kostlan Apr 5 '17 at 17:26
• @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. – DanielSank Apr 5 '17 at 17:46
• @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. – Kevin Kostlan Apr 5 '17 at 23:12
• @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. – DanielSank Apr 5 '17 at 23:15

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, and thus the only thermally populated excitations, be they phonons or anything else, do not have enough energy to break a Cooper pair.
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.