Question

Why does not the bare interaction potential appear in the Bogoliubov theory?

Answer 1

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Why does not the bare interaction potential appear in the Bogoliubov theory?

quantum-mechanics quantum-field-theorycondensed-matter renormalization superfluidity

They use some effective potential defined by the s-wave scattering length, but not the bare atom-atom interaction V(r)V(r).

Why?

It is standard practice in second quantization to use the bare interaction and then transform to the momentum space.

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askedJul 16 '15 at 13:47



kaiser
504●4●12

What is the "bare" atom-atom interaction? Of course at large distances it will look like V(r)∼1/r6V(r)∼1/r6 for neutral atoms. But what about when rr is comparable to the Bohr radius? – Mark Mitchison Jul 16 '15 at 15:32

It should be a repulsive core. – kaiser Jul 16 '15 at 18:05

1

Until the electron orbitals start to overlap, at which point the scattering problem is a many-electron problem and the concept of an atom-atom potential no longer makes sense. My point is that there does not exist any "bare" potential V(r)V(r), there are only effective potentials valid beyond a certain radius. If you want to use the Lennard-Jones potential, you find that its Fourier transform is divergent as k→∞k→∞, due to the singularity as r→0r→0. But if you don't care about physics at this scale, you replace it with a pseudo-potential whose Fourier transform is simple to work with. – Mark Mitchison Jul 16 '15 at 20:09 

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