# Tag Info

## Hot answers tagged superconductivity

4

As a general rule adding thermal energy doesn't cause electronic transitions. That's because typical electronic transition energies are a few electron volts or around 100kT at room temperature. In a metal the electrons aren't in discrete energy levels but instead reside in a continuous band of energy levels called the conduction band. While thermal energy ...

3

The super-current is carried by the gradient of the phase of the condensate, and there is a finite energy cost associated with this. If the gradient energy is larger than the BCS condensation energy (the energy gained by forming Cooper pairs), then superconductivity will disappear. This is the critical current. In order to be more quantitative, consider ...

2

Chiral $p$-wave superconductor and He A phase can be considered being equivalent phases of matter, for the following reason: The fact that the superconductor is charged does not make too much difference in this regard, because it only affects the electromagnetic response (one has Meissner effect, the other does not), but electromagnetic field is an external ...

2

All phenomenology of superconductivity is described by solid state theory with spontaneously broken electromagnetic $U_{EM}(1)$ symmetry, so let's answer on your questions by using this idea. First question: where does this wavefunction form come from? So, EM symmetry group is spontaneously broken bu such VEV. Which VEV breaks this symmetry? It is the ...

1

so you are confusing a superconductor whose macroscopic wavefunction describing only the ground state could be described the way you did, with a superconducting qubit which has a totally different wavefunction and is a junction between two blobs of superconducting material. this introductory article might straighten things out: ...

1

Spin indeed plays a role in this. Cooper pairs are bosonic in nature, below the critical temperature they condense into the same state to form a superfluid phase responsible for superconductivity. This is why they do not "bump into eachother", they are all in the same state. Moreover, the condensation of Cooper pairs opens a gap in the excitation spectrum ...

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