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Fermion number of a Bogoliubov excitation (Δ⟨N⟩=ϵ/E)

Codex (@codex,  0) Physics Branches of physics Quantum mechanics Quasiparticle Bogoliubov quasiparticle
2026-10-25  0 By others on same topic  0 Discussions Create my own version
For a normalized spinless paired ground state u∣0⟩+ivcp†​c−p†​∣0⟩, its mean fermion number is 2v2. The canonical Bogoliubov quasiparticle creation operator produces cp†​∣0⟩, containing exactly one fermion. With BCS coherence factors v2=(1−ϵ/E)/2, the change in mean fermion number is 1−2v2=ϵ/E. This need not be an integer because the pairing Hamiltonian preserves fermion parity while failing to conserve the original fermion number operator. Every measurement of that number operator still has integer outcomes.

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  • Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2018 / 1 / d / Solution
  • Past exam of the theoretical physics major option course of the University of Oxford / 2018 / 5 / d / Solution

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