Answer Happy

Consider two interacting electrons near a completely filled Fermi surface. Assume that the momentum of each electron is limited to a spherical layer around the Fermi surface by condition EF < (p) < EF + hwp, where (p) is the energy of the electron, F = (PF) is the Fermi energy and wp is the Debye frequency. We choose the zero for one-particle energies such that F = 0. The density of one-particle states in the above window can be assumed to be constant and equal to v(EF). We further limit the consideration to the pair states with total momentum equal to zero. The Hilbert space of this two-electron system is spanned by the states |_p,p) = √(a²_,ª} + a²_mª;){ground state) The Hamiltonian acting in the above Hilbert space is H = =Σ = (p) (aapt + a apt) - VoΣ(apa pa-pitapit + apa² pata pitapit). (2) P1 P2 Here Vo> 0 is the attraction strength, and p₁ and p2 are limited to a half of the above- mentioned spherical layer in the momentum space. a.) From Hamiltonian (2), obtain explicitly the matrix elements HP₁ P2 = (-P₁, P₁|H|- P2, P2). Then assume that there is an eigenstate with an energy E, which a form Epp-p, p). From stationary Schrodinger equation EE, Cpl-P₂P) = p.p CpHp.pl - p',p') obtain that = 1 (1) Vov (EF) 2 Sharp de E-26 b.) Show that in the limit Vov(EF) < 1 the lowest energy of the two-electron system has energy E = -2hwpe Vortep) (4) which is separated from higher energy states by a gap and hence represents a bound state.