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4

Well, $1/2\otimes1/2=0\oplus1$, so a system with two fermions has integer spin. But it is still a two fermion system, and therefore its wavefunction must be antisymmetric, as usual. This is not specific to Cooper pairs, but is basic Quantum Mechanics... [what is specific to Cooper pairs is that their size is $\gg a_0$, which means they are highly ...


3

It is not true that all superconductors are gapped. For example, d-wave superconductors in cuprates are gapless. The energy gap in the superconductor arises from the fact that breaking the Cooper pair requires finite energy. The low-lying quasi-particle excitations are all pair breaking excitations, so they are gapped from the ground state by the amount of ...


2

The Landau criterion is not in itself a criterion for superfluidity, but a criterion for the breakdown of superfluidity. Indeed, if applied to insulators or ordinary fluids, it would tell you that all of these are also superfluid... What the Landau criterion tells you is the velocity of the superfluid flow at which excitations are created from the ...


2

You are probably aware that the underlying mechanism responsible for superconductivity in cuprates or Fe-based superconductors is still subject to intense debate and research. The usual BCS electron-phonon coupling (EPC) is too weak to account for superconductivity at temperatures of order $100$ K, and produces a gap of s-wave symmetry which is incompatible ...


2

One way to estimate electron-phonon coupling is to take a look on the hot-electron relaxation rate, which can be more or less directly probed by pump-probe spectroscopy. The values for $\lambda$ are around 0.2-0.5. There are few articles on this topic from C. Gadermaier. Here are the links http://dx.doi.org/10.1063/1.4726164 ...


1

So upon looking into this further, it looks like Cooper pair interactions are pretty weak, on the order of 10^-3 eV. As such, I'm guessing the reason that we see superconducting at low temps and not at higher temps (with a few "hot superconductors" making an exception) is because at higher temperatures there is enough energy in the system such that Cooper ...


1

Yes, pairing is possible from a repulsive interaction. The reason behind this is that pairing has to occur in a certain angular momentum channel : $l=0$ for s-wave superconductivity, $l=1$ for p-wave, and so on. To see this, you can expand the repulsive $k$-dependent interaction on Legendre polynomials. Check this review that deals with the Kohn-Luttinger ...


1

The short answer is this: each line on your plot is denoting one fermionic mode (i.e. a fermionic creation/annihilation operator $c$), or equivalently each line corresponds to two Majorana modes (since a fermionic mode defines two Majorana operators $\gamma_1 = \frac{1}{2} \left(c+c^\dagger \right)$ and $\gamma_2 = \frac{1}{2i} \left(c-c^\dagger \right)$). ...



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