The Bogolyubov model of liquid helium is considered. The validity of substituting a c-number for the k = 0 mode operatorâ0 is established rigorously. The domain of stability of the Bogolyubov's Hamiltonian is found. We derive sufficient conditions which ensure the appearance of the Bose condensate in the model. For some temperatures and some positive values of the chemical potential, there is a gapless Bogolyubov spectrum of elementary excitations, leading to a proper microscopic interpretation of superfluidity. Jp, 03.75.Fi,
The modelLet us consider a system of N spinless identical nonrelativistic bosons of mass m enclosed in a centered cubic box Λ ⊂ R 3 of volume V = |Λ| = L 3 with periodic boundary conditions for wave functions. The Hamiltonian of the system can be written in the second quantized form aŝHereâ # k = {â † k orâ k } are the usual boson (creation or annihilation) operators for the one-particle state ψ k (x) = V −1/2 exp(ikx), k ∈ Λ * , x ∈ Λ, acting on the Fock spacesymm is the symmetrized n-particle Hilbert space appropriate for bosons, andThe sums in (1) run over the dual set2m) is the one-particle energy spectrum of free bosons in the modes k ∈ Λ * (we propose = 1),N Λ = k∈Λ * â † kâ k is the total particle-number operator, µ is the chemical potential, ν(k) is the Fourier transform of the interaction pair potential Φ(x). We suppose that Φ(x) = Φ(|x|) ∈ L 1 (R 3 ) and ν(k) is a real function with a compact support such that 0 ν(k) = ν(−k) ν(0) for all k ∈ R 3 . Under these conditions the Hamiltonian (1) is superstable [1]. So long as the rigorous analysis of the Hamiltonian (1) is very knotty problem, Bogolyubov introduced the model Hamiltonian of superfluidity theory [2,3]. He proposed to disregard the terms of the third and fourth order in operatorsâ # k , k = 0 in the Hamiltonian (1),c N.N. Bogolyubov, Jr., D.P. Sankovich