Physics Asked by Kiji on July 15, 2021
Im looking into Quantum Computing, where the BCS Theory is used to build Qubits with a BEC.
Why does the Bose-Einstein Condensate not interact with other particles and hence has no dissipation?
In other words how does the BEC form a superconducting material. Why is there no dissipation for BEC’s?
So I assume you know in the BCS theory, fermions form Cooper Pairs (bound electrons) at very low temperatures. Since their paired state has a lower energy than the Fermi energy, they are bound. So the pair is now a composite boson with spin 0 or 1 instead of the fermions spin 1/2.
This allows them to "condense" into the same quantum state which is what we see in BECs. So really the fact that the pair essentially behaves like a boson allows for the superconducting effect to happen. Since electrons follow the Pauli Exclusion Principle, you cannot just break up one Cooper Pair without changing energies of all other pairs. This fundamentally causes an "energy gap" at very low temperatures, meaning that phenomena such as electron scattering and other single-particle excitations (small excitations) are forbidden. Hence why you would not see dissipation effects or other low energy interactions. You can then describe the system with a macroscopic wavefunction, as is done in BEC.
There are special cases (experimental conditions) that have been experimentally observed where Superconductivity and dissipation coexist but I'm not familiar with the theory or experimental results so I cannot really comment on that.
Correct answer by Alon Shoshan on July 15, 2021
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